Fly Back Diode Selection - Page 1
Forum section: electronics / beginners

author: jbeach
author_rank: Newbie
timestamp: April 25, 2018, 10:19:54 pm
content: Hello,I wish to build a circuit to control a DC motor. The motor pulls Approx 14 Amps at 12 VDC.When selecting a flyback diode, am I good just selecting one that has a surge rating of greater than 14 amps? Or is there other things that I need to consider when selecting.Also, I have a 409 Amp 12VDC motor that I occasionally use. The motor has it's own built in controller. However, In the power line, I installed a solenoid with contacts rated at 500 Amps. The motor is intermittent duty. The on time is less than 60 seconds. This motor runs the winch on my Jeep. The purpose of this solenoid is so that should something get wet and short out in the motor/winch controller the winch will not drive uncommanded. In normal circumstances the solenoid will only turn off after the winch stops driving, and thus only interrupt the current for the indicator light, however it is rated to interrupt 500 amps. However, if I do have to interrupt up to 500 amps while the winch motor is driving, what sort of flyback protection should I use? Should I find a diode with a surge rating of 500 Amps?Thanks,-Jonathan

author: Jwillis
author_rank: Super Contributor
timestamp: April 26, 2018, 04:29:23 am
content: Hey hows it goingWhen sizing a flyback diode you need to know the power being dissipated into the diode.That can be calculated first by knowing the inductance of the motor coil in micro Henrys .Do you have an LCR meter ? Then you can calculate the energy (Joules) stored in the coil withE= energy in Joules L=Is Inductance in micro Henerys I = AmpsE= L*I^2(squared)/2I have a motor that has similar characteristics 12 v 14 ampIt has an inductance of of 177 uHSo the energy stored in the coil E = 177 * 14^2/2 = 177 * 196 /2 = 17346 Micro Joules = 0.017346 JoulesSo we'll use a millisecond to calculate the watts because the inductive discharge actually decreases exponentially.So the energy decreases over time .That gives us 17.346 watts for the first millisecond of discharge .WOE! your thinking but don't freak out that,s not much because that's not continuous.The internal resistance of the diode will slow that discharge down over time .A half second discharge is around 0.034692 wattsYou could use a 5 watt diode with out any trouble. Even a 3 watt can handle that short burst. But I wouldn't go much smaller.Also a fast switching diode would preform better than say a rectifying diode. A 3 watt 1N4148 fast switching diode or a 1N400X would be fine.

author: jbeach
author_rank: Newbie
timestamp: April 26, 2018, 11:09:51 pm
content: Yes I have a (Cheapo) L.C.R. meter and can measure the inductance of the 12VDC 14 Amp motor.What I cannot measure is the inductance of the 12VDC 409 Amp motor. I wonder if I could just contact about it. (It's a Warn Zeon 10S Winch)I've never really looked at watt ratings on a diode before. I figured their watt rating would simply be their continuous forward current rating multiplied by their forward voltage drop.I brought up the datasheet from Digikey for a 1N4001-G diode and was unable to find a watt rating. The datasheet for the 4148 shows it rated for 1/2 watt (500mW)I think I'm missing something.Thank you for the great info. I think it gives me enough to go on though.-Jonathan

author: Ian.M
author_rank: Super Contributor
timestamp: April 27, 2018, 12:58:54 am
content: Worst case, the back-EMF suppression diode has to carry a surge that peaks at the motor stall current. That's easily calculated from the supply voltage and the winding resistance of the motor. Take a couple of readings, disconnecting and turning the shaft a bit in-between to get the lowest reading. For very high current applications you may want to also add in the cable resistance (from CSA, length and resistivity of copper), and unless care is taken to minimise the loop area (and thus the cable inductance), a TVS diode across the contactor may also be advisable.There's little benefit in using a fast diode unless the motor (or other inductive load) is being PWMed, as diode speed is usually a reference to its reverse recovery time, and in a non-PWMed back-EMF suppression application the dissipation during reverse recovery isn't generally an issue because of the very low frequency of polarity reversals. However the lower Vf drop of a Schottky diode may be benificial in high current applications to reduce the peak power dissipation.