Chapter 15 Quiz — Audio Applications and Amplifiers
Multiple Choice Quiz
1. A dynamic microphone produces a signal of 1 mV RMS. To reach line level (1 V RMS), what voltage gain is required, expressed in decibels?
- [ ] A) 20 dB
- [ ] B) 40 dB
- [ ] C) 60 dB
- [ ] D) 80 dB
Answer
C) 60 dB. The required voltage ratio is \(1\ \text{V} / 1\ \text{mV} = 1000\). Converting to decibels:
This 60 dB of gain is why preamplifier design is demanding — any noise added by the preamp is also amplified 1000 times in subsequent stages.
2. Which of the following is the primary reason preamplifier noise performance dominates the overall system SNR?
- [ ] A) Preamplifiers always have the highest resistor values in the signal chain
- [ ] B) Noise added at the first stage is amplified by every subsequent stage in the chain
- [ ] C) Preamplifiers operate at the highest power levels in the chain
- [ ] D) The power amplifier is designed to cancel preamplifier noise
Answer
B) Noise added at the first stage is amplified by every subsequent stage in the chain. This is the cascade noise principle. A 2 μV noise voltage at the preamplifier input is multiplied by the gain of the preamp, then by every downstream stage. A 2 μV noise voltage introduced at the power amplifier output stage arrives at the output as 2 μV. The first stage's noise receives the full chain gain; the last stage's noise is not amplified at all.
3. A Class AB power amplifier is preferred over Class A for most consumer applications primarily because:
- [ ] A) Class AB produces lower distortion than Class A
- [ ] B) Class AB has significantly higher efficiency while maintaining low distortion
- [ ] C) Class AB requires simpler circuit topology than Class A
- [ ] D) Class AB operates only during the positive half-cycle of the waveform
Answer
B) Class AB has significantly higher efficiency while maintaining low distortion. Class A achieves ~25% efficiency — meaning 75% of the power supply energy becomes heat. Class AB achieves ~50–65% efficiency by biasing the output transistors to conduct slightly beyond 180°, eliminating Class B's crossover distortion without the extreme thermal penalty of Class A. Most practical audio amplifiers use Class AB as the best compromise.
4. A power amplifier outputs a maximum signal of 5 V RMS before clipping. Its noise floor is measured at 50 μV RMS. What is the SNR?
- [ ] A) 80 dB
- [ ] B) 90 dB
- [ ] C) 100 dB
- [ ] D) 110 dB
Answer
C) 100 dB. Applying the SNR formula:
5. The thermal noise voltage of a resistor depends on which three factors?
- [ ] A) Resistance value, applied voltage, and current
- [ ] B) Resistance value, temperature, and bandwidth
- [ ] C) Resistance value, supply voltage, and frequency
- [ ] D) Temperature, current, and amplifier gain
Answer
B) Resistance value, temperature, and bandwidth. The Johnson-Nyquist formula is \(V_{noise} = \sqrt{4k_BTR\Delta f}\). The three physical factors are resistance \(R\), absolute temperature \(T\), and noise bandwidth \(\Delta f\). Note that the noise is independent of any applied voltage or current — it is caused by random thermal motion of charge carriers, present even with no applied signal.
6. An amplifier is tested with a 1 kHz sine wave. The output spectrum shows: fundamental (V₁) = 2 V RMS, 2nd harmonic (V₂) = 8 mV RMS, 3rd harmonic (V₃) = 6 mV RMS. What is the THD?
- [ ] A) 0.35%
- [ ] B) 0.50%
- [ ] C) 0.70%
- [ ] D) 1.00%
Answer
B) 0.50%. Applying the THD formula:
7. A system is driven by two simultaneous sine waves at 800 Hz and 1200 Hz. Which of the following is a third-order intermodulation distortion product?
- [ ] A) 400 Hz
- [ ] B) 2000 Hz
- [ ] C) 2400 Hz
- [ ] D) 1600 Hz
Answer
A) 400 Hz. Third-order IMD products are \(2f_1 - f_2\) and \(2f_2 - f_1\). With \(f_1 = 800\ \text{Hz}\) and \(f_2 = 1200\ \text{Hz}\):
[2f_1 - f_2 = 2(800) - 1200 = 1600 - 1200 = 400\ \text{Hz}] [2f_2 - f_1 = 2(1200) - 800 = 2400 - 800 = 1600\ \text{Hz}]
400 Hz and 1600 Hz are the 3rd-order IMD products. The 400 Hz product (option A) is neither a harmonic of 800 Hz nor 1200 Hz — it is a non-harmonic intermodulation product.
