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PICList Thread
'The relay circuit'
1999\01\04@071006 by Jaume Aragay Badia

picon face
This is the circuit I'm using to drive the relays:


SIGNAL             Optoisolator
FROM    47 ohm       4N25   +-------+---------o +5V
PIC    -----                |       |
o----+     +------+        |       |
      -----       |        |       |
                  V ---> |/        |
                  - ---> |         |
                  |      |\        |
                  |        V     |/
                  |        +-----|   BD139
                  |        |     |\
                  |        |      V
                  |        +      +-----------+------------+
                  |       | |    | |          |            |
                  | 220   | |    | | 220      -            +
                  |  ohm  | |    | |  ohm     /\ 2N4148   | | relay coil
                  |        +      +           -           | | v=5V three of
them.
                  |        |      |           |            +  v=6V the
other one.
                  |        |      |           |            |
                 ---      ---    ---         ---          ---


Do you think it would be better to hook the coil to +5V and use the PIC
signal to ground the other pin of the coil?

Jaume.

----------------------
 Jaume Aragay Badia
  spam_OUTaragayTakeThisOuTspamemail.com
----------------------

1999\01\04@074042 by Tjaart van der Walt

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Jaume Aragay Badia wrote:
>
> This is the circuit I'm using to drive the relays:

<snippety>

1) 1k resistor between base of NPN transistor and PIC output.
2) 22k resistor between PIC output and ground.
3) NPN emittor to ground.
4) NPN collector to negative side of relay.
5) Positive side of relay to +5V
6) Flyback Diode 1 : Anode to Collector, Cathode to +5V

Optional :
1) Split 1k up into two 470 ohm, with 10nF to ground from the middle.
2) Flyback Diode 2 : Anode to Ground, Cathode to Base.

>
> Do you think it would be better to hook the coil to +5V and use the PIC
> signal to ground the other pin of the coil?

Only if you will make lots of money on repairs. ;)
The PIC won't handle the spikes forever.


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1999\01\04@075118 by Thomas McGahee

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Jaume,
There are much simpler ways to drive the relay(s) that will lower
your parts count and cost considerably.



                                    O +5V (unregulated)
                                    |
                                    |
                                ----*
                                |   |
                                |   C Diode is usually 1N4001
                               ---  C
                                A   C relay coil
                                |   C
                                |   C
                                |   |
                                 ---*
                                    |
                         1K       |/
              PIC o--- /\/\/\-----|   gain>50
                                  |\
                                    V
                                    |
                                    |
                                   ---

This method also results in less current being drawn from the PIC, and ensures
that the relay coil gets full voltage. The circuit you drew would have a
relay coil voltage equal to V+ minus two diode drops due to the darlington
connection. That is marginal, considering that V+ is only 5 volts to
begin with. Note also that I recommend driving the relay via
the unregulated voltage rather than +5, as this helps isolate
switching noise from the PIC.

I hope this helps.
Fr. Tom McGahee
Electronics Instructor
Don Bosco Technical High School

{Quote hidden}

----------
{Quote hidden}

1999\01\04@083956 by paulb

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Jaume Aragay Badia wrote:

> This is the circuit I'm using to drive the relays:

 Somewhat better, but you haven't specified the coil resistances and
what sort of power supply you are using.

 What people are telling you is:

 1} There is no "magic" about an optocoupler.  It has a particular
application; to couple a logic (possibly analog) state between two
circuits *that have no common power supply connection*.  Your circuit
appears to have a common ground connection on both sides of the opto-
coupler and much worse, we suspect you are using the same 5V for PIC and
relays.  The optocoupler is worse than useless under these
circumstances.

 2} Assuming a 1.5V drop in the opto LED, the 47 ohm resistor will
tend to set the current at 3.5 V / 47 ohms = 70 mA or so.  You may note
the PIC is rated for only 25 mA per output.  You therefore need at least
180 ohms in this position but as has been said, the opto is useless
*unless* you are going to use a power supply on the "transistor" side
that is both electrically and physically separate.

 3} The 220 ohm emitter load of the opto is far too low.  15 k ohm
minimum.

 4} No evident purpose of a 220 ohm resistor across the relay when
you already have a "snubber" or "flyback" diode.

 5} As others said, don't use emitter follower circuits for logic, you
only waste voltage.  Unless of course, you *want* to waste voltage.  An
emitter follower with collector supply from an unregulated higher
voltage will regulate its output to the base voltage less 0.65V.  But
that's not really what you want here.

