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My goal is to reduce battery consumption of a op-amp to zero while a microchip sleeps.

Intend to do so with a transistor driven by a pin.

While sleeping, the chip would use little battery, and the op-amp would use none.

An interrupt would set the OPAMP pin as HIGH, which would allow the transistor to relay the power to the op-amp.

The PWM pin would do its thing, and get increase with the op-amp, and the resulting voltage would drive the speaker.

Once the chip is done with the sound, the chip sleeps and turns the OPAMP pin as LOW, stopping the use of the op-amp.

schematic

simulate this circuit – Schematic created using CircuitLab

Here are my questions:

  • Does the schematic make sense and am I hooking things the proper way?
  • Is a pin's output stable enough to drive a op-amp with a transistor to save on battery?
  • Would the added pin relay actually reduce significantly the op-amp's battery use?
  • Is there a better way to amplify a sound while also reducing battery power?
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    \$\begingroup\$ You can use a transistor like that, but you must use a PNP one not an NPN one. \$\endgroup\$
    – Majenko
    Sep 11 at 19:36
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    \$\begingroup\$ It works as an emitter follower, he will lose about 0.7 volts at the emitter. A PNP may give more voltage but requires a base resistor. \$\endgroup\$
    – Gil
    Sep 11 at 20:09
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    \$\begingroup\$ Why not just use an op amp with an enable pin? \$\endgroup\$ Sep 13 at 12:38
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    \$\begingroup\$ @B7th Look at something like the OPA341 with a shutdown pin: ti.com/lit/ds/sbos202a/sbos202a.pdf \$\endgroup\$ Sep 16 at 21:15
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    \$\begingroup\$ Quiescent current during shutdown is just a few nA, and almost nothing if the chip is below 30 deg C \$\endgroup\$ Sep 16 at 21:17
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You could try this - most op-amps are not well-suited to driving speaker loads directly, though some may be okay driving headphone loads.

In order to minimize the shutdown current you also likely need to drive the PWM output to 0V or the internal circuitry of the op-amp may draw current from the output, perhaps partially powering up.

As shown you’ll lose a bit (700mV+) of voltage in the transistor, but that may be okay. The base will only draw what it needs from the output.

A modern alternative might be to use an amplifier with a shutdown input such as the PAM8406 - a class D or AB amplifier with <1uA current draw in shutdown. There are other similar chips, so I suggest fully investigating if this is a viable alternative.

enter image description here

There are also op-amps with shutdown inputs, but an audio amplifier may be more appropriate.

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  • \$\begingroup\$ Thanks for the info on that, that does seem like a lot of capacitors can you explain why? \$\endgroup\$
    – B7th
    Sep 16 at 19:41
  • \$\begingroup\$ The input coupling caps are there to block DC. The supply bypass caps are there because (especially in class D operation) there are very fast-switching edges and effective high-frequency bypassing is desirable. The ceramic caps provide that and the bulk reservoir is the electrolytic cap. Keep in mind this little IC can produce several watts of output to the speakers, which implies relatively high peak currents. \$\endgroup\$ Sep 16 at 19:46
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Does the schematic make sense and am I hooking things the proper way?

Do it like this: -

enter image description here

You wouldn't use an NPN emitter follower to do this; use a PNP common-emitter circuit. I wouldn't recommend trying to switch the 0 volts to the amplifier with an NPN either if your amplifier is driving a loudspeaker.

Is a pin's output stable enough to drive a op-amp with a transistor to save on battery?

Yes.

Would the added pin relay actually reduce significantly the op-amp's battery use?

What's a pin relay?

Is there a better way to amplify a sound while also reduce battery power?

There are always better ways but, to decide that needs a full detailed specification of exactly what you want to do (not nearly enough room to be covered here possibly).

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  • \$\begingroup\$ Thank you for the info, can you explain why the PNP is better suited for this than a NPN? \$\endgroup\$
    – B7th
    Sep 16 at 19:41
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    \$\begingroup\$ They are opposite polarity devices. To use a pnp effectively you have it switching the positive supply to the audio chip. To use an NPN effectively you’d switch 0 volts or a negative supply but you don’t want to have any transistor in the audio return path hence switching 0 volts is a bad idea. That leaves using a pnp transistor as the only feasible BJT option. \$\endgroup\$
    – Andy aka
    Sep 16 at 19:53
  • \$\begingroup\$ Awesome! This is the most easily feasible response from the two, I will test it tonight and let you know! \$\endgroup\$
    – B7th
    Sep 16 at 20:39

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