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I have a power BJT NPN 2n3055 driven by 12V 25khz square pulses. The transistor is able to withstand 15A continuous current.

The capacitor is discharged when Q1 is open. Could it create a problem for the transistor?

Can the capacitor discharge be a potential problem for the pulse generator "V2"?

I actually built this cicruit and so far it's still alive.

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ About 42%, roughly. \$\endgroup\$ – PlasmaHH Feb 14 '17 at 15:24
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    \$\begingroup\$ When Q1 is open the capacitor is charging, it will not be discharged (except, perhaps, at the exact moment when Q1 opens) \$\endgroup\$ – DerStrom8 Feb 14 '17 at 15:29
  • \$\begingroup\$ Why do you think there may be a problem? \$\endgroup\$ – Chu Feb 14 '17 at 15:37
  • \$\begingroup\$ I still have issues understanding the DerStorm8's comment. I need to think about it. From what I understand, the Q1 in its open state sets its collector potential to the saturation voltage value, which is 0.9V in my circuit. The capacitor is discharged to that value via Q1. That creates the high current limited by the resistance of Q1 collector-emitter junction, which has dynamic nature. \$\endgroup\$ – Dmitry Pryadkin Feb 14 '17 at 15:45
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    \$\begingroup\$ Using "open" and "closed" to describe transistor states is confusing. We say a switch is open when it is not conducting, but many beginners seem to say that a transistor is open when it IS conducting. In any case, if the capacitor is only 68 pF, it can't contain enough energy to damage the smallest transitor. \$\endgroup\$ – Peter Bennett Feb 14 '17 at 17:27
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No, the 2N3055 is a relatively large power transistor. Its output capacitance alone is probably higher than 68pF.

A rough analysis: Because the current gain is rather low, with 120mA base current, the output current is limited to around 4 amps. The power dissipation due to discharging the capacitor, \$f CV^2 / 2\$ is around half a milliwatt.

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No and no.

68 pF is such a low value that even at 25V the energy stored is just about 21 nJ. Charge stored is 1.7 nC. For this charge to produce a > 15A current peak, it will need to go trough the BJT in just < 113 ps, which is ridiculously fast considering its transition frequency is 2.5 MHz and that the capacitor will have some ESR. Also, such a current pulse, if it could happen a all, would be well within the device SOA.

Summarising: it's perfectly fine, don't worry.

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Though the transistor act as a switch it collector current is restricted by the base current and ESR of the capacitor

With rough calculation the base current is approximately 120 mA, 2N3055 has a DC gain of 20 - 70, taking the maximum gain the max collector current can be 120mA X 70 = 8.4 Amps, the peak current of 2N3055 is 15Amps, so the transistor seems operating with in the safe limit

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