Question 9.1: Objective: Design an inverting amplifier with a specified vo...

The input current is given by
i_{1} = \frac{v_{I}}{R_{1}} = \frac{v_{s}}{R_{1}}

If i_{1}(max) = 5  μA, then we can write
R_{1} = \frac{v_{s} (max)}{i_{1}(max)} = \frac{0.1}{5  ×  10^{−6}} ⇒ 20  k\Omega
The closed-loop gain is given by
A_{v} = \frac{−R_{2}}{R_{1}} = −5

We then have
R_{2} = 5R_{1} = 5(20) = 100  k\Omega
Trade-offs: If the signal source has a finite output resistance and the desired output voltage is v_{o} = −0.5 \sin ωt , the circuit must be redesigned. Assume the output resistance of the source is R_{S} = 1  k\Omega.
Redesign Solution: The output resistance of the signal source is now part of the input resistance to the op-amp. We now write
R_{1} + R_{S} = \frac{v_{s} (max)}{i_{1}(max)} = \frac{0.1}{5  ×  10^{−6}} ⇒ 20  k\Omega
Since R_{S} = 1  k\Omega, we then have R_{1} = 19  k\Omega. The feedback resistor is then R_{2} = 5(R_{1} + R_{S} ) = 5(19 + 1) = 100  k\Omega.
Comment: The output resistance of the signal source must be included in the design of the op-amp to provide a specified voltage gain.
Computer Verification: Figure 9.11(a) shows the PSpice circuit schematic with the source resistance of 1 kΩ and an input resistance of 19 kΩ. Figure 9.11(b) shows the 100 mV sinusoidal input signal. Figure 9.11(c) is the output signal which shows that a gain of 5 (magnitude) has been achieved and also shows that the output signal is 180 degrees out of phase with respect to the input signal. Finally, the input current is shown in Figure 9.11(d) with a maximum value of 5 μA. The actual circuit characteristics are not influenced to any great extent by the nonideal parameters of the μA-741 op-amp used in the circuit simulation