 Because you don't want to operate the relays from the same 5V
regulated line as the PIC, you don't really need 5V relays, but if you
already have them, you can use a series resistor for *each* to drop
whatever unregulated voltage you have, to 5V for the relay.

 6}  The BD139 is overkill unless the relay draws more than 100mA.  If
it does, you *really* don't want to share the PIC supply line.

 I agree with the others; use a NPN transistor with emitter to ground
(note; ground at the relay supply, not the PIC supply!), fed with a
resistor of 1 k ohm or more from the PIC.  Put a flyback diode across
the relay coil, keep all the relay wiring together and away from all the
PIC wiring, power the relay from the unregulated supply or a separate
regulator if it varies too much.
--
 Cheers,
       Paul B.

1999\01\04@123424 by Jaume Aragay Badia

picon face
>   Somewhat better, but you haven't specified the coil resistances and
> what sort of power supply you are using.
>

This board adds to an existing machine. The machine transformer's output is
13VAC aprox. Then I use this circuit:

220Vac       13Vac
      +-----+    +----+                      +------+
   ---+Trafo+----+    +------------+---------+ 7805 +-----o 5V output
   ---+     +----+    +-----+      |         +--+---+
      +-----+    +----+     |     ---           |
                 Diode      |     --- 200uF     |
                       Bridge     |      |            |
                           ---    ---          ---



> power the relay from the unregulated supply or a separate
> regulator if it varies too much.

Just before the 7805 or using an other 7805?

The coil resistances are:

n.a. Ohms for the 5V relays. I use the SHINMEI RSA-5 relays here.
68 Ohms for the 6V relay. 530 mW opearting power. 4'2V Pick-up. 0'6V
Drop-out.


Thanks for tour time and effort, I really apreciate it. :)

Jaume.

----------------------
 Jaume Aragay Badia
  KILLspamaragayKILLspamspamemail.com
----------------------

1999\01\04@155403 by Michael J. Ghormley

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face
Here's my try (flame shields up!):


SIGNAL                      +-------+---------o Separate +5V from welder PS
FROM                        |       +--+
PIC   220 ohm               | Relay |  |
o----/\/\/\-------+        |      (  ---
                  | 4N25   |      (   ^
                  V ---> |/       (  /_\ 1N4001
                  - ---> |         |  |
                  |      |\        +--+
                  |        V       |
                  |        |     |/
                  |        +-----|   BD139
                  |        |     |\
                  |        |       V
                  |        \       |
                  |        /       |
                  |     1K \       |
                  |        /       |
                  |        |       |
                  |        |       |
                 ---      ---     ---
                  -        -       -

I am not aware of the Beta of the BD139, but if it is too low then a 1K resistor
may be needed
between the emittor or the 4N25 and the base of the BD139.

What's the use of an opto-iso if you are hooking both sides of the circuit to th
e same power supply?

In an electrically noisy welding environment I would keep the opto-iso, but make
a new +5VDC from
the welder's side of the circuit.  If your original voltage comes from the welde
rs PS, then I would
think about a wall-wart for the PIC.

I would also keep aware of ground loops.

That's my two pence.

Michael

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**********************************************
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**********************************************

1999\01\04@172111 by paulb

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Jaume Aragay Badia wrote:

> This board adds to an existing machine. The machine transformer's
> output is 13VAC aprox. Then I use this circuit:

> 220Vac       13Vac
>        +-----+    +----+                      +------+
>     ---+Trafo+----+    +------------+---------+ 7805 +-----o 5V output
>     ---+     +----+    +-----+      |         +--+---+
>        +-----+    +----+     |     ---           |
>                   Diode      |     --- 200uF     |
>                   Bridge     |      |            |
>                             ---    ---          ---

>> power the relay from the unregulated supply or a separate
>> regulator if it varies too much.
> Just before the 7805 or using an other 7805?

 I'd be quite suspicious about that 200 µF capacitor.  I'll bet if you
'scope it when the relay tries to pull in, the output voltage is
dropping beyond the 7.5V necessary for the 7805 to regulate, at least
part of the cycle.

 Your relay supply should be derived from a second pair of diodes from
the transformer feeding a second capacitor (probably about 1000 µF or
more - I trust the transformer is good for an amp or so).  You won't
really need a regulator - you can just use a series resistor for *each*
relay.  Or use 12V relays instead.

