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PICList Thread
'Low dropout regulators'
1996\06\25@222841 by Steve Hardy

1996\06\25@235751 by Mark K Sullivan

1996\06\26@010329 by Peter Homann

{Quote hidden}

> I am constructing a battery operated PIC application where I need to be
> able to handle 2 Lithium cells.  Since this produces up to 7.4V, this
> needs to be regulated down to 5V.  To maximise battery life, the
> regulator needs low dropout (100mV) and quiescent current in the low
> microamps range.  It should be able to supply up to 100mA peak.
>
> Maxim has regulators which are ideal, but made of unobtainium (they
> haven't even sent the 'free samples').  So in desperation I designed my
> own using two transistors and a power FET to regulate the negative
> side.  This works fine except that I end up using a zener diode with
> only microamps of current through it, thus its voltage is _way_ out of
> spec: about 6V for a 10V zener!
>
> Does anyone know of (obtainable) zeners etc. which are rated for
> ultra-low bias current (say 1-5uA)?  Tempco is not too much of a worry -
> say +/-5% over 0-40 deg C.
>
> Regards,
> SJH
> Canberra, Australia
>

Steve,

I've been going through exactly the same problem. Have a look at
the National LM2936Z5. Its a low dropout 5V regulator, supplies
up to 150mA (I think), and has a quiescent current of around
15uA.

Should do the trick.

I think RS Components have them.

Let me know how you get on.

Peter.
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1996\06\26@020111 by gary skinner

>I am constructing a battery operated PIC application where I need to be
>able to handle 2 Lithium cells.  Since this produces up to 7.4V, this
>needs to be regulated down to 5V.  To maximise battery life, the
>regulator needs low dropout (100mV) and quiescent current in the low
>microamps range.  It should be able to supply up to 100mA peak.
>
>
How about a Zetex ZM33064, TO92 package Supply Voltage Monitor designed for
uP systems. Guaranteed to operate over 1V. Threshold set to 4.6V, built-in
20mV hysteresis, open col o/p sinking 10mA. Power-on reset over 4.6V. Input
voltage to 10V.

I have 100 in stock plus the 6 page data sheet.

Let me know if you want to try one.

regards

peter



-----------------------------------------------------------------------------
                         Peter J. Crowcroft
               PO Box 88458, Sham Shui Po, Hong Kong
Voice: 852-2720 0255.     Fax: 852-2725 0610      Email: .....diykitKILLspam@spam@hk.super.net
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-----------------------------------------------------------------------------

1996\06\26@022727 by Philippe TECHER

1996\06\26@023134 by Peter J. Crowcroft

>>National LM2936Z5. Its a low dropout 5V regulator, supplies
>up to 150mA (I think), and has a quiescent current of around
>15uA.
>

I just tried to buy these last week from the Hong Kong NS agent & was told
these are now obsolete & discontinued. Typically (for HK) they could not
tell me what was the replacement part which I am sure there is because they
have been very hard to get these last months due to cellular phone manufacture.

-----------------------------------------------------------------------------
                         Peter J. Crowcroft
               PO Box 88458, Sham Shui Po, Hong Kong
Voice: 852-2720 0255.     Fax: 852-2725 0610      Email: diykitKILLspamhk.super.net
               Web: http://www.hk.super.net/~diykit
-----------------------------------------------------------------------------

1996\06\26@131901 by Dwayne Reid

Hi,
I see that NJR uses a SOT89 and Motorola uses an pin SOT
package for their 78Lxx series regulators.  Any suggestions
on which one to go with?

Regards,
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1998\09\09@153900 by David VanHorn

1998\09\09@153907 by Mike Sauve

1998\09\10@110729 by Tim Economu

Try the LM7805. It is overtemp and overcurrent proof. And cheap.
Kind regards,
Tim

Timothy Economu                                          360-579-2117
Design Engineer                                       fax 360-579-2117
Trace Engineering                email EraseMEeconomuspam_OUTTakeThisOuTwhidbey.com
7185 South Cultus Bay
Clinton, Washington 98236
{Original Message removed}

'Regulators and New PICS'
1998\09\22@163810 by Craig Lee

1998\09\22@174037 by Matt Bonner

1998\09\24@125401 by Matt Bonner

Craig Lee wrote:
>
> On a similar topic, when are the HV parts available?

Craig,

I wrote the other day that I'd keep you apprised of my attempts to
locate an HV part mentioned at the MChip seminar.  I just heard from
Gail at Pioneer that I've got a couple of 16HV540 samples on the way.
She also handles Edmonton, so give her a call.
 cell: 403-606-2601
 email: desjardinsgspam_OUTpios.com

--Matt

1998\09\24@125623 by Harrison Cooper

1998\09\24@132921 by Matt Bonner

2000\02\10@131916 by M. Adam Davis

2000\02\10@132128 by Matt Burch

Mark,

I have had good success with the Linear LT1121 series of low-dropout linear
regulators... I-sub-q is somewhere in the 50uA range, I recall.

For really low-power, long-life battery operations, consider going to a
step-down switching regulator with an automatic shut-down under low-load
conditions... for example, the LTC1735 would fit the bill nicely.

mcb



At 12:09 PM 02/10/2000 -0600, Mark Peterson wrote:
{Quote hidden}

>I am running a small circuit that includes a PIC off of a 24 VDC battery
>system.  I am minimizing my current draw by having the PIC sleep most of
>the time.  I wake it up every few seconds, have it power up the analog
>input conditioning circuitry, make its decisions, send an output signal if
>required, and then have it go back to sleep.  Current draw using this
>scheme is peanuts.  My problem is the quiescent current that a classic 7805
>regulator uses, roughly 3.5 mA.  I saw the earlier messages about the Maxim
>and TelCom regulators.  They look great but their maximum input voltage is
>around 11 volts.  I've considered various dropping resistor and zener
>schemes to use in conjunction with one of the low power regulators but they
>all end up sucking up as much current as the original 7805.  Any clever
>ides out there on this?
>
>Thanks.
>

-------------------------------------------------------------------
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 Project Engineer   | 619 E. 8th St. Suite D | fax: (785) 749-9214
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-------------------------------------------------------------------

2000\02\10@133202 by Matt Burch

In a similar vein, I once had to power a pic from a tiny lithium battery
with a puny max output current. (lithium coin cells aren't known for their
great ability to deliver current!)  Solution: trickle-charge a large
tantalum cap thru a resistor, and wake the pic up every once in a while to
do a furious bit of sampling, processing, and transmitting before going
back to sleep to let the capacitor charge back up again.

mcb



At 01:16 PM 02/10/2000 -0500, M. Adam Davis wrote:
>I saw a circuit recently where the PIC was powered off a zener @ about 4.8vdc
>from a higher supply (as you mention).  When the PIC woke up, it started
PWMing
>a transister which hooked up to the power supply and it became its own pwm
>voltage regulater.  Due to another diode in the circuit, when it powered
up to
>5v, the zener took no current.

-------------------------------------------------------------------
    Matt Burch      |  Pinnacle Technology   | tel: (785) 832-8866
 Project Engineer   | 619 E. 8th St. Suite D | fax: (785) 749-9214
KILLspammburchKILLspampinnaclet.com |  Lawrence, KS  66044   |  http://www.pinnaclet.com
-------------------------------------------------------------------

2000\02\10@151625 by Russell McMahon

2000\02\10@155926 by paulb

2000\02\10@161354 by Matt Burch

At 07:57 AM 02/11/2000 +1100, Paul B. Webster VK2BZC wrote:

>  The main question here:  *Why* would you put a resistor in series,
>when the lithium cell is by your own argument, the resistor?

IIRC, it worked better that way.  :)

mcb

-------------------------------------------------------------------
    Matt Burch      |  Pinnacle Technology   | tel: (785) 832-8866
 Project Engineer   | 619 E. 8th St. Suite D | fax: (785) 749-9214
RemoveMEmburchTakeThisOuTpinnaclet.com |  Lawrence, KS  66044   |  http://www.pinnaclet.com
-------------------------------------------------------------------

'[OT] Alternative Parts (voltage regulators)'
2000\04\12@024204 by David E Arnold

If you want high efficiency you might consider a switching regulator.
There are versions available w/ low quiescient current and features
like logic controlled shutdown, etc. But they're more complex than LDO's
and tend to use more space. But there's some quite simple ones
such as the LTC1503-2 from Linear Technology, which requires no
external inductors and just 4 small caps wired to it.

Disadvantage of an LDO is all the current must flow through the input;
hence, they exhibit lower efficiency.

-Dave






Dale Botkin <spamBeGonedalespamBeGoneBOTKIN.ORG> on 04/11/2000 10:50:16 PM

Please respond to pic microcontroller discussion list <TakeThisOuTPICLISTEraseMEspam_OUTMITVMA.MIT.EDU>

To:   RemoveMEPICLISTTakeThisOuTMITVMA.MIT.EDU
cc:    (bcc: David E Arnold/SYBASE)
Subject:  Re: [OT] Samsung Parts (voltage regulators)




On Wed, 12 Apr 2000, Russell McMahon wrote:

> Of particular interest to me was a TO92 regulator (KA76L05Z) with low
> dropout specs and max input voltage apparently similar to the LM2936 (which
> I use due to its very low quiescent current, very low dropout, higher than
> many max input voltage) and at a MUCH lower price from RS.

I just stumbled upon the ICL7663, programmable (2 resistors), low dropout,
<10uA quiescent current, logic controlled shutdown, and a tempo output
that looks like it could be used for generating LCD contrast voltage.
Works like a charm. Is there something else I should look at as well?
Typically I want to use either 4xAA or a 9V battery and have it last for
many months, so the 78L05 isn't a particularly good choice.  The 7663 cost
me a couple of bucks, but if there's something equivalent in performance
and cheaper or easier to get, I'd love to hear about it.

