Output waveform problem of integrator circuit

shaorc

Output waveform problem of integrator circuit [Copy link]

The figure is an ideal integrator circuit. Due to the inverse relationship between the output and the input, in case (1), when the input is a constant DC quantity, the output is as shown in the figure (1)? If the analysis in case (1) is correct, in case (2), when the input is a square wave signal, if you want to get a positive triangular wave output waveform, does the input square wave signal have to be a negative voltage input?

maychang

Waveform (2) is incorrect.

The input is a positive pulse from 0 to Uin. After the first Uin positive pulse, the input drops to zero, followed by the second pulse. During the period when the input is zero, the output should remain unchanged, that is, the output is a horizontal line segment. When the second pulse arrives, the output drops at a certain slope.

maychang

The output is shown as the red line in the figure below.

Obviously, the output cannot continue like this. After a few pulses, the output will saturate to the negative direction.

shaorc

maychang published on 2021-4-14 08:46 The output is shown as the red line in the figure below. Obviously, the output cannot continue like this. After a few pulses, the output will be negatively saturated.

Oh, I understand this now.

But in case (2), why does the output remain constant instead of decreasing when the input is zero? Is it because the feedback capacitor has no discharge path?

In addition, the integral circuit can indeed play a role in waveform conversion, that is, input a square wave and output a triangle wave. How can this effect be achieved in the first post?

maychang

shaorc posted on 2021-4-14 09:05 maychang posted on 2021-4-14 08:46 The output is shown in the red line in the figure below. Obviously, the output cannot continue like this. It will take more than a few pulses...

In mathematical integration, the result of integrating a constant (zero-power function) is a linear function (first-power function), the result of integrating a linear function is a quadratic function... The result of integrating the number zero is a constant.

maychang

shaorc posted on 2021-4-14 09:05 maychang posted on 2021-4-14 08:46 The output is shown in the red line in the figure below. Obviously, the output cannot continue like this. It will take more than a few pulses...

"Why does the output remain constant and not decrease when the input is zero? Is it because the feedback capacitor has no discharge loop?"

The input is zero, and the two input terminals of the op amp are "virtually shorted", so the potential of the inverting input terminal must be equal to the potential of the non-inverting input terminal, that is, the potential of the inverting input terminal is zero. Therefore, the current passing through the resistor R1 in the first circuit is zero, the current in the capacitor C1 is also zero, and the voltage across C1 remains unchanged.

shaorc

maychang published on 2021-4-14 09:31 "Why does the output remain unchanged instead of decreasing when the input is zero? Is it because the feedback capacitor has no discharge circuit? 』 The input is zero, by...

"The current in capacitor C1 is also zero, and the voltage across C1 remains unchanged"

If you want the first integrator circuit to output a triangular wave, do you need to meet the following conditions?

(1) The op amp needs positive and negative power supply

(2) The op amp input must be a sinusoidal wave without a DC component.

In this way, during the positive half cycle of the input, the voltage of the integrating capacitor C1 increases in the negative voltage direction. When the negative half cycle of the input comes, a reverse current flows through the integrating capacitor C1, and the integrating capacitor voltage rises from negative voltage to 0. This cycle repeats and a triangular wave is obtained?

In addition: For the op amp to meet the virtual short and virtual open conditions, does it have to operate in a deep negative feedback state?

maychang

shaorc posted on 2021-4-14 10:37 "The current in capacitor C1 is also zero, and the voltage across C1 remains unchanged" If you want the integrator circuit in the first post to output a triangular wave, is it...

{If you want the integrator circuit in the first post to output a triangular wave, do you need to meet the following conditions?}

That's right. Both conditions are indispensable. But the op amp input does not have to be a sine wave, other waveforms are also acceptable as long as there is no DC component.

If the output is required to be a triangular wave, the input must be a rectangular wave with no DC component. In other words, the input signal is symmetrical about the horizontal axis.

maychang

shaorc posted on 2021-4-14 10:37 "The current in capacitor C1 is also zero, and the voltage across C1 remains unchanged." If you want the output of the first post's integrator circuit to be a triangular wave, is it...

"In addition: For the op amp to meet the virtual short and virtual disconnect conditions, does it have to work in a deep negative feedback state?"

Yes.

Gen_X

@shaorc

Your output graph is correct!

If the output is not expected to be a triangle wave below "zero", the input needs to be negative.