Question 1 |
Using the incremental low frequency small-signal model of the MOS device, the Norton
equivalent resistance of the following circuit is
r_{ds}+R+g_mr_{ds}R | |
\frac{r_{ds} +R}{1+g_m r_{ds}} | |
r_{ds}+\frac{1}{g_m}+R | |
r_{ds}+R |
Question 1 Explanation:
\begin{aligned}v_{\pi }&=-V_{x} \\ V_{x}&=\left ( I_{x}-g_{m}V_{x} \right )r_{ds}+I_{x}R\\ V_{x}\left ( 1+g_{m} r_{ds}\right )&=\left ( r_{ds}+R \right )I_{x} \\ R_{N}&=\frac{V_{x}}{I_{x}}=\frac{R+r_{ds}}{1+g_{m}r_{ds}}\end{aligned}
Question 2 |
An enhancement MOSFET of threshold voltage 3 V is being used in the sample and
hold circuit given below. Assume that the substrate of the MOS device is connected
to -10 V. If the input voltage V_I lies between \pm 10 V, the minimum and the maximum
values of V_G
required for proper sampling and holding respectively, are
3V and -3 V | |
10 V and -10 V | |
13 V and -7 V | |
10 V and -13 V |
Question 2 Explanation:
for holding MOSFET should be OFF.
\begin{aligned}V_{1\, min}&\rightarrow -10\, V \\ V_{G}-V_{1\, min}& \lt 3 \\ V_{G}& \lt 3-10V\Rightarrow -7V \\ \text{For Sampling, } V_{G}-V_{i\, max}& \gt 3\\ V_{G}& \gt 3+V_{i\, max}\\ V_{G}&\gt 13\end{aligned}
Question 3 |
In the circuit shown, V_1=0 \; and \; V_2=V_{dd}. The other relevant parameters are mentioned in the figure. Ignoring the effect of channel length modulation and the body effect, the value of I_{out} is _____mA(rounded off to 1 decimal place).
4.2 | |
6 | |
8.3 | |
2.4 |
Question 3 Explanation:
\begin{aligned} \frac{I_{2}}{I_{1}} &=\frac{(W / L)_{2}}{(W / L)_{1}}=\frac{3}{2} \\ I_{2} &=\frac{3}{2} \times I_{1}=1.5 \mathrm{mA} \end{aligned}
M_{3} is OFF because V_{1}=0 \Rightarrow I_{3}=0
M_{4} is ON because V_{2}=V_{D D}
\begin{aligned} I_{5} &=I_{4}=I_{2}=1.5 \mathrm{mA} \\ \frac{I_{6}}{I_{5}} &=\frac{(W / L)_{6}}{(W / L)_{5}}=\frac{40}{10}=4 \\ I_{6} &=4 I_{5}=4 \times 1.5=6 \mathrm{mA} \\ \therefore \quad I_{\text {out }} &=I_{6}=6 \mathrm{mA} \end{aligned}
Question 4 |
In the circuit shown, the threshold voltages of the pMOS (|V_{tp}|) and nMOS (V_{tn}) transistors are both equal to 1 V. All the transistors have the same output resistance r_{ds} of 6 M\Omega. The other parameters are listed below:
\mu_n C_{ox}=60\mu A/V^2; \left ( \frac{W}{L} \right )_{nMOS}=5
\mu_p C_{ox}=30\mu A/V^2; \left ( \frac{W}{L} \right )_{pMOS}=10
\mu_n \; and \; \mu_p are the carrier mobilities, and C_{ox} is the oxide capacitance per unit area. Ignoring the effect of channel length modulation and body bias, the gain of the circuit is____ (rounded off to 1 decimal place).
\mu_n C_{ox}=60\mu A/V^2; \left ( \frac{W}{L} \right )_{nMOS}=5
\mu_p C_{ox}=30\mu A/V^2; \left ( \frac{W}{L} \right )_{pMOS}=10
\mu_n \; and \; \mu_p are the carrier mobilities, and C_{ox} is the oxide capacitance per unit area. Ignoring the effect of channel length modulation and body bias, the gain of the circuit is____ (rounded off to 1 decimal place).
-800 | |
-750.6 | |
-900 | |
-652.3 |
Question 4 Explanation:
M_{3} and M_{4} are identical PMOS transistor and they have equal current.
