# Ideal diode - PNP emitter-base breakdown voltage question (PMOSFET)

I'm hoping to create this ideal diode. With 12V.

However, I feel like this would require the two PNP transistors to have an emitter-base breakdown voltage of 12V or more. Is this correct? (Such PNP transistors do not seem to exist.)

(source: birds-are-nice.me)

This page seems to strongly suggest that this can be used with 12V.

https://birds-are-nice.me/ipfs/QmSYsdRafbq19UzocGmWnkt9Gn3CuZga2TXvDKXFGQtJvw/

Does this circuit require emitter-base breakdown voltage of 12V or more? If not why not? And if so, what alternative can I use for a 12V ideal diode?

I was also thinking of using 4.7K for both R2 and R3. Is there a reason that they are different values in the diagram?

• This page seems to strongly suggest that this can be used with 12V. Then you see more than I do, there's talk about 5 V but nothing higher. Commented Dec 14, 2017 at 15:02
• "If you're using one of these circuits to prevent a 12V lead-acid battery from discharging, which is the mot common application, that's not a problem". Which kind of suggests that this would work on 12V. However, is it your understanding that it wouldn't? Commented Dec 14, 2017 at 15:27
• That 12 V is mentioned in relation to the max. gate-source voltage of the MOSFETs. The author might have overlooked the fact that the Vbe breakdown is an issue with the circuit from your question. And Vbe breakdown will happen at 12 V when the "diode" is in reverse mode so 12 V at the right side. Then the input will be pulled up to about 6 V through the BE of the right transistor (in forward mode) and BE of the left transistor (in reverse mode and breakdown). If there is a load to ground on the left side then it will receive current this way due to the breakdown. Commented Dec 14, 2017 at 16:28
• If you doubt this, simulate the circuit in a simulator and check what happens. Commented Dec 14, 2017 at 16:29