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I'm trying to measure a differential voltage in LTSpice using the following structure: enter image description here

The simulation results are correct, the voltage at the output is very similar to the differential input voltage. My question is: Why are the 40k resistors R7 and R9 needed? When I remove them, the results are not correct (see second picture), and with some OpAmps the simulation will even fail.

correct results

These are the waveforms when I remove R7 and R9. But that is the typical differential amplifier structure. Can anyone tell me why this happens?

enter image description here enter image description here

Edit: this is the full schematic including Vcc and Vss: enter image description here

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  • \$\begingroup\$ Is the GND symbol to the right of D3 actually connected? If so, move it downward one or two grid points to show a net. Whether or not it is connected changes the DC relationship between V2+V3 and the diff amp. \$\endgroup\$
    – AnalogKid
    Commented Jan 24 at 14:04

2 Answers 2

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Can anyone tell me why this happens?

You are applying a sinewave of amplitude 48 volts p-p to a differential amplifier that has a gain of unity powered from a +/- 3.3 volt supply. The fact that you have "limited" the sinewave with an attenuator and diode doesn't automatically mean that the offset seen on Vdiff+ isn't going to push the op-amp into excessive saturation.

You need to make R8 and R10 at least ten times lower in value. The clue is that your output waveform is badly clipping.

To prove this to yourself, view the waveform Vdiff+.

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  • \$\begingroup\$ Thanks for answering, but I don't think the op-amp is saturating. We are supposed to be measuring a square wave, not a sine wave, since the bridge rectifier makes the voltage at the measuring point switch between 0 V and 9.6 V. I have checked the voltages at the inputs of the op-amp and they are within the supply levels. \$\endgroup\$
    – FernandoPT
    Commented Jan 24 at 11:40
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    \$\begingroup\$ Without the extra 40 kohm resistors (that you used to attenuate the inputs), plot the voltages on the inputs to the op-amp. They should be near identical and well within the power rails of the op-amp. If they are not then this is your problem (an offset as per the mechanism outlined in my answer). \$\endgroup\$
    – Andy aka
    Commented Jan 24 at 11:57
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    \$\begingroup\$ You are right, I measured the voltage at the inputs without R7 and R9 and it is outside of the power supply. Thank you! \$\endgroup\$
    – FernandoPT
    Commented Jan 24 at 12:12
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Notice the sensed voltage (at least Vdiff+) is clamped by D2/D3 between 0 and 9.6V (or rather a Vf beyond that range). Suppose it's at +9.6V, and follow the resistor divider path R6/R8 to the output. Even if the amp is saturated hard negative (-3.3V) (which is plausible, being that we're applying positive voltage to -IN), the 1M/1M divider puts -IN at ((9.6V) - (-3.3V)) * (1MΩ) / (1MΩ + 1MΩ) + (-3.3V) = 3.15V, dangerously close to the positive supply rail. But the input differential is not sufficient to do that, it's only about 1.2V and the gain is 1, so the output will be about 1.2V (or, try to be), and Vdiff+ will hit the ESD clamp diode. Note in the datasheet, the inputs are limited to 0.3V outside Vdd/Vss, usually implying clamp diodes to the respective supplies for ESD protection.

The common-mode shunt resistors R7, R9 act to increase the noise gain of the amplifier. They literally reduce the input (error voltage) to the amp, increasing noise as the name suggests, reducing loop gain and bandwidth; normally this is undesirable, but here, it reduces the common-mode voltage, allowing the amp to continue to function at all.

The values are given by the common-mode (average of both inputs) or normal-mode (each divider by itself) gain from the sensed nodes to the amp, and the design maximum value of V3.

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