The following circuit is part of a larger circuit and processes analog audio. The input comes from the top-right and the output is the rightmost terminal of the pot, which then feeds into a straightforward audio amp chip (LM386N to be precise) before finally going to a headphone jack.

(pre-)amplifier circuit

My basic read is that the signal comes in and

  1. gets high-pass filtered by C10/R10
  2. something ampey involving R11, Q1 and its base components, and C12
  3. gets low-pass filtered by R13/C13
  4. goes through the pot

What I really don't understand is what is going on with #2 (though a friend suggested the entire thing may be a single amplifier and the 4 parts can't functionally be separated).

  1. Is this a single circuit or is my breakdown more or less accurate?
  2. In either case what is #2 doing?
  3. How does a transistor work when the signal is at the collector and not the base? In looking at amplifiers I've been unable to find a single example of this.
  4. What is the function of the diode in all this?
  5. What is the function of the pot? Is this related to volume, or is it related to the beta of the transistor as friend suggested?

Keywords that can readily be googled or links will suffice for answers if you don't want to type it out.

  • \$\begingroup\$ You sure R10 is connected to ground, not VCC? \$\endgroup\$
    – Kaz
    Dec 30, 2019 at 2:03
  • 1
    \$\begingroup\$ This looks more like a turn-on 'thump' filter, to kill the audio when power is first turned on. Else it makes no sense at all. \$\endgroup\$
    – user105652
    Dec 30, 2019 at 2:35

1 Answer 1


This is not an amplifier. Q1 will clamp the signal at power-on to keep DC from propagating. If the input has +DC on it, the DC will propagate until C10 charges. Q1 will clamp the DC until C11 charges. After C11 charges, there is no base current path and Q1 is effectively out of the circuit. D1 keeps Q1's base from going negative on power-off.

  • \$\begingroup\$ Does that mean that the RC pairs really are acting as filters? And does that mean the pot is for attenuation and does not affect the transistor circuit? \$\endgroup\$ Dec 29, 2019 at 8:16
  • 3
    \$\begingroup\$ @JustinOlbrantz Yes, your analysis of the filters is accurate. The pot is probably an attenuator, but circuit isn't complete enough to tell for sure. \$\endgroup\$
    – Mattman944
    Dec 29, 2019 at 8:21
  • \$\begingroup\$ I'm guessing that C10 will not actually charge due to some DC return path for the signal (not shown in the circuit). I.e., it's just a coupling cap. Q1 and R11 form a voltage divider. We can imagine Q1 to be a resistance which starts low on power up and rises. The purpose might be to "fade in" the signal on power up to suppress any power up glitches from earlier circuitry from going to the amplifier. This happens very quickly though; the RC constant for 10 uF and 2200 ohms is some 22 ms. \$\endgroup\$
    – Kaz
    Dec 30, 2019 at 1:59

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