I'm fairly new to LTSpice and I'm having a hard time making this circuit to work. So I'm making an Operational Integrator but my Output signal doesn't seem to match my predictions. enter image description here

If I'm not mistaken Voltage at node Va should be zero since the Operational Amplifier is in a negative feed back configuration. I'm using the Universal Operational Amplifier and I was wondering if maybe the parameters of the Op Amps where the cause of this unusual signal at Va which in turn effects the Vout.

EDIT: Here are some more detailed images relating to my confusion. I understand that the resistors and the capacitors are unusually non ideal for practical use but I just wanted to know why the voltage across the input terminals differ by so much.
enter image description here

enter image description here

enter image description here

  • \$\begingroup\$ The saturation is not at all unreasonable, but the positive level is. Assuming the output starts at zero, what will be the slew rate of the output? So how long will it take for the output to saturate? What is your timebase? \$\endgroup\$ Nov 14, 2018 at 3:45
  • \$\begingroup\$ The slop of the output signal should be -2 volts/sec and I set the initial conditions of Va and Vout to be 0. My Vout signal also hits 2 V instantly which I didn't expect. \$\endgroup\$
    – Nick Yarn
    Nov 14, 2018 at 3:49
  • 1
    \$\begingroup\$ Even though it's a universal op-amp, it's still going to have somewhat realistic parameters inside. That means that your resistor is insanely small and the cap is insanely big. Try a 10k\$\Omega\$ resistor and a 100\$\mu\$F cap -- see if things get better. \$\endgroup\$
    – TimWescott
    Nov 14, 2018 at 4:08
  • \$\begingroup\$ There's not enough info in your picture. I can't see the .TRAN card, for example. So no clue what you are doing there. But looks like the default case of .TRAN 1. If so, the spice program will go through a process of calculating the steady state and initialize things to those values. That's not usually a good thing for integrators without shorting switches available and periodically active. It doesn't start where you think it starts. So it doesn't yield what you wish it would yield. \$\endgroup\$
    – jonk
    Nov 14, 2018 at 4:08
  • \$\begingroup\$ In addition to Tim's excellent suggestions about the values (an opamp can't drive amps at its output, except in a very few cases), try setting the UIC checkbox on the .TRAN card, as well. Those two things (value changes and the UIC checkbox) should give you something more as you expect to see. \$\endgroup\$
    – jonk
    Nov 14, 2018 at 4:17

1 Answer 1


Here's an example of how to set things up. I've used the LT1800 (which is a rail to rail in/out opamp) to get the maximum range for display purposes. (It's also in the library.)

You can set up your voltage rails using separate voltage supplies with drawn images on the schematic. But it really isn't necessary. As you see below, it's just three spice lines. So that's another way to go. (I usually start with 99 and work the numbers backwards as I go in order to avoid LTspice's desire to start with 1 and work upwards from there as you add pictorial voltage sources to a schematic.)

enter image description here

As you can see above, the rate of change in the voltage at \$V_\text{OUT}\$ is \$\frac{\text{d}V_\text{OUT}}{\text{d}t}=\frac{5\:\text{V}-\left(-5\:\text{V}\right)}{500\:\text{ms}-0\:\text{ms}}=20\:\frac{\text{V}}{\text{s}}\$.

Since you know \$I=\frac{2\:\text{V}-0\:\text{V}}{10\:\text{k}\Omega}=200\:\mu\text{A}\$ then it follows that \$\frac{\text{d}V_C}{\text{d}t}=\frac{I}{C}=\frac{200\:\mu\text{A}}{10\:\mu\text{F}}=20\:\frac{\text{V}}{\text{s}}\$.

And that is a match.

The .IC card is relatively important if you want to get the largest range. What it does is sets up constraints on the solver for the initial conditions that it is forced to apply before starting the run. By setting \$V_\text{OUT}=5\:\text{V}\$, I'm forcing it to work out the fact that there will be, initially, \$5\:\text{V}\$ across capacitor \$C_1\$.

Note that the .TRAN card uses "UIC" in order to get the run started correctly. The use of UIC means that Spice will not go through the "initial transient solution" step (the so-called "ITS") to find the DC solution at \$t=0\$.

When you use UIC, the initial value of every single energy storage (voltage and current) device is treated as zero, except for those which are explicitly provided using the .IC card. The .IC card can specify node voltages or else inductor currents.

When you don't use UIC, then everything you explicitly provide using the .IC card, gets added as a constraint. But the ITS is still permitted to otherwise find initial values based upon a steady state solution... consistent with the given constraints.

This last bit may not be what you want here. So I recommend using UIC.

If you know, a priori, all of the initial values for the energy storage devices in the circuit, you can use UIC to compute the steady state solution without the transient response leading up to it (that may occur if you instead allowed Spice to first perform the ITS step by not using UIC.)

Hopefully, you are able to interpret the above information into your situation with success.


Your Answer

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

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