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I am a noob with analog electronics, so pardon me for any mistakes. I recently started with Op-Amp and currently working on a project whereI am using Op-Amp in integrator mode to convert square wave to triangle wave and then fro there again an integrator to convert that triangle wave to sine wave.

I tried the circuit with single power supply that is +5V and ground, but was not able to make to work. So is it mandatory to use a dual power supply like +5V and -5V for an Op-amp to work as integrator? Because all the circuits I referred to were using a dual mode power supply.

Additional details: The Op-Amp is LM358 and the square wave is from an arduino board +5V and 0V obviously. The duty cycle of the wave is fixed at 50% and the freq. is controllable from 100Hz to 100KHz. I tested the whole set-up with 1KHz freq using my DSO.

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Can an Op-Amp work as an integrator without dual mode power supply?

Sure it can, the opamp doesn't care and also it doesn't know.

Here are two integrator circuits, on the left with a symmetrical supply. On the right using a single supply. Note how I created a 5 V "virtual ground" as a "working point" for the opamp.

schematic

simulate this circuit – Schematic created using CircuitLab

Do note that only the left circuit can output positive and negative voltages.

The right circuit's output signal will always be positive.

But both are integrators.

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  • \$\begingroup\$ Yes - both are integrators. HOWEVER, what is the input voltage which is integrated? In the second case, it is not simply Vin; you have forgotten the DC level at the pos. input. \$\endgroup\$ – LvW Nov 23 '18 at 10:21
  • \$\begingroup\$ @LvW You're right, I ignored the fact that the circuit on the right also has a shifted DC working point at the input. I did not want to clutter my answer with such details but focus on the big picture. \$\endgroup\$ – Bimpelrekkie Nov 23 '18 at 10:25
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No, an opamp wired as a MILLER integrator and with single supply does NOT integrate the input voltage applied at the input resistor - if it has a bias voltage (app. half the supply voltage Vs) at the non-inverting input. Of course, this voltage also acts as an input and is integrated.

This is because the DC operating point at the output is not at zero volts (instead it is Vs/2). Hence, at t=0 the voltage across the feedback capacitor is (Vs/2-Vin). Simple example: A dc input voltage of Vin=0.5Vs will not cause any change at the output (Offset parameters neglected).

As another example, consider Vin=0. In this case, the output will ramp (starting at Vs/2) up to the pos. supply rail.

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