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OpAmps

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drkidd22

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Hello,

I'm new to the omp amp world and would like some advice on a circuit I have. The objective is to be able to tweak the pot so that I can send a signal out the opamp to a microchip's analog input pin.

In the simulation when I tweak the pot pin 1 of the opamp never changer, but the voltage going into its pin 3 does.

Any advice will be appreciated.
 
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Probably need to connect -24V to the opamp's "Ground" pin. My guess the output is already at 0V and has nowhere to go.
 
It would help if we knew what you were trying to do with this circuit. What kind of microchip, there are literally millions. Did you mean a microcontroller? If so the op-amp needs a very strict +5V. R6 and R7 are not connected to anything, is there a reason for that? You have if configured in a very odd way, but BrownOut is right it is a dual rail op-amp and would definitely work better if it had a dual rail supply. If you want, http://www.datasheetcatalog.org/datasheet/texasinstruments/opa2604.pdf there is the data-sheet.
 
The voltage gain of your opamp is R1/R5 (+1)= 6. So the 10V input is trying to be amplified to +60V which is impossible. Then the output of the opamp is always as high as it can go.

The pot affects the voltage at pin 2 slightly but not at pin 3.
 
I think its a useless circuit .. R6& R7 arent even connected. I dont think that vo = Vin . ( 1 + R1/R5 ) because you also apply a voltage on the inverting input ( R11 & R9)
 
I think its a useless circuit .. R6& R7 arent even connected.
maybe they are the load. The opamp works fine without them.

I dont think that vo = Vin . ( 1 + R1/R5 ) because you also apply a voltage on the inverting input ( R11 & R9)
The (+) input is +10V and the (-) input is almost at 0V. The gain is 6 so the output is saturated near the positive supply voltage.
 
You are right audioguru, There is a purpose for R6 and R7, I will post the complete circuit today.
the gain at 6 is what I also calculated, but once I put in the rest of the circuit maybe ya can help me out understand it thanks.
 
Here is the updated circuit.
All I'm trying to do is understand how it works.
I have tested it on a bread board and it works ok, but I have no idea how it really works.
 

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When you turn on the power supply, the output of the opamp goes high which turns on the LED in the opto-isolator strongly which turns on the transistor in the opto, which causes the (-) input of the opamp to go positive which reduces the output of the opamp which reduces the currents in the opto.
 
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