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Chapter 13 Quiz — Operational Amplifiers

Multiple Choice Quiz

1. Which of the following best describes an ideal operational amplifier?

  • [ ] A) Finite voltage gain, infinite input impedance, zero output impedance
  • [ ] B) Infinite voltage gain, zero input impedance, infinite output impedance
  • [ ] C) Infinite voltage gain, infinite input impedance, zero output impedance
  • [ ] D) Finite voltage gain, finite input impedance, finite output impedance
Answer

C) Infinite voltage gain, infinite input impedance, zero output impedance. The three defining characteristics of the ideal op-amp are \(A_{OL} \to \infty\), \(Z_{in} \to \infty\), and \(Z_{out} \to 0\). These idealizations make circuit analysis tractable while remaining accurate enough for most practical designs.

2. In a negative-feedback op-amp circuit, the "virtual short" rule states:

  • [ ] A) The inputs are physically connected together with a wire
  • [ ] B) The output voltage is always zero
  • [ ] C) The voltage difference between the two inputs is essentially zero
  • [ ] D) No current flows through the feedback resistor
Answer

C) The voltage difference between the two inputs is essentially zero. With negative feedback and very high open-loop gain, the feedback loop forces \(V_+ \approx V_-\). This is not a physical short circuit — no wire connects the inputs — but the feedback mechanism makes them appear shorted for analysis purposes.

3. An inverting amplifier has \(R_i = 5\ \text{k}\Omega\) and \(R_f = 100\ \text{k}\Omega\). What is the closed-loop voltage gain?

  • [ ] A) \(+20\)
  • [ ] B) \(-20\)
  • [ ] C) \(+21\)
  • [ ] D) \(-21\)
Answer

B) \(-20\). The inverting amplifier gain is:

\[A_V = -\frac{R_f}{R_i} = -\frac{100\ \text{k}\Omega}{5\ \text{k}\Omega} = -20\]

The negative sign indicates phase inversion. A non-inverting configuration would give \(+21\), but this is an inverting amplifier.

4. A non-inverting amplifier is configured with \(R_f = 47\ \text{k}\Omega\) and \(R_i = 10\ \text{k}\Omega\). What is the voltage gain?

  • [ ] A) 4.7
  • [ ] B) −4.7
  • [ ] C) 5.7
  • [ ] D) −5.7
Answer

C) 5.7. The non-inverting amplifier gain is:

\[A_V = 1 + \frac{R_f}{R_i} = 1 + \frac{47}{10} = 1 + 4.7 = 5.7\]

The gain is always positive (no phase inversion) and always greater than or equal to 1 for the non-inverting configuration.

5. A voltage follower (unity-gain buffer) is used primarily because:

  • [ ] A) It provides very high voltage gain with low noise
  • [ ] B) It provides very high input impedance and very low output impedance
  • [ ] C) It inverts the phase of the signal by 180°
  • [ ] D) It has a gain-bandwidth product much higher than other configurations
Answer

B) It provides very high input impedance and very low output impedance. The voltage follower has \(A_V = 1\) — it provides no voltage gain. Its value is impedance transformation: the extremely high input impedance does not load the source, and the extremely low output impedance can drive heavy loads. It isolates circuit stages from each other.

6. An op-amp has a gain-bandwidth product (GBW) of 5 MHz. When configured as a non-inverting amplifier with \(R_f = 90\ \text{k}\Omega\) and \(R_i = 10\ \text{k}\Omega\), what is the maximum bandwidth of the circuit?

  • [ ] A) 5 MHz
  • [ ] B) 1 MHz
  • [ ] C) 500 kHz
  • [ ] D) 50 kHz
Answer

C) 500 kHz. First calculate the closed-loop gain:

\[A_{CL} = 1 + \frac{90}{10} = 10\]

Then use the GBW relationship:

\[BW = \frac{GBW}{A_{CL}} = \frac{5\ \text{MHz}}{10} = 500\ \text{kHz}\]

7. Which configuration is produced when a capacitor replaces \(R_f\) in an inverting amplifier?

  • [ ] A) Differentiator
  • [ ] B) Voltage follower
  • [ ] C) Summing amplifier
  • [ ] D) Integrator
Answer

D) Integrator. Placing a capacitor in the feedback position (where \(R_f\) normally sits) creates an integrator. The output is proportional to the time integral of the input: \(V_{out} = -(1/RC)\int V_{in}\,dt\). Placing a capacitor at the input (where \(R_i\) normally sits) would create a differentiator.

8. An op-amp's slew rate is 2 V/µs. What is the maximum frequency at which a sinusoidal output of 8 V peak can be reproduced without distortion?

  • [ ] A) 250 kHz
  • [ ] B) 159 kHz
  • [ ] C) 39.8 kHz
  • [ ] D) 25 kHz
Answer

C) 39.8 kHz. Using the slew rate frequency limit:

\[f_{max} = \frac{SR}{2\pi V_{peak}} = \frac{2 \times 10^6}{2\pi \times 8} = \frac{2{,}000{,}000}{50.27} \approx 39{,}800\ \text{Hz} = 39.8\ \text{kHz}\]

Above this frequency, the output cannot keep up with the required rate of change and produces a triangular waveform instead of a sinusoid.