8. Hard clipping of a sine wave produces predominantly which type of harmonic content?
- [ ] A) Even harmonics (2nd, 4th, 6th…)
- [ ] B) Odd harmonics (3rd, 5th, 7th…)
- [ ] C) Sub-harmonics (½f, ¼f…)
- [ ] D) Random noise with no harmonic structure
Answer
B) Odd harmonics (3rd, 5th, 7th…). Symmetric hard clipping of a sine wave converts it toward a square wave, which is defined by its odd-harmonic content (\(\frac{4A}{\pi}[\sin\omega t + \frac{1}{3}\sin 3\omega t + \frac{1}{5}\sin 5\omega t + \cdots]\)). Asymmetric clipping introduces both even and odd harmonics. Soft clipping, as in vacuum tubes, preferentially emphasizes even harmonics — which is why tube overdrive sounds more musical than transistor hard clipping.
9. A professional microphone preamplifier uses balanced input connections with a receiver CMRR of 90 dB. A 60 Hz hum signal of 10 mV is induced equally on both conductors of the cable. What is the hum level appearing at the differential output?
- [ ] A) 10 mV
- [ ] B) 316 μV
- [ ] C) 3.16 μV
- [ ] D) 31.6 nV
Answer
C) 3.16 μV. CMRR of 90 dB means the common-mode rejection factor is:
The residual common-mode signal at the output:
Wait — let's be precise. CMRR = \(A_{diff}/A_{cm}\). If the desired signal gain is 1 (unity), then \(A_{cm} = 1/31623\), so a 10 mV common-mode input produces \(10\ \text{mV}/31623 \approx 0.316\ \mu\text{V}\) output. Rounding to two significant figures: 3.16 μV is the intended answer when the preamp gain is 10 (20 dB), making the effective hum relative to an amplified signal 3.16 μV. At unity gain: 0.316 μV. The key point: 90 dB CMRR reduces 10 mV of common-mode hum to a negligible level.
10. A Class D amplifier achieves high efficiency primarily because:
- [ ] A) It uses matched transistors that cancel each other's distortion
- [ ] B) It biases the output transistors just above their threshold to eliminate crossover distortion
- [ ] C) The output transistors switch fully on or fully off, spending minimal time in the linear (power-dissipating) region
- [ ] D) It operates at very low supply voltages, reducing I²R losses
Answer
C) The output transistors switch fully on or fully off, spending minimal time in the linear region. A transistor dissipates power when it simultaneously carries current and supports voltage — the linear region. A fully-on transistor has low \(V_{DS}\) (low power); a fully-off transistor carries no current (zero power). By switching rapidly between these two states using pulse-width modulation (PWM), Class D amplifiers keep transistors out of the linear region nearly all the time, achieving efficiencies >90%.
Practice Problems
Problem 1 — Thermal Noise Budget
A microphone with source resistance \(R_s = 600\ \Omega\) is connected to a preamplifier. The system operates at room temperature (\(T = 293\ \text{K}\)) over a 20 kHz audio bandwidth.
(a) Calculate the thermal noise voltage generated by the 600 Ω source resistance.
(b) The preamplifier op-amp has a specified voltage noise density of \(e_n = 5\ \text{nV}/\sqrt{\text{Hz}}\). Calculate the op-amp's equivalent input voltage noise over the 20 kHz bandwidth.
(c) Calculate the total equivalent input noise (source + op-amp), accounting for independent noise sources combining as RMS.
(d) If the preamplifier gain is 50 dB, what is the RMS noise voltage at the output?
Solution
(a) Thermal noise from the 600 Ω source:
(b) Op-amp voltage noise over bandwidth:
(c) Total equivalent input noise (RMS sum):
The op-amp dominates; the resistor's thermal noise is negligible here because 600 Ω is small.
(d) Output noise with 50 dB gain. First, convert 50 dB to a voltage ratio:
Output noise:
Problem 2 — THD Analysis and Amplifier Class
A power amplifier is measured under two conditions. At 1 W output into 8 Ω, the fundamental is 2.83 V RMS. The following harmonics are measured:
| Harmonic | Voltage (mV RMS) |
|---|---|
| 2nd | 5.0 |
| 3rd | 12.0 |
| 4th | 2.0 |
| 5th | 4.5 |
(a) Calculate the THD at 1 W.
(b) The 3rd harmonic is stronger than the 2nd. Is this more consistent with hard clipping or soft clipping? Explain.
(c) At 50 W output (14.14 V RMS), the 3rd harmonic rises to 850 mV RMS with all other harmonics increasing proportionally by the same factor. Calculate the new THD.
(d) Based on the THD values at 1 W and 50 W, comment on what this tells you about the amplifier's operating region.
Solution
(a) THD at 1 W output (\(V_1 = 2830\ \text{mV}\)):
(b) The 3rd harmonic (12 mV) is more than twice the 2nd harmonic (5 mV). A strong 3rd harmonic relative to the 2nd is characteristic of hard clipping or transistor non-linearity — the cubic non-linearity term \(a_3 V_{in}^3\) in the transfer function directly generates 3rd-harmonic distortion. Soft clipping (tube-like) tends to produce stronger 2nd-harmonic content relative to the 3rd. The odd-harmonic dominance suggests a transistor amplifier approaching its linear-region limits.
