I am new to this and I am trying to drive a set of 9V coils with the pic.
Should I use some type of driver or buffer and if I need to how would I
drive it?
The easiest way I have found is to drive the LED half of an optoisolator.
Connect the output side to the relay and its source. I use the H11G2 (just
because I have them around).
At 08:54 AM 1/12/2000 -0500, Bradshaw, William V wrote:
>I am new to this and I am trying to drive a set of 9V coils with the pic.
>Should I use some type of driver or buffer and if I need to how would I
>drive it?
>
>Thanks,
>Vern
>
>
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> I am new to this and I am trying to drive a set of 9V coils with the pic.
> Should I use some type of driver or buffer and if I need to how would I
> drive it?
>
> Thanks,
> Vern
I believe in power FETs to do this job.
(IRFD210, VP2206, etc)
You can "do" high side or low side switching with P and N channel FETs.
Good option! I use them to control a stepping motor. They could also remove
your neeed for the relays.... good call.
TAS
At 01:25 PM 1/12/2000 -0500, Thomas C. Sefranek wrote: {Quote hidden}
>"Bradshaw, William V" wrote:
>
>> I am new to this and I am trying to drive a set of 9V coils with the pic.
>> Should I use some type of driver or buffer and if I need to how would I
>> drive it?
>>
>> Thanks,
>> Vern
>
>I believe in power FETs to do this job.
>(IRFD210, VP2206, etc)
>
>You can "do" high side or low side switching with P and N channel FETs.
>
>--
>Thomas C. Sefranek WA1RHP
>ARRL Instructor, Technical Specialist, VE Contact.
>http://www.harvardrepeater.org
>http://hamradio.cmcorp.com/inventory/Inventory.html
>
>
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2N3904 collector to relay coil and annode of diode (1N4148 or 1N4001 or
similar .. whatever is cheap and/or readily available)
other side of relay coil and cathode of diode to +9V
To turn relay on: raise output to logic "1", Vdd
To turn relay off: drop output to logic "0", GND
The diode shunts voltage overshoot from the coil's inductance when the
transistor is switched off and prevents damage to the transistor.
You can substitute similar transistors for 2N3904 if available or cheaper:
2N2222 is a close substitute
Vital characteristics of 2N3904:
hFe 200 typ.
BVCEO 40
IC 600mA
Also, I am assuming that the coils are 250mA or less .. you can substitute
1k for the 2.2k to drive up to about 500mA.
Enjoy,
Ken
>I am new to this and I am trying to drive a set of 9V coils with the pic.
>Should I use some type of driver or buffer and if I need to how would I
>drive it?
>
>Thanks,
>Vern
William, I'm sure someone else will give pointers as to which transistor/fet
you can use to drive the coil. I just want to remind you to put a diode
across the coil to prevent damage to the electronics when it is switched
off. If you are not quite sure how or why, shout..
> I am new to this and I am trying to drive a set of 9V coils with the pic.
> Should I use some type of driver or buffer and if I need to how would I
> drive it?
>
> Thanks,
> Vern
>
Cute, but a bit of a waste of an opto-isolator, if it's not isolating
anything! A transistor will do the same job, a *lot* cheaper, with more
gain.
And you *did* mean to put a resistor in series with the opto input,
did you not - or has it one included? And the diode across the relay?
--
Cheers,
Paul B.
yes... I DID mean to have a resistor on the output. I made this with 'parts
on hand' and I have not claimed (I though it was clear) that it was not the
cheapest or easiest way to do what he wanted. I use optos just because I
control high voltages with the relays and do not want to take the chance
that I will short out something and it will get back to the PIC. Glad I did
too, because I had a metal plate contact my circuit and all I got was a
bunch of 'chattering relays'. I also said in another posting that driving
with a fet was his best bet and that is how I drive my stepper motors.
Also, yes, I do have diodes on my relays as well. I am also using two pole
relays that are wired in parallel as a redundant measure. Sorry I left that
off the ASCII drawing as well.
TAS
At 11:48 PM 1/13/2000 +1100, Paul B. Webster VK2BZC wrote: {Quote hidden}
>Terry A. Steen wrote:
>
>> The easiest way I have found is to drive the LED half of an
>> optoisolator.
