# NCEES practice exam problem 120



## grownupsara (Oct 17, 2007)

Okay, I'm stumped with NCEES practice problem 120. I have a background in power design, so I'll admit that amplifiers are a weak point for me, so I apologize if this is a dumb question.

Looking at the problem, it has an amplifier circuit with a resistor and capacitor attached. You have to figure out the bode plot for A=V1/V2 versus frequency, where V1 is the input voltage and V1 is output voltage.

From the what the solution says: A=(1/jwC)/®=1/RCw, it looks to me like they're saying that V1=1/jwC and V2=R. This is what's confusing me. Maybe the current is the same and cancels out in the numerator and denominator?

Thanks in advance to anyone who can help explain this! I have the EERM, but can't find anything in it that helps me with this.


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## Flyer_PE (Oct 17, 2007)

I'm not sure where you're difficulty lies but I'll take a shot at it.

They're finding the transfer function of the op-amp circuit using KCL.

The perfect op-amp rules are:

1. V+ = V-

2. Current into V+ and V- = 0

The current through the capacitor is equal to the current through the resistor. (V1/R)

The voltage across the capacitor (V2) is equal to the current multiplied by the impedance = (V1/R) * (1/jwC)

The result is V2/V1=1/jwCR

Hope this helps.

Jim


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## grownupsara (Oct 17, 2007)

Jim,

Thanks for your help. That does answer my question. I've been having trouble understanding exactly what an amplifier actually does, but your explanation makes sense. Thanks!


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## chicago (Oct 17, 2007)

grownupsara,

As a side note, if you look at Figure 44.8 p.44-5 in EERM, you will find some of the most common gains for op-amp circuit configurations. That is always a good place to start. As you can see, problem 120's circuit is identical to Figure 44.8(e) integrator.

I doubt the actual exam will have anything more fancier than these. There might be a slight variation thereof, but follow KCL's rule as Jim has outlined, and you can find the gain of almost any simple op-amp circuit.


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