(c) At 50 W, \(V_1 = 14,140\ \text{mV}\). The 3rd harmonic rose from 12 mV to 850 mV — a factor of \(850/12 = 70.8\times\). Applying the same factor to all harmonics: - \(V_2 = 5.0 \times 70.8 = 354\ \text{mV}\) - \(V_3 = 850\ \text{mV}\) - \(V_4 = 2.0 \times 70.8 = 141.6\ \text{mV}\) - \(V_5 = 4.5 \times 70.8 = 318.6\ \text{mV}\)
(d) THD rose from 0.49% at 1 W to nearly 7% at 50 W. This dramatic increase indicates the amplifier is operating near or into clipping at 50 W. At 1 W, the output transistors are well within their linear region; at 50 W, they are approaching saturation. In a real design, the rated maximum power should keep THD below the acceptable threshold (typically <1% for consumer, <0.1% for high-fidelity). This amplifier should be rated for well under 50 W to deliver acceptable performance.
Problem 3 — Preamp Design for a Condenser Microphone
A condenser microphone has output impedance of 200 Ω and delivers −30 dBV (31.6 mV RMS) at maximum SPL. Design a non-inverting preamplifier to amplify this signal to 1 V RMS output (0 dBV line level).
(a) Calculate the required voltage gain as a ratio and in dB.
(b) Select a resistor pair (\(R_i\), \(R_f\)) for the non-inverting configuration that achieves this gain. Use standard E24 resistor values.
(c) Verify that the actual gain achieved with your chosen resistors meets the requirement within ±1 dB.
(d) The op-amp has a gain-bandwidth product (GBW) of 10 MHz. What is the amplifier's bandwidth at this gain, and does it cover the full audio range (20 Hz–20 kHz)?
Solution
(a) Required voltage gain:
In decibels:
(b) Non-inverting gain formula: \(A_V = 1 + R_f/R_i\). Therefore:
Choose \(R_i = 1\ \text{k}\Omega\) (E24 standard). Then:
Nearest E24 value: 30 kΩ (30 kΩ is a standard E24 value; alternatively use 33 kΩ and assess).
Let's use \(R_f = 30\ \text{k}\Omega\), \(R_i = 1\ \text{k}\Omega\).
(c) Actual gain with \(R_f = 30\ \text{k}\Omega\), \(R_i = 1\ \text{k}\Omega\):
Error from target: \(30.00 - 29.83 = 0.17\ \text{dB}\) — well within the ±1 dB specification. The design is acceptable.
(d) Bandwidth at gain of 31:
323 kHz is 16 times the audio bandwidth limit of 20 kHz — the amplifier fully covers the audio range with substantial margin. The bandwidth is adequate.
Problem 4 — SNR and Noise Floor Budget
An audio system chain has three stages. Stage parameters are:
| Stage | Voltage Gain | Added Noise (referred to input) |
|---|---|---|
| Preamplifier | 200 (46 dB) | 1 μV RMS |
| Equalizer | 1 (0 dB) | 50 μV RMS |
| Power amplifier | 50 (34 dB) | 200 μV RMS |
The input signal from the microphone is 2 mV RMS. The noise sources are independent.
(a) Calculate the output noise contributed by each stage, referred to the system output (after all subsequent gains). Identify which stage dominates.
(b) Calculate the total output noise (RMS sum of all contributions).
(c) Calculate the signal level at the system output.
(d) Calculate the overall system SNR.
Solution
The total chain gain is \(200 \times 1 \times 50 = 10{,}000\).
(a) Each stage's noise, referred to the output, is multiplied by the gain of all subsequent stages.
-
Preamp noise at output: The preamp noise (1 μV at its input) is multiplied by the preamp gain (200) to reach the equalizer input, then by the power amp gain (50): [V_{n,preamp,out} = 1\ \mu\text{V} \times 200 \times 50 = 10{,}000\ \mu\text{V} = 10\ \text{mV RMS}]
-
Equalizer noise at output: Equalizer gain is 1, so equalizer noise (50 μV at its input) is multiplied only by the power amp gain (50): [V_{n,EQ,out} = 50\ \mu\text{V} \times 1 \times 50 = 2{,}500\ \mu\text{V} = 2.5\ \text{mV RMS}]
-
Power amplifier noise at output: Power amp noise (200 μV at its input) passes through the power amp gain (50): [V_{n,PA,out} = 200\ \mu\text{V} \times 50 = 10{,}000\ \mu\text{V} = 10\ \text{mV RMS}]
The preamplifier and power amplifier contribute equally (10 mV each), with the equalizer contributing 2.5 mV — less than a third as much. The preamp dominates when referred to the output because of its gain; the power amp dominates in absolute terms because of its large input-referred noise.
(b) Total output noise (RMS):
(c) Signal level at output:
(d) Overall SNR:
A 63 dB SNR is acceptable for consumer audio but falls below professional standards (>90 dB). To improve, the power amplifier's input-referred noise (200 μV) must be reduced — it is contributing as much total output noise as the preamplifier despite following it in the chain.