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relay coil that heats.
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Old 24th April 2013, 06:30 PM   #11
sofaspud is offline sofaspud  United States
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1W isn't adequate if it must drop 12V at ~100mA.
Could also just remove the 1N4001 and put a resistor on the collector of Q5, and turn the zener 180...
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Old 24th April 2013, 06:32 PM   #12
Project16 is offline Project16  France
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I performed the test with the resistance and now the coil does not heat but my resistance test was 1/2w and it was scorching hot!

Thank you all!
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Old 24th April 2013, 06:34 PM   #13
Rundmaus is offline Rundmaus  Germany
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Calculate. Though rumours say differently, Ohm's law does not mind being used from time to time.

Rundmaus
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Old 24th April 2013, 07:02 PM   #14
jaycee is offline jaycee  United Kingdom
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sofaspud, that would defeat the purpose of the catch diode, which is to prevent relay coil back-EMF destroying Q5.

I'd be surprised if the relay coil uses 100mA. I looked up a fairly typical relay for this job, a Finder 40.31 12V, which has a 220 ohm coil and that's 55mA. You could probably get away with 40mA of current and it'd pull in.

edit: Project16, what did you measure for the resistance of your relay's coil?

Last edited by jaycee; 24th April 2013 at 07:04 PM.
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Old 24th April 2013, 07:15 PM   #15
Project16 is offline Project16  France
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Quote:
Originally Posted by jaycee View Post
edit: Project16, what did you measure for the resistance of your relay's coil?
The relay is an Omron G2R 2, the resistance of the coil is 275 ohms (manufacturer specific).
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Old 24th April 2013, 07:26 PM   #16
jaycee is offline jaycee  United Kingdom
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OK, well because you are going from 24V to 12V that makes things easy - ideally, you want a 275 ohm resistor, but the closest are 270 ohms and 220 ohms.

If we go with 270 ohms, that is 270+275 = 545 ohms. At 24v this gives us 24/545 = 0.044A (44mA).

The power dissipated in our resistor will be P=I^2R = (0.044*0.044)*270 = 0.523W - as you can see, a half watt resistor will be rather hot ! A 2W resistor is best for safety.

A 220 ohm resistor will also work just as well, there will only be slightly more current flowing which will not damage the coil.
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Old 24th April 2013, 07:31 PM   #17
Project16 is offline Project16  France
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Thank you for all these explanations and calculations.

Regards!

Last edited by Project16; 24th April 2013 at 07:39 PM.
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Old 24th April 2013, 07:49 PM   #18
sofaspud is offline sofaspud  United States
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Quote:
Originally Posted by jaycee View Post
sofaspud, that would defeat the purpose of the catch diode, which is to prevent relay coil back-EMF destroying Q5.
I saw my use of the zener to be (while reverse-biased) holding the relay coil at 12V. It then becomes the catch diode for back-EMF (while forward-biased). It does double duty.
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Old 24th April 2013, 08:02 PM   #19
jaycee is offline jaycee  United Kingdom
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sofaspud, actually there's an interesting article on Rod Elliot's page about using a Zener to limit back EMF. The important thing is that it does not slow the opening speed of the contacts. However it's not used as a way to run a lower voltage coil on a higher voltage. I'm not sure what the implications of that are, but I wouldn't want to recommend a mod that might degrade the function of the DC protection.
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Old 24th April 2013, 08:20 PM   #20
Project16 is offline Project16  France
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I just thought of something .
One regulator 78L12 type might do the trick in the space of resistance, right?
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