I'm trying to make a simple boost pedal for my guitar using the simplest schematic I found. I'm wondering why the input is connected to the +9v battery in this schematic through a 430kohm resistance. There should be no such connection from what I understood in this video about transistors.

Here is the schematic for the boost pedal I'm trying to make.


What happens current-wise where the two different voltages meet at R1?

I found another schematic here that I understand better according to the video.

Schematic 2


1 Answer 1


You have found two different schematics for circuits that do two different things. The bottom schematic is for making a beep with no volume control, and no real sound reproduction ability. The top schematic is an actual amplifier, which needs to faithfully reproduce the guitar's output voltage, only bigger.

It's what's known as a "single stage amplifier" -- try Googling for it to get a description of the circuit theory.

In short, R1, R2 and R4 are there to make sure that with no sound there's some standing current in Q1, because the only way that Q1 can make its collector voltage increase is by turning off a little bit. By biasing Q1 into the center of its operating range, the amplifier can accommodate large (ish) voltage swings from the guitar.

C1 is a DC blocking capacitor. It lets the transistor base voltage ride where it needs to (and the input voltage ditto), while allowing audio-frequency AC to pass. The impedance looking into the transistor base is pretty high (somewhere around 30k\$\Omega\$, doing the math in my head), and the source impedance is pretty low (600\$\Omega\$ for a guitar???). So the AC voltage at that node is pretty much set by the guitar.

  • \$\begingroup\$ Without the bias current of R1 the sound would be distorted. \$\endgroup\$
    – user105652
    Jan 17, 2019 at 20:03
  • \$\begingroup\$ Thank you for the quick answer! I now understand that R1 is used for the bias. I have looked up single stage amplifier but I still don't understand what happens to the voltage where the end of C1 meets the end of R1 as we have two different voltages that meet. I also have another question if you don't mind, would it still work if I used a 10nf capacitor instead of a 100nf at C1? \$\endgroup\$
    – Servietsky
    Jan 17, 2019 at 22:43
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    \$\begingroup\$ You would be unhappy with a 10nF cap there. If I'm doing the math right (in my head, which means I'm probably screwing it up at least a little) that circuit starts ignoring low-frequency signals at about 60Hz. If you changed to the 10x smaller cap that'd be 600Hz, and your guitar would sound horribly tinny. If I were building that circuit I'd experiment with a bigger cap there (up to 470nF), especially if it were for a bass. \$\endgroup\$
    – TimWescott
    Jan 17, 2019 at 22:50
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    \$\begingroup\$ Oh -- I edited my answer to include your second question. If you haven't seen that already. \$\endgroup\$
    – TimWescott
    Jan 18, 2019 at 1:26
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    \$\begingroup\$ @Servietsky The capacitor pushes the voltage up and down with the input, and the resistors slowly charge or discharge the capacitor so that the average voltage is a certain voltage. (If the average voltage gets too high, it discharges more than it charges, and if the average voltage gets too low, it charges more than it discharges) \$\endgroup\$
    – user253751
    Jan 18, 2019 at 2:45

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