2
\$\begingroup\$

I want to connect a BJT as a zener diode, that is operating in the avalanche mode when the Base and Emitter is reverse biased. My question is if the data sheet states the forward current for Base and Emitter, is this current still relevant in reverse bias mode, or is it less for some reason?

Cheers

Kv

BJT connected with BE reverse biased

\$\endgroup\$
1
\$\begingroup\$

Forward current is very different from the allowed reverse current. Many datasheets don't list reverse breakdown voltage and even fewer list the allowable reverse current. It sometimes helps to check datasheets from different vendors when they produce the same transistor, but this is basically the vendor's way of saying: We don't support this mode, you are on your own.

Here is an example of a Motorola BC547 datasheet that lists emitter to base breakdown voltage V(BR)EBO as 6V minimum. The current at which it was measured is 10µA. If you use much larger currents, the transistor will likely be damaged. The absolute maximum rating for collector current is 100mA, maximum emitter current in forward mode will be similar in magnitude.

I don't know if zener diodes perform much better when it comes to noise, but I do know that a transistor used in avalanche breakdown is often used as a white noise generator.

\$\endgroup\$
  • \$\begingroup\$ I've also noticed not too much info on this topic as well. \$\endgroup\$ – kvresto Feb 24 '15 at 4:37
1
\$\begingroup\$

No, the maximum current will be less. Think about thermal considerations alone- the breakdown voltage will likely be in the 9V range for a modern Si NPN transistor so the power dissipation will be about 10x higher at the same current. A few mA should be okay.

You should know that there is degradation of forward beta when transistors are operated in this mode, so once you use it in this mode (for a length of time with significant current) you should not expect it to ever work quite as well as a transistor as it did originally.

\$\endgroup\$
  • \$\begingroup\$ Thanks for your input, valuable information, I know what to do now. \$\endgroup\$ – kvresto Feb 24 '15 at 4:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.