Dale
---
The most exciting phrase to hear in science, the one that heralds new
discoveries, is not "Eureka!" (I found it!) but "That's funny ..."
               -- Isaac Asimov

2000\04\12@063742 by paulb

2000\04\12@081229 by David VanHorn

2000\04\12@083141 by paulb

2000\04\12@084349 by David VanHorn

'[OT] Samsung Parts (voltage regulators)'
2000\04\12@102732 by mike

'[OT] Alternative Parts (voltage regulators)'
2000\04\16@052050 by Russell McMahon

2000\04\16@052054 by Russell McMahon

'[OT] Samsung Parts (voltage regulators)'
2000\04\16@052058 by Russell McMahon

'[OT] Alternative Parts (voltage regulators)'
2000\04\16@113807 by David VanHorn

'[OT] Samsung Parts (voltage regulators)'
2000\04\16@220932 by Dale Botkin

2000\04\16@221740 by l.allen

2000\04\17@025553 by Russell McMahon

2001\01\18@142018 by Chris Carr

2001\01\18@152131 by Robert Francisco

Check out Brooks Instruments

----- Original Message -----
From: John Pearson <xeroEraseME.....CMC.NET>
To: <EraseMEPICLISTMITVMA.MIT.EDU>
Sent: Friday, January 19, 2001 12:13 AM
Subject: [OT]: Electronic Air Valves, pressure regulators


{Quote hidden}

> I am looking for electronicly activated air pressure valves and pressure
> regulators. I want to control air pressure in the range of 10 to 50 psi.
>
> Also looking for help with my project.
>
> Anyone have experience, knowledge, or sell these things.
>
> Thanks
>
> John
>
> --
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>
>

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I was going to use a voltage regulator from Toko America, what Digi-Key
listed as a 260 ma 5.5 volt regulator (datasheet:
http://207.208.84.9/semiconductors/pdf/tk112xxb.pdf).

However, when it arrived and I started looking at the data sheet, it seemed
to be only a 150 ma regulator. What I'd like to know is, can I safely put
two of these together, both outputs tied to the same positive (circuit)
supply rail? Will I then be able to get 300 ma out of the two regulators?

If I can't tie them to the same supply rail, can I run some of my components
off a separate rail? For example, can I have one regulator power the LCD in
my circuit, and the other regulator power the MCU, but still have output
pins of the MCU connected to input pins of the LCD, and everyone sharing a
common ground?

As a side question, does anyone know of a 250 - 300 ma 5.5 volt
surface-mount regulator?

TIA,

--
Rick

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2002\01\26@020942 by David VanHorn

>
>If I can't tie them to the same supply rail, can I run some of my
>components off a separate rail? For example, can I have one regulator
>power the LCD in my circuit, and the other regulator power the MCU, but
>still have output pins of the MCU connected to input pins of the LCD, and
>everyone sharing a common ground?

This will definitely work.

As to putting active regs in parallel, probably not.
If their ref voltages aren't identical, then one ends up hogging the load,
and overheating I think.

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2002\01\26@025322 by 859-1?Q?Alexandre_Guimar=E3es?=

> As a side question, does anyone know of a 250 - 300 ma 5.5 volt
> surface-mount regulator?

   What is the input voltage ? That will be the basis to determine how much
heat you will generate on the linear regulator. If it is too high you will
have to give up on the smd part. If it is low enough take a look at linear
technology, they have some low drop regulators that should fit your needs.

Best regards,
Alexandre Guimaraes

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2002\01\26@063440 by Russell McMahon

> However, when it arrived and I started looking at the data sheet, it
seemed
> to be only a 150 ma regulator. What I'd like to know is, can I safely put
> two of these together, both outputs tied to the same positive (circuit)
> supply rail? Will I then be able to get 300 ma out of the two regulators?
>
> If I can't tie them to the same supply rail, can I run some of my
components
> off a separate rail? For example, can I have one regulator power the LCD
in
> my circuit, and the other regulator power the MCU, but still have output
> pins of the MCU connected to input pins of the LCD, and everyone sharing a
> common ground?

Two regulators will have slightly different Vouts - how different depend on
specified Vout tolerances.

You can definitely use separate regulators to each drive part of the load.
The small difference in voltages that result should generally be less than
you will get from other sources.

You can parallel two and expect reasonable results as long as the Vouts are
not vastly different and both have current limits.
The higher Vout regulator will take most of the load at low loads as it
keeps driving when the other turns off at set Vout. Once this has reached
its current limit it will start to sag and the other will then make up the
difference. This means one will TEND to run at near its max rated current.

You can put "spreading resistors" in the outputs to make Vout sag as current
is drawn but this destroys your regulation somewhat and is probably not
needed. At eg 100 mA a sag of 0.1 volt would be enough needing V/I = 1 ohm.
This may do nasty things to your regulator compensation and is probably best
avoided. Twill probably work well enough as is.

If you do NOT value the regulators over-current  protection and don't need
very low dropout you can try the olde days trick of a PNP transistor in
parallel with the regulator.
Vin feeds reg through a resistor Rb.
Emitter to Vin.
Base to Rb and regulator in.
Collector to Vout.

As long as drawn current through Rb is < Vbe turn on (about 0.6v) regulator
will be the sole source of Vout.
When Vrb exceeds Vbe the transistor turns on and Vout rises until the
regulator backs off to stop this thereby reducing Vrb.
Works well enough.
Short the output and transistor will still try to provide Vout.

Probably finding a single suitable SMD regulator would be better.


       Russell McMahon

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2002\01\26@085831 by Byron A Jeff

On Fri, Jan 25, 2002 at 09:46:08PM -0800, Rick Mann wrote:
> I was going to use a voltage regulator from Toko America, what Digi-Key
> listed as a 260 ma 5.5 volt regulator (datasheet:
> http://207.208.84.9/semiconductors/pdf/tk112xxb.pdf).
>
> However, when it arrived and I started looking at the data sheet, it seemed
> to be only a 150 ma regulator. What I'd like to know is, can I safely put
> two of these together, both outputs tied to the same positive (circuit)
> supply rail? Will I then be able to get 300 ma out of the two regulators?

Probably not a good idea. Linear bipolar regulators generally have a "feature"
called thermal runaway. Simply put the hotter it gets, the more current it
draws, making it even hotter. When you parallel once one becomes hotter than
the other, taking all of the current.

>
> If I can't tie them to the same supply rail, can I run some of my components
> off a separate rail? For example, can I have one regulator power the LCD in
> my circuit, and the other regulator power the MCU, but still have output
> pins of the MCU connected to input pins of the LCD, and everyone sharing a
> common ground?

Yes. That should be fine.

>
> As a side question, does anyone know of a 250 - 300 ma 5.5 volt
> surface-mount regulator?

Of course there is the third option: use a pass transistor in parallel with
the regulator. Then when the regulator starts passing too much current the
pass transistor kicks in and delivers the rest. A surface mount low sat
PNP or PMOS power transistor should be able to easily deliver a 1/2 amp.

Hope this gives you some ideas.

BAJ

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2002\01\26@090440 by info

2002\01\26@095735 by Dave Dilatush

Rick Mann wrote...

>I was going to use a voltage regulator from Toko America, what Digi-Key
>listed as a 260 ma 5.5 volt regulator (datasheet:
>http://207.208.84.9/semiconductors/pdf/tk112xxb.pdf).
>
>However, when it arrived and I started looking at the data sheet, it seemed
>to be only a 150 ma regulator. What I'd like to know is, can I safely put
>two of these together, both outputs tied to the same positive (circuit)
>supply rail? Will I then be able to get 300 ma out of the two regulators?

Looking at the data sheet for the TK11255BM (I'm assuming that's the
one), it would appear you might be able to do this.

If you do, be aware that the two regulators aren't going to share the
load current equally; in fact, one of them will end up supplying most of
the load current while the other one supplies only a small part of it,
or none at all.

The reason for this is that there's always some variation in the output
voltage setpoint from device to device; and if you have two devices in
parallel, one of them will be trying to set the output voltage just a
little bit higher than the other.  The one that wants the higher output
voltage (even if it's only higher by a few hundred microvolts) will end
up sourcing all the current, while the regulator that wants to set a
slightly lower voltage will simply shut off.

The only time you'll get both regulators conducting is if your load
current exceeds the short-circuit current limit of the regulator which
is carrying the lion's share of the load (or the regulator gets too hot,
activating its overtemperature protection), whereupon the other one will
kick in and start passing current.

>If I can't tie them to the same supply rail, can I run some of my components
>off a separate rail?  For example, can I have one regulator power the LCD in
>my circuit, and the other regulator power the MCU, but still have output
>pins of the MCU connected to input pins of the LCD, and everyone sharing a
>common ground?

Yes, you can do that.  The trick, however, is to somehow partition your
circuit and divide the load more or less evenly between the two
regulators.  I don't know how much current your LCD takes; if it's just
a character LCD module it probably doesn't draw more than a few
milliamps and, if so, putting it on a separate regulator wouldn't help
your overall situation much.

>As a side question, does anyone know of a 250 - 300 ma 5.5 volt
>surface-mount regulator?

No, though you could look around and see what National Semiconductor,
Maxim, and Linear Technology might offer.

Hope this helps...

Dave

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2002\01\26@101124 by Dave Dilatush

Byron A Jeff wrote...

>Linear bipolar regulators generally have a "feature"
>called thermal runaway. Simply put the hotter it gets, the more current it
>draws, making it even hotter.

Any examples?

All of the IC voltage regulators I've ever used or investigated--and
that's a LOT of them--not only have internal current limiting and output
short-circuit protection, but thermal overload protection and, in some
cases, pass transistor SOA protection as well.  Some go even further,
adding protection against input overvoltage and even protection against
being inserted into a PC board backwards.

