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My LTspice schematic

I am trying to create an integrator in LTspice that I will use as a subcircuit in a signal chain later, but can't get the simulation results to agree with my theoretical calculations. I derived the following transfer function:

$$ H(s)=\frac{R_2}{R_{1}R_{2}Cs+R_{1}} $$

When I used MATLAB linear simulator to test the response of this transfer function to a 5 Hz square wave, I got the expected result: a triangle wave at steady state. However, LTspice would not cooperate. LTspice simulation results

This looks almost like a differentiator. I double and triple checked to make sure I had the transfer function right. I also built the physical circuit, and measured a triangle wave output at steady state. Can someone help me make my LTspice simulation produce the desired results?

I'm somewhat of a noob with LTspice so I apologize if I left out any important information needed to diagnose the problem.

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    \$\begingroup\$ It is acting like Vcc and Vee aren't hooked up. LTSpice may need a bit of wire between the terminals of a part and the supplies -- I wouldn't know, because for stylistic reasons I always do this. Try moving the Vee and Vcc flags over, wire them up, and give it a whirl. \$\endgroup\$
    – TimWescott
    Dec 22, 2020 at 3:57

2 Answers 2

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The standard form for your transfer function is:

$$G_s=K\frac{\omega_{_0}}{s+\omega_{_0}}$$

where voltage gain \$K=-\frac{R_2}{R_1}\$ and \$\omega_{_0}=\frac1{R_2\,C_1}\$. This is a low-pass filter with voltage gain \$K=-2.7\$ and \$\omega_{_0}\approx 3.704\:\frac{\text{rad}}{\text{s}}\$ or \$f_{_0}\approx 589.5\:\text{mHz}\$.

From this, I'd expect integrator behavior, not differential. In particular, I'd expect an upward ramp followed by a downward ramp, etc. So it should look like a triangle wave at the output.

If you center your square-wave around \$0\:\text{V}\$ (which you didn't do), then you'd expect to see a charging current of about \$\frac{\pm 4.5\:\text{V}}{1\:\text{k}\Omega}=\pm4.5\:\text{mA}\$. At \$100\:\text{ms}\$ per half-cycle, this works out to \$\Delta\,V=\frac{4.5\:\text{mA}}{100\:\mu\text{F}}\cdot 100\:\text{ms}=4.5\:\text{V}\$. So that should be the peak-to-peak for your triangle, assuming you center your square-wave. (The triangle wave will also be centered around \$0\:\text{V}\$, given a little bit of time to "settle-in.")

With the square-wave you have, I'd expect to see the same triangle-wave (given enough time to settle-in, with the integrating capacitor developing a net quiescent charge), but the average value is now the voltage gain times the input average voltage, or \$-2.7\cdot 4.5\:\text{V}=-12.15\:\text{V}\$. For that, you'll need an opamp that preferably has rail-to-rail output and use a \$V_\text{EE}=-15\:\text{V}\$.

Let's perform a run in LTspice using \$\pm 15\:\text{V}\$ supply rails and an opamp whose output is close to rail-to-rail, the LT1800:

enter image description here

Just as predicted.

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You have supplied the opamp with a +9 V rail only, then given it a 9 V input when it has -2.7x gain. It will not be able to drive the output negative.

To get things working

  • Supply the opamp with +/- 15 Volts.

  • Reduce your input signal to 1 V amplitude.

Then play with the amplifier supply and signal amplitude while you explore what input common mode range and output swing means for that opamp. Hint, it's not a rail-to-rail opamp.

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