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I have a stereo potentiometer (10 kΩ) and a +5 VDC voltage (regulated by a 7805) with which I am trying to generate two "inverse" voltages, so at the same time one channel being high resistance and the other low resistance ie. channel one 5 V and channel two 0 V, or channel one 2.5 V and channel two 2.5 V. I hope you get it.

I then want to buffer these voltages with a TL072. The op-amp is overheating and I have a feeling that the power supply does not like what I do either. The load impedance is around 4.7 kΩ, which shouldn't be the problem. But what is?

This how I hooked it up:

circuit

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  • \$\begingroup\$ Welcome! From where did you buy the TL072? Are they oscillating by any chance? They are unity gain stable, but still... \$\endgroup\$
    – winny
    Commented Mar 21, 2023 at 16:47
  • \$\begingroup\$ Why don't you use an inverting op-amp to get the negative rail. \$\endgroup\$
    – Andy aka
    Commented Mar 21, 2023 at 16:54
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    \$\begingroup\$ Aren't you applying too high supply voltage to opamp? You are applying 34V whereas recommended is 30V max. What is the voltage connected to the potentiomenter? Hope it is lesser than the V+ of the opamp \$\endgroup\$
    – sai
    Commented Mar 21, 2023 at 16:57
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    \$\begingroup\$ @linsensuppe0815, +/-18V is absolute max if I am not mistaken. We should use the opamp based on the recommended conditions. \$\endgroup\$
    – sai
    Commented Mar 21, 2023 at 17:01
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    \$\begingroup\$ @sai It depends on the exact part number, some versions have ROC of +/- 20V max. \$\endgroup\$
    – GodJihyo
    Commented Mar 21, 2023 at 17:17

1 Answer 1

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You don't need a stereo potentiometer. I've built such a circuit on the breadboard just now and it works fine. Nothing's overheating. I've bought the op-amps in my "collection" directly from TI, though.

schematic

simulate this circuit – Schematic created using CircuitLab

TL072 is bypassed with 1uF from both either supply rail to ground.

OA1 is a buffer. OA2 inverts the voltage with respect to 2.5V. So, as A goes from 0V..5V, B goes from 5V to 0V. That's what you wanted, right?

The resistors R2-R5 can come from a DIP-packaged 4-resistor array. Their value is not critical. 10kΩ..100kΩ should be fine.

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  • \$\begingroup\$ Why C5? (half a microsecond with R3) \$\endgroup\$ Commented Mar 22, 2023 at 6:35
  • \$\begingroup\$ Thank you for providing the circuit! So R4 and R5 are for the biasing to 2.5V? What are C3 and C4 there for? \$\endgroup\$ Commented Mar 22, 2023 at 8:13
  • \$\begingroup\$ @DanielChisholm Can you explain what time constant you calculated? I know that capacitors like C5 in the feedback of an op-amp are for stability, but never knew which values are suitable. \$\endgroup\$ Commented Mar 22, 2023 at 9:33
  • \$\begingroup\$ R3*C5 = 484ns ("half a microsecond"). OA2 is set up as a voltage inverter with gain=1 (R2 = R3), but a a 22pF cap is paralleled with R3.... presumably to boost the frequency response of the feedback. I wouldn't have put it there, but knowing @Kubahasn'tforgottenMonica is no dummy he probably has a very good reason(s) that I am overlooking \$\endgroup\$ Commented Mar 22, 2023 at 10:56
  • \$\begingroup\$ It’s just a bit of compensation for stray capacitances on input. Will keep the circuit working if you replace the op-amp with something even faster. But for a TL07x that’s already good practice imho. Especially that I’ve said that the exact resistor value is no big deal but match is. TL08x, 100k resistor and poor layout and it may turn into a bell. So I’m conservative that way :) \$\endgroup\$ Commented Mar 22, 2023 at 16:17

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