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Let's consider this device I have built: It is essentially a power supply for a 24V DC motor. The power supply is switched on and off according to a proper control signal (which drives a relay) which is not important at the moment since it has always been put to ON while the device was working.

enter image description here

Let's focus on the final part of the circuit, with the motor, the fuse of 10A and the transistor BUV20.

The motor should require an average current of 8A at 24V and BUV20 can provide 50A. BUV20 is cooled through a fan (shown at left in the picture) and a metal heatsink (with thermal paste.) The motor's nominal voltage is 24V, but I have inserted a potentiometer in order to decide its value (generally I will use a voltage between 17-20 V since I do not need its maximum speed.)

The device was working at these values of voltage, then it suddenly started to rotate at an exaggerated speed. Then, the 10A fuse stopped it after some seconds. (Maybe since it is a delay fuse.)

After this event, I checked the fuse of 10A and it is broken. Then I have measured the collector - emitter resistance of BUV20, and I have seen it is 0.

So, my opinion is that BUV20 has broken while it was working, and its collector and emitter terminals shorted. So, 27Vpeak were provided to the motor, which started to run at exaggerated speed.

What might be the cause of this transistor problem? I may change it with another BUV20, but I think its accident may not be casual, and there may be a design problem. So, what may be the cause of collector and emitter short while the transistor is working? A motor spike, a transistor defect etc. Do you have any advice?

UPDATE:

enter image description here

Under your great advises, I have done the following modifications in my circuit (in red), with also other additional elements I have put by myself:

  1. A 100nF CAPACITOR on the biasing network of BDX53C, to avoid fast variations of the biasing voltage when moving the potentiometer.

  2. The original delayed fuse of 10A has been replaced by an INSTANT FUSE OF 10A. This because the delayed fuse has burnt only after some seconds, and that time has been enough to break the transistor.

  3. A REVERSE DIODE has been put in parallel to Base-Emitter Junction of the BUV20.

  4. A ZENER DIODE OF 24V, 5W has been put in parallel to the motor. This is not a complete protection, but follows this logic: if BUV20 breaks and collector and emitter become shortened (as happened the last time), the zener diode puts on the motor only 24V, and not potentially 27Vpeak of the collector voltage. This zener will not survive a lot in this condition, but it gives time to the other protection systems for activating (note that 24V is the maximum voltage of my motor, so in this time the motor will be safe).

  5. A COMPARATOR which detects a short between collector and emitter of BUV20 due to its break.I have designed the resistors so that, if the emitter voltage will be over 23V (it should not happen in my operation conditions), the comparator output will become low and a digital logic (precisely, an arduino system) will switch off the relay that supplies the BDX53C.

What do you think about these measurements? Should I modify them or insert other protections?

My main aims are, in decreasing order of priority:

  1. protecting the motor
  2. protecting the BUV20
  3. protecting other elements
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    \$\begingroup\$ Motors are an inductive load so when switching on/off there can be damaging voltage peaks. I would want to see a flyback diode in parallel with the motor to keep those peaks in check. But why aren't you using a switched/PWM type solution to control the motor's speed. Your design looks very "vintage" in that respect. \$\endgroup\$ Sep 4, 2020 at 8:24
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    \$\begingroup\$ This isn't a good circuit for so many reasons. But you just want to know why it is breaking. You have clues, good ones, here already. But let me just ask a question. Have you attempted to thevenize this into a simpler circuit and then analyze it? Is it possible that you could start to run out of beta? Assuming for a moment that's possible, how would the motor inductance react in response? \$\endgroup\$
    – jonk
    Sep 4, 2020 at 9:08
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    \$\begingroup\$ I would not even consider that circuit. So I can't tell you what I'd do to fix the beta. I was just asking you to imagine what might happen if the beta starts to collapse (it is not fixed vs collector current.) If the motor current rises, and the beta not only is exhausted but starts to move oppositely in response (which it can do), then the inductor will suddenly find itself in an impossible situation. How do you think it will react? \$\endgroup\$
    – jonk
    Sep 4, 2020 at 9:21
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    \$\begingroup\$ If the current must then decline, the only possible response by the inductance in the motor is to reverse it's applied voltage. That's the only way the current can decline in order to meet the beta available. But the reversal means the BUV20 emitter is yanked hard negative. This creates a very large VCE at high current, extremely high instantaneous dissipation, extremely rapid internal point junction heating with no time to spread out to the case. \$\endgroup\$
    – jonk
    Sep 4, 2020 at 9:28
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    \$\begingroup\$ Note that I didn't assume you turned things off. It's already a problem without that added issue. Adding that issue just complicates it further. Motors, especially 8 A motors at 24 V and so about 200 W, aren't a piece of cake to drive. Your circuit is way out of its league. \$\endgroup\$
    – jonk
    Sep 4, 2020 at 9:38

3 Answers 3

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which may be the cause of this transistor problem?

