Figure 541
the MOSFET in the Circuit of Fig$V_{t}=$ $1 \mathrm{~V}$

Question

Figure 541 
the MOSFET in the Circuit of Fig$V_{t}=$ $1 \mathrm{~V}$

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The Answer is $(b) Refer to Figure 5.4 .3 below. We assume that with R_{L} reduced to 10 \mathrm{k} \Omega , the MOSFET remains in saturation. Thus, I_{D} remains unchanged at 0.25 \mathrm{mA} . Figure 5.4.3 shows the circuit before and after Thévenin theorem is applied to simplify the drain circuit. Now, V_{D}=2.5-0.25 imes 5=1.25 \mathrm{~V} Thus, V_{D}>V_{G} , and saturation-mode operation is confirmed. The drain voltage becomes V_{D}=1.25 \mathrm{~V} (c) Figure 5.4.4 If R_{L} is disconnected, the circuit becomes as in Fig. 5.4.4. Assuming that the transistor remains in saturation, then I_{D}=0.25 \mathrm{~mA} and V_{D}=5-0.25 imes 10=2.5 \mathrm{~V} which is greater than V_{G} , thus confirming saturation-mode operation. The drain voltage is now V_{D}=+2.5 \mathrm{~V} (d) Figure 5.4.5 With R_{L} disconnected and the MOSFET operating at the edge of saturation, V_{D G}=-V_{t} , thus V_{D}=-1 \mathrm{~V} and the current remains unchanged, I_{D}=0.25 \mathrm{~mA} From the circuit in Fig. 5.4.5, we can write \begin{aligned} R_{D} & =\frac{5-V_{D}}{I_{D}}=\frac{5-(-1)}{0.25} \ & =24 \mathrm{k} \Omega \end{aligned}$
$(b) Refer to Figure 5.4 .3 below. We assume that with R_{L} reduced to 10 \mathrm{k} \Omega , the MOSFET remains in saturation. Thus, I_{D} remains unchanged at 0.25 \mathrm{mA} . Figure 5.4.3 shows the circuit before and after Thévenin theorem is applied to simplify the drain circuit. Now, V_{D}=2.5-0.25 imes 5=1.25 \mathrm{~V} Thus, V_{D}>V_{G} , and saturation-mode operation is confirmed. The drain voltage becomes V_{D}=1.25 \mathrm{~V} (c) Figure 5.4.4 If R_{L} is disconnected, the circuit becomes as in Fig. 5.4.4. Assuming that the transistor remains in saturation, then I_{D}=0.25 \mathrm{~mA} and V_{D}=5-0.25 imes 10=2.5 \mathrm{~V} which is greater than V_{G} , thus confirming saturation-mode operation. The drain voltage is now V_{D}=+2.5 \mathrm{~V} (d) Figure 5.4.5 With R_{L} disconnected and the MOSFET operating at the edge of saturation, V_{D G}=-V_{t} , thus V_{D}=-1 \mathrm{~V} and the current remains unchanged, I_{D}=0.25 \mathrm{~mA} From the circuit in Fig. 5.4.5, we can write \begin{aligned} R_{D} & =\frac{5-V_{D}}{I_{D}}=\frac{5-(-1)}{0.25} \ & =24 \mathrm{k} \Omega \end{aligned}$
(b) Refer to Figure 5.4 .3 below.
$(b) Refer to Figure 5.4 .3 below. We assume that with R_{L} reduced to 10 \mathrm{k} \Omega , the MOSFET remains in saturation. Thus, I_{D} remains unchanged at 0.25 \mathrm{mA} . Figure 5.4.3 shows the circuit before and after Thévenin theorem is applied to simplify the drain circuit. Now, V_{D}=2.5-0.25 imes 5=1.25 \mathrm{~V} Thus, V_{D}>V_{G} , and saturation-mode operation is confirmed. The drain voltage becomes V_{D}=1.25 \mathrm{~V} (c) Figure 5.4.4 If R_{L} is disconnected, the circuit becomes as in Fig. 5.4.4. Assuming that the transistor remains in saturation, then I_{D}=0.25 \mathrm{~mA} and V_{D}=5-0.25 imes 10=2.5 \mathrm{~V} which is greater than V_{G} , thus confirming saturation-mode operation. The drain voltage is now V_{D}=+2.5 \mathrm{~V} (d) Figure 5.4.5 With R_{L} disconnected and the MOSFET operating at the edge of saturation, V_{D G}=-V_{t} , thus V_{D}=-1 \mathrm{~V} and the current remains unchanged, I_{D}=0.25 \mathrm{~mA} From the circuit in Fig. 5.4.5, we can write \begin{aligned} R_{D} & =\frac{5-V_{D}}{I_{D}}=\frac{5-(-1)}{0.25} \ & =24 \mathrm{k} \Omega \end{aligned}$
We assume that with

