1
\$\begingroup\$

I want to drive a DC shunt motor like in the picture below. The PWM circuit with optocoupler, mosfet driver etc. is not shown. This circuit works fine on 12V DC motors. It even worked when tested on the DC shunt motor at 200V. But when tested at more than 300V, the IGBT broke. Does anyone have some advice on what I can improve on this circuit? After reading the question at Controlling a very small DC motor with PWM? I realised that I should move C1 into parallel with R2 and that D1 should be fast switching. Do you think this would solve the problem? I noted that when increasing the PWM, there are big spikes momentarily during the adjustment. I believe that one of those spikes broke the IGBT. I am going to implement a controller on the nucleo that I use for the PWM that allows the PWM to go up gradually, even if the potentiometer is instantly turned to 100% duty cycle. But I don't know if that would fix the problem? Thanks! Note: The diode at D2A is a BYV34X-600

enter image description here

\$\endgroup\$
  • \$\begingroup\$ Please ask a good specific question: A better question than "How do I solve the problem?" would be "How can I stop the IGBT from exceeding its maximum voltage rating?" The more speciific you are the better answers you will get to your questions: electronics.stackexchange.com/help/how-to-ask \$\endgroup\$ – Voltage Spike May 6 '16 at 20:39
  • \$\begingroup\$ What part number is D1? What is the purpose of C1, and is it really 22uF? \$\endgroup\$ – Bruce Abbott May 6 '16 at 21:18
  • \$\begingroup\$ D1 is a P600K diode, so I should it replace it with a fast switching one. C1 is really 22uF, I don't believe that it serves any purpose. I think it should be like the C1 in this answer electronics.stackexchange.com/questions/172948/… \$\endgroup\$ – Greig Swanepoel May 8 '16 at 13:39
2
\$\begingroup\$

Remove C1, because it just makes current to flow into IGBT and heating it without any benefit (don't know where you got such idea). Next replace R2,R1 with IGBT gate driver IC and other passive elements. D2A can stay if you want, D1 should be a fast/power diode and it has to remain there.

\$\endgroup\$
  • \$\begingroup\$ Yes, am surprised C1 didn't smoke. He was probably thinking he was trying to achieve a smoothed voltage, but for that he would have to have a choke etc. \$\endgroup\$ – www-0av-Com Jan 14 '19 at 17:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.