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'[EE]: Using parallel 7805 regulators'
2003\03\30@224749 by Ricardo D. Medina

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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@225703 by Jai Dhar

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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" <spam_OUTrickymTakeThisOuTspamMONTEVIDEO.COM.UY>:

{Quote hidden}

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2003\03\30@230743 by Des Bromilow

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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

>>> .....jdharKILLspamspam@spam@ENGMAIL.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" <rickymspamKILLspamMONTEVIDEO.COM.UY>:

{Quote hidden}

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2003\03\30@233645 by Marc Joffe

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Check out the website
http://www.mitedu.freeserve.co.uk/Circuits/Power/boosti.htm

It shows how to use an PNP transistor to boost the current. If you have
the parts lying around this may be easy for you to do.

Marc

On Sun, 2003-03-30 at 23:09, Des Bromilow wrote:
> 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
>

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2003\03\31@011047 by Daniel Dourneau

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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

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__________________
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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'[EE]: Using parallel 7805 regulators'
2003\04\01@075154 by Rick C.
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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

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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

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At 07:54 AM 01/04/2003 -0500, you wrote:
>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.)

Another common solution is to put a smallish resistor on the input to the
7805, and a PNP transistor across the whole lot.

Emitter goes to resistor input,
Base goes to resistor, regulator input junction,
Collector goes to regulator output.

Not sure what size of resistor you need to put there, but it will make a
difference, and depends on the Hfe of the transistor.

--Brendan

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