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=100V_{B E} (on) = 0.6  V, β = 100, and VA=V_{A} = ∞. The designed output IOI_{O} is to be 200 µA. The bias voltages are to be V+=5 VV^{+} = 5  V and V=0V^{−} = 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.

10.2
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The reference current can be written as
IREF=IO(1+2β)=(200)(1+2100)=204 µAI_{REF} = I_{O} \left(1 + \frac{2}{β} \right) = (200) \left(1 + \frac{2}{100} \right) = 204  µA
From Equation (10.1), the resistor R1R_{1} is found to be

IREF=V+ − VBE − VR1I_{REF} = \frac{V^{+}  −  V_{B E}  −  V^{−}}{R_{1}}      (10.1)
R1=V+ − VBE(on)IREF=5 − 0.60.204=21.6 kΩR_{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 VBE(on)=0.7 VV_{B E} (on) = 0.7  V is assumed, the value of IREFI_{REF} or R1R_{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 IOIREFI_{O} \cong I_{REF}.

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