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.

  • 1
    \$\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


The standard form for your transfer function is:


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.


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.