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Low Side for BLDC circuit

I am trying to spin a BLDC using discrete components with an STM32F030C8T6 and using its TIM1 to control the high and low sides. High and low sides are controlled by NPN transistors that each control a pair of PNP transistors for MOSFET control. The problem I am facing is that the Q9 transistor in the above circuit doesn't turn on when applying a 20kHz 50% duty PWM signal to its base. The Vge is always stuck at 0.2 volts. I tried lowering the resistance to 500 ohms and the Vge is still at 0.5 volts. Why am I not getting enough current to turn the transistor on?

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  • \$\begingroup\$ What is "Vge"? A BJT dose not have a 'g' pin. Are you actually referring to "Vbe" - the voltage measured across the base-emitter junction? \$\endgroup\$ Commented Jun 23 at 1:00
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    \$\begingroup\$ Does it turn on without PWM? Simplify your problem, eliminating the unknowns. \$\endgroup\$ Commented Jun 23 at 10:36

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Reset to default
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Q8 is drawn as a PNP, yet S8050 is an NPN transistor. Which is it? Did you mean S8550?

In any case, Q9 appears to be reversed E-C. Presumably it's supposed to level-shift the logic level signal to the 15V level. 15V is more than the E-B breakdown voltage of most transistors, but, if you're lucky, not enough to destroy the GPIO etc.

I'm not sure what Q8 et. al are supposed to accomplish. Maybe you have a reference design you are trying to implement. There's no path to discharge the IGBT gate capacitance quickly. Perhaps a ground is missing somewhere. R12 has no obvious purpose.

Maybe this is what you had in mind:

schematic

simulate this circuit – Schematic created using CircuitLab

enter image description here

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  • \$\begingroup\$ I tried drawing the circuit in Kicad using a reference circuit. It is very likely that I messed up the drawing after comparing yours and the reference circuit. Could you please explain why the transistor is not saturating if I have the circuit similar to one you shared? Why is Vge stuck at 0.2 volts? \$\endgroup\$ Commented Jun 22 at 19:14
  • \$\begingroup\$ Assuming no measurement errors, the circuit is incorrectly built and does not match my schematic, or one or more parts are damaged. \$\endgroup\$ Commented Jun 22 at 19:26
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Pin 2 and 3 of Q9 are switched - your emitter should be connected to ground.

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  • \$\begingroup\$ Connecting Q9 emitter to ground would cause other problems as there would be excessive Q9D base current. \$\endgroup\$ Commented Jun 23 at 14:27
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    \$\begingroup\$ Ok sorry, not directly to ground but to the 1k5 emitter resistor. \$\endgroup\$ Commented Jun 23 at 15:51

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