# Why doesn't my operational integrator work?

Here is the schematic I'm following, to build an operational integrator.

I'm using a LM358N IC which has 2 op amps. Below is a crude drawing of my connections.

I don't seem to get what I want, when I feed a square wave signal 0 to 5 V with a 20 ms period. I expect a negative sloped signal, but just get 0 V output.

What am I doing wrong?

• What are the component values R and C? – Stefan Wyss Nov 10 '18 at 6:52
• How do you reset the cap? – jonk Nov 10 '18 at 6:54
• R = .4 MOhms C = 47 MicroF – Nick Yarn Nov 10 '18 at 7:23
• I didn't. Do I have to?? I also tried to make a Inverting Amplifier and it also didn't work... – Nick Yarn Nov 10 '18 at 7:25
• If I change my input voltage signal to -1 to 1 V square wave would I need one? – Nick Yarn Nov 10 '18 at 7:26

simulate this circuit – Schematic created using CircuitLab

Figure 1. The op-amp output cannot go negative unless there is a negative supply to the op-amp.

• Ok, but why do I see a positive output when I applied a positive input signal to the invertinv terminals – Nick Yarn Nov 10 '18 at 17:55
• If you are feeding the input with +5 V and with no DC path in the feedback then you are probably pushing the input outside the common-mode input range (0 V to V+ −1.5 V) at 25°C. When you do that anything can happen. – Transistor Nov 10 '18 at 19:51

There are two basic problems with that circuit.

• You are applying a positive voltage to an inverting configuration. The op-amp needs to provide a negative output value n response for which it needs a negative supply.
• you don’t have a dc feedback pathway. Without DC feedback (1) your op-amp operating point is not defined and (2) op-amp bias currents will be integrated driving the output even with no input.

To fix the former, either use a negative supply, or bias the positive input above ground.

If this is the totality of your circuit, to fix the latter, add a large resistor in parallel with the capacitor which effectively turns the configuration into a low-pass filter. This behaves as an integrator for signals above the cut-off frequency.

DC integrators require additional circuitry to set their DC bias. Either switches that periodically discharge the capacitor, or external circuitry that reacts to the output and changes the input accordingly (e.g., in an oscillator configuration).