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In the below mentioned circuit which is used for automotive EMR braking. We are facing freewheeling diode burn out (FR607 is currently being used).

enter image description here

We are still clueless about the cause. But as per design this is a 6 A rated diode with surge value of 300 A which according to my design team is well enough. But we suspect that this ampere rated diode is not OK. We need to find the maximum freeweeling current in this circuit. How to calculate this current value. The coil resistance is 0.8 amps. Load of the circuit is 30 amps and the supply is a 24 V DC battery source 200 Ah.

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    \$\begingroup\$ 1. Resistance is not measured in amps. 2. Coil inductance matters so I don't know why you bothered to measure coil resistance without trying to find its inductance. 3. Battery amp hours does not matter. I think your design team needs to do some studying on what happens when you interrupt current through an inductor. Figure out the current devay waveform. Find out what the current at the beginning of the decay, at the end, and how long it takes with the L/R time constant. Calculate, simulate, or solve the differential equation. \$\endgroup\$
    – DKNguyen
    Commented Dec 27, 2019 at 4:40
  • \$\begingroup\$ Welcome to EE.SE. I have inlined the image for you and fixed the missing capitalisation. You could improve your question by adding a properly formatted link to the diode datasheet. How long and how frequently can the diode handle that current? \$\endgroup\$
    – Transistor
    Commented Dec 27, 2019 at 4:40
  • \$\begingroup\$ Dear Nguyen yes it's actually 0.8 ohms. Just a typo sorry. \$\endgroup\$
    – Gowthaman
    Commented Dec 30, 2019 at 7:16

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Since inductors work against changes in current. The maximum (peak) freewheeling current is 30A same as the load current. When the switch opens this current will flow through the freewheel diode. internal resistance in the coil and the voltage drop of the diode will cause this voltage to drop fairly rapidly.

One sure way to kill that FR607 is contact chatter, because it's only rated for 6A and a 50% duty cycle could put average 15A (or more) through it, so make sure you're not getting accidental PWM from the relay.

Perhaps use an automotive rectifier diode (like are used for altenators), depending on how often it is switched you may not need a heatsink

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  • \$\begingroup\$ Thanks Jasen. By any chance regarding the first sentence of your comment is there any theory or formula to calculate this free wheel current or the negative potential that developed across the inductive load in this circuit?? \$\endgroup\$
    – Gowthaman
    Commented Dec 30, 2019 at 7:14
  • \$\begingroup\$ I reasoned from "inductors work against changes in current" to get "the peak current will be the same", The voltage will be whatever voltage drop the diode produces. but the actual EMF produced includes that needed to overcome the coils internal resistance, so the current tails off fasten than just the reduction caused by the external voltage drop. \$\endgroup\$
    – Jasen Слава Україні
    Commented Jan 1, 2020 at 22:56

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