What is the best way to power off a dual supply op amp (LM741)? Should I cut the positive, negative, or both? Thanks!

I’m building an audio oscillator using four op amps, and I’d like to be able to turn the oscillator on/off by cutting power to the entire circuit. The four op amps are configured in a feedback loop, so I believe that when I cut power there is no input signal to any of the op amps.

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  • \$\begingroup\$ Depends on the remainder of the circuit and why you want it turned off. \$\endgroup\$
    – jippie
    Jun 22 '15 at 7:28
  • \$\begingroup\$ What I meant to say is: It is bad practise to disconnect either or both power rails from a semiconductor if your input signals aren't "safe". Input signals can potentially damage the device if one or more power rails are disconnected, hence why it is important to be more specific in your question. \$\endgroup\$
    – jippie
    Jun 22 '15 at 8:53
  • \$\begingroup\$ OKAY, thanks! Just added detail on the rest of the circuit to the post. \$\endgroup\$ Jun 22 '15 at 19:30
  • \$\begingroup\$ You'll have to cut both +V and -V, because the non-inverting inputs are connected to ground (which is a suboptimal design but doesn't change the answer for this question). There are other options, like creating virtual ground from the regular power supply using extra circuitry. \$\endgroup\$
    – jippie
    Jun 22 '15 at 20:00
  • \$\begingroup\$ Did you build the oscillator already? It is a rather "uncommon" circuit - and I doubt if it can oscillate as desired. \$\endgroup\$
    – LvW
    Aug 11 '15 at 9:51

What you need is to is to either make the voltage difference between two nodes zero or make open circuiting so that they can not reference each other in order to actually power off.

Cutting one would be open circuiting the internal circuit therefore powering off the op-amp.

As for the "best way": the rest of the circuit is of importance here, since probably powering the op-amp off change other stuff.


You DO need to disconnect both batteries when powering the circuit down.

There are a couple of options to consider.

1) Modify the circuit to run from a single-ended power supply. This is easy to do in your circuit because the only nodes that connect to ground are the 4- (+) inputs on the op-amps and the load.

The changes needed are simple: add a voltage divider reference from V+ to V- with the mid-point going to the 4- (+) inputs on the op-amps. Add a bypass capacitor across the lower resistor for stability.

You will also need to add an output coupling capacitor. The (+) side of the cap goes to the op-amp output pin, the (-) side goes to the load. I normally also add a bleeder resistor from the (-) side of the capacitor to ground (V-) so that you don't get a thump when connecting the load.

2) Use a DPST or DPDT switch to disconnect both batteries.

3) There is a simple trick that can work well under certain circumstances: connect the batteries in series with blocking diodes. You then use a SPST switch to short the diodes.


simulate this circuit – Schematic created using CircuitLab

The downside of this approach is that the ground terminal will bounce around by one diode drop as current consumption changes from (+) to (-). Depending on the PSRR rating of the op-amps, this can lead to distortion at the output.

However, I have successfully used this technique in the past.

  • \$\begingroup\$ I like your Diodes because they simplify the switch to SPST.You could connect capacitors across the diodes to deal to PSSR.Generaly PSSR is good at low frequency and gets progressively worse as frequency increases so the caps can be small and cheap. \$\endgroup\$
    – Autistic
    Aug 20 '15 at 2:17

As an alternative: There also are op amps with a built in shutdown function, have you considered using those? This example consumes <1uA when you pull the dedicated pin low. Perhaps you can find one that suits your needs.


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