The picture below is named as headphone amplifier on the internet. My goal is not to do "headphone amplifier". I want to design an analog frontend with a bandwidth of 10 MHz for ADC input. This circuit will form the amplifier part.

What is the detailed analysis of the circuit in this picture?

Let me explain the circuit as far as I understand. I would be glad if you correct both the points I do not understand and my mistakes.

  • Input voltage is adjusted with the original RP1.1 potentiometer.

  • The C1 capacitor required for AC coupling and together with the ground resistance (R1 47k) creates a "high pass filter". The purpose here is to prevent noise frequencies (0.33 Hz) from passing. From there it was connected to the noninverting pin of the opamp.

  • R3 and R2 resistors that determine the gain were connected to the inverting pin.

What I don't understand here is the C2 (47uF) capacitor. I don't know what this capacitor does.

Also, R4 and C3 components are circuit elements that need to be explained for me.

enter image description here

  • \$\begingroup\$ Would you explain what kind of detailed analysis you want that can be accepted as answer? NE5532 has 10 MHz typical gain-bandwidth product. \$\endgroup\$
    – Justme
    Apr 26, 2020 at 18:51
  • \$\begingroup\$ The 5532 has an AC gain of 6(ish), courtesy of R3 and R2. R4 is redundant. R1 is a route for the amplifier input bias current. C3 is a surprise, at best it does nothing but filter RF, so it would be better at the input, at worst it destabilises the opamp, presumably it doesn't destabilise a 5532. \$\endgroup\$
    – Neil_UK
    Apr 26, 2020 at 19:12
  • \$\begingroup\$ A good starting point for you would be to study up on the inverting and non-inverting op-amp configurations, and then come back to this circuit. \$\endgroup\$
    – rpm2718
    Apr 26, 2020 at 19:18
  • \$\begingroup\$ @Neil_UK: C3 is a minimalistic version of a Boucherot cell AKA the Zoebel network. \$\endgroup\$
    – frr
    Apr 26, 2020 at 19:57
  • \$\begingroup\$ @frr duh, yes, I suppose it is intended to be a headphone amp, I would have recognised it with a series resistor. \$\endgroup\$
    – Neil_UK
    Apr 26, 2020 at 20:00

1 Answer 1


I'm not quite sure what you trying to achieving .. NE5532 has a unity gain on 10M, and drop from 1K frequency. https://www.onsemi.com/pub/Collateral/NE5532-D.PDF, page-5, top left graph. and phase margin around 10-15 degree on 10M.

That headphone are simple non-inverting amplifier, with some strange implementation. at first there is no 10 Megahertz you can apply on input as C3 would be effective short on that frequency.

R1 defined '0' potential, so power should be bipolar. And as it mention before a weird back-feeding loop. it set as multiply by 2 on DC (R3, R4) and 5.5 on AC (R3, R2)

Finally, a power output of NE5532 absolutely not enough for "headphone amplifier"

  • \$\begingroup\$ sorry pal, i have a feeling that complexity for 10Mhz ADC would be beyond your level, Starting from, level of the signal, impedance, differential input of ADC, voltage reference, and in the end the correct layers of PSB. Even for me it it faaar complex to achieve the correct result. If you wish to play with ADC, start at some datasheet on that particular ADC it will give some basic idea of schematics how it run, and then devil in details .. \$\endgroup\$
    – JamesBrown
    Apr 28, 2020 at 4:50

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