Op amp circuit waveform analysis problem

shaorc

As shown in Figure 1, the input of the op amp is 220VAC (there is a string of voltage-dropping resistors before R1 and R2 that are omitted)

The waveform measured at point A is indeed the same as the theory, a perfect sine wave with decreasing amplitude.

[1] The waveform measured at point B after passing through R5 is shown in Figure 2. The negative half cycle of the waveform is smoothed out. It only passes through one resistor, so why does the negative half cycle disappear?

[2] The waveform output from point C is shown in Figure 3. I am even more confused as to why the waveform of the negative half cycle is like this?

【3】Is the UB part of LM324 an inverting amplifier circuit or an integrator circuit?

Figure 1

Figure 2

Figure 3

gmchen

The three problems are actually caused by the same reason, that is, UB is neither an inverting amplifier nor an integrating circuit, but a precision rectifier circuit.

gmchen

When point A is in the negative half cycle, UB is in the inverting amplification state, and point B is the virtual ground, which is the half cycle in which the voltage is 0 in Figure 2. UB inverts and amplifies, outputting half a sine wave, which is the part in Figure 3 where the voltage is greater than 0.

When point A is in the positive half cycle, UB should have output a negative voltage, but the output current path is blocked by D1, and its output cannot be fed back to the input. The op amp is actually in an open-loop state, and there is no current in R5, so the potential of its inverting input terminal (point B) is consistent with that of point A, and the output terminal (point C) is directly swung to the vicinity of the negative power supply.

gmchen

You can measure the waveform on the right end of D1, which is the waveform of a full-wave rectifier output.

wupan1981

Learn

scxuchenglong

1. The waveform at point B is only half a wave because of the feedback signal after rectification. If the feedback loop is disconnected, it will become a sine wave.

2. It is also related to feedback superposition. The operational amplifier will change the phase. You can analyze it carefully for details.

3. It is a rectifier circuit, and the capacitor in the feedback loop is just an anti-self-excitation