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I am trying to build a circuit which is controlled from an analog input signal Vin to produce a -24V signal with >50mA output. My analog card can only produce ~10mA, therefore I generate the signal using an Op-Amp. I used the analog simulation tool from Microchip (MINDI) and the simulation result differs from my hand calculation. Simulation of inverting opamp

The circuit represents the standard layout of a inverting amplifier. I removed most of filter capacitors and other resistors from the design guideline. Building a non-inverting amplifier with the same principle gives the correct results. Is there a bug in the calculation of the simulation or does my circuit has a significant error?

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I am trying to build a circuit which is controlled from an analog input signal Vin to produce a -24V signal with >50mA output.

To produce -24 volts from an op-amp means that it has to have a negative power rail because an op-amp cannot conjure up any voltage that is outside the range of the power rails and, your negative rail is in fact ground (0 volts): -

enter image description here

The MCP6V51 is a rail-to-rail output device so, if the negative rail were (say) -25 volts you could expect -24.9 volts at best.

To produce circa 50 mA requires a more specialized op-amp because the MCP6V51 cannot normally be expected to deliver more than a few mA - it has a short circuit output current of -36 mA and, under normal load conditions, the data sheet implies about 5 mA in the table on page 4.

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  • \$\begingroup\$ I used the simulation for verification of my calculation. The opamp in the circuit is one of the MCP devices supported from the MINDI tool. \$\endgroup\$
    – v3xX
    Commented Jan 27, 2020 at 11:11
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    \$\begingroup\$ I don't believe what you have said affects my answer. \$\endgroup\$
    – Andy aka
    Commented Jan 27, 2020 at 11:12
  • \$\begingroup\$ My pick would be ti.com/lit/ds/symlink/opa192.pdf which should provide the necessary current \$\endgroup\$
    – v3xX
    Commented Jan 27, 2020 at 11:14
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    \$\begingroup\$ That's the short circuit current (into a zero ohm load). If you look at figure 20 it implies that the voltage dropped in the output stage when driving 50 mA is about 3 volts. You need to account for this and give your negative supply enough headroom (maybe make it -28 volts). However, on some loads you may overheat the device. If for instance you wanted to deliver 50 mA at (say) -10 volts on the output, the op-amp would be burning 0.05 x (28 - 10) watts = 0.9 watts and would go into thermal shutdown. \$\endgroup\$
    – Andy aka
    Commented Jan 27, 2020 at 11:36
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    \$\begingroup\$ If your load is always connected to ground (positive rail) then you can buffer the op-amp output with a BJT emitter follower and take negative feedback from the emitter (R1). Then you choose the BJT that has the appropriate thermal characteristics and copper surrounding it. \$\endgroup\$
    – Andy aka
    Commented Jan 27, 2020 at 12:16

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