What should be considered when picking a flyback diode?

Catching a flyback pulse is actually a rather easy application for a diode. Many of the diode parameters don't matter much. The basics are:

1. Reverse voltage. This is simply the maximum voltage the coil will be driven with, which is 24 V in your example. There is nothing magic about the flyback catching role here.

2. Forward current capability. This is just the maximum current there will ever be in the coil when it is suddenly turned off. Your relay coil dissipates 700 mW with 24 V across it. (700 mW)/(24 V) = 30 mA. The current can't somehow magically be higher than that.

Note that for a relay that isn't regularly pulsed, you can use the maximum pulse current spec for the diode instead of the continuous current spec. Every time the relay is switched off, there is a finite amount of energy that will be dissipated by the diode. At most, the diode will dissipate only the energy stored in the coil. In practice, the coil resistance actually dissipates most of that energy.

One spec you didn't mention that sometimes does matter for flyback-catching applications is reverse recovery time. That is important when the coil may be switched on again before the current from the previous pulse has fully died down yet. This is common in PWM applications. In such cases, during the time the diode is still on but the new pulse has already started, the diode becomes a short across the coil. This can damage the diode, damage the switch, and/or draw too much supply current. At the low voltages in your example, a Schottky diode would be an obvious solution to this.

posted almost 4 years ago

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Olin Lathrop‭

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a coil with plain on/off functionality, for example a 24VDC relay coil with 700mW max coil power.

That means a coil current of 0.7 watts / 24 volts = ~30 mA.

A relay might have a coil inductance of anything from 1 henry upwards but this can be estimated by the relay activation time in the data sheet. It's approximate but you could assume L/R equals the activation time. Let's say it's 20 ms.

What does R equal? It equals voltage squared divided by power (700 mW): -

$$R = \dfrac{V^2}{P}$$

So, with 700 mW and 24 volts, the resistance is 823 Ω. Hence inductance equals ~16.5 henries based on L/R = 20 ms.

Energy stored is half of $0.03^2 \times 16.5$ = 7.425 mJ.

So, if the relay was toggled on and off continuously (20 ms to activate and 20 ms to deactivate) that is a frequency of 25 Hz and the worst case power dissipated in everything due to magnetic energy stored is 186 mW.

It's looking fairly trivial for virtually any diode that has a reverse voltage rating of 30 volts or above because most of that power would be eaten by the relay's internal resistance of 823 Ω.

But, if in doubt, you could easily simulate this circuit.

Q2: What about using TVS diodes as flyback?

When the relay is deactivated, the diode becomes forward biased so a TVS is ineffective in the normal position for a flyback diode. If, however, you wish to reverse the direction of the TVS in order to dissipate the power more quickly in order to turn the relay off more quickly, then you will still need a diode in addition to the TVS to avoid the TVS conducting when the relay is activated.

This time, the power dissipated will be higher than the previously calculated 186 mW because, in the worst case scenario, the relay could be energized and rapidly deactivated nearly twice as fast. So assume 372 mW (worst case) and assume also that most of this will be "spent" in the TVS (if you want to be cautious).

I'd still use a simulator because the time taken to set it up and get numbers is about the same length of time I took to write this answer.

posted almost 4 years ago

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4y ago

Andy aka‭

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