Objective: An op-amp with a T-network is to be designed as a microphone preamplifier. Specifications: The circuit configuration to be designed is shown in Figure 9.12. The maximum microphone output voltage is 12 mV (rms) and the microphone has an output resistance of 1 kΩ. The op-amp circuit is to

Question

Objective: An op-amp with a T-network is to be designed as a microphone preamplifier.

Specifications: The circuit configuration to be designed is shown in Figure 9.12. The maximum microphone output voltage is 12 mV (rms) and the microphone has an output resistance of 1 kΩ. The op-amp circuit is to be designed such that the maximum output voltage is 1.2 V (rms). The input amplifier resistance should be fairly large, but all resistance values should be less that 500 kΩ.
Choices: The final design should use standard resistor values. In addition, standard resistors with tolerances of ±2 percent are to be considered.

Accepted Answer

We need a voltage gain of
[latex]|A_{v}| = \frac{1.2}{0.012} = 100 [/latex]
Equation (9.18) can be written in the form

[latex]A_{v} = \frac{v_{O}}{v_{I}} = − \frac{R_{2}}{R_{1}} \left(1 + \frac{R_{3}}{R_{4}} + \frac{R_{3}}{R_{2}} \right) [/latex] (9.18)
[latex]A_{v} = − \frac{R_{2}}{R_{1}} \left(1 + \frac{R_{3}}{R_{4}} \right) − \frac{R_{3}}{R_{1}} [/latex]
If, for example, we arbitrarily choose [latex]\frac{R_{2}}{R_{1}} = \frac{R_{3}}{R_{1}} = 8 [/latex] , then
[latex]−100 = −8 \left( 1 + \frac{R_{3}}{R_{4}} \right) − 8[/latex]

which yields
[latex]\frac{R_{3}}{R_{4}} = 10.5 [/latex]
The effective [latex]R_{1}[/latex] must include the [latex]R_{S}[/latex] resistance of the microphone. If we set [latex]R_{1} = 49 k\Omega[/latex] so that [latex]R_{1,eff} = 50 k\Omega[/latex], then
[latex]R_{2} = R_{3} = 400 k \Omega[/latex],
and
[latex]R_{4} = 38.1 k \Omega[/latex],
Design Pointer: If we need to use standard resistance values in our design, then, using Appendix C, we can choose [latex]R_{1} = 51 k\Omega[/latex], so that [latex]R_{1,eff} = 52 k\Omega[/latex], and we can choose [latex]R_{2} = R_{3} = 390 k\Omega[/latex]. Then, using Equation (9.18), we have
[latex]A_{v} = −100 = \frac{−R_{2}}{R_{1,eff}} \left(1 + \frac{R_{3}}{R_{4}} \right) − \frac{R_{3}}{R_{1,eff}} = \frac{−390}{52} \left(1 + \frac{390}{R_{4}} \right) − \frac{390}{52}[/latex]
which yields [latex]R_{4} = 34.4 k \Omega[/latex]. We may use a standard resistor of [latex]R_{4} = 33 k \Omega[/latex]. This resistance value then produces a voltage gain of [latex]A_{v} = −103.6 [/latex].
Trade-offs: If we consider ±2 percent tolerances in the standard resistor values, the voltage gain can be written as
[latex]A_{v} = \frac{−R_{2}(1 ± 0.02)}{1 k \Omega + R_{1}(1 ± 0.02)} \left[ 1 + \frac{R_{3}(1 ± 0.02)}{R_{4}(1 ± 0.02)} \right] − \frac{R_{3}(1 ± 0.02)}{1 k \Omega + R_{1}(1 ± 0.02)} [/latex]
or

[latex]A_{v} = \frac{−390(1 ± 0.02)}{1 + 51(1 ± 0.02)} \left[ 1 + \frac{390(1 ± 0.02)}{33(1 ± 0.02)} \right] − \frac{390(1 ± 0.02)}{1 + 51(1 ± 0.02)} [/latex]
Analyzing this equation, we find the maximum magnitude as [latex]|A_{v}|_{max} = 111.6 [/latex] or +7.72 percent, and the minimum magnitude as [latex]|A_{v}|_{min} = 96.3[/latex] or −7.05 percent.
Comment: As required, all resistor values are less than 500 kΩ. Also the resistance ratios in the voltage gain expression are approximately equal. As with most design problems, there is no unique solution. We must keep in mind that, because of resistor value tolerances, the actual gain of the amplifier will have a range of values

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