> The coil resistances are:
> n.a. Ohms for the 5V relays.

 That's a funny resistance!

> I use the SHINMEI RSA-5 relays here.

> 68 Ohms for the 6V relay. 530 mW operating power.

 100 mA.  OK, 100mA load on the 200 µF capacitor over 1/100 second ...
CV=It V=IT/C = 0.001/2*10e-4 = 0.5*10 = 5V ripple drawn by the relay
alone.  But adding the current draw of the PIC and various over-small
resistors probably made it critical.

 The capacitor can stay at 200 µF for operating the PIC *alone*.

> 4'2V Pick-up. 0'6V Drop-out.

 No need for regulation here, but 4.2 V is a bit close in terms of the
emitter follower circuit...
--
 Cheers,
       Paul B.

1999\01\05@055040 by Jaume Aragay Badia

picon face
Thanks for your input. Finally there are two different circuit proposals:


SIGNAL                      +-------+---------o Separate +5V from welder PS
FROM                        |       +--+
PIC   220 ohm               | Relay |  |
o----/\/\/\-------+        |      (  ---
                  | 4N25   |      (   ^
                  V ---> |/       (  /_\ 1N4001
                  - ---> |         |  |
                  |      |\        +--+
                  |        V       |
                  |        |     |/
                  |        +-----|   BD139
                  |        |     |\
                  |        |       V
                  |        \       |
                  |        /       |
                  |     1K \       |
                  |        /       |
                  |        |       |
                  |        |       |
                 ---      ---     ---

using a 1K resistor between 4N25 emitter and BD139 base if needed.

And this other one:



                                    O +V (unregulated) being 12Vac
                                    |
                                    |
                                ----*
                                |   |
                                |   C Diode is usually 1N4001
                               ---  C
                                A   C relay coil
                                |   C
                                |   C
                                |   |
                                 ---*
                                    |
                         1K       |/
              PIC o-*- /\/\/\-*---|   gain>50
                    |         |   |\
                   +++       ---    V
               22K | |        A     |
                   +++        |     |
                    |         |     |
                   ---       ---   ---

A couple questions on this second circuit:

1.- Is the second diode a 1N4001?
2.- Would a 2N2222A a good choice for the NPN transistor?




The values of the DOCUMENTED relays ("n.a." was "not available", Paul) from
FINDER I'll be using are:

1.- For the big one, FINDER ref. # 40.52:

       Coil: 12Vdc, 220ohm, 55mA.

2.- For the other 3 of them, FINDER ref. # 30.22:
       Coil: 12Vdc, 360 ohm, 33mA.



Once again, Thanks for your time and patience.

Jaume.


----------------------
 Jaume Aragay Badia
  RemoveMEaragayTakeThisOuTspamemail.com
----------------------

1999\01\05@062120 by Michael J. Ghormley

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Jaume Aragay Badia wrote:

<snip>

{Quote hidden}

Looks to me like a garden variety 1N4148 would do, but a 1N4001 will also work.

> 2.- Would a 2N2222A a good choice for the NPN transistor?

The 2N2222A is rated for 150mA which ought to work, but I always get a funny
feeling when punking in a little TO-18 cased transistor to drive a 12V relay.  I
would spend $0.15 more and get a TO-220 cased transistor like a TIP-29A or some
such unless space or component costs are at a premium.

<snip>

> Once again, Thanks for your time and patience.

T'aint'n a thang.

One last thing:  I don't think that the +V at the top of this second circuit is
supposed to be AC -- just unregulated DC, methinks.  Perhaps the original poster
will have some thoughts...

Michael

* TAKE THE '.NOSPAM' OUT OF MY ADDRESS TO REPLY
**********************************************
Outside of a dog, a book is man's best friend.
Inside of a dog, it's too hard to read anyway!
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**********************************************

1999\01\05@071949 by Jaume Aragay Badia

picon face
> One last thing:  I don't think that the +V at the top of this second
circuit is
> supposed to be AC -- just unregulated DC, methinks.  Perhaps the original
poster
> will have some thoughts...

> Michael


You're right Michael, that voltage is unregulated, not ac. My fault, not the
sender's. ThanX.

Jaume.