9. CMRR (Common-Mode Rejection Ratio) is defined as:

  • [ ] A) The ratio of output voltage to input voltage in dB
  • [ ] B) \(20\log_{10}(A_{differential}/A_{common})\) in dB
  • [ ] C) The difference between the positive and negative supply voltages
  • [ ] D) The ratio of GBW to slew rate
Answer

B) \(20\log_{10}(A_{differential}/A_{common})\) in dB. CMRR measures how much better the op-amp amplifies the desired differential signal compared to the unwanted common-mode signal. A CMRR of 100 dB means the differential gain is 100,000 times larger than the common-mode gain — virtually all the common-mode interference is rejected.

10. "Virtual ground" at the inverting input of an op-amp occurs when:

  • [ ] A) The inverting input is directly connected to ground with a wire
  • [ ] B) The non-inverting input is grounded and negative feedback is applied
  • [ ] C) Both input terminals are connected to a common supply rail
  • [ ] D) The output voltage is exactly zero volts
Answer

B) The non-inverting input is grounded and negative feedback is applied. When \(V_+ = 0\) V (grounded non-inverting input), the virtual short rule (\(V_+ = V_-\)) forces \(V_- = 0\) V as well. The inverting input sits at 0 V without being physically connected to ground — it is virtually grounded through the feedback action. This is the key insight for analyzing the inverting amplifier, summing amplifier, integrator, and differentiator.

Practice Problems

Problem 1 — Design an Inverting Amplifier for Gain −25

Design an inverting amplifier with a closed-loop voltage gain of exactly −25. Use standard resistor values.

(a) Select an appropriate \(R_i\) value and calculate the required \(R_f\).

(b) If the nearest standard E24 resistor value for \(R_f\) is 240 kΩ, what is the actual achieved gain?

(c) Calculate the bias-compensation resistor \(R_{bias}\) to minimize errors from input bias current. Connect it from the non-inverting input to ground.

(d) If the op-amp has an input offset voltage \(V_{OS} = 2\ \text{mV}\), what is the resulting DC error at the output?

Solution

(a) Choose \(R_i = 10\ \text{k}\Omega\) (good compromise between source loading and noise). Calculate \(R_f\):

\[R_f = |A_V| \times R_i = 25 \times 10\ \text{k}\Omega = 250\ \text{k}\Omega\]

(b) With \(R_f = 240\ \text{k}\Omega\) (nearest standard E24 value):

\[A_V = -\frac{R_f}{R_i} = -\frac{240}{10} = -24\]

The actual gain is −24 rather than −25 (4% low). Use a 240 kΩ resistor in series with a 20 kΩ trim pot if precise gain is required.

(c) Bias-compensation resistor \(R_{bias} = R_i \| R_f\):

\[R_{bias} = \frac{R_i \cdot R_f}{R_i + R_f} = \frac{10 \times 240}{10 + 240} = \frac{2400}{250} = 9.6\ \text{k}\Omega \approx 10\ \text{k}\Omega\]

Connect a 10 kΩ resistor from the non-inverting input to ground.

(d) Output DC error from offset voltage (using actual gain magnitude of 24):

\[V_{OS,out} = V_{OS} \times \left(1 + \frac{R_f}{R_i}\right) = 2\ \text{mV} \times \left(1 + \frac{240}{10}\right) = 2\ \text{mV} \times 25 = 50\ \text{mV}\]

The 2 mV input offset appears as a 50 mV DC offset at the output.

Problem 2 — Gain-Bandwidth Product Analysis

An op-amp datasheet specifies GBW = 8 MHz and slew rate SR = 3 V/µs.

(a) You need to amplify a signal with gain \(A_V = +40\) (non-inverting configuration). What is the bandwidth limit imposed by the GBW?

(b) The same amplifier must faithfully reproduce a 5 V peak sinusoidal output. What is the maximum frequency allowed by the slew rate?

(c) If the signal to be amplified has frequency content up to 150 kHz, does this design meet both requirements? If not, what gain and/or slew rate would be required?

Solution

(a) Bandwidth limit from GBW. For non-inverting gain of +40:

\[BW = \frac{GBW}{A_{CL}} = \frac{8\ \text{MHz}}{40} = 200\ \text{kHz}\]

(b) Maximum frequency from slew rate with 5 V peak output:

\[f_{max} = \frac{SR}{2\pi V_{peak}} = \frac{3 \times 10^6}{2\pi \times 5} = \frac{3{,}000{,}000}{31.42} \approx 95.5\ \text{kHz}\]

(c) The signal requires 150 kHz bandwidth. The GBW allows 200 kHz (adequate), but the slew rate only allows 95.5 kHz (inadequate — the slew rate is the binding constraint).