R_{L}

reduced to

10 \mathrm{k} \Omega

, the MOSFET remains in saturation. Thus,

I_{D}

remains unchanged at 0.25

\mathrm{mA}

. Figure 5.4.3 shows the circuit before and after Thévenin theorem is applied to simplify the drain circuit. Now,

V_{D}=2.5-0.25 imes 5=1.25 \mathrm{~V}

Thus,

V_{D}>V_{G}

, and saturation-mode operation is confirmed. The drain voltage becomes

V_{D}=1.25 \mathrm{~V}

(c)
$(b) Refer to Figure 5.4 .3 below. We assume that with R_{L} reduced to 10 \mathrm{k} \Omega , the MOSFET remains in saturation. Thus, I_{D} remains unchanged at 0.25 \mathrm{mA} . Figure 5.4.3 shows the circuit before and after Thévenin theorem is applied to simplify the drain circuit. Now, V_{D}=2.5-0.25 imes 5=1.25 \mathrm{~V} Thus, V_{D}>V_{G} , and saturation-mode operation is confirmed. The drain voltage becomes V_{D}=1.25 \mathrm{~V} (c) Figure 5.4.4 If R_{L} is disconnected, the circuit becomes as in Fig. 5.4.4. Assuming that the transistor remains in saturation, then I_{D}=0.25 \mathrm{~mA} and V_{D}=5-0.25 imes 10=2.5 \mathrm{~V} which is greater than V_{G} , thus confirming saturation-mode operation. The drain voltage is now V_{D}=+2.5 \mathrm{~V} (d) Figure 5.4.5 With R_{L} disconnected and the MOSFET operating at the edge of saturation, V_{D G}=-V_{t} , thus V_{D}=-1 \mathrm{~V} and the current remains unchanged, I_{D}=0.25 \mathrm{~mA} From the circuit in Fig. 5.4.5, we can write \begin{aligned} R_{D} & =\frac{5-V_{D}}{I_{D}}=\frac{5-(-1)}{0.25} \ & =24 \mathrm{k} \Omega \end{aligned}$

Figure 5.4.4
If

R_{L}

is disconnected, the circuit becomes as in Fig. 5.4.4. Assuming that the transistor remains in saturation, then

I_{D}=0.25 \mathrm{~mA}

and

V_{D}=5-0.25 imes 10=2.5 \mathrm{~V}

which is greater than

V_{G}

, thus confirming saturation-mode operation. The drain voltage is now

V_{D}=+2.5 \mathrm{~V}

(d)
$(b) Refer to Figure 5.4 .3 below. We assume that with R_{L} reduced to 10 \mathrm{k} \Omega , the MOSFET remains in saturation. Thus, I_{D} remains unchanged at 0.25 \mathrm{mA} . Figure 5.4.3 shows the circuit before and after Thévenin theorem is applied to simplify the drain circuit. Now, V_{D}=2.5-0.25 imes 5=1.25 \mathrm{~V} Thus, V_{D}>V_{G} , and saturation-mode operation is confirmed. The drain voltage becomes V_{D}=1.25 \mathrm{~V} (c) Figure 5.4.4 If R_{L} is disconnected, the circuit becomes as in Fig. 5.4.4. Assuming that the transistor remains in saturation, then I_{D}=0.25 \mathrm{~mA} and V_{D}=5-0.25 imes 10=2.5 \mathrm{~V} which is greater than V_{G} , thus confirming saturation-mode operation. The drain voltage is now V_{D}=+2.5 \mathrm{~V} (d) Figure 5.4.5 With R_{L} disconnected and the MOSFET operating at the edge of saturation, V_{D G}=-V_{t} , thus V_{D}=-1 \mathrm{~V} and the current remains unchanged, I_{D}=0.25 \mathrm{~mA} From the circuit in Fig. 5.4.5, we can write \begin{aligned} R_{D} & =\frac{5-V_{D}}{I_{D}}=\frac{5-(-1)}{0.25} \ & =24 \mathrm{k} \Omega \end{aligned}$
Figure 5.4.5
With

R_{L}

disconnected and the MOSFET operating at the edge of saturation,

V_{D G}=-V_{t}

, thus

V_{D}=-1 \mathrm{~V}

and the current remains unchanged,

I_{D}=0.25 \mathrm{~mA}

From the circuit in Fig. 5.4.5, we can write

\begin{aligned}R_{D} & =\frac{5-V_{D}}{I_{D}}=\frac{5-(-1)}{0.25} \& =24 \mathrm{k} \Omega\end{aligned}

Figure 541
the MOSFET in the Circuit of Fig $V_{t}=$ $1 \mathrm{~V}$

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Figure 541 the MOSFET in the Circuit of Fig Vt=V_{t}=Vt​= 1 V1 \mathrm{~V}1 V

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Figure 541
the MOSFET in the Circuit of Fig $V_{t}=$ $1 \mathrm{~V}$

Correct Answer:
Verified