Objective: Design a driver amplifier to supply current to an LED. Specifications: The available voltage source is variable from 0 to 5 V and has an output resistance of 200 Ω. The required diode current is 10 mA when the maximum input voltage is applied. The required closed-loop transconductance

Question

Objective: Design a driver amplifier to supply current to an LED. Specifications: The available voltage source is variable from 0 to 5 V and has an output resistance of 200 Ω. The required diode current is 10 mA when the maximum input voltage is applied. The required closed-loop transconductance gain is then [latex]A_{g f} = I_{o}/V_{i} = (10 × 10^{−3})/5 → 2 mS[/latex].
Choices: An op-amp with the characteristics described in Example 12.8 and a BJT with [latex]h_{F E} = 100[/latex] are available.

Accepted Answer

(Design Approach): To minimize loading effects on the input, an amplifier with a large input resistance is required; to minimize loading effects on the output, a large output resistance is required. For these reasons, a series–series feedback configuration, or transconductance amplifier, is selected.
The closed-loop gain is
[latex]A_{g f} = 2 × 10^{−3} \cong 1/β_{z}[/latex]
and the resistance feedback transfer function is
[latex]β_{z} = 500 \Omega[/latex]
The dependent open-loop voltage source of the op-amp, as shown in Figure 12.17, can be transformed to an equivalent dependent op-loop transconductance source for the transconductance amplifier, as shown in Figure 12.12. We find that
[latex]A_{g} = A_{v}/R_{o} [/latex]
The parameters specified for the op-amp yield
[latex]A_{g} = 100 A/V[/latex]
The loop gain for the series–series configuration is
[latex]A_{g} β_{z} = (100)(500) = 5 × 10^{4} [/latex]
Referring to Table 12.1, the expected input resistance is

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} = (10)(5 × 10^{4}) k \Omega → 500 M\Omega[/latex]
and the expected output resistance is
[latex]R_{o f} = (100)(5 × 10^{4}) \Omega → 5 M\Omega[/latex]
These input and output resistances should minimize any loading effects at the amplifier input and output.
For this example, we may use the amplifier configuration shown in Figure 12.27, in which the load resistor [latex]R_{L}[/latex] is replaced by an LED. In the ideal case,
[latex]β_{z} = R_{E} = 500 \Omega[/latex]

Computer Simulation Verification: Figure 12.32 shows the circuit used in the computer simulation. Again, a standard µA-741 op-amp was used in the circuit and a standard diode was used in place of an LED. When the input voltage reached 5 V, the current through the diode was 10.0 mA, which was the design value. The input resistance [latex]R_{i f}[/latex] was found to be approximately 2400 MΩ and the output resistance [latex]R_{of}[/latex] was found to be approximately 60 MΩ. Both of these values are larger than predicted because of the differences in the assumed op-amp parameters and those of the µA-741 op-amp.
Comment: Again, an almost ideal feedback circuit can be designed by using an op-amp.

Question 12.12: Objective: Design a driver amplifier to supply current to an...

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