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Using LTSpice I have simulated the following circuit. It is a circuit integrator using the op-amp model: LF356. The simulation has a saturation voltage of 12 and -12 V and a square wave input source. Since the input is a square wave, I expect the voltage at the output to be a triangular wave.

This is reflected in the image below. However I do not understand what is happening to the circuit during the transient. Furthermore, what is the need of the large resistor R1? When I try to simulate it without that resistor it produces an error.

In the image V(n003) is the input voltage and V(n005) is the output voltage (Vo) of the amplifier.

LTSpice Simulation

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3 Answers 3

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The simulator solves for the steady-state output voltage first- since you have +2V applied initially, the output voltage would be about -8V after a long time with +2V at the voltage source V1. Vout = -2V*(R1/R2)

Then the square wave starts, and the average voltage is zero, so the op-amp output will start to approach 0V with a time constant of 1ms.

The purpose of R1 is to prevent the output from slowly drifting towards a supply rail because of op-amp offset voltage, for example. Even a tiny DC voltage integrated over sufficient time will eventually exceed the available op-amp output swing.

By the way, it's better to name the nets of interest (F4), otherwise if you make some change to the circuit the numbering can change resulting in you getting results from unexpected nets in your circuit.

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The textbook, or "pure" integrator, doesn't have an R1 there, but its presence is explained in the other two answers. Adding it, however, makes the pole at 0 to be shifted to a finite frequency, which means the circuit becomes a lowpass:

$$\dfrac1s \rightarrow \dfrac{\omega_p}{s+\omega_p}$$

If you'll apply the inverse Laplace to find the impulse responses for the two, you'll find that they will be \$1\$ and \$\omega_p\exp(-t\omega_p)\$, where \$\omega_p=1/\tau\$. Which means that a step reponse will be \$t\$ for the pure integrator (a stairway to heaven), and \$1-\exp(-t/\tau)\$ for the 1st order lowpass -- that's what you see superimposed over the triangular waveform.

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Furthermore, what is the need of the large resistor R1?

If you didn't have the feedback resistor the op-amp would saturate to one rail or the other because there is massive DC gain, possibly as much as 100,000.

However I do not understand what is happening to the circuit during the transient.

You not understanding something is not enough information to figure out what might be happening in your little grey cells. It all looks pretty normal to me.

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