I've never seen any integrated circuit voltage regulator do what you've
described, or anything even remotely resembling it; not ever.  My
experience has been that most IC voltage regulators are very rugged.

If you know any that do have this thermal runaway "feature", I'd be
interested in getting the manufacturer and part numbers so I know what
not to buy in the future.

Dave

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2002\01\26@103649 by Byron A Jeff

On Sat, Jan 26, 2002 at 03:10:04PM +0000, Dave Dilatush wrote:
{Quote hidden}

> Byron A Jeff wrote...
>
> >Linear bipolar regulators generally have a "feature"
> >called thermal runaway. Simply put the hotter it gets, the more current it
> >draws, making it even hotter.
>
> Any examples?
>
> All of the IC voltage regulators I've ever used or investigated--and
> that's a LOT of them--not only have internal current limiting and output
> short-circuit protection, but thermal overload protection and, in some
> cases, pass transistor SOA protection as well.  Some go even further,
> adding protection against input overvoltage and even protection against
> being inserted into a PC board backwards.
>
> I've never seen any integrated circuit voltage regulator do what you've
> described, or anything even remotely resembling it; not ever.  My
> experience has been that most IC voltage regulators are very rugged.
>
> If you know any that do have this thermal runaway "feature", I'd be
> interested in getting the manufacturer and part numbers so I know what
> not to buy in the future.

I explained myself poorly. Sorry. Of course they have current limiting. But
the net effect is that one of the regulators will always be maxed out
because it runs warmer than the others.

All of the dicussions I've seen on parallel regulators indicates that their
outputs should never be tied directly together.

If you have an example of where regulator outputs are tied diectly togther
and they equally share the load, I'd sure like to see it because that's a
part I would use.

BAJ

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2002\01\26@122300 by Dave Dilatush

Byron A Jeff wrote...

>But the net effect is that one of the regulators will always be maxed out
>because it runs warmer than the others.

Yes, one of the regulators will end up sourcing most of the load
current; but it will do so because it has the higher of the two output
voltage setpoints, not specifically due to its being warmer.

However, you could also have some interesting thermal dynamics going on
with two regulators in parallel like this.  What would happen, for
instance, if you have two regulators connected thus, whose output
voltage setpoints were very close at room temperature but they have
slightly different reference voltage temperature coefficients and/or
error amplifier gain tempcos?  The regulators might share current
differently at different load levels or temperatures; or they could even
swap current back and forth, heating and cooling alternately as some
kind of funky "thermal oscillator"; or, if both regulators had rather
strong negative output voltage tempcos they could settle into an
arrangement where they actually would share the load more or less
evenly.  Not likely, but possible.

The bottom line, though, is what Russell McMahon said: provided the
voltage outputs of the two regulators aren't vastly different and they
are both current-limited, there shouldn't be any huge problems.

The only "should never" I'm aware of is, don't ever just connect two
regulators in this fashion and casually assume they'll somehow
automagically share the load equally; they won't.  
Also, there are a few possible complexities that must be taken into
account:

One possible "gotcha" involves low-dropout ("LDO") regulators; these
tend to be rather picky in their input/output filter capacitor
requirements, with respect to both capacitance and ESR.  With the wrong
capacitors these things can become unstable and end up oscillating.
Putting two of them in parallel could end up making the input/output
filtering requirements even more complex.

A second possible complication involves the particular overcurrent
protection mechanism that ends up limiting the current in the regulator
that initially supplies the bulk of the load current.  If its output
current is limited by the internal current-limit circuitry, all is
probably well and the second regulator will simply step in and start
supplying the remainder of the load current.  But if the first regulator
ends up tripping its overtemperature protection BEFORE it goes into
current limit (for instance, because the package size is small and
there's a large input-output voltage differential), funny things can
happen.  Some regulators may implement this overtemperature protection
by simply cranking back on the internal current limiter when the die
temperature goes too high; but others may operate by actually shutting
down the regulator entirely and waiting for it to cool off, whereupon it
turns on again.  Obviously such on/off/on/off behaviour wouldn't be too
helpful in this situation, so a careful study of the datasheet is in
order.

>If you have an example of where regulator outputs are tied diectly togther
>and they equally share the load, I'd sure like to see it because that's a
>part I would use.

Like I said, both here and in my earlier reply to the OP, this business
of equal load sharing simply isn't too likely to happen.  
Frankly, I think it's a whole lot easier--and cleaner--just to get a
heftier voltage regulator if you need more current.  
Probably be cheaper, too.

Dave

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2002\01\26@125429 by andy n1yew

or you can use a pass transistor..

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2002\01\26@131053 by jim

2002\01\26@134055 by Thomas C. Sefranek

2002\01\27@024717 by Roman Black

2002\01\27@044103 by Alan Shinn

2002\04\03@183758 by Pic Dude

Yep, I know how harsh an automobile can be, voltage-wise.
Thanks.

----- Original Message -----
From: "Matt Pobursky" <spamBeGonepiclistKILLspamMPS-DESIGN.COM>
To: <.....PICLISTspam_OUTMITVMA.MIT.EDU>
Sent: Wednesday, April 03, 2002 5:19 PM
Subject: Re: [EE]: Automotive 5V Regulators Was: 7805 heating up...


{Quote hidden}

> Dropping the input voltage to a linear regulator for automotive
> applications is a bad idea. Why?
>
> The nominal battery voltage may be 12V, but it will dip as low as
> 8-9V when cranking the engine and as high as 14V+ when charging
> on a lightly loaded electrical system. There will also be some
> variation on the top end due to alternator voltage regulator
> variations.
>
> The two good choices for bulletproof designs are:
>
> 1. A low dropout linear regulator like National's LM29XX series.
> Specifically made for automotive applications and will maintain
> regulation even during low voltage (engine cranking) situations.
> Available in TO-200 packages (as well as SMD and TO-92
> varieties), heatsink accordingly.
>
> 2. A DC-DC converter (for reasons discussed previously). Lowers
> your power losses and works well over the entire voltage range.
> Can also eliminate electrical noise on output which is present on
> the 12V input due to the PWM/output filter. Lots of alternatives
> here depending on your output current needs, cost, packaging,
> etc.
>
> I've been down this road a lot of times and always end up at one
> of the two mentioned destinations...
>
> Matt Pobursky
> Maximum Performance Systems
>
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'[EE]: Swithing Voltage Regulators (cheap)'
2002\04\12@082056 by Edson Brusque

2002\04\12@175114 by 859-1?Q?Alexandre_Guimar=E3es?=

2002\04\12@211130 by Edson Brusque

2002\04\12@230600 by Russell McMahon

2002\04\12@233125 by 859-1?Q?Alexandre_Guimar=E3es?=

'[EE]: Cheap Swithing Voltage Regulators x SwitchC'
2002\04\12@235234 by 859-1?Q?Alexandre_Guimar=E3es?=

2002\04\13@082444 by Peter L. Peres

'[EE]: Swithing Voltage Regulators (cheap)'
2002\04\13@104357 by miked

'[EE]: Cheap Swithing Voltage Regulators x SwitchC'
2002\04\13@113451 by 859-1?Q?Alexandre_Guimar=E3es?=

2002\04\13@120103 by Sean H. Breheny

2002\04\14@145423 by Peter L. Peres

2002\04\14@145437 by Peter L. Peres

2002\04\14@145441 by Peter L. Peres

2002\04\14@151020 by Sean H. Breheny

2002\04\16@013919 by Peter L. Peres

>Hi Peter,
>
>I'm sure you are right in general, but I looked at the spice model for
>the 2N2222, for example, in SwitchCad, and I couldn't see any noise
>parameters. Perhaps I missed them.
>
>Sean

Hi Sean,

here is some help from Berkeley Spice 3f5 online help pages (note: NF is
the noise factor; note2: sitchcad3 seems to have Berkeley spice origins
from what I can tell; note3: If a parameter is not given you can add it
yourself (edit the relevant library file with an editor);  note4: the
table below is wrapped. The asterisks and other unexpected scribblings in
some lines on column 0 belong to the previous line):

-- quote start:

3.4.4.  BJT Models (NPN/PNP)


     The bipolar junction transistor model in  SPICE  is  an
adaptation  of  the  integral charge control model of Gummel
and Poon.  This modified Gummel-Poon model extends the  ori-
ginal  model to include several effects at high bias levels.
The model automatically simplifies to the simpler Ebers-Moll
model when certain parameters are not specified.  The param-
eter names used in the modified Gummel-Poon model have  been
chosen to be more easily understood by the program user, and
to reflect better both physical and circuit design thinking.


     The dc model is defined by the parameters IS,  BF,  NF,
ISE,  IKF,  and  NE which determine the forward current gain
characteristics, IS, BR, NR, ISC, IKR, and NC  which  deter-
mine  the  reverse current gain characteristics, and VAF and
VAR which determine the output conductance for  forward  and
reverse regions.  Three ohmic resistances RB, RC, and RE are
included, where RB can  be  high  current  dependent.   Base
charge  storage  is  modeled  by forward and reverse transit
times, TF and TR, the forward transit  time  TF  being  bias
dependent  if desired, and nonlinear depletion layer capaci-
tances which are determined by CJE, VJE, and MJE for the B-E
junction  ,  CJC, VJC, and MJC for the B-C junction and CJS,
VJS, and MJS for  the  C-S  (Collector-Substrate)  junction.
The temperature dependence of the saturation current, IS, is
determined by the energy-gap, EG, and the saturation current
temperature  exponent,  XTI.  Additionally base current tem-
perature dependence  is  modeled  by  the  beta  temperature
exponent  XTB  in  the  new model.  The values specified are
assumed to have been measured at the temperature TNOM, which
can  be specified on the .OPTIONS control line or overridden
by a specification on the .MODEL line.

