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schematic

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.

EDIT 1

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?

EDIT 2

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

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    \$\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

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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.

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