Question 1 |
Select the correct statement(s) regarding CMOS implementation of NOT gates.
Noise Margin High (NM_H) is always equal to the Noise Margin Low (NM_L) irrespective of the sizing of transistors. | |
Dynamic power consumption during switching is zero. | |
For a logical high input under steady state, the nMOSFET is in the linear regime of
operation. | |
Mobility of electrons never influences the switching speed of the NOT gate. |
Question 1 Explanation:
(A) NM_H will not be always equal to NM_L
because it depends on transistors parameters
like size
NM_H=V_{IL}-V_{OL}
NM_L=V_{OH}-V_{IH}
Condition for NM_H=NM_L : when V_{TN}=|V_{TP}|\; and \; V_{IP}=\frac{V_{DD}}{2}
If \frac{K_p}{K_n}< > 1 then NM_H\neq NM_L
(B) Due to capacitive leading of stage, dynamic power consumption during switching will not be zero.
(C) For V_{DD}-|V_{TP}|\leq V_{in}\leq V_{DD} [logic high input]
PMOS \rightarrow cut off
NMOS \rightarrow Linear
(D) Mobility of electrons influences the switching speed because
Propagation delay,
\tau _p=\frac{\tau _{PLH}+P_{PHL}}{2}
\tau _{PLH}=\frac{C_LV_{DD}}{\mu _p C_{ox} \frac{W}{L} (V_{GS}\;\; V_{TP})^2}
\mu _p dependent on mobality
Therefore (C) is only correct.
NM_H=V_{IL}-V_{OL}
NM_L=V_{OH}-V_{IH}
Condition for NM_H=NM_L : when V_{TN}=|V_{TP}|\; and \; V_{IP}=\frac{V_{DD}}{2}
If \frac{K_p}{K_n}< > 1 then NM_H\neq NM_L
(B) Due to capacitive leading of stage, dynamic power consumption during switching will not be zero.
(C) For V_{DD}-|V_{TP}|\leq V_{in}\leq V_{DD} [logic high input]
PMOS \rightarrow cut off
NMOS \rightarrow Linear
(D) Mobility of electrons influences the switching speed because
Propagation delay,
\tau _p=\frac{\tau _{PLH}+P_{PHL}}{2}
\tau _{PLH}=\frac{C_LV_{DD}}{\mu _p C_{ox} \frac{W}{L} (V_{GS}\;\; V_{TP})^2}
\mu _p dependent on mobality
Therefore (C) is only correct.
Question 2 |
In the circuits shown, the threshold voltage of each Nmos transistor is 0.6 V. Ignoring the effect of channel length modulation and body bias, the values of Vout1 and Vout2, respectively, in volts, are
1.8 and 1.2 | |
2.4 and 2.4 | |
1.8 and 2.4 | |
2.4 and 1.2 |
Question 2 Explanation:
V_{\text {out } 1}=3-0.6-0.6=1.8 \mathrm{V}
Question 3 |
In the circuit shown, A and B are the inputs and Fis the output. What is the functionality of the circuit?
Latch | |
XNOR | |
SRAM Cell | |
XOR |
Question 3 Explanation:
So, the given logic circuit acts as an XNOR gate
Question 4 |
In the circuit shown, what are the values of F for EN=0 and EN=1, respectively?
0 and D | |
Hi-Z and D | |
0 and 1 | |
Hi-Z and \bar{D} |
Question 4 Explanation:
When E N=0
x_{1}=(\overline{D \cdot 0})=1 \Rightarrow PMOS is in OFF state
x_{2}=(\overline{1+D})=0 \Rightarrow NMOS is in OFF state
Both the transistors are in OFF state, which offers high impedance.
\begin{aligned} \text { When } E N=1: x_{1}&=(\overline{D \cdot 1})=\bar{D} \\ x_{2}&=(\overline{0+D})=\bar{D} \\ F &=D \end{aligned}
Question 5 |
A standard CMOS inverter is designed with equal rise and fall times (\beta _n=\beta _p). If the width of the pMOS transistor in the inverter is increased, what would be the effect on the LOW noise margin (NM_L) and the HIGH noise margin NM_H?
NM_L increases and NM_Hdecreases | |
NM_L decreases and NM_H increases. | |
Both NM_L and NM_H increase. | |
No change in the noise margins. |
Question 5 Explanation:
Making PMOS wider, shifts input transition point (V_{IT}) towards V_{DD}
Making NMOS wider, shifts input transition point (V_{IT}) towards zero.
So, as PMOS made wider, NML increases and NMH decreases.
