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schematic I'm trying to build a sensors motor controller for a 500 watt bicycle hub motor. The problem I am having with it is when I connect power (4500 mAh 12s lipo) I usually destroy a transistor or two immediately.

At first, I figured the problem was too much current in the MOSFETs so I got bigger ones capable of 120 amps, 480 amps pulse (much more than the battery can even do) and it still broke the transistors, although not quite as loudly.

Next I figured that timing could be the problem, so I left the MCU disconnected when I applied Vcc - issue still occurred.

I then added pull-up/pull-down resistors on the bridge to make sure they weren't in some kind of floating phase - issue remained.

I also tried hooking up things in a variety of orders such as with/without the motor, with/without the MCU, but I still get the same problem.

Let me clarify a few things about the schematic:

High side are P-channel, low side are N-channel, 100k resistors on the pull-up/pull-down, schmidt trigger mainly to invert signal to P-channel so they are active high from MCU (ATmega328). No gate resistors (from what I read this only slows down switching which could make matters worse).

I really am lost as to what the problem could be here. I still haven't tried putting pull-up/pull-down resistors on the drivers or messing with the code too much, because I figure the code doesn't have much to do with it at this point since the issue occurs without hooking up motor or MCU. I wanted to stop for a minute and see if I'm missing something obvious before I go on, because I feel like I'm just rapid firing trivial changes at this point.

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  • \$\begingroup\$ What is your bridge deadtime during commutation? \$\endgroup\$
    – Tony Stewart EE75
    Mar 31, 2019 at 21:33
  • \$\begingroup\$ Gate diode//R is designed to speed up Turn Off to create deadtime. This is most likely your issue \$\endgroup\$
    – Tony Stewart EE75
    Mar 31, 2019 at 21:35
  • \$\begingroup\$ What would a gate diode look like? Is that a diode pointed towards the gate? There is no programmed dead time. However, I wouldn't think there needs to be because consecutive states would never allow for a shoot through. \$\endgroup\$
    – thisissparzo
    Mar 31, 2019 at 22:18
  • \$\begingroup\$ Could it have something to do with bootstrapping the p channel mosfet? Although I thought p channel was for high side switching, some would make me believe otherwise. \$\endgroup\$
    – thisissparzo
    Mar 31, 2019 at 22:19
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    \$\begingroup\$ BLDC drivers usually use N channel pairs (not complementary pairs); when the bottom N channel device (in your circuit) is turned on, it will conduct via the top P channel body diode which will result in large currents. I think the architecture is the issue. See digikey.com/en/articles/techzone/2016/dec/… \$\endgroup\$
    – Peter Smith
    Apr 1, 2019 at 8:08

2 Answers 2

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There's a lot amiss here. The TC4427 is only rated to 18V on the Vdd pin. You don't show what the Vd supply is set at, but it needs to be at the Vcc that is applied to the top FETs so that it can pull the gate up on the P channel FET to turn it off. Vcc with a 12s battery is going to be 40-45V... If Vdd is connected to Vcc, you'd then (assuming that the TC4427 hadn't failed) be applying the full 40+V to the gates - these are only good for 20V. You'd need to generate a high side gate drive supply at Vcc less a suitable drive voltage (10-15V). Lastly, and certainly not least, look at the body diodes on the output FETs - the high side one will always be conducting. The source and drain pins should be swapped.

I'd suggest looking up a reference design for a bridge and gate driver circuit. You can find ones that use drivers that don't need the inverters, have shoot-through protection (that prevent both FETs being on at the same time) and incorporate a bootstrap supply for the high side FET). Most use N channel only, since the Rds of the N channel FETs is a lot lower.

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  • \$\begingroup\$ I had a separate supply for the drivers. It was 18 volts. I think you found it with those body diode directions. I never questioned them but now that makes a lot of sense. I had more than a half dozen people look at this and no-one noticed that either. Awesome! I'll try swapping those around. \$\endgroup\$
    – thisissparzo
    Apr 13, 2019 at 15:41
  • \$\begingroup\$ You're still not going to be able to turn the top FETs off if you have 18V on the drivers - you need to get Vgs well below Vth, i.e. above Vcc-Vth relative to the ground of the drivers if you stick with p-channel FETs. I can't find any drivers that are configured for p-channel gate drives that are rated above 36V. \$\endgroup\$
    – Phil G
    Apr 13, 2019 at 16:58
  • \$\begingroup\$ Oh, what if I can just use external power supply for it and invert output? \$\endgroup\$
    – thisissparzo
    Apr 13, 2019 at 19:39
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enter image description here

As soon as you turn the N channel MOSFET at the low side of your h bridge on, the body diode of the P MOSFET becomes forward biased, and a large undefined current will flow, probably taking them both out.

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  • \$\begingroup\$ Yes, I think you are right, as someone else also said. Thank you for your time. Will build another. \$\endgroup\$
    – thisissparzo
    Apr 13, 2019 at 15:42

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