----------------------
 Jaume Aragay Badia
  spamBeGonearagayspamBeGonespamemail.com
----------------------

1999\01\05@082116 by Ian Rozowsky

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I've always used a BC337 (TO92) or BC817 (SOT23), one side of 12V
relay to 12V, other side to collector, emitter to ground. Base
driven from PIC through 10k. 1N4007 across coil. I've got at least
150,000 instances of this circuit in the field, so it must work.

regards

Ian Rozowsky
Development Engineer
Centurion Systems
Box 506 Cramerview 2060 South Africa
Tel   : +27-11-708-2680
Fax   : +27-11-708-2630
e-mail: TakeThisOuTrozEraseMEspamspam_OUTcentsys.co.za

1999\01\05@082151 by Tjaart van der Walt

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face
Jaume Aragay Badia wrote:

{Quote hidden}

1N4001 will work fine.

> 2.- Would a 2N2222A a good choice for the NPN transistor?
Just about any NPN will do. I have a bunch of BC337's - so I
use them, but a 2n2222A will also work fine.

{Quote hidden}

These currents shouldn't pose any problem for a 2N2222A

It is a bit of overkill, but if you use it in some
mission critical application, I'd also add a cap to
limit the (remote-ish) possibility of a pike inducing
the transistor to switch :

{Quote hidden}

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1999\01\05@091015 by Thomas McGahee

flavicon
face
----------
> From: Jaume Aragay Badia <EraseMEaragayspamEMAIL.COM>
> To: RemoveMEPICLISTEraseMEspamEraseMEMITVMA.MIT.EDU
> Subject: Re: The relay circuit
> Date: Tuesday, January 05, 1999 5:43 AM
>
> Thanks for your input. Finally there are two different circuit proposals:
>
SNIP
{Quote hidden}

Please note that the proper voltage is unregulated DC (not AC).

> 1.- Is the second diode a 1N4001?

It can be a 1N4001 or a 1N4148. It is not critical.
It is used to clamp the base-emitter voltage to
a maximum negative swing of -.7 volts to prevent
accidental breakdown when the transistor switches
off and the coil produces a nasty pulse. It also
prevents this yucky stuff from getting back to the PIC.
As someone else mentioned, you can split the 1K resistor
into two 470 ohm units in series, and tie a 10 nf cap
to ground in case you have any additional noise
problems. Probably over-kill in most cases, but
quite useful for those really bad cases.

Make sure that your grounding paths are properly
designed, as the relays generate a fair amount
of noise into the ground system. If possible the
heavy currents should be handled by wider traces
and should be routed such that the ground noise
does not affect the PIC's ground system. There
has to be a common connection between the ground
system for the PIC and the ground system for the
transistor/relay section, but avoid running
the transistor/relay current through the PIC
section.

> 2.- Would a 2N2222A a good choice for the NPN transistor?

Yes, a 2N2222A will work. I usually use MPSA05 or MPS6566
transistors because I have thousands of them that I picked
up dirt-cheap a few years ago. The transistor type is not
critical. It must be able to withstand the voltage and be
able to carry the necessary current. By using a 1K resistor
we guarantee that the transistor is operated in the
saturated mode where power dissipation is small.

I hope this helps.
Fr. Tom McGahee

{Quote hidden}

1999\01\05@175106 by Michael J. Ghormley

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Ian Rozowsky wrote:

> I've always used a BC337 (TO92) or BC817 (SOT23), one side of 12V
> relay to 12V, other side to collector, emitter to ground. Base
> driven from PIC through 10k. 1N4007 across coil. I've got at least
> 150,000 instances of this circuit in the field, so it must work.

You have given me courage Ian.  So the 2N2222A should work just fine,
yes?  I will try a TO-92 next time.

Is the 1N4007 better than the 1N4001?  I thought that their current
ratings were about the same, but their PIV was just higher (no databook
handy).  Am I wrong or is there another reason for the 1N4007?

Michael

* TAKE THE '.NOSPAM' OUT OF MY ADDRESS TO REPLY
**********************************************
Outside of a dog, a book is man's best friend.
Inside of a dog, it's too hard to read anyway!
                                 Groucho Marx
**********************************************

1999\01\06@013610 by John Smith

picon face
From: Michael J. Ghormley

>Is the 1N4007 better than the 1N4001?  I thought that their current
>ratings were about the same, but their PIV was just higher (no databook
>handy).  Am I wrong or is there another reason for the 1N4007?
>
>Michael

None of the my suppliers here in South Africa stock the 4001, they say it is
more expensive than the 4007 ( which has always been the bread and butter
diode for TVs etc. ) because of the volumes.

