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I was studying some stuff here and I came across a question that I couldn't find on the internet. I created this circuit for demonstrating my problem:

Circuit under analysis

The transistor Q6 is a normal NPN transistor, the only difference is that it's maximum Vce is 30V. Now here is the situation: If I oscillate it, when it's ON the voltage drop will be on the resistors and there will be very low voltage between Colletor and Emitter. But when it's turned OFF the Collector wil be at 48V and the Emitter at 0V.

As long as the breakdown voltage is not reached we are good, but does the turning on of the transistor (that inicial drop between the 48V-30V) damages the transistor? I assume that the 48V drop on the turning on takes some time and in that time amount the transistor will be operating at a voltage higher than it should, is that true? Does it damage the transistor?

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  • \$\begingroup\$ There's a problem with the picture you have provided. The link appears to be expired. \$\endgroup\$ – KingDuken Jun 23 '17 at 22:41
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But when it's turned OFF the Collector wil be at 48V and the Emitter at 0V.

Is 48 V more than 30 V? Yes, then you have a problem.

The off state is exactly the state that's normally expected to cause the highest \$V_{ce}\$ and where the maximum \$V_{ce}\$ spec is most likely to become important.

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  • \$\begingroup\$ I know I'm late, bate thank you very much for your help! \$\endgroup\$ – Augusto Mattos Feb 19 '18 at 14:01

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