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Chapter 12: Problem 8

A transformer-coupled class A amplifier drives a \(16-\Omega\) speaker through a \(3.87: 1\) transformer. Using a power supply of \(V_{C C}=36 \mathrm{~V}\), the circuit delivers \(2 \mathrm{~W}\) to the load. Calculate: a. \(P(\) ac \()\) across transformer primary. b. \(V_{L}(\mathrm{ac})\). c. \(V(\mathrm{ac})\) at transformer primary. d. The rms values of load and primary current.

Short Answer

Expert verified
a. The AC power across transformer primary is \(2W\). b. The AC load voltage is \(8V_{ac}\). c. The AC voltage at transformer primary is \(30.96V_{ac}\). d. The rms value of load current is \(0.5A_{ac}\), and the rms value of primary current is \(0.129A_{ac}\).

Step by step solution

01

Determine the AC power across transformer primary

Given that the amplifier delivers 2W to the 16-Ω load, it means that the AC power, \( P_{ac} \), across the transformer primary is also 2W. This is due to the principle that power is conserved in a transformer (ignoring power loses). Thus \( P_{ac} = 2W \).
02

Calculate the AC load voltage

Using the formula \( P = V^2/R \), we can rearrange to solve for voltage and get \( V = \sqrt{PR} \). Substitution of known values gives \( V = \sqrt{(2)(16)} \), yielding \( V = 8V_{ac} \) as the AC load voltage. This is \( V_L(ac) \).
03

Compute the AC voltage at transformer primary

Using the transformer ratio of 3.87:1, represents the relation between primary and secondary voltages. Since we know the secondary voltage \( V_S \) is 8V (from step 2), we can get the primary voltage \( V_P \) by multiplying the secondary voltage by 3.87. Hence \( V_P = 3.87 * 8V_{ac} = 30.96V_{ac} \) is the AC voltage at transformer primary.
04

Calculate the rms values of the load and primary current

Using Ohm's law (V = IR), we can change it to \( I = V/R \) to find the AC load current \( I_{L(ac)} \), which would be \( I_{L(ac)} = 8V_{ac} / 16Ω = 0.5A_{ac} \). In the transformer, the primary current \( I_{P(ac)} \) can be calculated using the transformer turns ratio, which is inverse for currents. Thus, \( I_{P(ac)} = I_{L(ac)} / 3.87 = 0.5A_{ac} / 3.87 = 0.129A_{ac} \).

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Most popular questions from this chapter

Calculate the efficiency of a class \(\mathrm{B}\) amplifier for a supply voltage of \(V_{C C}=22 \mathrm{~V}\) driving a 4- \(\Omega\) load with peak output voltages of: a. \(V_{L}(\mathrm{p})=20 \mathrm{~V}\). b. \(V_{L}(\mathrm{p})=4 \mathrm{~V}\)

Draw the circuit diagram of a class A transformer-coupled amplifier using an \(n p n\) transistor.

A \(160-\mathrm{W}\) silicon power transistor operated with a heat sink \(\left(\theta_{S A}=1.5^{\circ} \mathrm{C} / \mathrm{W}\right)\) has \(\theta_{J C}=0.5^{\circ} \mathrm{C} / \mathrm{W}\) and a mounting insulation of \(\theta_{C S}=0.8^{\circ} \mathrm{C} / \mathrm{W}\). What maximum power can be handled by the transistor at an ambient temperature of \(80^{\circ} \mathrm{C}\) ? (The junction temperature should not exceed \(200^{\circ} \mathrm{C}\).)

Draw the circuit diagram of a class B npn push-pull power amplifier using transformercoupled input.

Sketch the circuit diagram of a quasi-complementary amplifier, showing voltage waveforms in the circuit.
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