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 VBE(on)=0.6 V,β=100, and VA=∞. The designed output IO 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.

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The reference current can be written as
IREF=IO(1+β2)=(200)(1+1002)=204 µA
From Equation (10.1), the resistor R1 is found to be
IREF=R1V+ − VBE − V− (10.1)
R1=IREFV+ − VBE(on)=0.2045 − 0.6=21.6 kΩ
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 VBE(on)=0.7 V is assumed, the value of IREF or R1 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 IO≅IREF.