     The  BJT parameters used in  the  modified  Gummel-Poon
model are listed below.  The parameter names used in earlier
versions of SPICE2 are still accepted.



     name   parameter                               units   default
example
 area

1    IS     transport saturation current            A       1.0e-16
1.0e-15
 *
2    BF     ideal maximum forward beta              -       100
100
3    NF     forward current emission coefficient    -       1.0        1
4    VAF    forward Early voltage                   V       infinite
200
5    IKF    corner for forward beta
            high current roll-off                   A       infinite
0.01
 *
6    ISE    B-E leakage saturation current          A       0
1.0e-13
 *
7    NE     B-E leakage emission coefficient        -       1.5        2
8    BR     ideal maximum reverse beta              -       1
0.1
9    NR     reverse current emission coefficient    -       1          1
10   VAR    reverse Early voltage                   V       infinite
200
11   IKR    corner for reverse beta
            high current roll-off                   A       infinite
0.01

12   ISC    B-C leakage saturation current          A       0
1.0e-13
 *
13   NC     B-C leakage emission coefficient        -       2
1.5
14   RB     zero bias base resistance               Z       0
100
 *
15   IRB    current where base resistance
            falls halfway to its min value          A       infinite
0.1
 *
16   RBM    minimum base resistance
            at high currents                        Z       RB         10
 *
17   RE     emitter resistance                      Z       0          1
 *
18   RC     collector resistance                    Z       0          10
 *
19   CJE    B-E zero-bias depletion capacitance     F       0
2pF
 *
20   VJE    B-E built-in potential                  V       0.75
0.6
21   MJE    B-E junction exponential factor         -       0.33
0.33
22   TF     ideal forward transit time              sec     0
0.1ns
23   XTF    coefficient for bias dependence of TF   -       0
24   VTF    voltage describing VBC

12   ISC    B-C leakage saturation current          A       0
1.0e-13
 *
13   NC     B-C leakage emission coefficient        -       2
1.5
14   RB     zero bias base resistance               Z       0
100
 *
15   IRB    current where base resistance
            falls halfway to its min value          A       infinite
0.1
 *
16   RBM    minimum base resistance
            at high currents                        Z       RB         10
 *
17   RE     emitter resistance                      Z       0          1
 *
18   RC     collector resistance                    Z       0          10
 *
19   CJE    B-E zero-bias depletion capacitance     F       0
2pF
 *
20   VJE    B-E built-in potential                  V       0.75
0.6
21   MJE    B-E junction exponential factor         -       0.33
0.33
22   TF     ideal forward transit time              sec     0
0.1ns
23   XTF    coefficient for bias dependence of TF   -       0
24   VTF    voltage describing VBC

            dependence of TF                        V       infinite
25   ITF    high-current parameter
            for effect on TF                        A       0
 *
26   PTF    excess phase at freq=1.0/(TF*2PI) Hz    deg     0
27   CJC    B-C zero-bias depletion capacitance     F       0
2pF
 *
28   VJC    B-C built-in potential                  V       0.75
0.5
29   MJC    B-C junction exponential factor         -       0.33
0.5
30   XCJC   fraction of B-C depletion capacitance   -       1
            connected to internal base node
31   TR     ideal reverse transit time              sec     0
10ns
32   CJS    zero-bias collector-substrate
            capacitance                             F       0
2pF
 *
33   VJS    substrate junction built-in potential   V       0.75
34   MJS    substrate junction exponential factor   -       0
0.5
35   XTB    forward and reverse beta
            temperature exponent                    -       0
36   EG     energy gap for temperature
            effect on IS                            eV      1.11
37   XTI    temperature exponent for effect on IS   -       3
31   TR     ideal reverse transit time              sec     0
10ns
32   CJS    zero-bias collector-substrate
            capacitance                             F       0
2pF
 *
33   VJS    substrate junction built-in potential   V       0.75
34   MJS    substrate junction exponential factor   -       0
0.5
35   XTB    forward and reverse beta
            temperature exponent                    -       0
36   EG     energy gap for temperature
            effect on IS                            eV      1.11
37   XTI    temperature exponent for effect on IS   -       3
       -- hit return for more, ? for help --

38   KF     flicker-noise coefficient               -       0
39   AF     flicker-noise exponent                  -       1
40   FC     coefficient for forward-bias
            depletion capacitance formula           -       0.5
                                                    o
41   TNOM   Parameter measurement temperature        C      27         50

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2002\04\18@145552 by Peter L. Peres

I am new to electronics and have some questions about the 7800 series
voltage regulators. What happens when the supply voltage goes below the
regulated output voltage? Does the regulator stop giving an output or does
it just output the supply voltage? Any help would be greatly appreciated.

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2002\06\29@143144 by Olin Lathrop

> I am new to electronics and have some questions about the 7800 series
> voltage regulators. What happens when the supply voltage goes below the
> regulated output voltage?

It starts a runaway nuclear reaction that obliterates the regulator, the
circuit board, and everything else within 20Km in a large fireball.  This is
generally considered "bad".

That failing, the output voltage drops below spec.

> Does the regulator stop giving an output or does
> it just output the supply voltage?

The regulator will continue trying to regulate.  In normal operation it
drives its pass element as hard as it has to in order to maintain the
desired output voltage.  When that output voltage goes low, the pass element
is turned on harder.  If the input voltage is below the output voltage set
point, it obviously can't maintain the desired input voltage, so that pass
element will be driven as hard as it can manage.  However, there will always
be some voltage drop accross the regulator, so the output voltage will be a
bit less than the input voltage.

Note also that the input voltage must be kept at or above the output
setpoint plus some margin for the regulator to work properly.  This margin
is often called the "dropout voltage", which is a rather high 2V for the
LM7805.  Therefore, the input voltage for a LM7805 must be at least 7V for
it to maintain 5V output over the full current range.


*****************************************************************
Embed Inc, embedded system specialists in Littleton Massachusetts
(978) 742-9014, http://www.embedinc.com

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2002\06\29@152630 by Dale Botkin

On Sat, 29 Jun 2002, Olin Lathrop wrote:

> It starts a runaway nuclear reaction that obliterates the regulator, the
> circuit board, and everything else within 20Km in a large fireball.  This is
> generally considered "bad".
>
> That failing, the output voltage drops below spec.

Could I be seeing things?  I'd swear Olin just made a joke...  8-)

Dale

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2002\06\29@180156 by Joe Farr

2002\06\30@054438 by Roman Black

Richard Mellina wrote:
>
> I am new to electronics and have some questions about the 7800 series
> voltage regulators. What happens when the supply voltage goes below the
> regulated output voltage? Does the regulator stop giving an output or does
> it just output the supply voltage? Any help would be greatly appreciated.

Some of the low-dropout 5v regulators i've tried
are ok with this. At least under 100mA or so.
When the supply volts are less than 5v the output
is fixed at the supply volts minus about 50mV.
-Roman

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2002\06\30@130458 by Shawn Mulligan

Since your're new to electronics it might be fun to answer this question
experimentally. You can build a simple adjustable power supply and test the
behaviour of the 7800 series regulators directly.

The power supply can be built using the LM317 adjustable voltage regulator
and 4 external components. The actual circuit is available in the datasheet
(and you should start becoming familiar with datasheets) or through an
online search (try Google, for example).

Connect the power supply to the 7800 to be tested and the 7800 to the load
to be driven. If you don't have the actual load/device built yet, mock it
with a resistor or similiar device drawing the expected amount of current.
This is important because the voltage regulator may 'drop out' or loose
regulation at different points depending on current draw.

Then you're ready to experiment. Adjust the input voltage while measuring
the output voltage and observe the behaviour of the regulator when it
approaches the critical point where Vin < Vout.

Just as an aside, you don't want Vin to be too high either (e.g. 12V into a
7805, supplying 5V) because voltage regulators are very inefficient and
you'll simply waste power, which will be apparent by the heating of the
voltage regulator.

Hope this was helpful.

Shawn


>I am new to electronics and have some questions about the 7800 series
>voltage regulators. What happens when the supply voltage goes below the
>regulated output voltage? Does the regulator stop giving an output or does
>it just output the supply voltage? Any help would be greatly appreciated.


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'[OT]: Voltage Regulators and hair loss'
2002\06\30@133921 by Shawn Mulligan

You are losing all of your hair simply because of evolutionary improvements
on the species homo sapiens.

Early humans had sloping postures, underdeveloped intellect and hairy
frames. Modern man has risen above his ape-like ancestors as a result or
several tiny improvements/changes in the genetic code. These changes
represent evolutionary advances whereby the strong survive and the weak
perish.

These microscopic changes are hidden within human DNA; however, we can
clearly witness their manifestations on the human being. The frame has
straighened. Intellectual abilities have increased. And lastly, the once
protective coating of hair, no longer necessary, is disappearing. First on
the hands and feet, then buttocks, the legs, arms, chest and finally the
head. Indeed baldness is the highest form of evolution. You are not losing
your hair because of stress or age. You are simply ahead of your time.

Also ahead of his time, Shawn



>Seemed like a good reason why I was loosing all  my hair. Maybe it's old
>age after all - doh !
>


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On Sat, 29 Jun 2002 11:14:41 -0500 Richard Mellina <RemoveMErsillemEraseMEspam_OUTFLASH.NET>
writes:
> I am new to electronics and have some questions about the 7800 series
> voltage regulators. What happens when the supply voltage goes below
> the
> regulated output voltage? Does the regulator stop giving an output
> or does
> it just output the supply voltage? Any help would be greatly
> appreciated.

