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I'm trying to convert a 0 - -500 mV signal to 0 - 1.1 V. I came up with the following circuit in LT spice. Is there anything wrong with it or can it be improved? Any way to get rid of the negative reference or create it simply, or have I messed it up in some other way? I don't have any knowledge about transistor circuits.

This is for a PSU voltmeter.

edit: I should have clarified: input should draw/supply less than 100 nA

so, requirements are

  1. high impedance input
  2. invert the signal
  3. scale from 0 to 1.1 V
  4. hopefully better load regulation than what I have

I have a few bjt and 1 single opamp (1/4 of a LM324) to spare.

enter image description here

edit: Here's a simplified circuit which uses an opamp for the gain and output. It seems the bjt inverter works just like an opamp inverter except the voltage at emitter, not at the base, gets inverted at collector. Thus, with no extra input bjt, there is a single pn junction increase in the output which can be lowered by an output bjt with similar current across it (and maybe gluing the two bjt will help minimize temperature difference). [With 2 pn drops this becomes 2 pn junction increase in output which is probably why I needed the darlington output in the schematic above. Also, every increase in gain will require an offset for Re close to Vcc/gain, to keep Vee in range. Thus, for gain of 1, no offset is required. That's my rudimentary understanding of the electronics.]

Input current is less than 37 nA. Output gain can be set at the opamp without disturbing the input.

enter image description here

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    \$\begingroup\$ Do you have to make it with BJTs? Have you considered using opamps instead? I should also say that the circuit seems transistor-dependent, which means that the circuit may give different outputs even for the same model transistors in practice. PS: I simulated the circuit in ISIS and I got 1.3VDC output for -500mVDC input. \$\endgroup\$ – Rohat Kılıç Sep 19 at 8:35
  • \$\begingroup\$ @RohatKılıç I can't spare more than 1 opamp, and I couldn't figure out how to do it with just 1. It doesn't matter to me if it is bjt model dependent. I'm only concerned with the load regulation which looks bad... but I don't know if the ADC load will vary during operation. \$\endgroup\$ – Indraneel Sep 19 at 8:51
  • \$\begingroup\$ @RohatKılıç with some more understanding of what is going on, I removed 2 bjt and added an opamp. Should work, I think, and no offset voltage to mess with. \$\endgroup\$ – Indraneel Sep 20 at 7:04
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Is there anything wrong with it or can it be improved?

As I stated in the comments, the circuit appears to be transistor-dependent. If you build up this circuit in practice, you may exchange the transistors until you find the required behavior.

I can't spare more than 1 opamp, and I couldn't figure out how to do it with just 1.

Well, all you need is a DC amplifier with a gain of approximately -2V/V. It's quite easy with only one opamp:

schematic

simulate this circuit – Schematic created using CircuitLab

I used an LM358 here but you can use any other opamps you have at hand.

The input stage with complementary transistors acts as a buffer with restoration (to compensate the VBE drop). So the voltage at the emitter of 3906 is nearly equal to the input voltage.

Without C4, the following stage is just an inverting amplifier. And the gain is \$\mathrm{A_V=-(R2/R1) = 2.2V/V}\$. The addition of C4 in parallel to R2 makes the amplifier a 1st order active low-pass filter to reject any AC components (above \$\mathrm{(2\pi \ 150k\ 1\mu)^{-1} = 1Hz}\$). So the circuit amplifies only DC.

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  • \$\begingroup\$ ah thanks ! but I'm sorry I forgot to mention, I also need a high impedance input (<100 nA). Your circuit is supplying 0-7.3 uA which is too much for me. So this will need another opamp for the buffer, unless you can do it with a bjt buffer as input.... then it may need a voltage offset, and it gets too complicated for me. \$\endgroup\$ – Indraneel Sep 19 at 9:53
  • \$\begingroup\$ @Indraneel then see the updated circuit. Note that an LM358 chip has two independent, common-supply opamps. If you mean "only one opamp IC" with "do it with just 1" then this may work for you. \$\endgroup\$ – Rohat Kılıç Sep 19 at 10:06
  • \$\begingroup\$ I have just 1 single opamp to spare (not a chip.... 1 LM324, 3 opamps already used up, 1 left)....... otherwise it wouldn't be a problem and no fun ! \$\endgroup\$ – Indraneel Sep 19 at 10:09
  • \$\begingroup\$ @Indraneel okay then. See the final circuit. The input stage with complementary transistors acts as a buffer with restoration (to compensate the 0.6V drop). So the voltage at the emitter of 3906 is nearly equal to the input voltage. Note that the input bias current is under 100nA. Try this one. If fails, I can't help because I'm out of options :) \$\endgroup\$ – Rohat Kılıç Sep 19 at 10:29
  • \$\begingroup\$ Thanks! It seems to be called a voltage mirror. I will check it out. Right now it still has a -100 mV offset (in LTSpice). I'll see if it can be fixed reliably. \$\endgroup\$ – Indraneel Sep 19 at 12:57

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