Question 12.8: Objective: Design a feedback amplifier to amplify the output...
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 R_{S} = 5 k \Omega and the nominal input resistance of the power amplifier is R_{L} = 75 \Omega .
Choices: An op-amp with parameters R_{i} = 10 k\Omega, R_{o} = 100 \Omega , and a low-frequency gain of A_{v} = 10^{4} is available. [Note: In this simple design, neglect frequency response.]
(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 A_{v f} = 0.5/0.01 = 50. For the ideal case, A_{v f} = 1/β_{v} , so the feedback transfer function is β_{v} = 1/50 = 0.02. The loop gain is then
T = β_{v} A_{v} = (0.02)(10^{4}) = 200
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 | A_{vf} = \frac{V_{o}}{V_{i}} = \frac{A_{v}}{(1 + β_{v} A_{v})} | R_{i}(1 + β_{v} A_{v}) | \frac{R_{o}}{(1 + β_{v} A_{v})} |
| Shunt–series (current amplifier) | Current | Current | A_{if} = \frac{I_{o}}{I_{i}} = \frac{A_{i}}{(1 + β_{i} A_{i})} | \frac{R_{i}}{(1 + β_{i} A_{i})} | R_{o}(1 + β_{i} A_{i}) |
| Series–series (transconductance amplifier) | Voltage | Current | A_{gf} = \frac{I_{o}}{V_{i}} = \frac{A_{g}}{(1 + β_{z} A_{g})} | R_{i}(1 + β_{z} A_{g}) | R_{o}(1 + β_{z} A_{g}) |
| Shunt–shunt (transresistance amplifier) | Current | Voltage | A_{zf} = \frac{V_{o}}{I_{i}} = \frac{A_{z}}{(1 + β_{g} A_{z})} | \frac{R_{i}}{(1 + β_{g} A_{z})} | \frac{R_{o}}{(1 + β_{g} A_{z})} |
R_{i f} \cong (10)(200) k\Omega → 2 M \Omega
and the output resistance to be
R_{of} \cong (100/200) \Omega = 0.5 \Omega
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
\frac{1}{β_{v}} = 1 + \frac{R_{2}}{R_{1}} = 50
and
\frac{R_{2}}{R_{1}} = 49
The feedback network loads the output of the amplifier; consequently, we need R_{1} + R_{2} to be much larger than R_{o}. However, the output resistance of the feedback network is in series with the input terminals, so extremely large values of R_{1} and R_{2} will reduce the actual signal applied to the op-amp because of voltage divider action.
Initially, then, we choose R_{1} = 1 k\Omega and R_{2} = 49 k\Omega.
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 R_{i f} was found to be approximately 580 MΩ and the output resistance R_{of} 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.