       Linear regulators have a "drop out voltage." For the 7800 series, it's
about 2V. If the input voltage is less than the rated output voltage +
2V, the output falls out of regulation and the output voltage starts to
decrease. Once out of regulation, the output voltage is about Vin - the
dropout voltage. If you are using a line powered supply, AC line ripple
shows up in the regulated output when the "valley" of the incoming DC +
ripple falls below the Vout+Vdo . I typically look at the regulator Vin
on a DC coupled scope to make sure the bottom of the ripple is still
above the input dropout voltage under low line and high load. You can
also by low drop out regulators ("LDO's") that reduce this dropout
voltage. These are good for battery operated products since the equipment
can operate longer before the battery voltage is low enough for the
regulator to drop out.


Harold


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'[EE] Alternator Regulators.'
2002\07\13@213721 by John Dammeyer

2002\07\13@224119 by Roman Black

2002\07\13@234949 by John Dammeyer

2002\07\14@134000 by Peter L. Peres

The phase control scheme may work but imho do not allow the alternator to
free run, provide a resistor to limit the voltage at all times. 200V may
be too much for the insulation of that device. Also, the frequency could
be 400Hz-ish or more so you need a SCR scheme that works at that voltage.
Me I'd rectify and filter the voltage directly and use a buck dc/dc
converter to regulate rail voltage and a second with current and voltage
settings for charging, with switchover through FETs used as switches. I
suppose that 25A is @24Vac or such, not at 200.

Peter

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2002\07\14@153820 by Peter L. Peres

On Sun, 14 Jul 2002, Roman Black wrote:

>Hi John, I remember when you posted about this before.
>My suggestion is to go to a motorcycle wrecker and find
>a shunt regulator from pretty much any small modern motorcycle
>that has a single phase alternator. They are mostly perm
>magnet, the bigger bikes use 3 phase, the smaller bikes
>generally single phase.
>
>There are only a couple of wires to connect, usually very
>straightforward. Cheap, quick and should be reliable. :o)

Always assuming that there are small bikes with 25A alternators. Else it
will be a shunt *ignition* regulator.

Peter

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2002\07\14@174332 by Mike Singer

Peter L. Peres wrote:
.
.
> Me I'd rectify and filter the voltage directly and use a buck dc/dc
> converter to regulate rail voltage and a second with current
> and voltage settings for charging, with switchover through FETs
> used as switches. I suppose that 25A is @24Vac or such, not
> at 200.
.
.
  I think, it's a good idea of using single regulated DC/DC converter. It should produce 13 to 15V DC.    I doubt about alternator's 200VAC unloaded. It is diesel sailboat, not jet fighter. In order to be used effectively alternator should produce somewhat 14AC at min rpm. For a sailboat diesel max/min rpm is aprox 3, I think.    So 14-40V to 13-15V regulated switching DC/DC converters aren't rare things on a market. After all, for what heck do this PICs have PWM?
  Mike.

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2002\07\15@011055 by John Dammeyer

2002\07\15@035720 by Roman Black

John, you design should avoid the chance of the
alternator running open-circuit for any periods
of time, this is the main cause of alternator
insulation breakdown, and will kill your alternator.
The SCR type regulators are designed to keep the
alternator volts down, by loading the thing into the
battery/load or shorted through the SCRs.

If you are going to use a PIC to control SCR firing
it might be prudent to add a hardware safeguard
circuit, where a cap is charged through the cycle,
and if reaches a setpoint automatically turn on the
SCR. So if you get an open-circuit load, (blown fuse,
PIC glitch etc) the alternator has a minimum load.
Probably the last 1/3 of the cycle (60') should be
enough as your hardware safeguard.
-Roman

John Dammeyer wrote:
{Quote hidden}

>
> I pulled the 200VAC from the BMW D13 manual where they say the two phase
> alternator produces as much as 200VAC open circuit.  These little
> Shibaura alternators may be far less.  I'll connect up a 1/3HP electric
> motor to the alternator and run it at 1725RPM just to see what comes
> out.
>
> I'm leaning toward a two SCR plus two regular diode bridge setup where
> the SCRs are triggered with PWM from a PIC.  I'll start with a 6.3VAC
> transformer to test out the circuit and then migrate up to 36VAC just to
> ensure that things are working correctly.  I will post the circuit
> design when the testing is done.

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2002\07\15@041024 by Roman Black

Peter L. Peres wrote:
>
> On Sun, 14 Jul 2002, Roman Black wrote:
>
> >Hi John, I remember when you posted about this before.
> >My suggestion is to go to a motorcycle wrecker and find
> >a shunt regulator from pretty much any small modern motorcycle
> >that has a single phase alternator. They are mostly perm
> >magnet, the bigger bikes use 3 phase, the smaller bikes
> >generally single phase.
> >
> >There are only a couple of wires to connect, usually very
> >straightforward. Cheap, quick and should be reliable. :o)
>
> Always assuming that there are small bikes with 25A alternators. Else it
> will be a shunt *ignition* regulator.


Good point, the larger bikes are 25A+, not sure about
the smaller ones. Typical running current on a small bike
is well over 15A with lights and ignition. Most bikes now
are required/designed to have lights on at all times.
-Roman

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2002\07\15@042303 by Alan B. Pearce

>I pulled the 200VAC from the BMW D13 manual where they say
>the two phase alternator produces as much as 200VAC open
>circuit.  These little Shibaura alternators may be far less.

Don't count on it. Did you manage to run your original alternator open
circuit at some stage? This may well be why it has died. I believe that many
of the original alternators on motor vehicles would die if run without a
load because the rectifier diodes used did not have the Reverse Voltage
rating to withstand open circuit use. More modern alternators may well have
moved on from this situation now high current/high voltage diodes are more
common than they were then.

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2002\07\15@090808 by John Dammeyer

>
> >I pulled the 200VAC from the BMW D13 manual where they say
> >the two phase alternator produces as much as 200VAC open
> >circuit.  These little Shibaura alternators may be far less.
>
> Don't count on it. Did you manage to run your original alternator open
> circuit at some stage? This may well be why it has died.


The manual states the procedure on how to test the output of the
alternator and that's what I did.  At maximum engine RPM the alternator
produced about 3VAC.

It's possible that at some point I may have inadvertently shut off the
Master Battery switch while the engine was still running.  This would
have disconnected the battery from the alternator output and blown the
regulator which in turn took out the windings.   It's a common problem
on boats with master switches and Diesels that don't care if they have
electricity for operation.  This "Load Dump" can be prevented by putting
a hefty transorb on the output of the alternator regulator to keep the
voltage within some safe point.  Given that this diesel is from the mid
70's the chance of it surviving a Load Dump is pretty unlikely.  So more
than likely I shot myself in the foot.

That's one reason why I would like to have a system that has high
voltage diodes and regulator transistors to be capable of with standing
the Load Dump voltages.

Sigh....


John

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2002\07\15@134459 by Peter L. Peres

> load dumps

I think that you need to look at some surplus place for long line MOVs.
They are used for telephone lines and such. I have never used one but I
know you can get them in coffee mug size with stud terminals. That should
stop a load dump at least for a while. They are meant to prevent excessive
voltage from appearing at expensive equipment in phone switches I think.

Peter

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2002\07\15@173846 by Mike Singer

John,
test the alternator at that RPM, it is intended for. If you have 6v 3000 RPM motorcycle alternator then it is OK.
Find 12v alternator of low RPM engines or change if possible transmission ratio from engine to the alternator to increase RPMs.

  Cheer up. Mike.

John Dammeyer wrote:
{Quote hidden}

> The manual states the procedure on how to test the output of the
> alternator and that's what I did.  At maximum engine RPM the
> alternator
> produced about 3VAC.
>
> It's possible that at some point I may have inadvertently shut off the
> Master Battery switch while the engine was still running.  This would
> have disconnected the battery from the alternator output and blown the
> regulator which in turn took out the windings.   It's a common problem
> on boats with master switches and Diesels that don't care if they have
> electricity for operation.  This "Load Dump" can be prevented
> by putting
> a hefty transorb on the output of the alternator regulator to keep the
> voltage within some safe point.  Given that this diesel is
> from the mid
> 70's the chance of it surviving a Load Dump is pretty
> unlikely.  So more
> than likely I shot myself in the foot.
>
> That's one reason why I would like to have a system that has high
> voltage diodes and regulator transistors to be capable of
> with standing
> the Load Dump voltages.
>
> Sigh....

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2002\07\15@190532 by John Dammeyer

2002\09\24@140546 by Mike Singer

part 1 240 bytes content-type:text/plain; (decoded 7bit)

William Chops Westfield wrote:
> ...I'd also like to be able to get
> 5V from a 3V (or even 1.5V) battery...


Microchip TC125/126.

_Attached .gif._ from 21372b.pdf

Start up at 0.9v.

Mike


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part 2 5417 bytes content-type:image/gif; (decode)

---

part 3 154 bytes
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2003\03\30@225703 by Jai Dhar

Why not order some samples from either National Semi or Maxim? National's LM2679 (I think that's it) is a nice 5V regulator, or MAX787 from Maxim. Both of them send out free samples.

Quoting "Ricardo D. Medina" <rickymEraseME@spam@MONTEVIDEO.COM.UY>:

{Quote hidden}

> Hi List,
>
> I have to power a network Hub that requires 5v DC at 1A, with up to 2-3A
> peak currents. I was wondering, is there some way to parallel 7805
> regulators to fit this task? (The problem is that I live in Uruguay, an LM
> 350 costs U$S 7, and I can't just mail order them from Digikey beacuse of
> new Customs laws.)
>
> Thank you very much for your help,
>
> Ricardo Medina
>
> PD: I´m sorry if this isn't Quoted Printable format, but I'm stuck with
> Outlook Express. I selected Quoted printable, but I'm not too sure it will
> work....
>
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2003\03\30@230743 by Des Bromilow

Why not do that trick with the transistor, and the base voltage set point.