Hence their V_{SG} should be equal.
\begin{aligned} V_{S G 3} &=V_{S G 4}=\frac{V_{D D}}{2}=2 V \\ I_{S D} &=\frac{\mu_{p} C_{o x}}{2}\left(\frac{W}{L}\right)_{p}\left(V_{S G}-\left|V_{T}\right|\right)^{2} \\ &=\frac{30}{2} \times 10(2-1)^{2}=150 \mu \mathrm{A} \end{aligned}
now, by using current mirror property all transistor should have equal current.
\begin{aligned} I_{\mathrm{DSN}}&=I_{\mathrm{SDP}}=150 \mu \mathrm{A} \\ \text { For } M_{1} \quad g_{m 1}&=\sqrt{2 \mu_{n} C_{o x} \frac{W}{L} \times I_{D S}} \\ &=\sqrt{2 \mu_{n} C_{0 x}} \frac{W}{L} \times I_{D S}\\ & =\sqrt{2 \times 60 \times 5 \times 150}=300 \mathrm{m} \end{aligned}
M_{2}, M_{3} and M_{4} from active load for M_{1}. This active load in equivalent to resistance r_{d s 2} i.e. 6 \mathrm{M} \Omega
M_{1} is common source amplifier.
\begin{aligned} \therefore \quad \frac{V_{\text {out }}}{V_{\text {in }}} &=A_{v}=-g_{m 1} \times\left(r_{\text {ds2 }} \| r_{\text {ds1 }}\right) \\ &=-300 \mathrm{M} \mathrm{C} \times 3 \mathrm{M} \Omega=-900 \end{aligned}
Hence their V_{SG} should be equal.
\begin{aligned} V_{S G 3} &=V_{S G 4}=\frac{V_{D D}}{2}=2 V \\ I_{S D} &=\frac{\mu_{p} C_{o x}}{2}\left(\frac{W}{L}\right)_{p}\left(V_{S G}-\left|V_{T}\right|\right)^{2} \\ &=\frac{30}{2} \times 10(2-1)^{2}=150 \mu \mathrm{A} \end{aligned}
now, by using current mirror property all transistor should have equal current.
\begin{aligned} I_{\mathrm{DSN}}&=I_{\mathrm{SDP}}=150 \mu \mathrm{A} \\ \text { For } M_{1} \quad g_{m 1}&=\sqrt{2 \mu_{n} C_{o x} \frac{W}{L} \times I_{D S}} \\ &=\sqrt{2 \mu_{n} C_{0 x}} \frac{W}{L} \times I_{D S}\\ & =\sqrt{2 \times 60 \times 5 \times 150}=300 \mathrm{m} \end{aligned}
M_{2}, M_{3} and M_{4} from active load for M_{1}. This active load in equivalent to resistance r_{d s 2} i.e. 6 \mathrm{M} \Omega
M_{1} is common source amplifier.
\begin{aligned} \therefore \quad \frac{V_{\text {out }}}{V_{\text {in }}} &=A_{v}=-g_{m 1} \times\left(r_{\text {ds2 }} \| r_{\text {ds1 }}\right) \\ &=-300 \mathrm{M} \mathrm{C} \times 3 \mathrm{M} \Omega=-900 \end{aligned}
Question 5 |
A CMOS inverter, designed to have a mid-point voltage V_I equal to half of V_{dd}, as shown in the figure, has the following parameters:
V_{dd}=3V
\mu_n C_{ox}=100\mu A/V^2; V_{tn}=0.7 V for nMOS
\mu_p C_{ox}=40\mu A/V^2; |V_{tp}|=0.9 V for pMOS
The ratio of \left ( \frac{W}{L} \right )_n to \left ( \frac{W}{L} \right )_p is equal to ____(rounded off to 3 decimal places).
V_{dd}=3V
\mu_n C_{ox}=100\mu A/V^2; V_{tn}=0.7 V for nMOS
\mu_p C_{ox}=40\mu A/V^2; |V_{tp}|=0.9 V for pMOS
The ratio of \left ( \frac{W}{L} \right )_n to \left ( \frac{W}{L} \right )_p is equal to ____(rounded off to 3 decimal places).
