I am trying to recreate a circuit that is mentioned here, here and in various other sources.

For the project I intend on using this in it would be beneficial to be able to switch output voltage from GND to 8V. Maximum expected current on this output will be ~20mA.

What I came up with is this:

enter image description here

(544UD1 is a dual supply not rail-to-rail op amp that I have available,2T3117a is an NPN bjt: UcbMax=60V;UceMax=60V;UebMax=4V, Vref is done with voltage divider instead of Zener as in source schematic, switching on/off is done with analog switch placed on reference voltage)

LtSpice simulation gives following results: enter image description here

What I am worried about is that Qbase drops to -15V while QEmitter is in range from GND to 8V. This seems to violate UebMax rating.

I can also use a single supply op amp (for example OP777): enter image description here

This however does not fully address the problem (note time ~28ms) as Qbase can drop faster than Qemitter:

enter image description here

Simulation does not seem to mind this spec violation but I am not sure what happens in real life at this point.

Is there a simple way to address this issue?

  1. Should a different BJT be used? (Though many small BJTs have Ueb in range of 4V to 6V as far as I am aware)
  2. Should a different Op-Amp be used? Strictly single supply or dual supply can be used as well?
  3. Should an analog switch be placed elsewhere?
  • 1
    \$\begingroup\$ Just as a matter of style, most circuit designers usually show current as flowing from the top of the page or block to the bottom. Reversing this convention can lead to errors when people misread your schematics. \$\endgroup\$
    – TimWescott
    Commented Jul 6, 2022 at 22:54
  • \$\begingroup\$ What are the desired rise and fall times of this output? \$\endgroup\$
    – Jens
    Commented Jul 6, 2022 at 23:25
  • \$\begingroup\$ @Jens as this is a power supply and not a signal the rise and fall are not critical. Several ms should be fine. A device this is supposed to be powering requires a sequential power up. \$\endgroup\$ Commented Jul 7, 2022 at 5:18

2 Answers 2


A simple fix is to add a diode in reverse parallel with the Q6 base-emitter. This will clamp the reverse voltage to 0.7 V.

Adding this diode means that the opamp is discharging the load cap through the diode and R14. I'm not familiar with the opamp in the schematic, and some opamps have a problem with stability when driving a capacitive load.

Because the output current is only 20 mA, consider driving it with the opamp output directly. Again, this depends on the characteristics of the opamps available.

  • \$\begingroup\$ Thank you! I've increased R14 to 2k, reduced C3 to 1u and added 1N5819 as you suggested. Simulation seems to work great. Remains to be seen how this performs when built. \$\endgroup\$ Commented Jul 7, 2022 at 5:41

Much depends on how fast you need the output V8V to change when you switch off the reference. As it stands, Q6 will charge C3 (10 uF) very quickly (and perhaps cause excess current in Q6), but when the reference is removed, the discharge path for C3 will be through R18 (1k) and through the E-B junction of Q6, base resistor R14 (100R), and the Op-Amp. You could add a diode from emitter to base, as suggested, and then the discharge will be through R14. This could be as much as 80 mA (and 230 mA if you use the -15V supply), which is too much for the device. I would suggest increasing R14 to at least 1k, and perhaps reduce C3 to 1 uF, which will result in a TC of 1 mS. If you need faster discharge, you might be able to use an NPN/PNP pair to drive C3 between the 15V rail and ground. A simulation may help:

Capacitor driver simulation


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