Question 10.10: Objective: Determine the currents and voltages in a simple J...
Objective: Determine the currents and voltages in a simple JFET circuit biased with a constant-current source.
Consider the circuit shown in Figure 10.24. The transistor parameters are: I_{DSS1} = 2 mA, I_{DSS2} = 1 mA, V_{P1} = V_{P2} = −1.5 V, and λ_{1} = λ_{2} = 0.05 V^{−1} Determine the minimum values of V_{S} and V_{I} such that Q_{2} is biased in the saturation region. What is the value of I_{O}?
In order for Q_{2} to remain biased in the saturation region, we must have v_{DS} ≥ |V_{P}| = 1.5 V, from Equation (10.69). The minimum value of V_{S} is then
v_{DS} ≥ v_{DS} (sat) = v_{G S} − V_{P} = |V_{P}| (10.69)
V_{S} (min) − V^{−} = v_{DS} (min) = 1.5 V
or
V_{S} (min) = 1.5 + V^{−} = 1.5 + (−5) = −3.5 V
From Equation (10.70), the output current is
i_{D} = I_{DSS} \left( 1 − \frac{v_{G S}}{V_{P}} \right)^{2} (1 + λ v_{DS} ) = I_{DSS} (1 + λ v_{DS}) (10.70)
i_{D} = I_{O} = I_{DSS2}(1 + λ_{vDS} ) = (1)[1 + (0.05)(1.5)] = 1.08 mA
As a first approximation in calculating the minimum value of V_{I}, we neglect the effect of λ in transistor Q_{1}. Then, assuming Q_{1} is biased in the saturation region, we
have
i_{D} = I_{DSS1} \left(1 − \frac{v_{G S1}}{V_{P1}} \right)^{2}
or
1.08 = 2 \left(1 − \frac{v_{G S1}}{(−1.5)} \right)^{2}
which yields
v_{G S1} = −0.40 V
We see that
v_{G S1} = −0.40 V = V_{I} − V_{S} = V_{I} − (−3.5)
or
V_{I} = −3.90 V
Comment: Since Q_{1} is an n-channel device, the voltage at the gate is negative with respect to the source