So we use a 1000v diode on a circuit with 5v supply :-)

John Smith

JS Electronics
Cape Town
South Africa

1999\01\11@180241 by John Payson

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|  Because you don't want to operate the relays from the same 5V
|regulated line as the PIC, you don't really need 5V relays, but if you
|already have them, you can use a series resistor for *each* to drop
|whatever unregulated voltage you have, to 5V for the relay.

What do you think of something like this circuit?

    +Volts (e.g. 12V unreg)
      |
      +------------------.
      |                  |
     ||D                 |
     ||D Relay           =
     ||D                 ^ Diode
      |                  |
      +-----.            |
      |     |            |
     RES   CAP           |
      |     |            |
      +-----+------------'
      |
  [switched ground]

Since relays need more voltage when closed than open, this would
give the relay a little extra "kick" when it first energizes.  In
addition, when the relay is de-energized the cap will be charged
in the other direction, allowing the magnetic field in the inductor
to collapse more quickly than would otherwise be the case.

The one weakness I can see with this circuit is that a small delay
would be necessary after releasing the relay before it could be en-
ergized again, since the cap would have to be discharged first.
Otherwise, though, I'd expect it to work quite well and save power
versus running the relay always at full rated coil voltage.

Anyone here done anything like that?

1999\01\12@082249 by paulb

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face
John Payson wrote:

> What do you think of something like this circuit?

 All points noted and agreed.  If you are going to use the capacitor,
I'd do this:

          +--------|<|--------------+
          |                         |
+Vunreg o-+--VVVVV----+---UUUUU-----+----+
               R      |   Relay          |
                     ===                 \|
                     --- C                |----O BASE DRIVE
                      |                  v|
                      |                  |
                      M   Ground         M

 As before, R limits the continuous relay voltage and current.  C
allows pull-in at a higher voltage than the maintaining voltage, and
the diode limits kick-back.

 What's really good about this version, is that C isolates the relay
from the power supply also, at least to some extent.
--
 Cheers,
       Paul B.

1999\01\12@082305 by Dwayne Reid

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face
My variant (horrible ASCII art below) has been used in many different
projects over the past 20 years or so.  This works great for larger devices
- solenoids and the like.  My other trick with larger solenoids is to diode
isolate from the DC supply, then use a AC driven voltage doubler (using a
smaller than normal pump capacitor) to charge the capacitor to double the
main supply.  This eliminates the need for the resistor (and its heat
generation).  The pump capacitor has to be able to withstand the ripple
current but that has not proven to be a problem.

Another trick that works well for smaller relays (P&B T90 series, Aromat
JS1E series) is to charge a smaller than normal capacitor from the AC
supply, then rely upon the ripple for the reduced supply voltage.  I've used
that for years also.  Its a handy trick when you have a 24Vac supply (giving
you about 33Vdc unregulated) and 24V relays - and don't want to regulate the
relay supply or generate heat from a series resistor.  Again, the capacitors
have to be rated for the ripple current - I've used Sanyo and Nichicon with
no problems.


+Volts (e.g. 12V unreg)

-----------res-----+------+------------------.
                  |      |                  |
                 cap    ||D                 |
                  |     ||D Relay           =
                  |     ||D                 ^ Diode
                 Gnd     |                  |
                         +------------------
                         |
                         |
                         |
                   [switched ground]

dwayne


Dwayne Reid   <RemoveMEdwaynerTakeThisOuTspamspamplanet.eon.net>
Trinity Electronics Systems Ltd    Edmonton, AB, CANADA
(403) 489-3199 voice     (403) 487-6397 fax

1999\01\12@082321 by kypros.vassiliou

flavicon
John Payson wrote:
{Quote hidden}

Hi,
Yes Ibuilt a cct like this many times.
If you want to drive a lower voltage relay by a higher supply voltage
you can use the above curcuit (cct), the only comment I have is about
the diode which must be connected across the relays coil.
The resistor is calculated using Ohms law. First measure the resistance
of the relay (coil), devide this value by the coils voltage to find the
current through it and then calculate the extra voltage drop needed by
the series resistor (you have current and volt drop) to find its value.
The poewr of the resistor is P=I*I*V (I is current found before, V is
voltage drop across resistor).
I use to use the same cct just to lower the poewr consumed by a relay in
battery powered equipments. Here the capacitor is an electrolitic one
(it is not necessary to use capacitor in the previus case unless you
want to consume less power as I'll describe now) usually more than 220uF
placed in parallel to the resistor. The value of the resistor must be
equal(or less in some cases) to the value of the coils resistance. The
relay must be rated as the supply voltage.
The capacitor is used to give an initial kick to the relay to energise
so in cases where the resistance of the coil is low enough you can
increase its value to make sure the relay is always activated.

