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I was trying to drive the gate of 25N120 IGBT (2800pF gate capacitance) with the help of transistors. The working voltage is 10V. I have tried with BC548 NPN and BC558 PNP.

Everytime I apply PWM, the voltage level just remain 8V - 10V. The LED at output stays on. With positive trigger at base it shows 8V (seems NPN conduct try to ground the OUTPUT) and with negative trigger it shows 10V (seems PNP conduct and does its job.) The problem with the NPN why doesn't ground to 0V.

Even though I checked the output by implementing a 10nF capacitor, but the response is the same whether the output is connected or not.

MOSFET)IGBT drivers are not available here. The whole project was almost done, and now got stuck at this last part. My NPN and PNP configuration is in the photo.

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    \$\begingroup\$ Draw a schematics of your gate driver. \$\endgroup\$
    – Marko Buršič
    Jan 18, 2021 at 17:15
  • \$\begingroup\$ I added sir. The output is through the 15 ohm resistor. Pwm is not more than 800Hz. Transistor used bc548 bc558 \$\endgroup\$
    – Honey Singh
    Jan 18, 2021 at 20:09
  • \$\begingroup\$ IGBT = insulated gate bipolar transistor. MOSFET= metal oxide semiconductor field effect transistor. The things you are simply calling "transistors" are bipolar junction transistors. \$\endgroup\$
    – JRE
    Jan 18, 2021 at 20:35
  • \$\begingroup\$ That is not push-pull circuit. It is emmiter rollover. Used to amplify current. Switch transistors, pnp on top, enjoyed to +, npn on bottom, enjoyed to -. \$\endgroup\$
    – user263983
    Jan 18, 2021 at 21:57

1 Answer 1

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With such big gate charge your transistors are not best suited, also the Vgs of 15V would be more appropriate. You could buffer with additional transistor pair. Now, as for the correct order of transistors, you would mostly find your way for small voltage, but for higher voltage NPN has to be upper, and PNP is lower.

schematic

simulate this circuit – Schematic created using CircuitLab

schematic

simulate this circuit

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  • \$\begingroup\$ So should i have to apply one pair of transistors and one pair of mosfet in order to finally activate the igbt?? What number of p and n channel mosfets should i have to choose. And yeah i have BD140 BD139 transistor. are these two transistor can do standalone without the need of additional mosfet? \$\endgroup\$
    – Honey Singh
    Jan 19, 2021 at 2:56
  • \$\begingroup\$ You could use only a pair of transistors, but not the BC548/558, they can't provide such current. \$\endgroup\$
    – Marko Buršič
    Jan 19, 2021 at 9:56
  • \$\begingroup\$ thanks mate. so could anyone here provide the final circuit of Gate driver with the pair of BD139 BD140 transistors or Tip41C Tip42C pair. Max voltage i can provide is stable 10V for the gate driving circuit. and one more thing i want to keep the turn on off close to 1uS. this is why i hesitating to add 2 pairs of switching for gate cause i think that more i add transistors the more the delay time will be. \$\endgroup\$
    – Honey Singh
    Jan 19, 2021 at 14:01
  • \$\begingroup\$ @HoneySingh Well not so sure you will get 1us, try to see the IGBT gate charge and then calculate the time, note that you'll get 0.6V drop on BJT, so 10V might not be enough. Also you need a pre-drive circuit that will drive the base with those 10V. \$\endgroup\$
    – Marko Buršič
    Jan 19, 2021 at 18:09
  • \$\begingroup\$ Im usimg LM358 op amp to convert an ac pulse into the smooth square form which i think is good cause square pulse shows fully off and on. But im not good with opamp too. Somehow i made it but pnp transistor dosent work on it however optocoupler work. Not sure whats the problem. Its output is +ve and -ve as it should be but transistor dosent work. Lm358 has 2 inbuilt opamp. Im using single supply just +ve and ground. In case im able to use both opamp for both transistor so their bases dont interact with each other and i can be stressed out but i dont know what to do to make that work \$\endgroup\$
    – Honey Singh
    Jan 19, 2021 at 20:03

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