The trick is based on knowing the Vbe is 0.7V and you set a point at the base to give you the regulated voltage at the output of the transistor.
I remeber a univeristy lecturer showing it to me to boost the output of a zener.. you had to use a 5.7VDC zener to get 5VDC output. YOU could do similar with a single 7805 and offset the common by a single power diode (to get the extra 0.7VDC and then go from there. The power capacity of the circuit is based on the power rating of the transistor.

Des

>>> RemoveMEjdharspamBeGoneENGMAIL.UWATERLOO.CA 31/03/03 1:55:42 pm >>>
Why not order some samples from either National Semi or Maxim? National's LM2679 (I think that's it) is a nice 5V regulator, or MAX787 from Maxim. Both of them send out free samples.

Quoting "Ricardo D. Medina" <spamBeGonerickymKILLspam@spam@MONTEVIDEO.COM.UY>:

{Quote hidden}

> Hi List,
>
> I have to power a network Hub that requires 5v DC at 1A, with up to 2-3A
> peak currents. I was wondering, is there some way to parallel 7805
> regulators to fit this task? (The problem is that I live in Uruguay, an LM
> 350 costs U$S 7, and I can't just mail order them from Digikey beacuse of
> new Customs laws.)
>
> Thank you very much for your help,
>
> Ricardo Medina
>
> PD: I m sorry if this isn't Quoted Printable format, but I'm stuck with
> Outlook Express. I selected Quoted printable, but I'm not too sure it will
> work....
>
> --
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>

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2003\03\30@233645 by Marc Joffe

2003\03\31@011047 by Daniel Dourneau

At 00:34 31/03/03 -0300, you wrote:
>Hi List,
>
>I have to power a network Hub that requires 5v DC at 1A, with up to 2-3A
>peak currents. I was wondering, is there some way to parallel 7805
>regulators to fit this task? (The problem is that I live in Uruguay, an LM
>350 costs U$S 7, and I can't just mail order them from Digikey beacuse of
>new Customs laws.)

Please check:
http://www.national.com/an/AN/AN-103.pdf
It is a LM340 with a PNP transistor for boosting the output current.
It should be sufficient for your requirements.

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2003\03\31@015416 by Russell McMahon

__________________
I have to power a network Hub that requires 5v DC at 1A, with up to 2-3A
peak currents. I was wondering, is there some way to parallel 7805
regulators to fit this task? (The problem is that I live in Uruguay, an LM
350 costs U$S 7, and I can't just mail order them from Digikey beacuse of
new Customs laws.)
__________________


The transistor method that other people have suggested works well enough BUT
you lose the current limiting and thermal protection of the 7805. You can
retain over current protection and thermal protection if you are happy to
accept somewhat worse regulation.

You can parallel 7805's by placing SMALL resistors in the OUTPUTS with a
smoothing capacitor both at the regulator outputs AND at the final combined
output. The resistors will make the output slightly lower and reduce the
regulation achieved. They should drop say about 0.1 volt at fill current so
R = 0.1 V/ 1 Amp = 0.1 ohm. Power dissipated is only 0.1 watt so almost any
0.1 ohm resistor would do. This scheme works by reducing the load on
regulators which supply higher currents thereby balancing the overall share
of the load to each. if the drop is too low the highest output regulator
will not have its output reduced enough to fall below that of the lowest
output regulator. test selecting regulators to have as close an output
voltage as possible will allow the lowest possible resistors to be used.
((Note that placing resistors in the INPUTS to the regulators does NOT
produce current sharing)). With the use of 0.1R load resistors the output
voltage will droop about 0.1v more from no load to full load as it would
without the resistors.

You can parallel 7805s directly without any resistors at all with some
success. The higher output voltage regulators will provide more current than
the lower output ones until they reach their load limit. at that stage they
will start to shut down and reduce their output voltages somewhat allowing
the others to take up excess load. This means that some will work harder
than others but this is probably acceptable in practice.

By the way, quoted printable is what you should AVOID if possible :-).
Perfectly readable for me but makes quoting your message more difficult.

.


       Russell McMahon

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2003\03\31@083826 by Olin Lathrop

> I have to power a network Hub that requires 5v DC at 1A, with up to 2-3A
> peak currents. I was wondering, is there some way to parallel 7805
> regulators to fit this task?

Do not parallel them directly.  Each regulator has its own idea of what 5V
is, so they will not share the current evenly.  If your circuit can
tolerate a few 100mV droop at peak current, then you can put a small
resistor on the output of each regulator.  200mohm should be enough to
balance the load reasonalby well.

A better answer is to build one 3A regulator, or better yet, use a
switching regulator.


*****************************************************************
Embed Inc, embedded system specialists in Littleton Massachusetts
(978) 742-9014, http://www.embedinc.com

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2003\03\31@212526 by Ricardo D. Medina

Hi again,

I'm going to try Russell's and Olin's method, as it's cheaper, and quicker
for me to implement.

snip
+AD4- You can parallel 7805's by placing SMALL resistors in the OUTPUTS with a
+AD4- smoothing capacitor both at the regulator outputs AND at the final
combined
+AD4- output. The resistors will make the output slightly lower and reduce the
+AD4- regulation achieved.
snip

Thanks to everyone who contributed, and to all of you who make this list so
enjoyable.

Ricardo Medina

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Another common practice, as long as your normal load is constant and the
source is somewhat stable, is to put a resistor across the input and
output to take some of the primary task off the 7805. The peak loads
will be handled by the regulator. This is only effective if you run the
numbers. Calculate your resistor value by taking your average
unregulated voltage and average load current. Increase the resistor
value somewhere between 10 and 30%. (This may vary and depend upon
unregulated voltage and load current.)

Ex. Unreg ~13v,  constant load ~1A. Voltage across unknown resistor,
E=8  R=E/I  =  8/1 = 8 ohms. Try a resistor between 15 and 10 ohms
starting at the higher value resistor. Power rating for resistor: P=E/I
= 8 watts. A 15 to 20 watt resistor will still get warm. Your regulator
should be able to handle the peak currents for short periods of time.
Regulation will suffer just a little.

Paralleling 7805's as suggested below will work too, but you will still
be dissipating the same amount of heat.
Rick
>
>__________________
I have to power a network Hub that requires 5v DC at 1A, with up to
2-3A
peak currents. I was wondering, is there some way to parallel 7805
regulators to fit this task? (The problem is that I live in Uruguay, an
LM
350 costs U$S 7, and I can't just mail order them from Digikey because
of
new Customs laws.)
__________________


The transistor method that other people have suggested works well enough
BUT
you lose the current limiting and thermal protection of the 7805. You
can
retain over current protection and thermal protection if you are happy
to
accept somewhat worse regulation.

You can parallel 7805's by placing SMALL resistors in the OUTPUTS with a

smoothing capacitor both at the regulator outputs AND at the final
combined
output. The resistors will make the output slightly lower and reduce the

regulation achieved. They should drop say about 0.1 volt at fill current
so
R = 0.1 V/ 1 Amp = 0.1 ohm.

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2003\04\01@081437 by Marcelo Puhl

On 1 Apr 2003 at 7:54, you wrote:

>  I have to power a network Hub that requires 5v DC at 1A, with up to 2-3A peak
> currents. I was wondering, is there some way to parallel 7805 regulators to
> fit this task? (The problem is that I live in Uruguay, an LM 350 costs U$S 7,
> and I can't just mail order them from Digikey because of new Customs laws.)

I would suggest the good old MC34063 Switching regulator.
It is not too difficult to find, and it is cheap.

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2003\04\02@192253 by Brendan Moran

Hi all,

Can anyone tell me whether 78xx series regulators are stable with negative
resistance loads (i.e., where the current decreases as voltage increases,
even though the sign of the voltage and current is the same)? If so, what
are the constraints on the load impedance (i.e., how negative can it be?
How much capacitive load in addition to the negative resistance?).

I'm asking because I have a circuit where I was planning on using a 7824 to
feed the input of a switching power supply. I need to be able to power a
device which expects an input voltage between 20 and 32 volts from a
battery supply that can vary between 12 and 18 volts. The device is
somewhat sensitive to noise on the input (so says the manufacturer,
although it seems doubtful given that it has a switching supply internally)
so I was planning on using one of national's "simple switcher" ICs to step
up the battery voltage to 28v, and then regulate it down to a clean 24
using the 7824.

Thanks,

Sean

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Hello,

I might be disturbing you with a very simple question, but I ran out of
solutions for this problem.
I have designed a power supply which gives 13.8V, with the adjustable
LM338, and tied to the output, the input of an LM7805, in order to
obtain 5V also (I need both 13.8V and 5V). Is it ok to do this ? Because
when I power the board, the 7805 gets very hot quickly and I had to pull
it off (there is no power consumption as I haven't planted all the IC's
yet).
I am wondering what could be the cause of this, as all the connections
are ok.
Please give me a suggestion. Thank you !

Lucian

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2004\05\14@084718 by Aadu Adok

> this ? Because when I power the board, the 7805 gets very hot
> quickly and I had to pull it off (there is no power
> consumption as I haven't planted all the IC's yet). I am
it has to drop 13.8 - 5 = 8.8 volts.
if you IC-s are consuming, let's say 200mA (only you know the exact
value) then the power dissipation on 7805 would be 8.8 * 0.2 = 1.76
watts. I guess it will get quite hot without heatsink.

aadu

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2004\05\14@084924 by Jinx

> Is it ok to do this ?

Yes

> Because when I power the board, the 7805 gets very hot quickly
> and I had to pull it off (there is no power consumption as I haven't
> planted all the IC's yet)

Something is wrong. There is power consumption, possibly a short
or pins swapped. Have you double-checked the 7805 pin-out and
then checked it again ? How about the voltage out from the 7805 ?
Sure there's less than 30V on the input to the 7805 ? Looked for
solder bridges or a problem with the wiringtracks ? It'll be something
simple

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2004\05\14@085959 by Lucian

2004\05\14@090005 by Lucian

2004\05\14@091424 by Jinx

> The pinout of the 7805 is IN-GND-OUT, when looking from its
> front, isn't it ? (at least that's what I've found on the datasheets

Correct. Maybe you've a bad 7805 (possible) or capacitor (unlikely).
Do you have spares to substitute to find out ?

