# Op Amp circuit

Discussion in 'Homework Help' started by peter_morley, Mar 12, 2011.

1. ### peter_morley Thread Starter Member

Mar 12, 2011
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0
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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2. ### hgmjr Moderator

Jan 28, 2005
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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

3. ### peter_morley Thread Starter Member

Mar 12, 2011
179
0
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

Dec 26, 2010
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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.

5. ### steveb Senior Member

Jul 3, 2008
2,433
469
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.

Last edited: Mar 12, 2011
6. ### peter_morley Thread Starter Member

Mar 12, 2011
179
0
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.

7. ### peter_morley Thread Starter Member

Mar 12, 2011
179
0
Hey here is what ive done for this problem i know its not right...attachment.

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8. ### peter_morley Thread Starter Member

Mar 12, 2011
179
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Great this new post got added to my original post now nobody wants to help...boohoo.

9. ### Jony130 AAC Fanatic!

Feb 17, 2009
4,167
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Look at diagram that I drew