Question 10.1: Objective: Design a two-transistor current source to meet a ...
Objective: Design a two-transistor current source to meet a set of specifications.
Specifications: The circuit to be designed has the configuration shown in Figure 10.2(b).Assume that matched transistors are available with parameters V_{B E} (on) = 0.6 V, β = 100, and V_{A} = ∞. The designed output I_{O} is to be 200 µA. The bias voltages are to be V^{+} = 5 V and V^{−} = 0 .
Choices: The circuit will be fabricated as an integrated circuit so that a standard resistor value is not required and matched transistors can be fabricated.
The reference current can be written as
I_{REF} = I_{O} \left(1 + \frac{2}{β} \right) = (200) \left(1 + \frac{2}{100} \right) = 204 µA
From Equation (10.1), the resistor R_{1} is found to be
I_{REF} = \frac{V^{+} − V_{B E} − V^{−}}{R_{1}} (10.1)
R_{1} = \frac{V^{+} − V_{B E} (on)}{I_{REF}} = \frac{5 − 0.6}{0.204} = 21.6 k \Omega
Trade-offs: The design assumes that matched transistors exist. The effect of mismatched transistors will be discussed later in this section.
Comment: In this example, we assumed a B–E voltage of 0.6 V. This approximation is satisfactory for most cases. The B–E voltage is involved in the reference current or resistor calculation. If a value of V_{B E} (on) = 0.7 V is assumed, the value of I_{REF} or R_{1} will change, typically, by only 1 to 2 percent.
Design Pointer: We see in this example that, for β = 100, the reference and load currents are within 2 percent of each other in this two-transistor current source. In most circuit applications, we can use the approximation that I_{O} \cong I_{REF}.