0.225 | |
0.124 | |
0.324 | |
0.624 |
Question 5 Explanation:
I_{D n}=I_{D p} and both will be in saturation
\begin{aligned} \text{If }\quad V_{\mathrm{IN}}&=\frac{V_{D D}}{2}=1.5 \mathrm{V}=V_{\mathrm{GSN}}=V_{S G P} \\ \Rightarrow \frac{1}{2}&\left(\mu_{n} C_{O x}\right)\left(\frac{W}{L}\right)_{n}\left[V_{\mathrm{GSN}}-V_{T N}\right]^{2} \\ =& \frac{1}{2}\left(\mu_{p} C_{Q x}\right)\left(\frac{W}{L}\right)_{p}\left[V_{G S P}+V_{T P}\right]^{2} \\ 100 \times 10^{-6}\left(\frac{W}{L}\right)_{n} &[1.5-0.7]^{2} \\ &=40 \times 10^{-6}\left(\frac{W}{L}\right)_{p}[1.5-0.9]^{2} \\ \Rightarrow \quad \frac{\left(\frac{W}{L}\right)_{n}}{\left(\frac{W}{L}\right)_{p}}&=\frac{40}{100} \times \frac{(0.6)^{2}}{(0.8)^{2}}=0.225 \end{aligned}
\begin{aligned} \text{If }\quad V_{\mathrm{IN}}&=\frac{V_{D D}}{2}=1.5 \mathrm{V}=V_{\mathrm{GSN}}=V_{S G P} \\ \Rightarrow \frac{1}{2}&\left(\mu_{n} C_{O x}\right)\left(\frac{W}{L}\right)_{n}\left[V_{\mathrm{GSN}}-V_{T N}\right]^{2} \\ =& \frac{1}{2}\left(\mu_{p} C_{Q x}\right)\left(\frac{W}{L}\right)_{p}\left[V_{G S P}+V_{T P}\right]^{2} \\ 100 \times 10^{-6}\left(\frac{W}{L}\right)_{n} &[1.5-0.7]^{2} \\ &=40 \times 10^{-6}\left(\frac{W}{L}\right)_{p}[1.5-0.9]^{2} \\ \Rightarrow \quad \frac{\left(\frac{W}{L}\right)_{n}}{\left(\frac{W}{L}\right)_{p}}&=\frac{40}{100} \times \frac{(0.6)^{2}}{(0.8)^{2}}=0.225 \end{aligned}
Question 6 |
In the circuit shown below, the (W/L) value for M_{2} is twice that for M_{1} . The two nMOS
transistors are otherwise identical. The threshold voltage V_{T} for both transistors is 1.0V.
Note that V_{GS} \; for M_{2} must be \gt 1.0 V.
The voltage (in volts, accurate to two decimal places) at V_{x} is _______.
The voltage (in volts, accurate to two decimal places) at V_{x} is _______.0.33 | |
0.42 | |
0.55 | |
0.75 |
Question 6 Explanation:
\begin{aligned} \text{Let},\quad K_{n}&=\frac{\mu_{n} C_{o x}}{2}\left(\frac{W}{L}\right)\\ \text{Given that,}\\ \left(\frac{W}{L}\right)_{2} &=2\left(\frac{W}{L}\right)_{1} \\ \text{So,}\quad K_{n 2} &=2 K_{n 1}\\ \text{For }M_{1}, \quad V_{G S 1}-V_{T}&=2-1=1 V \\ \text{For }M_{2}, \quad V_{G S 2}-V_{T}&=2-V_{x}-1=1 \mathrm{V}-V_{x}<1 \mathrm{V} \\ V_{D S 2}&=\left(3.3-V_{x}\right)>\left(V_{G S 2}-V_{T}\right) \end{aligned}
So, M_{1} will be in linear region and M_{2} will be in saturation region.
