Analyzing a Circuit with an Op-Amp

In summary, the homework statement is trying to solve for the potential at the V+ input of the op-amp.
  • #1
Temp0
79
0

Homework Statement


http://i.imgur.com/TBrFMIT.png


Homework Equations





The Attempt at a Solution


So I basically start with node voltages at the positive terminal of the op-amp, and I label that node E, so the node voltage goes:

VE/80 + (VE - VD)/20 = 0 (I labelled the node at the bottom of the diamond D)

Continuing from this equation, I then analyze the circuit at node D, and I know the left node in the diamond is 10V, because it is connected to a voltage source which is connected to the ground. I also know the right node is 0V, because it's directly connected to the ground, so then I get the equation:

(VD - 10)/40 + (VD - VE)/20 + VD/60

From the node voltage equation at node E, I learn that VE = 0.8 VD, which means I can then substitute that value into my equation to solve for VD. Skipping the calculations, I get that VD = 4.84V.

From there, I analyze the node on the right of the diamond, which I know to be 0V, this gives me the equation:
-VD/60 + (-VC)/30 = 0 (I name the top node C)
Simplifying this equation and substituting in the number for VD gives me a VC value of -2.42V
I can also solve for VE using the equation I had earlier, VE = 0.8VD, so I get that VE = 3.87V.

With these values, I use the summing point constraints, and I know that the voltage in the node connected to the negative terminal of the op-amp is the same as the voltage in the node connected to the positive terminal, giving me VE' = 3.87V (Naming the negative terminal E')

Doing node voltage analysis on the node E' gives me an equation to solve for Vo, but I get a ridiculously high number when I try this. Could someone identify where I am going wrong with my work? Thank you in advance.
 
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  • #2
Temp0 said:

Homework Statement


http://i.imgur.com/TBrFMIT.png


Homework Equations





The Attempt at a Solution


So I basically start with node voltages at the positive terminal of the op-amp, and I label that node E, so the node voltage goes:

VE/80 + (VE - VD)/20 = 0 (I labelled the node at the bottom of the diamond D)

Continuing from this equation, I then analyze the circuit at node D, and I know the left node in the diamond is 10V, because it is connected to a voltage source which is connected to the ground. I also know the right node is 0V, because it's directly connected to the ground, so then I get the equation:

(VD - 10)/40 + (VD - VE)/20 + VD/60

From the node voltage equation at node E, I learn that VE = 0.8 VD, which means I can then substitute that value into my equation to solve for VD. Skipping the calculations, I get that VD = 4.84V.

From there, I analyze the node on the right of the diamond, which I know to be 0V, this gives me the equation:
-VD/60 + (-VC)/30 = 0 (I name the top node C)
Simplifying this equation and substituting in the number for VD gives me a VC value of -2.42V
I can also solve for VE using the equation I had earlier, VE = 0.8VD, so I get that VE = 3.87V.

With these values, I use the summing point constraints, and I know that the voltage in the node connected to the negative terminal of the op-amp is the same as the voltage in the node connected to the positive terminal, giving me VE' = 3.87V (Naming the negative terminal E')

Doing node voltage analysis on the node E' gives me an equation to solve for Vo, but I get a ridiculously high number when I try this. Could someone identify where I am going wrong with my work? Thank you in advance.

Your value for the potential at the V+ input of the op-amp looks okay.

Can you show your work for the node voltage analysis to determine Vo?
 
  • #3
Nevermind, I figured out why I was wrong, it's because I can't use the 0V node for node analysis, because there's an unknown current coming out of the ground.
 

What is an op-amp?

An op-amp, short for operational amplifier, is an electronic component that amplifies the voltage difference between its two input terminals. It has a high gain and can be used in a variety of circuit configurations for different purposes.

How does an op-amp work?

An op-amp works by taking the voltage difference between its two input terminals and amplifying it to a much higher voltage at the output. This is achieved through the use of transistors and other electronic components inside the op-amp.

What are the key characteristics of an op-amp?

The key characteristics of an op-amp include its high open-loop gain, high input impedance, low output impedance, and high bandwidth. It also has a differential input and a single-ended output, meaning it amplifies the difference between its two inputs.

What is the ideal op-amp model?

The ideal op-amp model assumes that the op-amp has infinite open-loop gain, infinite bandwidth, infinite input impedance, and zero output impedance. It also has no input offset voltage or bias currents, and its output voltage can change instantaneously in response to changes in the input voltage.

How do you analyze a circuit with an op-amp?

To analyze a circuit with an op-amp, you must first determine the circuit configuration and the input and output voltages. Then, you can use the ideal op-amp model and basic circuit analysis techniques to calculate the output voltage. You may also need to consider the effects of the op-amp's non-ideal characteristics, such as input offset voltage and bias currents, on the circuit's performance.

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