Diode snubber for 12v relay

boji

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I looked at the datasheet for 12v panasonic TQ relays and it draws 5.8mA with 2k of coil resistance during latching and then voltage falls (I believe) to 75% of that to maintain latch. How do you go about calculating the spike when the inductive 'flywheel' starts turning**?

(**I'd like to learn, but if you want to kindly reply and yet keep it simple, I need to buy some more snubbers and I just wanted to be sure 100v 150mA rated diodes would be sufficient.)
 

Bo Deadly

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I actually don't know the official answer for that but I would imagine that the answer is something like peak current times coil resistance. So 5.8mA * 2k = 11.6V. But of course capacitance and leakage inductance is going to crush that.

You can use just about any diode. People say even a signal diode is ok.
 

ruffrecords

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Inductors resist current change (like capacitors resist voltage change). 12V into 2K is 6mA so that will be the ultimate current through the relay when you turn it on. But at turn on the inductor resists this change from 0 mA to 2mA. Instead the current rises exponentially. It is an RL circuit. It acts like an RC circuit except current rises at a rate determine by the L/R  time constant instead of voltage rising determined by the RC time constant.

https://www.electronics-tutorials.ws/inductor/lr-circuits.html

At switch off the inductor has a chunk of stored magnetic energy. It resists the attempted from in current from 6mA to 0mA by raising the voltage across its terminals. The power supply is held rigidly at 12V so the other end of the coil rises in voltage. How much it rises depends on the inductance and the rate of change of current. As in the case of turn on, the rate of change of current is determine by L/R except this time R is very large because the driving transistor or switch is essentially open circuit. This usually means the inductor produces a huge positive voltage spike which will easily blow the brains out of a driver transistor. What you need is a nice small R across the inductor for the current to flow through. Tbis is what the snubber diode does.

Cheers

Ian
 

CJ

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everybody seems to use 1N4148 devices across the terminals.    since these come on a roll of 5000 for about 5 bucks after tariff,  price per device is about .00003 micro cents. 

how good is QC on a device this cheap?  so maybe a 1N4007 to get you into HP reliability.
 

Newmarket

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[silent:arts] said:
If you don't use a transistor or other driver circuit - is the diode still necessary?

You mean using a mechanical switch to operate the relay ?
In that case the 'back emf' voltage spike still occurs and could cause arcing across switch contacts so diode / snubber circuit still advised.
 

Newmarket

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CJ said:
everybody seems to use 1N4148 devices across the terminals.    since these come on a roll of 5000 for about 5 bucks after tariff,  price per device is about .00003 micro cents. 

how good is QC on a device this cheap?  so maybe a 1N4007 to get you into HP reliability.

FWIW I use 1N4002 on all mechanical relays.
 

buildafriend

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Newmarket said:
You mean using a mechanical switch to operate the relay ?
In that case the 'back emf' voltage spike still occurs and could cause arcing across switch contacts so diode / snubber circuit still advised.

stole there words from my mouth. there is mention of this in a book called the art of electronics. am I right by calling it a flyback diode?
 

Newmarket

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buildafriend said:
stole there words from my mouth. there is mention of this in a book called the art of electronics. am I right by calling it a flyback diode?

Yes you can call it a 'Flyback Diode'.
I tend to use the term 'back emf diode' as I think it better conveys the action.
'Flyback' is fine too but it's one of those words I tend to shy away from using in this context simply as the term 'flyback' is used in other contexts esp 'Flyback Converter' where the operation is rather more complex  :)
Same with 'Bootstrap'  :)

It's also worth noting that the diode alters the time to turn off - it will increase it. Unlikely to be a problem in switching audio or control signals but just as well to be aware. There are alternative 'snubbing' methods to get around it but for what I'd call 'normal' use the diode works fine.
 

abbey road d enfer

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boji said:
I looked at the datasheet for 12v panasonic TQ relays and it draws 5.8mA with 2k of coil resistance during latching and then voltage falls (I believe) to 75% of that to maintain latch. How do you go about calculating the spike when the inductive 'flywheel' starts turning**? 
As you may have gathered, the answer is not simple; assuming the switch has no residual resistance, the voltage spike depends on the actual resistance, inductance  and stray capacitance of the coil, which are, except for the former, unpublished parameters. The actual voltage can be quite high, up to several hundred volts.
Now if you use a non-mechanical switch such as a transistor or FET, they don't open like a mechanical switch, due to internal capacitances, so they would reduce the amplitude of the spike, but often will be destroyed as a consequence.

I need to buy some more snubbers and I just wanted to be sure 100v 150mA rated diodes would be sufficient.
Diodes are not the only way of snubbing back emf. As newmarket mentioned, diodes react in a way that can slow the relay's release. R-C snubbers can behave better in that case. Diodes clamp the back emf at about 0.7V, an R-C snubber may damp it at a larger voltage, that would be safe for the switch, but also limit the back current to a value that does not alter significantly the release.
 
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