integratormultisimoperational-amplifier
I am creating an Op-Amp Integrator circuit in Multisim. It seems to work for sinusoidal voltage inputs, square wave inputs, etc but when I connect it to a DC source the output voltage (channel A in the diagram below) just immediately goes to -11V. Based on the integrator equation (-1/RC)*integral(v(t)dt), the output voltage should be equal to -0.4t (-0.4 volts per second). I have tried using a bunch of different values for R, C and Vin. The Op-Amp is a 741 Op- Amp. Thank you for any advice.
I'm not entirely sure how to answer the why question, but simulation reproduces the same result; first the big picture:
But if you zoom in there are tiny little overshoot peaks in there...
which are actually pretty close to those measured. TL081 model is from TI, which is actually a very basic one... yet it predicted the overshoot pretty well.
Alas while I'm not completely sure what actually causes them, thanks to discussion below with the OP, all fingers point to the slew rate of the TL081. If I lower the slopes of the square wave by 3 orders of magnitude (from 1ns to 1us), then the output spikes go away (in simulation), but I also get some DC bias on the output (once it stabilizes):
Yeah, the datasheet slew rate for TL081 is 13 V/us. So in 1ns (first simulation) it would only be able to swing its output by 13mV, not enough to react to 25x greater input swing. Basically the output gets briefly pulled (up or down) by the input via the feedback resistor before the opamp can react to correct that. Actually it almost works like a voltage doubler during those spikes. Another way to check this is to measure the time span of the spikes. In [the first] simulation those spikes last 240ns (=4.15MHz), which is comparable to the 3-4MHz bandwidth of the TL081 (I note that the old datasheet said 4MHz but the new one says 3MHz.)
- If I apply at Vin a square wave with an amplitude from -5 V to +5 V, what happens to the voltage on V- node?
In a negative feedback arrangement the amplifier will do its best to keep the - input at the same potential as the + input. The + input is grounded and we refer to the - input as "virtual ground". The voltage there should always be zero unless we drive the output into saturation at which point the feedback won't be able to correct anymore.
- With a regular integrator at the output node I should have a triangular wave that goes from 0V to \$ -V_{in} \times \Large \frac{t}{RC} \$, but in this case what I get at the output? Is it a "rectified"square wave?
simulate this circuit – Schematic created using CircuitLab
Figure 2. (a) and (b) show how the circuit behaves on each alternate clock cycle. (A) and (B) show the result depending on which way the squarewave starts.
Consider the diode operation first: what output voltage will satisfy the virtual ground? Clearly when the input is positve and the output is at -0.7 V (a real diode - not an ideal one) the output will go to -0.7 to maintain 0 V at the - input.
When the input goes negative the diode will be reverse biased so the capacitor will run as a normal integrator and the output will be a triangle wave.
Finally, the output will depend on the starting conditions indicated by diagrams A and B. (Please excuse the arrows. CircuitLab doesn't have a free-style line symbol.
In practice, due to diode and capacitor leakage and any non-symmetry in the square wave the output may settle down somewhere in between the two conditions.
Best Answer
Upon initialization, MultiSim attempts to perform a DC balance. Since for DC a capacitor is the equivalent of an open circuit, that's what you get.
Instead, replace V1 with a step function, with the output at 0 for t = 0, and an output of 1 for t > 0. Alternatively, replace V1 with an offset square wave with amplitude 0.5 and a 0.5 offset - that is swinging from 0 to 1. Make the frequency 1/1000 Hz, and as long as you're only interested in t < 500 you'll be fine.