I am amplifying a signal for a current transformer (CT) with an op amp. This amplified output signal I need to compare using op amp comparator.

The supply voltage is 12V and the IC is an LM358. When I give the output of the amplifier to the comparator, the comparator gives an output signal of 7 volts irrespective of voltages at inverting and non inverting terminal, which should be either high Vcc or low 0V as it is a comparator. Can you help me with a solution?

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  • \$\begingroup\$ Hi! Without a schematic this will be impossible to answer. You can simply use the schematic editor that's built into this site's question editor to add one. \$\endgroup\$ Commented Sep 20, 2020 at 12:22
  • \$\begingroup\$ also, I removed the 741 and noise tags, because you mention neither in your question, and thus these tags seem unrelated. \$\endgroup\$ Commented Sep 20, 2020 at 12:30
  • \$\begingroup\$ Is your defective LM358 a cheap fake one from ebay? \$\endgroup\$
    – Audioguru
    Commented Sep 20, 2020 at 15:33
  • \$\begingroup\$ You've got three schematics now, @Marcus. Happy? \$\endgroup\$
    – Transistor
    Commented Sep 20, 2020 at 15:58
  • \$\begingroup\$ What do you want this circuit to do? (1) CT will input an AC signal to OA1. The input will be driven alternately positive and negative but you only have a single supply rail. (2) If the amplifier worked then OA3 would switch every time the input signal exceeded the threshold so your relay would be switched on and off at 50 or 60 times per second depending on your mains frequency. Is this what you intended? (3) You are switching your relay in emitter-follower mode so you will never get 12 V on the coil. (4) You have no snubber diode on the relay coil so your transistor won't last long. \$\endgroup\$
    – Transistor
    Commented Sep 20, 2020 at 16:01

2 Answers 2


There are many things wrong with this circuit. From the left:

The CT probably requires a load resistor. If it outputs 30mA at full scale (as your schematic appears to show), that should go into a resistor. For example, 100 ohms would give you 3V volts RMS if the CT can work with that high a resistance. The voltage will be AC and in that example it would be +/-4V peak from the tranformer with just the load resistor.

The circuit following it is a DC amplifier with gain 1+ (R2/R1). It cannot handle voltages below 0V and the op-amp will do unpleasant things (phase reversal at a minimum) if the input goes more than a few hundred mV below ground. You probably want some kind of precision rectifier circuit here rather than an amplifier, but that will depend on what exactly you are trying to do and how well you are trying to do it.

The following circuit is indeed a comparator but the common mode range of the LM358 does not include the power supply rail so the top part of the pot range will do nothing of value. R3 is unnecessary.

Your relay drive circuit will only give you about 9V (or less, depending on values) on the relay coil when 'on' because the LM358 output does not swing to the positive rail and the emitter follower transistor drops 700 or 800mV. There is also voltage drop across R6, which is unnecessary. It would be better to reverse the comparator and drive the transistor as a low-side driver, in which case you need the base resistor and you also would need a flyback diode across the relay coil.

It's a bad idea to drive a relay without hysteresis in the comparator, otherwise you will likely get chattering around the control point. That can be accomplished by adding a bit of positive feedback to the comparator.


I think the output of comparator seems to be floating. Try placing a buffer between amplifier and comparator.


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