Good morning, im new to electrical engineering and was given a school project to create a basic circuit that amplifies sound to a speaker. in the process of trying to solve this problem ive come across several different hurdles.

For one i designed a circuit after gathering information from youtube tutorials and i cant even test if its functioning properly cause im not sure if my speaker is working or the circuit is just wrong. I also dont know how to connect an epe speaker to a breadboard (trrs). Any kind of help would be appreciated, especially with regards to confirming if my circuit is correct.

I used to different op amps to get the two stage amplification. the output of the first op amp should be the input of the second. The voltage should be between 5-12v for the input. Did i ground the correct segment ? should I add a seperate battery for the second op amp or will that generate more noise? I put the capacitor there to reduce noise, should i add more?

enter image description here

  • 3
    \$\begingroup\$ The second op-amp is trying to have infinite gain and you'll probably end up turning all your sound into square waves, is that intended? \$\endgroup\$
    – user253751
    Apr 12 at 14:34
  • 2
    \$\begingroup\$ Most op-amps have difficulty driving a speaker, causing audio distortion starting at very low signal amplitude. The cause is insufficient current at the output pin...many op-amps limit current for self-protection. Where current-limiting sets in, distortion results. Headphones are often an easier load than a speaker, providing sufficient loudness before distortion begins. \$\endgroup\$
    – glen_geek
    Apr 12 at 14:36
  • \$\begingroup\$ An op-amp is not powerful enough to drive a speaker. Depends on speaker and op-amp though, but unlikely under normal circumstances. Also you can't use an inverting amplifier if you ground the positive input and provide only single-ended supply voltage. \$\endgroup\$
    – Justme
    Apr 12 at 14:36
  • \$\begingroup\$ What do you mean by "speaker"? A big-dia, powerful (e.g. 12-in 50W) speaker or a headphone speaker? With an op amp, you can drive only the latter. \$\endgroup\$ Apr 12 at 14:47
  • 1
    \$\begingroup\$ Can we also ask why you decided to use two op-amps instead of for example 0 or 1 or 3 or 4? Just out of curiosity. \$\endgroup\$
    – user253751
    Apr 12 at 15:10

2 Answers 2


Circuit is incorrect.
For the first opamp in the signal chain, grounding the non-inverting input (+ input) is wrong, if you intend to power the opamp with a single +9V DC supply. The +input should be half-way between +9V and GND, like the circuit below. A capacitor C2 is added so that this voltage remains fairly constant. R3, R4 have large resistance so that a battery supply is not stressed.


simulate this circuit – Schematic created using CircuitLab

For C1 (a polarized capacitor), its "+"end should face the opamp, not the input signal.
The potentiometer can have a larger value, like 10k so that the opamp needn't drive it with a lot of current...the limited current available at opamp output pin should mostly go to the headphone.
I have added an output resistor R5 so that the opamp is not stressed.
Capacitor polarity for C3 must face the opamp output pin. Headphones would be connected between "Out" and GND (bottom of +9V supply).
Some opamps may work better if headphones are connected between "Out" and +9V, in which case polarity of C3 should be reversed.


Your original (Falstaf?) schematic was a negative image so I changed it to a normal black on white positive. It had many errors: enter image description here

  • \$\begingroup\$ The 2nd op amp does not have to be an audio PA as the load is nothing different than an earphone i.e. a few tens of milliwatts and an op amp can drive it. \$\endgroup\$ Apr 12 at 18:39
  • \$\begingroup\$ The max output current of many opamps is only 15mA peak and many earphones are 32 ohms. then maximum output levels are low. \$\endgroup\$
    – Audioguru
    Apr 13 at 1:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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