Chapter 15 Practice Problems — Audio Applications and 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 — Preamplifier Gain and Signal Chain
A microphone produces a signal of 2 mV RMS. It is connected to a preamplifier followed by a power amplifier. The preamplifier has a voltage gain of 60 dB; the power amplifier has a voltage gain of 26 dB.
(a) Convert each stage's gain from dB to a linear voltage ratio.
(b) Find the signal voltage at the output of the preamplifier and at the output of the power amplifier.
(c) The power amplifier drives an 8 Ω speaker. Calculate the output power delivered to the speaker.
Hint
(a) To convert from dB: \(A_V = 10^{G_{dB}/20}\). For 60 dB: \(A_V = 10^{60/20} = 10^3 = 1000\). For 26 dB: \(A_V = 10^{26/20} \approx 20\).
(b) Apply gains in sequence: \(V_{preamp,out} = V_{mic} \times A_{V,pre}\). Then \(V_{power,out} = V_{preamp,out} \times A_{V,power}\).
(c) Power into a resistive load: \(P = V_{rms}^2 / R\). Use the RMS voltage at the power amplifier output and the speaker impedance of 8 Ω.
Problem 2 — Signal-to-Noise Ratio
A preamplifier has a maximum output voltage of 1.5 V RMS and a noise floor of 15 μV RMS measured over a 20 kHz audio bandwidth. A 10 kΩ source resistor is connected to the input.
(a) Calculate the SNR of the preamplifier in dB using \(SNR = 20\log_{10}(V_{signal}/V_{noise})\).
(b) Using the thermal noise formula \(V_n = \sqrt{4k_B T R \Delta f}\) with \(k_B = 1.38 \times 10^{-23}\) J/K, \(T = 300\) K, \(R = 10\) kΩ, and \(\Delta f = 20{,}000\) Hz, calculate the thermal noise generated by the source resistor alone.
(c) How does the source resistor's thermal noise compare to the amplifier's stated noise floor? Which dominates the overall system noise floor?
Hint
(a) Substitute \(V_{signal} = 1.5\) V and \(V_{noise} = 15 \times 10^{-6}\) V into the SNR formula. Compute the ratio first, then take the log.
(b) Compute the product \(4 \times 1.38 \times 10^{-23} \times 300 \times 10{,}000 \times 20{,}000\). Then take the square root. The result will be in the microvolt range.
(c) If the source resistor's thermal noise is comparable to or larger than the amplifier's noise floor, the resistor is the limiting factor and upgrading the op-amp will not improve SNR. Noise sources add in quadrature (root sum of squares): \(V_{total} = \sqrt{V_{resistor}^2 + V_{amp}^2}\).
Problem 3 — Power Amplifier Efficiency and Class
A Class AB power amplifier delivers 40 W of audio power to an 8 Ω speaker load. Its power supply provides ±35 V at a total supply current of 2 A (drawn equally from both rails).
(a) Calculate the total DC power drawn from the supply.
(b) Calculate the amplifier's efficiency \(\eta = P_{out}/P_{supply} \times 100\%\).
(c) A Class D amplifier with 92% efficiency delivers the same 40 W output. How much power does it draw from the supply, and how much power does it dissipate as heat?
Hint
(a) Total DC power is \(P_{supply} = V_{supply,total} \times I_{supply}\). With a dual supply ±35 V, the total supply voltage is 70 V (sum of both rails). Multiply by the total supply current.
(b) Efficiency \(\eta = (P_{out} / P_{supply}) \times 100\%\). For Class AB, expect 50–65%. Heat dissipated is \(P_{heat} = P_{supply} - P_{out}\).
(c) From the efficiency equation, \(P_{supply} = P_{out} / \eta\). Heat dissipated is \(P_{heat} = P_{supply} - P_{out}\). Compare the heat values for Class AB and Class D — this illustrates why Class D is preferred for battery-powered and high-power applications.
Problem 4 — Total Harmonic Distortion
A spectrum analyzer measures the output of a power amplifier driven by a 1 kHz sine wave at rated power. The measured harmonic amplitudes are: fundamental \(V_1 = 8.00\) V RMS, second harmonic \(V_2 = 24\) mV RMS, third harmonic \(V_3 = 16\) mV RMS, fourth harmonic \(V_4 = 4\) mV RMS.
(a) Calculate the THD using \(THD = \frac{\sqrt{V_2^2 + V_3^2 + V_4^2}}{V_1} \times 100\%\).
(b) The second harmonic is larger than the third. Does this suggest the distortion is more characteristic of a tube amplifier (even harmonics) or a transistor amplifier in hard clipping (odd harmonics)?
(c) The manufacturer specifies THD < 0.5%. Does this amplifier meet the specification? What would need to change to reduce THD to below 0.1%?
Hint
(a) Square each harmonic amplitude, sum them, take the square root, and divide by the fundamental. Keep units consistent (all in V RMS).
(b) Even harmonics (2nd, 4th) correspond to octave relationships and are characteristic of soft, tube-like distortion. Odd harmonics (3rd, 5th) are associated with hard clipping and sound harsher. Look at the ratio \(V_2/V_3\) to determine which character dominates.
(c) Compare your computed THD to 0.5% and 0.1%. To reduce THD, options include: operating at lower output power (more headroom), increasing negative feedback loop gain, improving output transistor linearity, or using a feed-forward distortion cancellation scheme.