>
>> +-------+
>> PIC OUT---| |----RELAY---- High voltage
>> | |
>> | H11G2 |
>> +--| |--+
>> | | | |
>> \|/ +-------+ \|/
>> GND GND
>
> Cute, but a bit of a waste of an opto-isolator, if it's not isolating
>anything! A transistor will do the same job, a *lot* cheaper, with more
>gain.
>
> And you *did* mean to put a resistor in series with the opto input,
>did you not - or has it one included? And the diode across the relay?
>--
> Cheers,
> Paul B.
>
>
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>Vital characteristics of 2N3904:
>hFe 200 typ.
>BVCEO 40
>IC 600mA
>
>Also, I am assuming that the coils are 250mA or less .. you can
substitute
>1k for the 2.2k to drive up to about 500mA.
My data book says the ABLOLUTE MAXIMUM allowable collector current for a
2N3904 is 200 mA.
It also says that the hFe of 200 typical (100 min/300max) is at 10 mA.
At 100 mA, the hFe is more like 30 than 200. The circuit Ken described
is fine if your relay will operate reliably with a current of 60-70 mA
with the 2.2K, 100 mA with the 1K resistor.
The 2N2222 is a little better, max about 200 mA. The 2N4401 is better
yet, but not quite as easy to find.
A better solution is one of the Zetex high-gain transistors available
from Digi-Key. A ZTX689B will sink 2 amps with 10 mA of base drive (but
be careful - this is measured in very short pulses!). I'd feel safe
planning on 500-600 mA with 10 mA of base drive. Assuming you have a 5V
supply, you need about 390 ohms to get 10 mA into a transistor because
you lose 0.7V across the base-emitter junction and 0.3 to 0.4 on the PIC
output. Check your databook and make sure the I/O pin can source 10 mA
or better at 4.5V output (they will specify it as Vcc-0.5V, I think).
This assumes you have a decent 5V supply.
Of course the easiest solution is to use an N-channel FET. An IRLD014 or
IRLD110 would work great. Just hook source to ground, gate to your PIC,
drain to the relay, relay to 9V. At Digi-Key, the IRLD110 is even
cheapter than the Zetex part. Don't forget the diode across the relay
coil, or you can destroy any of these transistors quick, especially the
Zetex ones.
part 0 3150 bytes >Ken Webster wrote:
>
>>Vital characteristics of 2N3904:
>>hFe 200 typ.
>>BVCEO 40
>>IC 600mA
>>
>>Also, I am assuming that the coils are 250mA or less .. you can
>substitute
>>1k for the 2.2k to drive up to about 500mA.
>
>My data book says the ABLOLUTE MAXIMUM allowable collector current for a
>2N3904 is 200 mA.
Ahh .. so it is. I was using an old NTE cross-reference and it has a
"Maximum collector current" column which lists 0.6 A for the 2N3904
equivelent .. consulting with a respectable databook (Motorolla) shows that
the continuous collector current is 200mA max. The NTE book must be showing
peak current rather than continuous current. Sorry about that!
>It also says that the hFe of 200 typical (100 min/300max) is at 10 mA.
>At 100 mA, the hFe is more like 30 than 200. The circuit Ken described
>is fine if your relay will operate reliably with a current of 60-70 mA
>with the 2.2K, 100 mA with the 1K resistor.
>The 2N2222 is a little better, max about 200 mA. The 2N4401 is better
>yet, but not quite as easy to find.
>
>A better solution is one of the Zetex high-gain transistors available
>from Digi-Key. A ZTX689B will sink 2 amps with 10 mA of base drive (but
>be careful - this is measured in very short pulses!). I'd feel safe
>planning on 500-600 mA with 10 mA of base drive. Assuming you have a 5V
>supply, you need about 390 ohms to get 10 mA into a transistor because
>you lose 0.7V across the base-emitter junction and 0.3 to 0.4 on the PIC
>output. Check your databook and make sure the I/O pin can source 10 mA
>or better at 4.5V output (they will specify it as Vcc-0.5V, I think).
>This assumes you have a decent 5V supply.
>
>Of course the easiest solution is to use an N-channel FET. An IRLD014 or
>IRLD110 would work great. Just hook source to ground, gate to your PIC,
>drain to the relay, relay to 9V. At Digi-Key, the IRLD110 is even
>cheapter than the Zetex part. Don't forget the diode across the relay
>coil, or you can destroy any of these transistors quick, especially the
>Zetex ones.
