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 $I_{O}$ ?

Question

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: [latex]I_{DSS1} = 2 mA, I_{DSS2} = 1 mA, V_{P1} = V_{P2} = −1.5 V[/latex], and [latex]λ_{1} = λ_{2} = 0.05 V^{−1}[/latex] Determine the minimum values of [latex]V_{S}[/latex] and [latex]V_{I}[/latex] such that [latex]Q_{2}[/latex] is biased in the saturation region. What is the value of [latex]I_{O}[/latex]?

Accepted Answer

In order for [latex]Q_{2}[/latex] to remain biased in the saturation region, we must have [latex]v_{DS} ≥ |V_{P}| = 1.5 V[/latex], from Equation (10.69). The minimum value of [latex]V_{S}[/latex] is then

[latex]v_{DS} ≥ v_{DS} (sat) = v_{G S} − V_{P} = |V_{P}| [/latex] (10.69)
[latex]V_{S} (min) − V^{−} = v_{DS} (min) = 1.5 V[/latex]
or
[latex]V_{S} (min) = 1.5 + V^{−} = 1.5 + (−5) = −3.5 V[/latex]
From Equation (10.70), the output current is

[latex]i_{D} = I_{DSS} \left( 1 − \frac{v_{G S}}{V_{P}} \right)^{2} (1 + λ v_{DS} ) = I_{DSS} (1 + λ v_{DS}) [/latex] (10.70)
[latex]i_{D} = I_{O} = I_{DSS2}(1 + λ_{vDS} ) = (1)[1 + (0.05)(1.5)] = 1.08 mA[/latex]

As a first approximation in calculating the minimum value of [latex]V_{I}[/latex], we neglect the effect of λ in transistor [latex]Q_{1}[/latex]. Then, assuming [latex]Q_{1}[/latex] is biased in the saturation region, we
have
[latex]i_{D} = I_{DSS1} \left(1 − \frac{v_{G S1}}{V_{P1}} \right)^{2} [/latex]
or
[latex]1.08 = 2 \left(1 − \frac{v_{G S1}}{(−1.5)} \right)^{2} [/latex]
which yields
[latex]v_{G S1} = −0.40 V[/latex]
We see that
[latex]v_{G S1} = −0.40 V = V_{I} − V_{S} = V_{I} − (−3.5)[/latex]
or
[latex]V_{I} = −3.90 V[/latex]
Comment: Since [latex]Q_{1}[/latex] is an n-channel device, the voltage at the gate is negative with respect to the source

Question 10.10: Objective: Determine the currents and voltages in a simple J...

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