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Chapter 13 Practice Problems — Operational Amplifiers

These problems are meant to build your problem-solving intuition. A hint is provided for each — try the problem on your own first before reading it.

Problem 1 — Inverting Amplifier Analysis

An inverting op-amp circuit has \(R_i = 10\) kΩ and \(R_f = 150\) kΩ, powered by ±15 V supplies.

(a) Calculate the closed-loop voltage gain \(A_V\).

(b) If the input is \(V_{in} = 0.4\) V DC, find \(V_{out}\). Will the output clip?

(c) What is the maximum input voltage that keeps the output in the linear region (assuming the op-amp saturates at ±13 V)?

Hint

(a) For the inverting configuration, \(A_V = -R_f/R_i\). Substitute the resistor values directly.

(b) Apply \(V_{out} = A_V \cdot V_{in}\). Compare the result to the ±13 V saturation limits to determine if clipping occurs.

(c) The output saturates at ±13 V. Set \(|A_V \cdot V_{in}| = 13\) V and solve for \(V_{in,max}\). Because the gain is negative, a positive input saturates the output negative and vice versa.

Problem 2 — Non-Inverting Amplifier Design

Design a non-inverting amplifier with a voltage gain of exactly \(+11\).

(a) Using the gain formula \(A_V = 1 + R_f/R_i\), find the required resistor ratio \(R_f/R_i\).

(b) Choose standard resistor values (common values: 1 kΩ, 2.2 kΩ, 4.7 kΩ, 10 kΩ, 22 kΩ, 47 kΩ, 100 kΩ) that achieve exactly this ratio.

(c) Using your chosen resistors, confirm the gain and calculate \(V_{out}\) when \(V_{in} = 0.25\) V. What is the input impedance seen by the source?

Hint

(a) Rearranging: \(R_f/R_i = A_V - 1 = 10\). Any pair of resistors whose ratio is 10:1 works.

(b) Convenient choices include \(R_i = 10\) kΩ with \(R_f = 100\) kΩ, or \(R_i = 1\) kΩ with \(R_f = 10\) kΩ. Verify the ratio gives exactly 10.

(c) Substitute into the gain formula to confirm \(A_V = 11\), then compute \(V_{out} = A_V \cdot V_{in}\). The input impedance of a non-inverting amplifier is essentially infinite because the signal drives the non-inverting (+) input directly.

Problem 3 — Summing Amplifier

A summing amplifier has three inputs: \(V_1 = 2\) V, \(V_2 = -1\) V, \(V_3 = 0.5\) V. The input resistors are \(R_1 = 10\) kΩ, \(R_2 = 20\) kΩ, \(R_3 = 10\) kΩ, and the feedback resistor is \(R_f = 40\) kΩ.

(a) State the two golden rules that make the summing amplifier analysis straightforward.

(b) Calculate the current from each input source to the virtual ground node.

(c) Apply KCL at the virtual ground node to find the output voltage \(V_{out}\).

Hint

(a) Golden Rule 1 (virtual short): \(V_+ = V_-\). Since \(V_+ = 0\), the inverting input is a virtual ground (\(V_- = 0\)). Golden Rule 2 (no input current): \(I_- = 0\), so all currents entering the virtual ground node must flow through \(R_f\).

(b) With the inverting node at 0 V, the current through each input resistor is \(I_n = V_n / R_n\). Compute \(I_1\), \(I_2\), and \(I_3\) individually — watch the sign of \(V_2\).

(c) KCL at the virtual ground: \(I_1 + I_2 + I_3 + I_f = 0\), where \(I_f\) flows through \(R_f\) toward the output. Then \(V_{out} = -I_f \cdot R_f = -R_f (V_1/R_1 + V_2/R_2 + V_3/R_3)\). Notice different input resistors give different weights to each input.

Problem 4 — Voltage Follower and Gain-Bandwidth

An op-amp has a gain-bandwidth product (GBW) of 5 MHz.

(a) The op-amp is configured as a voltage follower (unity-gain buffer). What is the maximum usable bandwidth in this configuration?

(b) The same op-amp is now used in a non-inverting amplifier with \(A_V = 50\). What is the maximum usable bandwidth?

(c) An audio signal at 15 kHz must be amplified by a factor of 20. Is this op-amp suitable? What minimum GBW would a replacement op-amp need?

Hint

(a) For a single-pole op-amp, \(\text{GBW} = A_{CL} \times BW\). With \(A_{CL} = 1\), \(BW = \text{GBW}/1 = 5\) MHz.

(b) With \(A_{CL} = 50\), \(BW = \text{GBW}/A_{CL} = 5\,\text{MHz}/50 = 100\) kHz.

(c) The required bandwidth must extend to at least 15 kHz (the signal frequency), so check whether \(A_{CL} \times f_{signal} < \text{GBW}\). Compute \(20 \times 15\,\text{kHz} = 300\) kHz. If this exceeds the available GBW, the op-amp cannot do the job — choose a replacement with GBW exceeding 300 kHz.