Op Amp circuit

Thread Starter

peter_morley

Joined Mar 12, 2011
179
Hello i've been trying to figure out how this circuit works for about an hour and I would appreciate some help considering I have a test for it soon. It seems like a simple circuit to me but when I get to the part where I have a 20kΩ resister in series with the 60kΩ and 240kΩ combination I get lost. I say combination because I believe they aren't in parallel but not sure how to describe it. I've tried node-voltage method and I get very close to the right answer but at this point its just guess and check...not fun anymore. Here is the circuit...
 

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hgmjr

Joined Jan 28, 2005
9,027
You could use Thevenin's Equivalent on R2 and R11 and Vout. That would allow you combine R2 and R11 into a single resistance. The Thevenin Equivalent voltage equation would then be in series with feedback path.

hgmjr
 

Thread Starter

peter_morley

Joined Mar 12, 2011
179
We have learned that in class but I'm not quite sure what you mean by that. Are you saying I should take the 60 and the 240 resistors in parallel so RTH equals 48kΩ. But how am I able to find VTH if I don't know the current going through? Should the short circuit current be 120mV/60kΩ which is my initial current into the branch part. Then take 48kΩ * 120mV/60kΩ to get VTH? sorry if that was incoherent. haha
-peter
 

Adjuster

Joined Dec 26, 2010
2,148
You are not trying to get a numerical value for VTH straight away. If you do as hgmjr suggested and find VTH as a function of Vout, you will end up with a circuit which is easy to solve for Vout.
 

steveb

Joined Jul 3, 2008
2,436
It's definitely good to develop the ability to use shortcuts like Thevenin's equivalents and other tricks. However, it is also important to be able to derive any relation from basic analysis.

In this case, since you are having trouble visualizing, even after given a shortcut approach, you should endeavor to discover this circuit on your own.

Here you can see the following relations if you assume an ideal opamp.

Vo=Vn+I11*R11
Vn=I1*R1
I11=I1+I2
I2=Vn/R2
I1=-V5/R7

Often, you need to do a formal solving of linear equations to get a solution, but look at these equations carefully. If you just do simple substitution to find Vo in terms of V5 and resistor values, you easily get the following.

Vo/V5=-(R1+R11*(1+R1/R2))/R7

Look what happens if R11 is zero. >>> Vo/V5=-R1/R7

Look what happens if R2 goes to infinity. >>> Vo/V5=-(R1+R11)/R7

Both equations are what you would get for a simple inverting amplifier.
 
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Thread Starter

peter_morley

Joined Mar 12, 2011
179
Thanks for clarifying that steveb. Understanding that Vn was the voltage drop -20kΩ * 120mV/6kΩ was key to understanding the problem. I got -6.8 volts which matched the multisim analysis thanks alot.
 

Jony130

Joined Feb 17, 2009
5,488
Look at diagram that I drew



I hope that diagram will help you solve you problem.
The only think you have to do is write down the equation for Va voltage.
And next if you solve for Va you can easily find Vb

Vb = Va * 1+R8/R12
 

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