\begin{aligned} I_{D_{1}}&=I_{D_{2}} \\ K_{n 1}\left[2\left(V_{G S 1}-V_{T}\right) V_{D S 1}-V_{D S 1}^{2}\right]&=K_{n 2}\left(V_{G S 2}-V_{T}\right)^{2} \\ K_{n 1}\left[2(2-1) V_{x}-V_{x}^{2}\right]&=2 K_{n 1}\left(2-V_{x}-1\right)^{2} \\ 2 V_{x}-V_{x}^{2} &=2\left(1+V_{x}^{2}-2 V_{x}\right) \\ &=2 V_{x}^{2}-4 V_{x}+2\\ 3V_{x}^{2}-6 V_{x}+2 &=0 \\ V_{x}^{2}-2 V_{x}+\frac{2}{3} &=0 \\ V_{x} &=1 \pm \sqrt{\frac{4-\frac{8}{3}}{4}}=1 \pm \sqrt{\frac{1}{3}} V \\ V_{GS2}&=\left(2-V_{x}\right) \geq V_{T}\\ \Rightarrow \quad \left(1-V_{x}\right) &\geq0\\ \text{So, valid value,} \quad V_{x}&=1-\sqrt{\frac{1}{3}}=0.4226 \mathrm{V} \end{aligned}
So, M_{1} will be in linear region and M_{2} will be in saturation region.
\begin{aligned} I_{D_{1}}&=I_{D_{2}} \\ K_{n 1}\left[2\left(V_{G S 1}-V_{T}\right) V_{D S 1}-V_{D S 1}^{2}\right]&=K_{n 2}\left(V_{G S 2}-V_{T}\right)^{2} \\ K_{n 1}\left[2(2-1) V_{x}-V_{x}^{2}\right]&=2 K_{n 1}\left(2-V_{x}-1\right)^{2} \\ 2 V_{x}-V_{x}^{2} &=2\left(1+V_{x}^{2}-2 V_{x}\right) \\ &=2 V_{x}^{2}-4 V_{x}+2\\ 3V_{x}^{2}-6 V_{x}+2 &=0 \\ V_{x}^{2}-2 V_{x}+\frac{2}{3} &=0 \\ V_{x} &=1 \pm \sqrt{\frac{4-\frac{8}{3}}{4}}=1 \pm \sqrt{\frac{1}{3}} V \\ V_{GS2}&=\left(2-V_{x}\right) \geq V_{T}\\ \Rightarrow \quad \left(1-V_{x}\right) &\geq0\\ \text{So, valid value,} \quad V_{x}&=1-\sqrt{\frac{1}{3}}=0.4226 \mathrm{V} \end{aligned}
Question 7 |
Two identical nMOS transistors M_{1} and M_{2} are connected as shown below. The circuit is used as an amplifier with the input connected between G and S terminals and the output taken between D and S terminals. V_{bias} and V_{D} are so adjusted that both transistors are in saturation. The transconductance of this combination is defined as g_{m}=\frac{\partial i_{D}}{\partial V_{GS}} while the output resistance is r_{0}=\frac{\partial V_{GS}}{\partial i_{D}} , where i_{D} is the current flowing into the drain of M_{2}. Let g_{m1} , g_{m2} be the transconductances and r_{01} , r_{02} be the output resistances of transistors M_{1} and M_{2} , respectively.
Which of the following statements about estimates for g_{m} and r_{0} is correct?
Which of the following statements about estimates for g_{m} and r_{0} is correct?
g_{m}\approx g_{m1}\cdot g_{m2}\cdot r_{02} \;and \; r_0 \approx r_{01}+r_{02}. | |
g_{m}\approx g_{m1}\ + g_{m2} \; and \; r_{0} \approx r_{01}+r_{02}. | |
g_{m}\approx g_{m1} \; and \; r_{0}\approx r_{01} \cdot g_{m2}\cdot r_{02}. | |
g_{m}\approx g_{m1} \; and \; r_{0}\approx r_{02}. |
Question 7 Explanation:
g_{m}=\frac{\Delta I_{D}}{\Delta V_{\text {in }}}=\frac{i_{D}}{v_{g s}}=\frac{i_{D 1}}{v_{g s}}=g_{m 1}
To calculate r_{o} :
\begin{aligned} v_{\pi 2} &=-I_{x} r_{01} \\ I_{x} &=g_{m 2} v_{\pi 2}+\frac{\left(V_{x}-I_{x} r_{01}\right)}{r_{02}} \\ I_{x} &=-g_{m 2} r_{01} I_{x}+\frac{V_{x}}{r_{02}}-I_{x} \frac{r_{01}}{r_{02}} \\ V_{x} &=r_{02}\left[1+r_{01} g_{m 2}+\frac{r_{01}}{r_{02}}\right] I_{x} \\ r_{0} &=\frac{V_{x}}{I_{x}}=r_{01}+r_{02}+r_{01} r_{02} g_{m 2} \\ & \approx r_{01} r_{02} g_{m 2} \end{aligned}
Question 8 |
Assuming that transistors M_{1} and M_{2} are identical and have a threshold voltage of 1V, the state of transistors M_{1} and M_{2} are respectively.
