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I have this circuit that amplifies an electric guitar signal of -1.5 to 1.5 V to 0 to 3 V. Op-amp The circuit worked perfectly in my Multisim simulation, but the signal doesn't change with the input it only stays at 3.3 Vs. The only thing I'm doing different from the the circuit above is I'm using two op amps on the same chip(MCP6002). I tried using a decoupling capacitor on first positive input, but I had the same problem.

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  • \$\begingroup\$ make sure that Vee of your protoboard is connected to +3.3V: the convention is that Vcc is a positive voltage and Vee is a negative voltage of the power supply; also make sure V1 AC has a zero DC; and correct your biasing network (R2/R1) to guarantee a positive signal \$\endgroup\$
    – V.V.T
    Nov 16, 2020 at 6:40
  • \$\begingroup\$ The way you are biasing the input is not correct, as mentioned, you are depending on the guitar to short R1 to ground, and some DC will flow through the guitar coil. Also, 1K input resistance is extremely small for a guitar. \$\endgroup\$
    – S.s.
    Nov 17, 2020 at 1:25

3 Answers 3

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From your own answer to your question:

The problem is that the guitar has it's own resistance from the potentiometer. After measuring that resistance and recalculating R2 using the resistance of the guitar as R1, the amp works as it should.

My comment:

That means you have omitted the input decoupling capacitor and are running a small DC current through the guitar's pickup. You need to add the capacitor and one on the output too.

schematic

simulate this circuit – Schematic created using CircuitLab

Figure 1. The input buffer.

The input impedance seen by the guitar is 100k || 100k = 50 kΩ which is possibly quite low for a guitar pickup. You can choose the de-coupling capacitor value based on the cut-off frequency as it acts as a high-pass filter. To avoid losing any bass from the guitar we'll aim for 20 Hz cut-off.

According to a random high-pass filter calculator you'll need 160 nF so anything from 220 nF to 1 μF would be fine.

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The problem is that the guitar has it's own resistance from the potentiometer. After measuring that resistance and recalculating R2 using the resistance of the guitar as R1, the amp works as it should.

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  • \$\begingroup\$ That means you have omitted the input decoupling capacitor and are running a small DC current through the guitar's pickup. You need to add the capacitor and one on the output too. \$\endgroup\$
    – Transistor
    Nov 16, 2020 at 21:58
  • \$\begingroup\$ What value of capacitor should I be using for this or how can I calculate this? When I added a .1 uF capacitor, the voltage is stuck at 3.3 again. \$\endgroup\$
    – Jacob Lara
    Nov 17, 2020 at 1:30
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Remember guitar amplifiers that used tubes? Their input resistance was one million ohms. Yours is only 4300 ohms.

The input coupling capacitor is supposed to feed a voltage divider of two series resistors so that the input of the first opamp is at half the supply voltage so that its output can swing equally up and down.

Your second opamp has a DC gain of 2.2 times which will amplify the +1.65V of the output of the first opamp. The output of the second opamp will be a saturated DC voltage as high as it can go. Then the second opamp will do almost nothing.

Do guitars play DC?? Coupling capacitors are used for audio and the 8.2k resistor needs a coupling capacitor to ground so that its 2nd opamp's DC gain is 1.

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