You haven't mistakenly used a 7905 have you ? (just a long shot)

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2004\05\14@092042 by Hazelwood Lyle

>-----Original Message-----
>From: Jinx [spam_OUTjoecolquittspam_OUTspam_OUTCLEAR.NET.NZ]
>Sent: Friday, May 14, 2004 9:15 AM
>To: PICLISTspam_OUTMITVMA.MIT.EDU
>Subject: Re: [EE:] Cascading Regulators
>
>
>> The pinout of the 7805 is IN-GND-OUT, when looking from its
>> front, isn't it ? (at least that's what I've found on the datasheets
>
>Correct. Maybe you've a bad 7805 (possible) or capacitor (unlikely).
>Do you have spares to substitute to find out ?
>
>You haven't mistakenly used a 7905 have you ? (just a long shot)

Since this is a machine-made board, is this a tested design? Is there
a short from Vcc to GND somewhere?
Oh, and regarding that "Long Shot", I've made that mistake recently
myself. I now keep my regulators sorted into separate parts bins to
reduce the chances of doing that again.

Lyle

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2004\05\14@092249 by Lucian

2004\05\14@093050 by Jinx

> Thank you for taking the time to reply to my stupid question :)

No worries. The problem's always in the last place you look eh ?

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2004\05\14@094742 by Russell McMahon

> I have a heatsink to be mounted, but for the moment, there should be no
> power consumption, so I thought that the power to be dissipated is
> small.

Make sure that the 7805 is not oscillating at a very high frequency. When
this happens, as it can, many strange things can happen. A DC meter is not a
good enough check if this is happening - the mean DC level is probably still
5v. An oscilloscope is the best means of testing this.

Also check the datasheet for the 7805 and be sure to add input and output
capacitors of the values it says and WHERE it says. In problem cases, having
short lead lengths between capacitor and regulator may matter. On some newer
regulators the capacitors must not have an ESR (equivalent series
resistance) that is either too high OT too low. AFAIR the 7805 does not
specify this.

As a rule a large capacitor at input and output should settle it down if
stability is questionable. Normally a very large output cap should not be
needed and can cause damage at shut down under certain circumstances due to
back discharge through the regulator, unless suitable protection diodes are
used.

Who said 3 terminal regulators were simple ??? ;-)
(They are actually, compared to the alternatives, but there are still things
to be known).


       Russell McMahon

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2004\05\14@100657 by Herbert Graf

> Hello,
>
> I might be disturbing you with a very simple question, but I ran out of
> solutions for this problem.
> I have designed a power supply which gives 13.8V, with the adjustable
> LM338, and tied to the output, the input of an LM7805, in order to
> obtain 5V also (I need both 13.8V and 5V). Is it ok to do this ? Because
> when I power the board, the 7805 gets very hot quickly and I had to pull
> it off (there is no power consumption as I haven't planted all the IC's
> yet).
> I am wondering what could be the cause of this, as all the connections
> are ok.
> Please give me a suggestion. Thank you !

       Three things I can think of off hand:

       1. You have something wired wrong. Triple check (or get someone else check)
your wiring. If you've got no load you shouldn't have a noticeable heating
of those components).

       2. A rare one, but one I have seen: did you follow the datasheet on input
and output caps for both regulators? I've seen a case where if you don't put
the output cap on a 7805 it can sometimes start oscillating, drawing a huge
amount of current.

       3. You have a short somewhere, relates to number one, but since wiring is
the likely cause of the problem it's good to make this one a separate point!

       It could be a bad component, but I doubt that very much.

       On top of all this, how much current do you plan to draw on the 5V rail?
You're 7805 will very likely need a heatsink unless it's supplying a very
small current. TTYL

----------------------------------
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http://repatch.dyndns.org:8383/pic_stuff/

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2004\05\14@120021 by Lucian

2004\05\14@120813 by Morgan Olsson

Russell McMahon 15:46 2004-05-14:
>capacitors must not have an ESR (equivalent series
>resistance) that is either too high OT too low.

The input cap: *always* the lower ESR the better

> AFAIR the 7805 does not specify this.

Varies between manufacturers.
I have never seen anybody specify more than 220nF input cap.  This must be foil or ceramic, electrolytes have too high internal series resistance (ESR).

Very important is that the total track lengths from the poles of this cap to the regulator is short, not more than 2cm.

I usually use 100nF foil close, and a electrolytic at board power input.  (as to keep it away from the hot regulator)  Where board area is small and i do not need large capacitance i have used one 10uF tantalum instead.

I once used a "professional" to make my layout.  The design was dead, and at analysis i saw she placed the output cap ten cm away from the 78L08, and GND and plus was routed 7cm apart, making a loop inductor, thus creating a nice LC cirquit.  It really was a stable OSCILLATOR!

/Morgan
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2004\05\15@004640 by Russell McMahon

> ICs. Anyway, th maximum current is around 500mA through the 7805.

(13.8 - 5 ) x 500 mA =~4.5w
Not vast but heatsink accordingly.

       RM

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2004\05\15@022638 by Lucian

2004\05\15@041036 by Russell McMahon

> I put it next to a heatsink, but not tightly coupled, as it is isolated.

Whether this is OK depends on what you mean by "tightly coupled". As I noted
before - at 500 mA you will dissipate about 4.5 watts in the 7805. This is
FAR too much for no heatsink but does not require an especially large one.


       RM

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2004\05\15@044850 by Jinx

> Could the potentiometer be too bad ? As it jumps suddenly from
> 13.3 to 14.6+.

Sorry, I missed that last line when replying. R2 (the 5k pot) is meant
to provide a variable output. IOW you wouldn't expect a jump from
13.3 to 14.6. But as I mentioned in the other reply, 120R is stated
for the LM338 in the example. Maybe that has some bearing on what
you're seeing, but it could just as likely be a dirty pot. Try turning it
around (ie keep the wiper connection and use the other end terminal)
to see if that makes a difference. Unless you're dead unlucky and the
jump, if there is one, happens to be in exactly the same place ;-)

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2004\05\15@081044 by gtyler

You got caps on all inputs and outputs? could be oscillating.


----- Original Message -----
From: "Lucian" <KILLspamlucifferspamBeGoneHOME.RO>
To: <@spam@PICLISTSTOPspam@spam@MITVMA.MIT.EDU>
Sent: Saturday, May 15, 2004 8:27 AM
Subject: Re: [EE:] Cascading Regulators


> I put it next to a heatsink, but not tightly coupled, as it is
isolated.
> I hope it will not get too hot. My mistake in designing the circuit
and
> connecting the LM338 and LM7805 on the same heatsink.
>
> Now my problem is solved and the circuit is working nice, apart from
the
> 13.8V problem: I can't get 13.8V out of the LM338 regulator. The
adjust
> resistors are as follows: 220R between ADJ and VOUT and 2.2K plus a 5K
> potentiometer between ADJ and GND. Am I mistaking the values of the
> resistors ? If so, what would be some other better values ? (I have
smd
> resistors and I don't know how could I change them, but to know for
the
> future). Could the potentiometer be too bad ? As it jumps suddenly
from
> 13.3 to 14.6+.
>
> Lucian
> {Original Message removed}

2004\05\15@090802 by PicDude

Yes.  This is expected, unless you're drawing really low current.  Just add a decent heatsink.  The usual 7805 will protect itself (by shutting off) if it overheats.

Since you mention cascading, I have this situation frequently and instead of adding a (relatively) large heatsink, I use 2 regulators in "series" -- a 7810 to drop the 13.8V to 10V, then send that to the 7805 to drop it to 5V.  Cost is another 35c or so for the 7810 regulator.

Cheers,
-Neil.


On Friday 14 May 2004 07:39 am, Lucian scribbled:
{Quote hidden}

> Hello,
>
> I might be disturbing you with a very simple question, but I ran out of
> solutions for this problem.
> I have designed a power supply which gives 13.8V, with the adjustable
> LM338, and tied to the output, the input of an LM7805, in order to
> obtain 5V also (I need both 13.8V and 5V). Is it ok to do this ? Because
> when I power the board, the 7805 gets very hot quickly and I had to pull
> it off (there is no power consumption as I haven't planted all the IC's
> yet).
> I am wondering what could be the cause of this, as all the connections
> are ok.
> Please give me a suggestion. Thank you !
>
> Lucian

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2004\05\15@095359 by Russell McMahon

> Since you mention cascading, I have this situation frequently and instead
of
adding a (relatively) large heatsink, I use 2 regulators in "series" -- a
7810 to drop the 13.8V to 10V, then send that to the 7805 to drop it to 5V.
Cost is another 35c or so for the 7810 regulator.
>

A word of warning - while there may be some advantage in doing this, there
is no free lunch.

This practice reduces the dissipation in any one component but the overall
dissipation is the same.
If the regulators are on the same heatsink, the heatsink has to dissipate as
much energy as it would have had to if there had only been one regulator.


       Russell McMahon

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2004\05\15@221657 by PicDude

On Saturday 15 May 2004 08:53 am, Russell McMahon scribbled:
> A word of warning - while there may be some advantage in doing this, there
> is no free lunch.
>
> This practice reduces the dissipation in any one component but the overall
> dissipation is the same.
> If the regulators are on the same heatsink, the heatsink has to dissipate
> as much energy as it would have had to if there had only been one
> regulator.



Absolutely agreed/understood/known.  The intention here was to eliminate the need for a heatsink, or in the case of my breadboard, I do this to reduce the heat on each device to the point where I don't burn my fingers when I occassionally touch it.  Of course in other circumstances, a heatsink is needed still, and then doing this is not as beneficial.