Question 6 |
The logic gates shown in the digital circuit below use strong pull-down nMOS transistors for
LOW logic level at the outputs. When the pull-downs are off, high-value resistors set the
output logic levels to HIGH (i.e. the pull-ups are weak). Note that some nodes are
intentionally shorted to implement "wired logic". Such shorted nodes will be HIGH only if
the outputs of all the gates whose outputs are shorted are HIGH.
The number of distinct values of X_{3}X_{2}X_{1}X_{0} (out of the 16 possible values) that give Y = 1 is _______.
The number of distinct values of X_{3}X_{2}X_{1}X_{0} (out of the 16 possible values) that give Y = 1 is _______.
5 | |
6 | |
8 | |
7 |
Question 6 Explanation:
\begin{aligned} A&=\left(X_{1} \oplus X_{2}\right) \bar{X}_{3} \\ B&=\left[\left(X_{1} \oplus X_{2}\right) \bar{X}_{3} X_{0}\right] \cdot \bar{X}_{0}=0\\ Y&=B+X_{3}=0+X_{3}=X_{3} \end{aligned}
Out of 16 possible combinations of x_{3} x_{2} x_{1} x_{0} x_{3} , will be high for 8 combinations. So, Y will be high for 8 combinations.
Question 7 |
The logic function f(X,Y) realized by the given circuit is
NOR | |
AND | |
NAND | |
XOR |
Question 7 Explanation:
From pull-down network,
\begin{aligned} \overline{f(X, Y)}&=\bar{X} \bar{Y}+X Y=X \odot Y \\ f(X, Y)&=\overline{X \odot Y}=X \oplus Y \end{aligned}
\begin{aligned} \overline{f(X, Y)}&=\bar{X} \bar{Y}+X Y=X \odot Y \\ f(X, Y)&=\overline{X \odot Y}=X \oplus Y \end{aligned}
Question 8 |
For the circuit shown in the figure, P and Q are the inputs and Y is the output.
The logic implemented by the circuit is
The logic implemented by the circuit is
XNOR | |
XOR | |
NOR | |
OR |
Question 8 Explanation:
As per GATE Answer key Marks to All.
Question 9 |
The logic functionality realized by the circuit shown below is
OR | |
XOR | |
NAND | |
AND |
Question 9 Explanation:
The output Y will be logic 1, when
A=1, B=1 and \bar{B}=0
\Rightarrow Y=A \cdot B \cdot \bar{B}=A \cdot B
\Rightarrow AND operation
A=1, B=1 and \bar{B}=0
\Rightarrow Y=A \cdot B \cdot \bar{B}=A \cdot B
\Rightarrow AND operation
Question 10 |
The functionality implemented by the circuit below is
2-to-1 multiplexer | |
4-to-1 multiplexer | |
7-to-1 multiplexer | |
6-to-1 multiplexer |
Question 10 Explanation:
When the outputs \left(O_{0^{\prime}} O_{1}, O_{2}, O_{3}\right) of the decoder are at logic 1, the corresponding tristate buffer is activated. In that case, whatever data is applied at the input of a buffer, becomes its output.
Hence, when
\begin{aligned} \Rightarrow \quad C_{1} C_{0}&=00, \quad \text{ Then } O_{0}=1, \\ \therefore \quad Y&=P \\ \Rightarrow \quad C_{1} C_{0}&=01, \quad \text{ Then } O_{1}=1 \\ \therefore \quad Y&=Q \\ \Rightarrow \quad C_{1} C_{0}&=10, \quad \text{ Then } O_{2}=1 \\ \therefore \quad Y&=R \\ \Rightarrow \quad C_{1} C_{0}&=11, \quad \text{ Then } O_{3}=1 \\ \therefore \quad Y&=S \end{aligned}
\therefore \quad the circuit effectively behaves as a 4 to 1 multiplexer.
Hence, when
\begin{aligned} \Rightarrow \quad C_{1} C_{0}&=00, \quad \text{ Then } O_{0}=1, \\ \therefore \quad Y&=P \\ \Rightarrow \quad C_{1} C_{0}&=01, \quad \text{ Then } O_{1}=1 \\ \therefore \quad Y&=Q \\ \Rightarrow \quad C_{1} C_{0}&=10, \quad \text{ Then } O_{2}=1 \\ \therefore \quad Y&=R \\ \Rightarrow \quad C_{1} C_{0}&=11, \quad \text{ Then } O_{3}=1 \\ \therefore \quad Y&=S \end{aligned}
\therefore \quad the circuit effectively behaves as a 4 to 1 multiplexer.
There are 10 questions to complete.