Question 12.12: Objective: Design a driver amplifier to supply current to an...
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 A_{g f} = I_{o}/V_{i} = (10 × 10^{−3})/5 → 2 mS.
Choices: An op-amp with the characteristics described in Example 12.8 and a BJT with h_{F E} = 100 are available.
(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
A_{g f} = 2 × 10^{−3} \cong 1/β_{z}
and the resistance feedback transfer function is
β_{z} = 500 \Omega
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
A_{g} = A_{v}/R_{o}
The parameters specified for the op-amp yield
A_{g} = 100 A/V
The loop gain for the series–series configuration is
A_{g} β_{z} = (100)(500) = 5 × 10^{4}
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 | 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} = (10)(5 × 10^{4}) k \Omega → 500 M\Omega
and the expected output resistance is
R_{o f} = (100)(5 × 10^{4}) \Omega → 5 M\Omega
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 R_{L} is replaced by an LED. In the ideal case,
β_{z} = R_{E} = 500 \Omega
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 R_{i f} was found to be approximately 2400 MΩ and the output resistance R_{of} 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.
