simulate this circuit – Schematic created using CircuitLab

I am trying a current sensing circuit and I noticed something unexpected. Connecting both the positive and negative inputs to ground the op-amp gets very hot (30˚C - 86˚F in a few seconds).

At the power up the output voltage is about -1 mV but it gets always more negative.

The three unused amplifiers are connected as shown.

I simulated the circuit using LTspice and it provides: Vout = 1.25 mV, IR4 = 25 nA, IR2 = -4.9 pA as expected. No heating from these currents.

All my resistors are 1% tolerance.

The chip is good because a simple inverting amplifier with gain = 1 works as expected.

I am not able to figure out what I am doing wrong.


I have connected the oscilloscope to the output. Apparently, the circuit is oscillating.

enter image description here

I changed the power supply from dual to single 12 V. Now:

  • Temperature is stable at about 22.8˚C
  • Output is stable at 8.05 mV
  • The output doesn't show any oscillation

enter image description here

Applying a differential voltage at the inputs, the output follows the inputs with a gain of about 43 instead of 50.

The only explanation I have for this behavior is that because the circuit is mounted on a breadboard, the stray capacitances make it oscillate.

Any better ideas?

How do I keep the double supply voltage without oscillations?


As suggested by tobalt, I added the two caps in parallel to the 50 kΩ resistors (values ready to hand):

  • 47 pF: They don't make much difference
  • 100 nF: They drastically reduce output oscillations but the output offset still drifts a lot (-50 mV in a few seconds) and never stops.

enter image description here

  • 2
    \$\begingroup\$ 30˚C isn't what I'd call hot let alone very hot unless your ambient local temperature is -10˚C. \$\endgroup\$
    – Andy aka
    Jan 8, 2023 at 10:49
  • \$\begingroup\$ it's fine. the amp has some quiescent power dissipation \$\endgroup\$
    – tobalt
    Jan 8, 2023 at 11:53
  • \$\begingroup\$ I don't think it'a fine: 1) Temperature doesn't stabilize. The room temperature is 20˚C and the chip reaches 35˚C in less than a minute. 2) Output voltage increases with temperature. @35˚C it is -60mV which is quite larger of the opamp's offset voltage. I turn the power off at 35˚C to avoid damaging the chip. \$\endgroup\$
    – Fab
    Jan 8, 2023 at 14:28
  • \$\begingroup\$ 35°C is body temperature, barely warm. But anyway if the temperature is increasing at a rapid rate maybe you have a short somewhere (perhaps to the output of one of the unused op-amps). I don't think your schematic reveals any possibilities. \$\endgroup\$ Jan 8, 2023 at 15:02
  • \$\begingroup\$ @Fab see edit about oscillation \$\endgroup\$
    – tobalt
    Jan 8, 2023 at 17:15

1 Answer 1


4 amps consume 4 mA quiescent current. At 24 V, that dissipates alread 96 mW while not doing anything.

Thermal impedance to ambient is about 100 K/W, so a rise of 10 K is absolutely expected.

Regarding the oscillation: This happens when you use a large value for the feedback. You can place a small cap e.g. 22 pF across the two 50kOhm caps, to remove it. Still you will still see a warmup of approximately 10 K or a bit more.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.