Can anyone tell me what the 10uF capacitor does in the feedback circuit of this amplifier? How do you calculate it? Would a 100uF cap behave drastically different in that location?

 

As the signal frequency increases, the 10µF capacitance will present a lower impedance – resulting in an increased gain of the circuit at higher frequencies. Replacing the 10µF with one of 100 µF will result in an increased gain at an otherwise lower frequency.

If your reason for wanting to use a 100µF capacitor is because that is all you have to hand, you could try increasing the 27KΩ and 1k5Ω resistance values to compensate for the increased capacitor value to achieve somewhere near the original frequency response.

 

As the signal frequency increases, the 10µF capacitance will present a lower impedance – resulting in an increased gain of the circuit at higher frequencies. Replacing the 10µF with one of 100 µF will result in an increased gain at an otherwise lower frequency.

If your reason for wanting to use a 100µF capacitor is because that is all you have to hand, you could try increasing the 27KΩ and 1k5Ω resistance values to compensate for the increased capacitor value to achieve somewhere near the original frequency response.

Thank you! It is intended to be used as a microphone preamp, so I'm guessing that the 100uF capacitor would increase the bass response. Is there a formula to find the cutoff frequency where a noticeable rolloff would occur? If it's below 20hz then it shouldn't really matter.

 

Edit: The capacitor is so that the 1/2 power supply voltage bias at the plus input is amplified by only a gain of 1 and thus not saturating the output.
Thus the output is also biased at 1/2 the supply voltage so ±AC signals can be amplified with only a single supply voltage.

Even with a 10µF capacitor with the switch closed so the capacitor is seeing 1.5kΩ, the low frequency corner (-3dB point) is 10.6Hz.
A larger capacitor and/or the 27kΩ resistor connected will lower that further.
The corner frequency equals (1 / 2πRC).

 

f=1/(2πRC)
The circuit shown (1.5k/10uF) rolls off at 10Hz. The capacitor reduces the gain to unity at DC. The main reason to do that is to prevent the amplifier offset voltage being amplified.
By the way, that circuit is for a condensor microphone which requires a bias current. Don't use it with a dynamic microphone.

 

If your reason for wanting to use a 100µF capacitor is because that is all you have to hand, you could try increasing the 27KΩ and 1k5Ω resistance values to compensate for the increased capacitor value to achieve somewhere near the original frequency response.

decrease, not increase.

 

f=1/(2πRC)
The circuit shown (1.5k/10uF) rolls off at 10Hz. The capacitor reduces the gain to unity at DC. The main reason to do that is to prevent the amplifier offset voltage being amplified.
By the way, that circuit is for a condensor microphone which requires a bias current. Don't use it with a dynamic microphone.

Thank you! I am using it for an electret microphone built into my headset. I get a lot of noise using a 20 foot audio cable from my PC's built-in mic port, so I'm going to try using an amplifier with a short cable and then run a long cable from the amplifier output to the line-level input on my PC to get a better signal to noise ratio.

 

Thank you! I am using it for an electret microphone built into my headset. I get a lot of noise using a 20 foot audio cable from my PC's built-in mic port, so I'm going to try using an amplifier with a short cable and then run a long cable from the amplifier output to the line-level input on my PC to get a better signal to noise ratio.

That should work, but don't use the pot on the output. You need a low impedance drive to the cable. If the gain is too high, reduce the 33k resistor. But don't delete the 100Ω resistor, because the op-amp needs it to prevent the capacitive load (of the cable) sending the op-amp unstable.