Need help to port a single transistor negative voltage regulator, from -15to-12

[workbench]

I have a schematic for a Negative Voltage Regulator.
However, the plan assumes an input voltage of -15V.
My power supply delivers only -12V.

Can someone tell me which values I have to adjust?


I think I have to change the values of R4 and R49. I do not know how to proceed.

[MarkF]

You are already using a LM7812 regulator for the positive voltage.
Just replace all the negative side with a LM7909.  Or use a LM7809 and LM7909 for +/-9VDC.

   

[workbench]

That's my plan too.
I already ordered some. But they will arrive in a few days.
I would like to test the circuit today.
All I have here are resistors Zener diodes and transistors.

[Zero999]

It's not a proper voltage regulator, but a potential divider (R50 & R49) with an emitter follower or buffer transistor (Q13).

V_OUT = R49*V_IN/(R50+R47) - V_BE

V_BE is the base-emitter voltage of the transistor, which will vary from 0.6V unloaded to around 0.8V, when it's passing a decent load current.

The above formula gives around 9.4V fully loaded. Changing R50 to 1k8 should give a similar output voltage with 12V in.

If you want to use the LM7809, then Q13, R50, R49, R4 and C4 will need to be removed. You might want to consider the LM78L09, which has a lower current limit of 100mA, which might be more suitable, than the 1A offered by the LM7809. Another option is the LM317L, which does require a couple of resistors, but gives better performance than the LM78L09.

[mikerj]

R4 and R49 form a potential divider, and the transistor provides current gain.  This means the output is a fixed fraction of the input so there is no voltage regulation i.e. if the input voltage changes, so will the output.

The output voltage can be approximately calculated as follows:
Vout = (R49/(R49+R4) * Vin) - Vbe

Vbe is the base-emitter voltage of the transistor, which will be approximately 0.6v.  Using the current values:

10k/(10k+4k7) * -15 + 0.6 = ~-9.6v

If you aim for the same voltage with a -12v supply you need a divider ratio of:

(-9.6-0.6)/-12 = 0.85

You could change either or both resistors to achieve this, but you should keep the impedance of the divider similar or lower (impedance=R49 in parallel with R4)

e.g. (10K/0.85)-10k = 1.76k so replacing R4 witk a 1.8k resistor would give you a very similar output voltage with -12v supply.

Edit: Sorry, a bit late in posting this!

[workbench]

thanks for the technical competence