Engineering Op-Amp Resistive Circuit Homework: Finding Vo in a Given Circuit

AI Thread Summary
The discussion focuses on solving for the output voltage (Vo) across a 6k resistor in an op-amp circuit. The user has established that Vo equals the output voltage of the op-amp and is attempting to find the input voltage (Vin) using voltage and current laws. They express frustration over having multiple unknown currents and feel overwhelmed by the equations derived from Kirchhoff's Voltage Law (KVL) and Kirchhoff's Current Law (KCL). A suggestion is made to use the conservation of current to relate V+ and Vout, simplifying the problem. The circuit can be further reduced using a resistive voltage divider approach, indicating that the solution is manageable.
sojo
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Homework Statement


The attached picture is the circuit I have been given and I am asked to find Vo (the voltage drop across the 6k resistor.
Screen Shot 2018-10-28 at 9.01.31 PM.png


Homework Equations


v_out/v_in = (R_1 + R_2)/R_1
V+ = V-

The Attempt at a Solution


So I know that the voltage drop across the 6k resistor is just the output voltage of the op-amp since there is a ground there.
I used the v_out/v_in = 3+9/3 = 4, so I know V_out = 4V_in
The thing I am having issues with is finding the V_in. I feel like I have 7 unknown currents and I've written out a ton of equations using the KVL and KCL but I feel like I just keep running around in circles with all the unknowns.
Am I going to have to use all the KVL KCL equations to solve this, or is there some simpler method I am missing to find the Vin so I can find the Vout?
 

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sojo said:
v_out/v_in = (R_1 + R_2)/R_1
V+ = V-

You used "conservation of current" to establish V– in terms of Vout. So now you use that same "conservation of current" law to find V+ in terms of Vout and the 4V. The current into the resistor network equals the current that flows from the resistor network.
 
Just a hint: All you need is the positive feedback factor k+ (setting the 4V dc source equal to zero).
In this case, we have Rp=4k||5k=2.222k and the problem reduces to two simple resistive voltage divider in series.
This shouldn`t be a big problem.
 
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