Objective: Design a feedback amplifier to amplify the output signal of a microphone to meet a set of specifications. Specifications: The output signal from the microphone is 10 mV and the output signal from the feedback amplifier is to be 0.5 V in order to drive a power amplifier that in turn will

Question

Objective: Design a feedback amplifier to amplify the output signal of a microphone to meet a set of specifications.

Specifications: The output signal from the microphone is 10 mV and the output signal from the feedback amplifier is to be 0.5 V in order to drive a power amplifier that in turn will drive the speakers. The nominal output resistance of the microphone is [latex]R_{S} = 5 k \Omega[/latex] and the nominal input resistance of the power amplifier is [latex]R_{L} = 75 \Omega[/latex] .
Choices: An op-amp with parameters [latex]R_{i} = 10 k\Omega, R_{o} = 100 \Omega[/latex] , and a low-frequency gain of [latex]A_{v} = 10^{4} [/latex] is available. [Note: In this simple design, neglect frequency response.]

Accepted Answer

(Design Approach): Since the source resistance is fairly large, an amplifier with a large input resistance is required to minimize loading at the input. Also, since the load resistance is low, an amplifier with a low output resistance is required to minimize loading at the output. To satisfy these requirements, a series–shunt feedback configuration, or voltage amplifier, should be used.

The closed-loop voltage gain must be [latex]A_{v f} = 0.5/0.01 = 50[/latex]. For the ideal case, [latex]A_{v f} = 1/β_{v} [/latex] , so the feedback transfer function is [latex]β_{v} = 1/50 = 0.02[/latex]. The loop gain is then
[latex]T = β_{v} A_{v} = (0.02)(10^{4}) = 200 [/latex]
Referring to Table 12.1, we expect the input resistance to be

Table 12.1	Summary results of feedback amplifier functions for the ideal feedback circuit
Feedback amplifier	Source signal	Output signal	Transfer function	Input resistance	Output resistance
Series–shunt (voltage amplifier)	Voltage	Voltage	[latex]A_{vf} = \frac{V_{o}}{V_{i}} = \frac{A_{v}}{(1 + β_{v} A_{v})} [/latex]	[latex]R_{i}(1 + β_{v} A_{v}) [/latex]	[latex]\frac{R_{o}}{(1 + β_{v} A_{v})} [/latex]
Shunt–series (current amplifier)	Current	Current	[latex]A_{if} = \frac{I_{o}}{I_{i}} = \frac{A_{i}}{(1 + β_{i} A_{i})} [/latex]	[latex] \frac{R_{i}}{(1 + β_{i} A_{i})} [/latex]	[latex]R_{o}(1 + β_{i} A_{i}) [/latex]
Series–series (transconductance amplifier)	Voltage	Current	[latex]A_{gf} = \frac{I_{o}}{V_{i}} = \frac{A_{g}}{(1 + β_{z} A_{g})} [/latex]	[latex]R_{i}(1 + β_{z} A_{g}) [/latex]	[latex]R_{o}(1 + β_{z} A_{g}) [/latex]
Shunt–shunt (transresistance amplifier)	Current	Voltage	[latex]A_{zf} = \frac{V_{o}}{I_{i}} = \frac{A_{z}}{(1 + β_{g} A_{z})} [/latex]	[latex] \frac{R_{i}}{(1 + β_{g} A_{z})} [/latex]	[latex] \frac{R_{o}}{(1 + β_{g} A_{z})} [/latex]

[latex]R_{i f} \cong (10)(200) k\Omega → 2 M \Omega[/latex]
and the output resistance to be
[latex]R_{of} \cong (100/200) \Omega = 0.5 \Omega[/latex]
These input and output resistance values will minimize any loading effects at the amplifier input and output terminals.
If we use the noninverting amplifier configuration in Figure 12.16, then we have
[latex]\frac{1}{β_{v}} = 1 + \frac{R_{2}}{R_{1}} = 50 [/latex]
and
[latex]\frac{R_{2}}{R_{1}} = 49 [/latex]
The feedback network loads the output of the amplifier; consequently, we need [latex]R_{1} + R_{2}[/latex] to be much larger than [latex]R_{o}[/latex]. However, the output resistance of the feedback network is in series with the input terminals, so extremely large values of [latex]R_{1}[/latex] and [latex]R_{2}[/latex] will reduce the actual signal applied to the op-amp because of voltage divider action.
Initially, then, we choose [latex]R_{1} = 1 k\Omega[/latex] and [latex]R_{2} = 49 k\Omega[/latex].
Computer Simulation Verification: The circuit in Figure 12.19 was used in a PSpice analysis of the voltage amplifier. A standard 741 op-amp was used in the circuit. For a 10 mV input signal, the output signal was 499.6 mV, for a gain of 49.96. This result is within 0.08 percent of the ideal designed value. The input resistance [latex]R_{i f}[/latex] was found to be approximately 580 MΩ and the output resistance [latex]R_{of}[/latex] was determined to be approximately 0.042 Ω. The differences between the measured input and output resistances compared to the predicted values are due to the differences between the actual µA-741 op-amp parameters and the assumed parameters. However, the measured input resistance is larger than predicted and the measured output resistance is smaller than predicted, which is desired and more in line with an ideal op-amp circuit.
Comment: An almost ideal feedback voltage amplifier can be realized if an op-amp is used in the circuit.

Question 12.8: Objective: Design a feedback amplifier to amplify the output...

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