Saturation, Saturation | |
Linear, Linear | |
Linear, Saturation | |
Saturation, Linear |
Question 8 Explanation:
\begin{array}{l} V_{G S 1}=2 V \\ V_{G S 2}=2.5-V_{x} \\ V_{D S 1}=V_{x} \\ V_{D S 2}=3-V_{x} \end{array}
Assume both MOSFETs in saturation and equate their currents
\begin{aligned} I_{D S 1} &=I_{D S 2} \\ \frac{k_{n}}{2}(2-1)^{2} &=\frac{k_{n}}{2}\left(2.5-V_{x}-1\right)^{2} \end{aligned}
After solving, V_{x}=0.5 \vee, 2.5 \mathrm{V}
V_{x} cannot be 2.5 \mathrm{V} . Because this will make M_{2} OFF.
Hence V_{x} may be 0.5 \mathrm{V}
\Rightarrow M_{1} is in linear region.
\Rightarrow M_{2} is in saturation region.
To verify further,
\begin{aligned} I_{D S 1} &=I_{D S 2} \\ k_{n}\left[(2-1) V_{n}-\frac{V_{x}^{2}}{2}\right]&=\frac{k_{n}}{2}\left(2.5-V_{x}-1\right)^{2} \end{aligned}
V_{G S 2}=2.5 \mathrm{V}-V_{x} must be greater than V_{T h}=1 V
After solving, V_{x}=0.588 \vee, 1.91 \mathrm{V}
Hence correct value of V_{x}=0.588 \mathrm{V}
This varifies above conclusion i.e.
M_{1} \rightarrow Linear region
M_{2} \rightarrow Saturation region
Question 9 |
An n-channel enhancement mode MOSFET is biased at V_{GS}>V_{TH} \; and \; V_{DS}>(V_{GS}-V_{TH}), where V_{GS} is the gate-to-source voltage, V_{DS} is the drain-to-source voltage and V_{TH} is the threshold voltage. Considering channel length modulation effect to be significant, the MOSFET behaves as a
voltage source with zero output impedance | |
voltage source with non-zero output impedance | |
current source with finite output impedance | |
current source with infinite output impedance |
Question 9 Explanation:
The small signal equivalent circuit of MOSFET in
saturation is as given below.
So, when the channel length modulation effect is significant, the MOSFET can be modelled as a current source with finite output impedance.
So, when the channel length modulation effect is significant, the MOSFET can be modelled as a current source with finite output impedance.
Question 10 |
For the circuit shown, assume that the NMOS transistor is in saturation. Its threshold voltage V _{tn}=1V and its transconductance parameter \mu_{n}C_{OX}(\frac{W}{L})=1mA/V^{2} . Neglect channel length
modulation and body bias effects. Under these conditions, the drain current I_{D} in mA is ___________.
4 | |
1 | |
2 | |
3 |
Question 10 Explanation:
\begin{aligned} V_{G S}&=\frac{8 \times 5}{8}-1 \times I_{D}=5-I_{D} \\ &(\text{Here }I_{D}\text{ is numerically in mA }) \\ I_{D} &=5-V_{G S} \quad \ldots(i)\\ I_{D} &=\frac{\mu_{n} C_{o x}}{2} \frac{W}{L}\left(V_{G S}-V_{T}\right)^{2} \\ 5-V_{G S} &=\frac{1}{2}\left(V_{G S}-1\right)^{2} \\ 10-2 V_{G S} &=V_{G S}^{2}+1-2 V_{G S} \\ V_{G S}^{2} &=9 \Rightarrow V_{G S}=3 V \\ I_{D} &=5-3=2 \mathrm{mA} \end{aligned}
There are 10 questions to complete.