Cheers,
-Neil

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2004\05\16@134827 by gtyler

'[EE]: DC:DC regulators - what effects efficiency?'
2004\05\31@033220 by Peter Mcalpine

'[EE]:Re: DC:DC regulators - what effects efficienc'
2004\05\31@061727 by K. Zeehuisen

'[EE]: DC:DC regulators - what effects efficiency?'
2004\05\31@082448 by Russell McMahon

2004\05\31@110119 by Martin Klingensmith

2004\05\31@152132 by John N. Power

{Quote hidden}

> From:         Peter Mcalpine[SMTP:RemoveMEpetemca1spamBIGPOND.NET.AU]
> Sent:         Monday, May 31, 2004 3:31 AM
> To:   TakeThisOuTPICLISTspamRemoveMEMITVMA.MIT.EDU
> Subject:      [EE]: DC:DC regulators - what effects efficiency?

> Hi All,
> Trying to get my head around DC:DC buck regulators..
> Have a circuit that generals runs from 12V input
> and has 3.5V output at 150ma.
> At present am using an ON Semi MC33063 chip. Although
> cheap it uses transistor switches inside, and would like
> to be able to have a bit more head room with a FET style
> unit..

> Anyway, at present with 12Vin, 3.5Vout@ 150ma I am
> getting 69% efficiency.

> What components most effect this efficiency number?

> Also, it seems that the DC:DC solution likes a bit of
> current going through it, as when I switch off things
> my efficiency drops to 48% (13ma output current).
> Should I do things differently?

As the output current decreases, the switching transistor
stays on for shorter times. Eventually, the transition times
(rise and fall) become comparable to the nominal
conduction time. Since efficiency is lower while the switch
is in its linear region, the average efficiency drops. Some
regulators have what is called "burst mode" to deal with
this. If the on time falls below a certain minimum, the
regulator switches to a mode in which all pulses have that
minimum length. Regulation then consists in producing
bursts of those pulses with variable numbers of pulses in
each burst. The burst length controls the output voltage,
but each pulse by itself is long enough to keep efficiency up.
The pulse interval within the burst is the same as in
normal mode, so all magnetics calculations remain valid.

> Thanks!

> Regards,
> Peter Mcalpine

John Power

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2004\06\01@041412 by Russell McMahon

2004\06\02@201608 by Peter Mcalpine

2004\06\02@202648 by Richard.Prosser

2004\06\02@211501 by Peter Mcalpine

2004\06\02@220158 by Richard.Prosser

2004\06\02@223606 by Russell McMahon

2004\06\03@031157 by hael Rigby-Jones

{Quote hidden}

>-----Original Message-----
>From: Russell McMahon [KILLspamapptechspamspam_OUTPARADISE.NET.NZ]
>Sent: 01 June 2004 09:14
>To: PICLISTRemoveMEMITVMA.MIT.EDU
>Subject: Re: [EE]: DC:DC regulators - what effects efficiency?
>
>
>> Inductor we are using is Tyco brand 220uh shielded ferrite
>core style
>> pn 3631b221k, value 220uh 10%
>> max RDC 0.460ohms
>> max IDC 0.70A
>> LQ test frequency 1Khz
>>
>> I have built another version using a LM2674, and on vero board it
>> performs similar to the MC33063, so guess with a good layout and
>> better selected components will get even better.
>
>Yes. I think that's rather low for the LM2674.
>If by LQ test frequency you mean the switching fequency of the
>converter, then I suspect that the 1 KHz frequency is far too
>low for the inductance. Lets see: I =~ t.V/L or t = LI/V For
>220 uH, 0.7A & 12-3.3 = 8v say. t = 220E-6 x 0.7/8 = 20 usec
>!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
>Your inductor is being well and truly saturated if you have a
>1 kHz duty cycle, unless you have a suitable cycle by cycle
>current trip operating (which the 33063 supports). Try a
>substantially higher frequency for the 33063.
>
>*** You don't say what catch/flywheel diode you are using. It
>MUST be a Schottky if you wish to minimise losses.
>

Oddly enough I have designed/simulated some low power switchers using Linear
Technology's "Switcher Cad" that have actually shown somewhat better
efficiency using a 1N4148 in place of the equivalent Schottky.  Was never
sure if this was genuine or an artifact of the Spice simulator.

Regards

Mike




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2004\06\03@192245 by Peter Mcalpine

Hi Russel,
The efficiency was a little better with a high load (150ma) by about
2%, but at low loads was worse by about 5%.
but having said all that, there was not a lot of thought put into the
components and there was a fair bit of HF noise on the output.
Interesting thing between the 2 circuits was the MC33063 no load Iinput
was 2ma, where the LM2674 was 4.5ma.. so when I do go to a low load
state the efficiency is worse due to quiescent current.

On a side issue, a module on my board calls for low ESR electros
of 1000uf or higher, as it can draw some high currents when it
transmits (GSM module). so my overall capacitance on the output
of the DC:DC reg is 10uf + 470uf + 2x1000uf - I would assume this would
not effect the DC:DC operation as the calculation are more for
a minimum filter cap... (also has various small nf ceramic on it)

Regards,
Peter


FLASHBACK -
You said
> I have built another version using a LM2674, and on vero
board it performs similar to the MC33063,

I assumed you meant that the efficiency was similar. If so, then there's
still something strange. If efficiency is higher then maybe all is OK.


   Russell McMahon

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part 1 5106 bytes content-type:text/plain; format=flowed; charset=Windows-1252; (decoded 7bit)

As the "GSR" has featured peripherally in discussions in the "is this going
to die" thread I decided to start a new thread related to it and others like
it.

I'll make the following observation at the start. If people want to ask
about it or comment on it please do so offlist to avoid annoying people who
get annoyed by such things. While I am notionally the "inventor" of this
circuit the design was actually proposed to me by God. Add a smiley to that
if you want. The basic circuit fell fully formed into my mind essentially
instantaneously immediately after I had provided a detailed written
requirement. The design was to meet a very specific need, which it met and
continues to meet very well indeed. This doesn't mean that it will meet all
other needs as well, divinely inspired or not :-). GSR stands for 'God's
Switching Regulator". End of discourse. Comments offlist please.

There are about 10,000 of these circuits in use at present. If it is used in
a new project, which it probably will be, there will be 100,000+ of them in
the next year or two.

The GSR is a buck regulator (maybe it should have been GBR :-) ) - so output
voltage must be lower than input voltage (unless additional coils are added,
which I have done).

It is usually not the most efficient circuit available.

It's attributes include
   Simplicity / low component count / low cost
   Tolerance to wide range of inductors.
   Good regulation for load and input variation, considering its
simplicity.
   Operation over a VERY wide range of input voltages (20:1 in original)
   Extremely strange mode of operation (which probably helps reduce EMI
substantially).

If you want a high performance circuit and cost is not important then you
probably want to use one of the many many IC switching regulators available.
If you want an ultra low cost design that performs surprisingly well then it
may meet your need. As an amateur 9v battery to 5 volt converter it could do
very well.

The "Black Regulator", refined by Roman Black from a circuit originally
provided by Richard Prosser (if my memory serves) has less components and an
even lower cost. It has worse regulation, requires larger inductors, tends
not to like wide ranges of input voltages and it is much easier to
understand its operation.

The attached GIF is subset from a larger circuit. It uses a high side P
Channel FET as the main switch BUT most low power designs would use a small
PNP transistor as the main high side buck regulator switch. In such cases
zener ZBUK2 is not required. Resistors RBUK3 & RBUK4 would need to be chosen
to suit the circumstances/

The original circuit had an output voltage of about 9 volts AFAIR and
"switched" for input voltages from about 10 volts to 200 volts. The P
Channel FET was rated at 200 volts and this set the upper (un)safe limit of
operation. When Vin is below target output voltage the regulator acts as a
pass through circuit with about 1v or less drop. This is important in the
original application.

In operation the switching waveform of  the FET (or high side PNP or
whatever) is "chaotic". No nice clean waveforms here. Switching can be made
far more regular and efficiency improved by adding a series
resistor-capacitor from the FET-Coil junction to the base of transistor
QBUK2. You can expect that EMI levels may rise when you do this.

OPERATION:

It *probably* works like this ;-)

All off, Vout = 0, ZBUK1 doesn't conduct.
QBUK1 off so RBUK2 turns QBUK2 on which turns BUKFET on via RBUK3.
Voltage is applied to inductor via FET and current increases.
CBUK2 charges until ZBUK1 starts to conduct.
QBUK1 turns on (driven by ZBUK1) turning off QBUK2 turning off FET.

The following is standard buck regulator stuff:
Inductor 'rings" as current must continue to flow.
Right end of inductor now goes positive relative to left end.
DBUK2 conducts and CBUK2 voltage *RISES* as energy is transferred from
inductor.
QBUK1 is held off as long as inductor has energy in it that drives output
and can transfer energy to load fast enough to maintain ZBUK1 in
conduction..
When inductor can't keep up the fight, CBUK2 falls *slightly*, ZBUK1 stops
conducting and cycle repeats.

If there is an RC added as abovem, when inductor left end starts to fall,
QBUK2 is driven off more quickly adding regenerative turn off. The reverse
happens at turn on. This adds "formal" hysteresis and makes the circuit more
understandable. There is in fact formal hysteresis present due to the eenrgy
stored in the inductor, but some can't see this.

If the inductor is shorted out the design becomes a linear regulator. I have
NEVER seen this circuit drop into a linear regulation mode but i have heard
that reported by someone else who played with it.

Apart from the inductor there are a few tens of cents of parts used (if a
small TO92 bipolar transistor is used instead of the FET. Inductor cost
depends on application.





           Russell McMahon


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____________________________________________

2004\12\21@042608 by Russell McMahon