  1. Too much heat generated and the transistor internals suffered melt down.

You didn't mention the amount of heat-sinking you had applied to the transistor so this is a definite maybe.

  1. Back emf from the DC motor exceeded the reverse bias for the base-emitter junction.

This is rated at -7 volts so be aware of this. Try putting a reverse diode across base and emitter. I would also put a 100 nF capacitor across the motor because if the potentiometer made a slight stutter and produced a wildly changing demand voltage at the emitter the back-emf generated from the motor's inductance will be high (potentially hundreds of volts).

Put a 100 nF capacitor across R2 as well (to keep the pot output steady).

Also check that the BDX53C hasn't suffered - this could be another clue - maybe the BDX53 went pop (due to reverse base-emitter voltage (it is only rated at 5 volts) and, in turn, destroyed the BUV20 output transistor.

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    \$\begingroup\$ You should be able to get away with a 1N400x diode across base and emitter. \$\endgroup\$
    – Andy aka
    Sep 4, 2020 at 12:15
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    \$\begingroup\$ It shouldn't need one - it would go across base and emitter and doesn't need a resistor. Maybe if you have a cunning idea you should draw a circuit and let me look? \$\endgroup\$
    – Andy aka
    Sep 4, 2020 at 18:27
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    \$\begingroup\$ @Kinka-Byo If you found this answer solved your problem then you should accept it as the correct answer.. \$\endgroup\$
    – user2121
    Sep 4, 2020 at 19:48
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    \$\begingroup\$ @Kinka-Byo there are several ways to do this and my recommendation across base and emitter directly address why I think the transistor broke. You may find that you do need both methods. Again, if you come up with a scheme, post it and I'll take a look. \$\endgroup\$
    – Andy aka
    Sep 5, 2020 at 12:47
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    \$\begingroup\$ To protect an NPN transistors base emitter from excessive reverse bias, the anode connects to the emitter. Remember that you may need to do other protections too and I did advise you to draw a circuit and let me see it. \$\endgroup\$
    – Andy aka
    Sep 11, 2020 at 16:35
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Usually the load is on the collector side when being switched. Now there is more losses as the load is driven with an emitter follower configuration.

Another thing is the darlingon is directly connecting the power transistor base to supply voltage without any current limiting. It is possible that turning the power transistor on exceeded the Vbe and Ib of the power transistor and it broke.

There is also nothing to prevent inductive kickback from destroying the power transistor when the load is turned off.

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Two obvious potential failure causes (assuming adequate heatsinking on the transistor. As there is Vbe * 3 or at least 2V across the transistor by design, you are wasting a lot of power here, at least 16W at 8A. In fact even at rated conditions, 25V in, 17V motor, 8A current, you are dissipating 64W in the transistor. With Theta(j-c) = 0.7C/W, the junction is 45C warmer than the transistor case, which is warmer than the (unknown) heatsink ... get the picture? And that's at 8A ... but see below)

  1. Reverse breakdown voltages during inductive spikes. This could be base-emitter reverse breakdown ... only a few V as Andy points out. A reverse diode across B-E only passes the problem back to the driver transistor. Put one to the supply as well, and note these must be fast diodes not 1N4007s.

  2. Excessive current (combined with Vce>2V and the SOA curve in the transistor datasheet). You assume the motor is only taking 8A yet when the transistor goes short circuit, the 10A fuse blows. So clearly the motor current is exceeding 10A. On starting, an 8A motor may even exceed 50A : check its stall current rating in the datasheet. It will momentarily take this, and if excessively loaded, will take more than its rated current while running.

So it may be momentarily taking more than 50A (pop!) or it may be taking enough while running that, combined with the rather high Vce (either from the driver, or the fact this is a linear regulator) the SOA is exceeded.

And there's no Safe Operating Area (SOA) curve in that datasheet, nor secondary breakdown specs, so ... how to determine SOA violation?

Measure the startup and actual operating current to determine if these may be the problem.

It may be better so simply move to a PWM motor controller, saving a lot of power, rather than fixing this one.

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  • \$\begingroup\$ Thank you for your answer. Can you tell me why should a pwm motor controller solve these issues? If I control the base voltage of BUV20 with a pwm signal, why should this solve these problems? \$\endgroup\$
    – Kinka-Byo
    Sep 4, 2020 at 13:32
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    \$\begingroup\$ Because it consumes about 60W less power under your rated conditions, running the transistor that much cooler., and it will be designed with driving motors in mind. If you PWM the base of the BUV20, make it a low-side switch. You'll get the power saving, but you'll still have to take care of the protection yourself. \$\endgroup\$ Sep 4, 2020 at 13:54

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