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I'm trying to build a simple voltage comparator using LM324N. Here is the circuit:

schematic

simulate this circuit – Schematic created using CircuitLab

The strange thing that happens is that after apply the input voltage, a small part of it gets clipped off. Here are the pictures of the circuit and the resulted voltages:

enter image description here enter image description here enter image description here

I've connected all of the grounds and negative terminals (i.e., CH1, CH2 and the negative terminals of the input voltage and power supply) together. Also using another LM324N and applying input voltage with a resistor didn't solve the problem. As soon as I disconnect the input voltage from LM324N, it gets back to its normal shape which is a sine wave with 3V amplitude.

How do I stop the voltage from clipping on the opamp input?

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    \$\begingroup\$ You are feeding a signal to the non inverting input that exceeds your opamp supply. As a result, the high input impedance property of an opamp is ruined. You need to have a DC bias on that input if you ground the negative supply of your opamp. Since you have a 3V amplitude sine wave, you'll also need to raise the supply voltage of the opamp to at least 6V \$\endgroup\$
    – LetterSized
    Commented Feb 20, 2021 at 22:32
  • \$\begingroup\$ @LetterSized Thank you so much. That solved the problem. \$\endgroup\$
    – S.H.W
    Commented Feb 20, 2021 at 22:44

2 Answers 2

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On many opamps, they have input protection diodes that prevent the input from exceeding -0.3V. The input goes to -3V so this is a problem and the diode is turning on to protect the input. Either make a bidirectional rail (so the v- terminal is -3V) or change the function generator to have a DC offset so the 3V sine wave has a DC offset of 1.5V and centered around that value.

enter image description here Source: LM124 Datashet

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  • \$\begingroup\$ Thanks. So when do these protection diodes turn on? How this depends on the voltage of the v-? \$\endgroup\$
    – S.H.W
    Commented Feb 20, 2021 at 22:52
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    \$\begingroup\$ When you go below or above the rails by 0.3V \$\endgroup\$
    – Voltage Spike ♦
    Commented Feb 20, 2021 at 22:57
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The datasheet of the chip you linked to has the answer.

In the nominal operating conditions, the inputs must be withing supply voltage limits, between 0V and 5V in your case. You are exceeding that by going below 0V and witnessing the results of it.

In the absolute maximum ratings, the inputs have a limit of -0.3V. Below that value, the chip does not have to work, but it also won't get permanent damage. Exceeding that can cause permant damage to the chip.

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  • \$\begingroup\$ common mode range includes -ve rail, and extends to (Vcc-1.7V). What I didn't see before was that you can drag inputs far above Vcc (up to +32V) without damage...anyone's guess where the output goes when inputs are w-a-y up above Vcc. \$\endgroup\$
    – glen_geek
    Commented Feb 20, 2021 at 23:44

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