# Inductive energy when 12V relay coil is released

Maybe this question is non-sense. If I have a commom 12V relay (12V, 400 ohms coil) WITHOUT having a flyback diode connected to it, and driven by a low-side protected transistor/IC like ZXMS6004FF (the internal over-voltage protection would do the diode job). The datasheet of the transistor shows it withstands to a 90mJ "Unclamped Single Pulse Inductive Energy".

This is for a very specific project question. Would the transistor get damaged?

Datasheet of ZXMS6004FF: https://www.diodes.com/assets/Datasheets/ZXMS6004FF.pdf

You need to know the inductance of the relay coil. Then you can calculate the total energy of the coil by $$\ E= \frac {1}{2}LI^2 \$$ and compare that to 90 mJ.

est.L=4H +/-50%, I=12/400=0.03A $$\ E= \frac {1}{2}LI^2 \$$ =1.8mJ +/-50% seems well under 90mJ

• only that 1.8mJ? So the transistor will be ok! – abomin3v3l Feb 2 '20 at 10:38
• @abomin3v3l Don't jump to comclusions, he estimated L to be 4 H, even if it would need to be considerably greater to do damage. Maybe the word should be emphasized so that a quick glance cannot be tricked. – a concerned citizen Feb 2 '20 at 22:24
• Yes, it will be OK and will electrically switch at a much faster rate than with a Vdd diode clamp for dV/dt. The 50% was an estimated tolerance on my estimate. – Tony Stewart EE75 Feb 3 '20 at 1:26