To meet the requirement:

Required slew rate: [SR_{min} = 2\pi f V_{peak} = 2\pi \times 150{,}000 \times 5 = 4.71\ \text{V/µs}]

Select an op-amp with SR ≥ 5 V/µs (with margin). For example, the TL071 (SR = 13 V/µs) or OPA2134 (SR = 20 V/µs) would work.

Problem 3 — Summing Amplifier Analysis

A summing amplifier has the following configuration: - \(V_1 = 0.5\ \text{V}\), \(R_1 = 10\ \text{k}\Omega\) - \(V_2 = -1.2\ \text{V}\), \(R_2 = 20\ \text{k}\Omega\) - \(V_3 = 2.0\ \text{V}\), \(R_3 = 40\ \text{k}\Omega\) - Feedback resistor: \(R_f = 40\ \text{k}\Omega\) - Op-amp supply: ±15 V

(a) Using the virtual ground assumption, find the current through each input resistor.

(b) Apply KCL at the virtual ground node to find the total current through \(R_f\).

(c) Calculate the output voltage \(V_{out}\).

(d) Does the output voltage remain within the linear operating range of the op-amp?

Solution

(a) With virtual ground at the inverting input (\(V_- = 0\)), each input current flows through its resistor from \(V_n\) to 0 V:

\[I_1 = \frac{V_1}{R_1} = \frac{0.5\ \text{V}}{10\ \text{k}\Omega} = 50\ \mu\text{A}\]
\[I_2 = \frac{V_2}{R_2} = \frac{-1.2\ \text{V}}{20\ \text{k}\Omega} = -60\ \mu\text{A}\]
\[I_3 = \frac{V_3}{R_3} = \frac{2.0\ \text{V}}{40\ \text{k}\Omega} = 50\ \mu\text{A}\]

(b) KCL at the virtual ground node (no current enters the op-amp input):

\[I_f = I_1 + I_2 + I_3 = 50 - 60 + 50 = 40\ \mu\text{A}\]

(c) Output voltage. The current \(I_f\) flows from the virtual ground (0 V) through \(R_f\) to the output:

\[V_{out} = 0 - I_f \cdot R_f = -40 \times 10^{-6} \times 40 \times 10^3 = -1.6\ \text{V}\]

Alternatively, using the formula directly:

\[V_{out} = -R_f\!\left(\frac{V_1}{R_1} + \frac{V_2}{R_2} + \frac{V_3}{R_3}\right) = -40\!\left(\frac{0.5}{10} + \frac{-1.2}{20} + \frac{2.0}{40}\right)\!\text{k} = -40(0.05 - 0.06 + 0.05)\ \text{mA} = -1.6\ \text{V}\]

(d) The output is −1.6 V. With ±15 V supply, the linear output range is approximately ±13 V. Since −1.6 V is well within this range, the op-amp remains in its linear region. The golden rules apply and the result is valid.

Problem 4 — Integrator Output for a Step Input

An op-amp integrator has \(R = 22\ \text{k}\Omega\) and \(C = 47\ \text{nF}\). At \(t = 0\), a step voltage of \(V_{in} = +3\ \text{V}\) is applied. Assume the initial condition \(V_{out}(0) = 0\) and the op-amp supply is ±12 V.

(a) Calculate the RC time constant.

(b) Write the expression for \(V_{out}(t)\) for \(t > 0\).

(c) How long does it take for the output to reach −6 V?

(d) The op-amp saturates at approximately ±11 V. How long until the output saturates?

(e) Sketch the shape of \(V_{out}\) versus time, labeling the saturation point.

Solution

(a) RC time constant:

\[RC = 22 \times 10^3 \times 47 \times 10^{-9} = 1{,}034 \times 10^{-6}\ \text{s} \approx 1.034\ \text{ms}\]

(b) For a constant input \(V_{in} = +3\ \text{V}\), the integrator output ramps linearly:

\[V_{out}(t) = -\frac{V_{in}}{RC}\cdot t = -\frac{3\ \text{V}}{1.034\ \text{ms}}\cdot t = -2{,}902 \cdot t\ \ (\text{with } t \text{ in seconds})\]

Or equivalently: \(V_{out}(t) = -2.902\,t/\text{ms}\ \text{V}\)

(c) Time to reach −6 V:

\[t = \frac{|V_{out}| \cdot RC}{V_{in}} = \frac{6\ \text{V} \times 1.034\ \text{ms}}{3\ \text{V}} = 2.068\ \text{ms}\]

(d) Time to saturation at −11 V:

\[t_{sat} = \frac{11\ \text{V} \times 1.034\ \text{ms}}{3\ \text{V}} = \frac{11.374\ \text{ms}}{3} = 3.79\ \text{ms}\]

After 3.79 ms, the output reaches −11 V and pins to the negative rail. The golden rules no longer apply once the op-amp saturates.

(e) Shape description: \(V_{out}\) is a straight line with slope \(-2{,}902\ \text{V/s}\) starting at 0 V at \(t = 0\), crossing −6 V at \(t = 2.07\ \text{ms}\), and clamping at −11 V at \(t = 3.79\ \text{ms}\). After saturation, the output remains flat at −11 V regardless of the input.