Regards
Kypros Vassiliou

1999\01\12@084844 by Eisermann, Phil

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face
{Quote hidden}

       [snip]

> Anyone here done anything like that?
>
       something quite similar, as a snubber
       circuit for transformers (as used in
       a SMPS). Except that the diode and
       resistor are in parallel, and the cap
       in series to supply rail:


             o   o
             |   |
            ---  |
            ---  |
             |   C
             |   C
         *---*   C
         |   |   C
         /   |   |
         \   =   |
         /   ^   |
         \   D   |
         |   |   |
         *---*---*
                 |
              (switch)


       you trade power dissipation in the switch
       for power dissipation in the resistor. it
       also slows the dv/dt of the (inductive)
       leakage spike.

1999\01\12@113612 by John Payson

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face
|  All points noted and agreed.  If you are going to use the capacitor,
|I'd do this:
|
|           +--------|<|--------------+
|           |                         |
| +Vunreg o-+--VVVVV----+---UUUUU-----+----+
|                R      |   Relay          |
|                      ===                 \|
|                      --- C                |----O BASE DRIVE
|                       |                  v|
|                       |                  |
|                       M   Ground         M

| As before, R limits the continuous relay voltage and current.  C
|allows pull-in at a higher voltage than the maintaining voltage, and
|the diode limits kick-back.

|  What's really good about this version, is that C isolates the relay
|from the power supply also, at least to some extent.

This version and a similar one posted with the diode directly across
the relay have different design pros and cons vs mine.  I think I
like this one the best, though.

The one potential pitfall I can see is that if the switch is released
after the coil inductance is "charged" but before the cap has reached
equilibrium, the circuit will try to send voltage out the supply rail
in the amount of (cap voltage)+(coil current)*(resistor).  After the
cap has reached equilibrium, however, the potential kickback voltage
would simply equal the supply voltage.  Further, if the capacitor and
resistor are small relative to the filter caps and loading of the sup-
ply, however, even the worst-case potential voltage spike will be mit-
igated pretty well.

Placing the diode across the relay coil directly would eliminate any
voltage spike risks (and if the diode is placed close to the coil, it
would also minimize any transients that could couple to nearby circ-
uits).  It could also, however, increase the time required for the
relay to drop (since the magnetic field decays at a rate proportional
to the induced voltage).  For a demonstration of this effect, take a
solenoid-operated door chime and wire it with a DC power supply and
brush two wires together to form the "button".  Without a clamp diode
installed, the contacts will arc noticeably, but the solenoid will
operate cleanly.  If you add a kickback diode, the contacts will no
longer arc, but the release of the plunger will be "mushy".  A cir-
cuit like the one above, though, should solve both problems.

1999\01\13@064749 by paulb

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face
John Payson wrote:

> Placing the diode across the relay coil directly would eliminate any
> voltage spike risks (and if the diode is placed close to the coil, it
> would also minimize any transients that could couple to nearby circ-
> uits).

 I've always felt this was "right", but on closer examination, there is
no justification for it!  The function of the diode at the instant of
switch-off is to take over the flow of current, which it does.  While
the diode current itself rises very rapidly to do so, the whole point is
that the coil current decays relatively slowly.

 If the diode is across the coil, then what happens is that the current
in that part of the circuit stays the same, but the current in the *lead
to the driver* suddenly drops to zero.  Inductive radiation from this
lead would likely be *worse* than if the diode were associated with the
driver as it is in the ULN devices.

 As to the capacitive impulse from the wire connecting driver and coil,
the voltage on which rises suddenly from a fraction of a volt to 0.7V or
so *above* the supply rail, this effect is entirely independent of the
location of the diode.

 Interesting?
--
 Cheers,
       Paul B.

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