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I built the H bridge shown in the schematic below for use with an unknown motor I pulled off of a broken CD player. I was seeing how hard I could drive the motor using large values for Vcc, and inadvertently applied well over the transistors' max collector/emitter voltage, 40V, albeit for a very short period of time. When running the same Arduino code I was using before I toasted the transistors (which ran the motor in the pattern left, off, right, off, repeat), the motor now seems to only run in one direction. It runs at full capacity for 1/4 cycle, and at low capacity for the other 3/4 of the cycle. I used an ammeter to confirm it was truly running at low capacity rather than just coasting for the 3/4 cycle.

In my research to solve the problem, I've become aware this is a poor H bridge circuit, but I would like to save it if possible, since soldering takes me a very long time. Given the behavior described above, is there an easy way to narrow down which transistor(s) are broken without unsoldering them and using the diode function on a multimeter?

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    \$\begingroup\$ High-side drive when using NPN or NMOS for the high-side transistors is your problem. You can look this up everywhere. It has to do with the transistor caring what is applied between base/gate and emitter/source, but your source pin is not at a fixed voltage while the signal you are applying to base/gate is always referenced to GND. It is like tossing potatoes onto a lift that is moving up and down as if it were always at a fixed height above where you are standing. \$\endgroup\$
    – DKNguyen
    Aug 23, 2020 at 20:33
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    \$\begingroup\$ Shoot-through (short-circuit between low and high side transistors in the same side between switching due to slow turn on/off times causing both to be on at the same time) is also an issue due to shared inputs, but first things first. Easily resolved with independent control of all 4 transistors. \$\endgroup\$
    – DKNguyen
    Aug 23, 2020 at 20:40
  • \$\begingroup\$ Add a PNP for high side drivere \$\endgroup\$
    – Tony Stewart EE75
    Aug 23, 2020 at 21:59
  • \$\begingroup\$ I know its a poor design but I'm more interested in whether given the behavior I described above there I can narrow down whether it is a high or low side transistor that malfunctioned \$\endgroup\$
    – Walden Marshall
    Aug 23, 2020 at 22:07
  • \$\begingroup\$ @WaldenMarshall You'll probably want a design more like this. Also, just so you know they are fairly safe to use, those BJTs mentioned there are spec'd at Vceo=150 V. So they should withstand some voltage abuse. That said, given Arduino I/Os, it's probably not good for more than about 200 mA (and you'd need to change resistor values shown there in order to achieve that.) And how fast you expect to change things also matters, too. But it's worth looking at a design, in broad strokes, that has a few added details to it. (You'd need two.) \$\endgroup\$
    – jonk
    Aug 23, 2020 at 22:25

2 Answers 2

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If you want a better bridge , use complementary power darlington emitter followers inside Op amp feedback loop on each side but invert the other side. Then use 12V to drive 24V-4 * 0.65V if using CMOS Op Amp to drive 10ma*hfe^2 but watch out for resonance with long leads.

Shootthru will not occur with this.

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I know its a poor design but I'm more interested in whether given the behavior I described above there I can narrow down whether it is a high or low side transistor that malfunctioned

Chances are it's the high side emitter followers that are damaged (one or both) and the reason is because fundamentally, this simplistic and naive design will never be able to fully turn on the high-side transistors because they are emitter followers and will only ever deliver a voltage at their emitters that is Vb minus 0.7 volts (irrespective of how high you make +Vcc).

So, when you raised +Vcc to 40 volts, and drew current through them, they would have a massive volt drop across them and fry. If the logic control levels were (say) 5 volts, then the emitter can only muster an output of 4.3 volts hence, there would be 35.7 volts dropped across them at your load current. If the load current is 1 amp (a pure guess), the power dissipated will be 35.7 watts in the high-side transistor that happens to have been activated. This means they/it would fry in an instant.

They should survive 40 volts but they won't survive the power/heat.

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