Chapter 11: Problem 14
An amplifier has an open-circuit voltage gain of 1000 , an input resistance of \(20 \mathrm{k} \Omega\), and an output resistance of \(2 \Omega\). A signal source with an internal resistance of \(10 \mathrm{k} \Omega\) is connected to the input terminals of the amplifier. An \(8-\Omega\) load is connected to the output terminals. Find the voltage gains \(A_{v s}=V_{o} / V_{s}\) and \(A_{v}=V_{o} / V_{i}\). Also, find the power gain and current gain.
Short Answer
Step by step solution
Understanding the Voltage Gain Definitions
Calculate the Input Voltage, \(V_i\)
Calculate \(A_v\), Voltage Gain of the Amplifier
Calculate \(A_{v_s}\), Voltage Gain from Source to Output
Calculate the Power Gain
Calculate the Current Gain
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Understanding Voltage Gain
- \( A_v = \frac{V_o}{V_i} \)
Additionally, the voltage gain from the source to the output (\( A_{v_s} \)) is crucial. It considers the voltage drop across the resistor network at the input as well as the gain provided by the amplifier. Through this, we find:
- \( A_{v_s} = \frac{V_o}{V_s} \)
Exploring Power Gain
- \( G_p = \frac{(V_o^2/R_L)}{(V_s^2/R_s)} \)
- \( G_p = A_{v_s}^2 \cdot \frac{R_s}{R_L} \)
Investigating Current Gain
- \( A_i = \frac{I_o}{I_s} \)
- \( A_i = \frac{V_o/R_L}{V_s/R_s} = A_{v_s} \cdot \frac{R_s}{R_L} \)
This boost in current demonstrates the amplifier's capability to drive a load effectively, even if the input current is weak. Current gain is crucial in scenarios where a small input current must control a larger load or further stages of amplification.
Resistor Network Analysis and Amplification
Comprehending these interplays helps discern true performance beyond just looking at an amplifier's open-circuit gain, factoring in realistic operation conditions that the amplifier will face in use.