Question 6.15: Using the circuit of Figure 6.50 assume the following values...
Using the circuit of Figure 6.50 assume the following values: V_{DD} = 10 V, R_1 = 40 kΩ, R_2 = 100 kΩ, R_S = 200 Ω, and MOSFET parameters: V_{TH} = 0.5 V, μ_nC_{ox} = 100 µA/V² , W/L = 50, and λ = 0. Calculate the maximum allowable value of R_D for the MOSFET to remain on the edge of saturation. Assume V_{GS} = 3 V.
From inspection of Figure 6.50, we can state that
V_{GG}=V_{GS}+I_DR_{DS} (6.128)
In Equation (6.128) V_{GG} is the gate voltage to ground. Since the gate current is negligible
V_{GG}=\frac{R_2}{R_1+R_2}V_{DD} =7.14 V (6.129)
And since
I_D=\frac{1}{2}\mu _nC_{ox}\frac{W}{L}(V_{GS}-V_{TH})^2 (6.130)
using the values provided by the example, Equation (6.130) yields
I_D=15.63 mA
The condition for the MOSFET to be on the edge of saturation is
V_{DS}=V_{GS}-V_{TH}
and since V_{GS} = 3 V and V_{TH} = 0.5 V, it yields
V_{DS}=2.5 V
By inspection of Figure 6.50, we see that V_{DD} = R_DI_D + V_{DS} + R_SI_D. Using the given values in the above equation, 10 = R_D 0.01563 + 2.5 + 200 0.01563. From the above equation we find the value of R_D to be
R_D ≅ 280 Ω.