Question

the transistor parameters are (v_{tn}=0.8 v) and (mu_{n}c_{ox}=100 mu a/v^{2},w/l = 50), and (lambda=0.02 v^{-1}). the circuit parameters are (v^{+}=5 v,v^{-}=-5 v,i_{q}=0.5 ma), and (r_{d}=6 komega). determine (v_{gsq}) and (v_{dsq}). select one: a. (v_{gsq}=3.48 v) and (v_{dsq}=2.25 v) b. (v_{gsq}=2.56 v) and (v_{dsq}=5.23 v) c. (v_{gsq}=2.25 v) and (v_{dsq}=4.46 v) d. (v_{gsq}=1.247 v) and (v_{dsq}=3.25 v)

Explanation:

Step1: Recall the drain - current formula for MOSFET in saturation

The drain - current formula for a MOSFET in saturation is $I_D=\frac{1}{2}\mu_nC_{ox}\frac{W}{L}(V_{GS}-V_{TN})^2(1 + \lambda V_{DS})$. In the DC - analysis, $I_D = I_Q=0.5\ mA$. We first assume the MOSFET is in saturation.
Given $\mu_nC_{ox}=100\ \mu A/V^2$, $\frac{W}{L}=50$, $V_{TN}=0.8\ V$, and $I_Q = 0.5\ mA = 500\ \mu A$.
Substitute into the formula $I_Q=\frac{1}{2}\mu_nC_{ox}\frac{W}{L}(V_{GSQ}-V_{TN})^2$.
$500=\frac{1}{2}\times100\times50\times(V_{GSQ}-0.8)^2$.

Step2: Solve for $V_{GSQ}$

First, simplify the right - hand side of the equation: $\frac{1}{2}\times100\times50\times(V_{GSQ}-0.8)^2 = 2500(V_{GSQ}-0.8)^2$.
Then, we have the equation $2500(V_{GSQ}-0.8)^2=500$.
$(V_{GSQ}-0.8)^2=\frac{500}{2500}=0.2$.
$V_{GSQ}-0.8=\pm\sqrt{0.2}$. Since $V_{GSQ}>V_{TN}$ for the MOSFET to be in conduction, we take the positive root.
$V_{GSQ}=0.8+\sqrt{0.2}\approx0.8 + 0.447=1.247\ V$.

Step3: Use the voltage - current relationship in the circuit to find $V_{DSQ}$

In the DC - equivalent circuit, applying KVL in the drain - source loop: $V^+-I_QR_D - V_{DSQ}-V^- = 0$.
Given $V^+=5\ V$, $V^-=-5\ V$, $I_Q = 0.5\ mA$, and $R_D = 6\ k\Omega$.
$V_{DSQ}=V^+-I_QR_D-V^-$.
$V_{DSQ}=5-0.5\times10^{- 3}\times6\times10^{3}+5$.
$V_{DSQ}=5 - 3+5=3.25\ V$.

Answer:

D. $V_{GSQ}=1.247\ V$ and $V_{DSQ}=3.25\ V$