I don't know what the circuit is called; I found it in the suggested schematics of a Resolver to Digital Converter IC.

The circuit is an amplifier circuit that has two push-pull amplifiers and a bunch of op-amps. The LTspice simulations give out clipped waveforms. How can I fix this?

How do I go about debugging this circuit? The output waveforms are getting clipped.

I'm pretty sure the problem is with the op-amps, even though I use the manufacturer supplied SPICE model of the LM2902 op-amp. When I use the universal OpAmp2 model, I get a 180° phase shifted sinewave with a lower amplitude. There is still some clipping occurring in the output waveform.

Waveform with clippings, the desired output is 14 V sine wave: Waveform with clippings, the desired output is 14 V sine wave

LTspice schematic, I have used the manufacturer supplied op-amp model: LTspice schematic, I have used the manufacturer supplied op-amp model

Reference schematic: Reference schematic

Desired output: Desired output

EDIT: For context I'm simulating a section of PCB of an RTDC Evaluation board, the circuit was suggested in the official datasheet of the AU6805 IC [page 17]. The IC would give a sinusoidal excitation signal which would be amplified and would be given to the Resolver Circuit. The Circuit will be integrated onto a Motor Controller. My job is to debug the circuit, starting by simulating it.

KiCAD schematic export

  • 2
    \$\begingroup\$ The load resistors are way too small. You're pushing 1A peaks through those poor transistors. Why are there 10 ohm (and lower) load resistors? Other than that it looks OK I think. Push-pull amplifiers that use follower stages are not rail-to-rail. Are op-amps you use rail-to rail?. Decrease the gain and there won't be clipping. I'm not quite sure what you want to achieve with this circuit. And, please. make the schematics readable. They are very fuzzy, especially the second one is unreadable. In KiCad, go to File->Export->Schematic to clipboard and it should make a nicer image I'd hope. \$\endgroup\$ Commented Jan 27 at 19:38
  • \$\begingroup\$ found it in the suggested schematics of a Resolver to Digital Converter IC Why do you think the particular circuit will do what you need it to do? Please, first of all, state what you're trying to do with this circuit. At the moment it is not clear at all. Please address these comments by editing the question, not replying to comments. \$\endgroup\$ Commented Jan 27 at 19:40
  • \$\begingroup\$ @Kubahasn'tforgottenMonica I've updated it now, let me know if I need to modify anything. \$\endgroup\$ Commented Jan 31 at 6:31

2 Answers 2


Look at this data sheet for the LM2902. Search on "output swing". Note that it only pulls the output to \$1.5 \mathrm V\$ of the supply rails.

Your simulation is accurate for those op-amps.

The circuit is just an op-amp with a discrete-transistor output stage: there's no special name for it.


This design has several problems that prevent it from being a serious power amplifier. This would be ok as a 600 ohm line driver, but would still have a output voltage several volts less than the supply rails.

A) As already mentioned, the op-amps output will be 1.5 volts less than each supply rail. Keep the load above 100 ohms, or even 1,000 ohms with these small transistors.

B) The output diodes partially cancel out the Vdrop (b-e) of the output transistors, but the resistors on their bases would normally be replaced with constant-current sources and sinks. As the op-amp output approaches V++ (or V--) the resistor supplies less and current to drive the transistor.

Remember that the diodes prevent the op-amp from directly driving the outputs transistors with current. They provide a somewhat constant bias voltage to the base so that they have an idle current of several milliamps, else you would get severe distortion.

C) A more common design would have Darlington transistors for outputs so less drive current is needed (each diode would become 2 in series), plus use of an op-amp with higher voltage supply rails to offset the Vdrop in the outputs.

D) The Darlington outputs can drive a much lower impedance load, but will consume more current, up to several amps. Large aluminum heatsinks would be required.

NOTE: If this is only meant to be a simulation, you are limited to the parts in the simulators library, and their voltage and current limits. If you can track watts dissipated at the outputs, please do so.


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