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During a test of a BJT totem pole driver I noticed that smoke began to come from my circuit. I made the assumption that it was from the BJT as the input current began to slowly climb - indicating thermal runaway. I reduced the frequency and it seems that the BJT again seems to work fine. However, the input power to the circuit seems to be somewhat higher than what is usual.

My question is, as a BJT begins to enter thermal runaway, is it basically dead and worth replacing? The power dissipation is still constant at low frequency, but as frequency climbs, the thermal runaway starts again.

Is there anything in particular experiences during a thermal runaway that could permanently damage the device in regards to the power dissipation it consumes, while it seemingly maintaining good performance at certain conditions?

I plan to replace the parts on Monday and run the tests again to see it this is the case, but I wanted a bit of a sanity check first.

Best, Jven

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2 Answers 2

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Power transistors under secondary breakdown focus heat on the amplifying channel which gets more conductive under additional heat. The area in question loses its properties and/or outright melts leaving a hole. The main result is a loss of hFE. If the thermal runaway does not result in a total loss of conductivity, it still can burn out the active area to a degree where only two independent diodes are left.

You should measure hFE if you can. If it is below spec, your power transistor is likely hosed to a degree where it will offload more of the workload to the driver transistor than it is good for longterm. If it blows its driver transistor, this will likely take out significant parts of the control circuitry as well. A power transistor alone should not be that expensive to replace.

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  • \$\begingroup\$ Thanks for your response. How do I test the HFE? In regards to switching performance, it would seem that it is working fine. I am driving the base hard with around 1A of current, and the rise and fall-times on the output capacitor would agree with a HFE of around 10 which is to be expected with the application (due to my current limiting resistor in the emitters). So it still seems to be switching on and off OK. But the power dissipation seems to be quite a bit higher than I would expect. \$\endgroup\$
    – jvnlendm
    Aug 18, 2023 at 23:37
  • \$\begingroup\$ Well, there are multimeters that do hFE but in-circuit you can just measure emitter current and divide by base current (and subtract 1): more likely than not there are resistors in the path (or you can put one there for measuring purposes) that allow you to guess at the current. \$\endgroup\$
    – user107063
    Aug 18, 2023 at 23:51
  • \$\begingroup\$ You can measure Hfe by applying a known current into the base and then measuring the current that flows into the collector. One way to do this is to use let’s say a 12V power supply; connect the emitter of the device under test to ground, a larger (1 to 10k) resistor between +12 and the base, then a smaller (100R) resistor from +12 to the collector. You can measure the voltages across each resistor and then calculate the base and collector currents. The ratio collector/base current is the Hfe value. \$\endgroup\$
    – Frog
    Aug 18, 2023 at 23:55
  • \$\begingroup\$ If the voltage across the collector’s resistor is close to 12V then the BJT is saturated and you won’t get a correct measurement, you’ll need to increase the base or decrease the collector resistance value. \$\endgroup\$
    – Frog
    Aug 18, 2023 at 23:57
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Just like any other electronic component, there is a Data sheet that has absolute maximums on it. Temperature is usually in that list of specs.

As with everything on the absolute maximum list, exceeding any value on it means under that condition, the manufacturer can no longer guarantee the part will perform to spec.

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    \$\begingroup\$ The temperature limit is usually not expressed in terms of "smoke begins to come". \$\endgroup\$
    – user107063
    Aug 19, 2023 at 8:41

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