Given the circuit diagram attached....
you have a microphone input to the positive terminal of the opamp, and the middle pin of a potentiometer connected to the negative terminal (note that the potentiometer isn't charged and is just grounded....
How is this circuit expected to provide an output; when one terminal is grounded and the other is a small microphone input which is around 100mv...
I tried making the circuit, the output from the opamp I'm finding is 0V
Am I perhaps making a mistake somewhere??

This is the circuit diagram
https://cdn.instructables.com/F1R/8DZI/H7KZ1PNJ/F1R8DZIH7KZ1PNJ.MEDIUM.jpg

 

Circuit in question for the convenience of members who might want to help:


It's a terrible schematic. There isn't sufficient information to tell you anything.

Which opamp? Input signal characteristics? AC or DC coupled? Battery voltages?

Show the schematic of the circuit you tested. Include supply voltages, opamp part number, and input coupling/amplitude/frequency.

 

Circuit in question for the convenience of members who might want to help:
View attachment 127970

It's a terrible schematic. There isn't sufficient information to tell you anything.

Which opamp? Input signal characteristics? AC or DC coupled? Battery voltages?

Show the schematic of the circuit you tested. Include supply voltages, opamp part number, and input coupling/amplitude/frequency.

It's a tl-082 opamp, battery voltages are +/- 9V
I don't have a schematic right now, I tested it on a breadboard
would the information be enough now?

 

would the information be enough now?

Still need to know how the signal is being coupled and what it's magnitude is.

How are you measuring the output? Do you have an oscilloscope? You can't use a DVM to measure audio frequency signals.

 

The circuit is a non-inverting microphone preamp with a minimum gain of 11 and a maximum gain of whatever the opamp open loop gain is.

To limit the maximum gain, add a resistor between the bottom of the pot and GND (the center junction of the two batteries). Note that this resistor will decrease the minimum gain.

Because there is no input coupling capacitor, the circuit will work with a dynamic microphone but not with an electret.

To properly terminate the unused opamp section, connect the + input to GND and the - input to the output.

ak

 

Still need to know how the signal is being coupled and what it's magnitude is.

How are you measuring the output? Do you have an oscilloscope? You can't use a DVM to measure audio frequency signals.

I'm measuring it through an arduino, checking the voltage readings, I honestly don't understand what signal coupling is

 

If the input is a microphone or some other signal with 0 Vdc offset, then the output at the opamp will be that signal amplified by the gain of the circuit. That goes through a coupling capacitor, which allows the DC level on the other side to be different. So the average DC level is 0 V at the opamp output, and 2.5 V at the Arduino input. The 2.5 V DC level is established by the two 100K resistors. This centers the audio signal within the overall range of the A/D converter in the Arduino.

ak

 

If the input is a microphone of some other signal with 0 Vdc offset, then the output at the opamp will be that signal amplified by the gain of the circuit. That goes through a coupling capacitor, which allows the DC level on the other side to be different. So the average DC level is 0 V at the opamp output, and 2.5 V at the Arduino input. The 2.5 V DC level is established by the two 100K resistors. This centers the audio signal within the overall range of the A/D converter in the Arduino.

ak

I understand thank you, and the arduino only measures DC voltage so that's all I was reading......
how could I read an AC signal though??

 

I understand thank you, and the arduino only measures DC voltage so that's all I was reading......
how could I read an AC signal though??

Read it more often. When you command the Arduino to take an A/D reading, it takes a snapshot of whatever is on the analog input pin being sampled. If the signal is audio (which changes constantly), the A/D output will be the voltage value at the instant it was sampled. To capture continuous audio you have to take thousands, or tens or thousands of samples *per second*, depending on how accurately you want the samples to represent the audio. Digitizing audio is tricky, and takes *lots* of memory.

If you are testing the audio to see if it is above or below some loudness level, that is an easier task.

ak