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

#### Solution:

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: 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:

$\dfrac{V_0}{V_1}=21$

$\dfrac{V_0}{V_1}=21$