Question about G9 PSU

Hi everyone !


I'm currently learning how bipolar transistors work.

In the same time, I'm building a tube preamp and the PSU of my preamp is based on the G9 PSU.

Right now, I'm working on the phantom power. I plan to use the same circuit as written on the schematics of the G9. A voltage multiplier which will be "regulated" by a zener diode (56V) and a NPN.

My question deals with the NPN of this PSU. What will EXACTLY happen if the phantom power is short-circuited. What if I "forget" the 10k (R42) output resistance ? In those cases, what will happen to the transistor (blocked / saturated / other ) ? And concerning the current flowing into the zener ?

Please, give me all the details you can

Thank you in advance !

Then the transistor burns out. This simple circuit is not short-circuit-protected, as it's ment to work internally in-the-box.

Jakob E.

Thanks for your REALLY fast reply :grin:

But for example, if I put a 10ohm resistor in parallel with R42 (10kohm), is their a way to know the value of the current flowing into the 10ohm resistor ? What will happen to the 48V output ? Will it be kept at this value or will it decrease as the output resistance decrease ?

How can I calculate the critical resistance I can plug on the 48V without burning anything ?

An other question concerning the power dissipation of the zener diode (D6):
the maximum current which will flow into the diode is the same current that flows into R40 (2.2kohm) if nothing else is connected to the zener diode.
In this case, the current flowing in R40 will be:
(60 - 56) / 2200 = 1.81mA

This current is the same flowing into D6 so the power dissipated by D6 will be
56 * 1.81.10^-3 = 0.1 W

Does it mean that a 1/4W zener would be enough ?