Determine the voltage gain of the circuit in Fig. P4.9


Assuming an ideal op-amp, determine the voltage gain of the circuit in Fig. P4.9.

determine the voltage gain of the circuit in Fig. P4.9

Image from: J. D. Irwin and R. M. Nelms, Basic engineering circuit analysis, 10th ed. Hoboken, NJ: John Wiley, 2011.

Solution:

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As we are treating this to be an ideal op-amp, remember the following:

-No current flows in or out of the the positive and negative terminals of the op-amp

-The voltage at the negative terminal is equal to the voltage at the positive terminal (V_-=V_+)

 

Looking at the diagram, we see that the positive terminal has a value of V_1 because there is a direct source feeding a voltage of V_1, therefore, the value at the negative terminal must also be V_1. Let us illustrate this as follows:

determine the voltage gain of the circuit in Fig. P4.9

Notice how the node attached to the negative terminal (colored purple) has a value of V_1. Also note the arbitrarily chosen currents flowing in and out of the node. Remember that I_3=0 A because as mentioned, no current flows in or out of the positive and negative terminals in the op-amp. Let us now focus on the orange colored node. The voltage at that point is 0 V, since the bottom node is grounded (this is required to maintain Kirchhoff’s current law KCL) .

Let us now write KCL at the purple node:

I_1-I_2=0

(Remember, I_3=0 A)

Substituting the nodal voltage values, we get:

\dfrac{0-V_1}{1}-\dfrac{V_1-V_0}{20}=0

-V_1=\dfrac{V_1-V_0}{20}

-20V_1=V_1-V_0

V_0=21V_1

Voltage gain is voltage out over voltage in:

\dfrac{V_0}{V_1}=21

 

Final Answer:

\dfrac{V_0}{V_1}=21

 

This question can be found in Basic Engineering Circuit Analysis, 10th edition, chapter 4, question 4.9.

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