0
\$\begingroup\$

I have a circuit which produces a (not high speed) signal between 0 and 9 volts with an adjustable slope, and I would like to increase this output voltage to between 0 and 200 volts, which keeping it ground referenced.

Currently, I am using a PNP transistor connected as a common emitter, which is controlled by an NPN transistor and the output of the lmv358 op amp (the spice model I'm using comes from this website. Here is the schematic in LTSPICE:

LTSpice Schematic

Everything works as expected when I run an operating point simulation, but when I try a transient simulation, I get huge oscillations and the simulation rarely has time to finish:

Huge oscillations on LTSpice transient simulation (this is the voltage accross R3; it should be 4 volts).

What have I done wrong? I haven't been able to find similar circuits online, which suggests that maybe I'm missing something sort of obvious. How is something like this usually done?

Thanks!

| improve this question | |
\$\endgroup\$
  • 2
    \$\begingroup\$ With 200V split across R2+Q3 , They share 40 Watts and are burnt up. Don't ever design a generator or power supply without power & load ratings \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Jan 17 at 1:15
  • \$\begingroup\$ What will happen when 200 V are directly applied to a 0.7 V base-emitter junction? \$\endgroup\$ – Circuit fantasist Jan 17 at 4:09
  • \$\begingroup\$ @Circuitfantasist When it gets to Vbe gets to 0.6V (600uA) the feedback is supposed to throttle it, so the maximum dissipation is ~120mW in Q3 and diddlysquat in R2. But it's not stable, that's the main problem. In reality you'd add an emitter resistor to Q3 to limit the current if you used that bit of the OP's circuit. \$\endgroup\$ – Spehro Pefhany Jan 17 at 4:33
  • \$\begingroup\$ @Spehro Pefhany, However, some resistor before the Q5 base would not be redundant... \$\endgroup\$ – Circuit fantasist Jan 17 at 4:48
  • \$\begingroup\$ But it seems R3 is redundant... it can be replaced by the R4-R5 voltage divider. \$\endgroup\$ – Circuit fantasist Jan 17 at 7:36
2
\$\begingroup\$

You’ve added a whole lot of gain and some phase shift, both of which negatively affect closed-loop stability, notwithstanding the internal compensation in the op-amp. How it is done depends on the requirements, frequency response and load driving capability in particular. Driving piezo actuators, for example, is more challenging because they are capacitive and add even more phase shift.

Speaking of the op-amp, you are applying almost double the absolute maximum supply voltage of 5.5V to it, so it would likely fail instantly in real life.

| improve this answer | |
\$\endgroup\$

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.