This would work nicely if you have a particulary power-thirsty relay.
Fortunately most small 9V relays take well under 100mA. For such
applications you just can't beat the price of a 2N3904 ($1.60 for 10, $12.80
for 100 in DigiKey).
BTW .. since this thread is alive again and since I've learned (thanks to
Jinx) how easy it is to draw a diagram using Paintbrush, here is a schematic
for the cheap 9V relay driver (good for 12V and 24V also).
I also attached a schematic for a really super-cheap 5V relay driver. It
uses an emitter follower so it actually delivers only 4.4V or so to the
relay but this has always worked fine for the 5V relays I have used. It
does not require a protection diode because the emitter follower won't allow
any undershoot.
Thanks to Don Burdette's post I think we finally have a good set of
(hopefully bug free) relay driver circuits. I made some schematics for them
and put up a web page summarizing them here:
I did add one additional circuit which uses the capacitor voltage doubling
pull-in for relays of any voltage (i.e. drive a 24V relay with a 12V
supply). Feel free to copy the schematics (or anything else you find
useful).
> Thanks to Don Burdette's post I think we finally have a good set of
> (hopefully bug free) relay driver circuits. I made some schematics for them
> and put up a web page summarizing them here:
>
> http://www.scooby.flysaturn.com/cktlib/relaydrv
>
> I did add one additional circuit which uses the capacitor voltage doubling
> pull-in for relays of any voltage (i.e. drive a 24V relay with a 12V
> supply). Feel free to copy the schematics (or anything else you find
> useful).
>
> Cheers,
>
> Ken
Was it not working?
It should be fine.
Try www.scooby.flysaturn.com/cktlib/relaydrv/index.html
(though I don't think there should be any difference)
>Ken Webster schrieb:
>
>> Thanks to Don Burdette's post I think we finally have a good set of
>> (hopefully bug free) relay driver circuits. I made some schematics for
them
>> and put up a web page summarizing them here:
>>
>> www.scooby.flysaturn.com/cktlib/relaydrv
>>
>> I did add one additional circuit which uses the capacitor voltage
doubling
>> pull-in for relays of any voltage (i.e. drive a 24V relay with a 12V
>> supply). Feel free to copy the schematics (or anything else you find
>> useful).
>>
>> Cheers,
>>
>> Ken
> >Please check the link !
> >Peter
>
>
> Was it not working?
> It should be fine.
> Try www.scooby.flysaturn.com/cktlib/relaydrv/index.html
> (though I don't think there should be any difference)
>
>
> >Ken Webster schrieb:
> >
> >> Thanks to Don Burdette's post I think we finally have a good set of
> >> (hopefully bug free) relay driver circuits. I made some schematics for
> them
> >> and put up a web page summarizing them here:
> >>
> >> www.scooby.flysaturn.com/cktlib/relaydrv
> >>
> >> I did add one additional circuit which uses the capacitor voltage
> doubling
> >> pull-in for relays of any voltage (i.e. drive a 24V relay with a 12V
> >> supply). Feel free to copy the schematics (or anything else you find
> >> useful).
> >>
> >> Cheers,
> >>
> >> Ken
>
> >Please check the link !
> >Peter
>
> Was it not working?
> It should be fine.
> Try www.scooby.flysaturn.com/cktlib/relaydrv/index.html
> (though I don't think there should be any difference)
>
> >Ken Webster schrieb:
> >
> >> Thanks to Don Burdette's post I think we finally have a good set of
> >> (hopefully bug free) relay driver circuits. I made some schematics for
> them
> >> and put up a web page summarizing them here:
> >>
> >> www.scooby.flysaturn.com/cktlib/relaydrv
> >>
> >> I did add one additional circuit which uses the capacitor voltage
> doubling
> >> pull-in for relays of any voltage (i.e. drive a 24V relay with a 12V
> >> supply). Feel free to copy the schematics (or anything else you find
> >> useful).
> >>
> >> Cheers,
> >>
> >> Ken
Ken, the first URL is technically invalid. It's your fault. The
trailing slash is actually necessary. Most web servers respond with
a redirection to the correct URL. Most web browsers will accept the
redirection and go to the second URL. So the effect is transparent
to most of us. Just costs an extra second or so. Clearly for Peter
this does not work.
I recommend using the second URL. However that does mean you are
relying on the web server to perform the translation to index.html.
marinapower.com
