USN
1 P E
PESIT Bangalore South Campus
Hosur road, 1km before Electronic City, Bengaluru -100
Department of Electronics & Communication Engineering
Scheme and Solution of First Internals
Subject & Code : Microelectronics/10EC63
Name of faculty : M Raghavendra
Answer any 5 Questions
Section : 6th “C”
Time : 1Hr 30Mins
Q.No
Questions
a. Consider a process technology for which Lmin=0.4µm, tox=8nm, µn=450 cm2/V and
1
Vt=0.7 V
i. Find Cox and k’n.
ii. For a MOSFET with W/L=8µm/0.8µm, calculate the values of VGS and VDSmin
needed to operate the transistor in the saturation region with a dc current
ID=100µA
iii. For the device in (b), find the value of VGS required to cause the device to operate
as a 1000Ω resistor for very small VDS.
Solution :
1
Marks
07
USN
1 P E
PESIT Bangalore South Campus
Hosur road, 1km before Electronic City, Bengaluru -100
Department of Electronics & Communication Engineering
2
USN
1 P E
PESIT Bangalore South Campus
Hosur road, 1km before Electronic City, Bengaluru -100
Department of Electronics & Communication Engineering
2
b. Sketch a CMOS realization for the function
Solution:
03
Design a fixed bias circuit with and without source resistance for drain current of
0.5mA, k’n(W/L)=1mA/V2, Vt=1V, VDD=10V. Calculate percentage change in drain
current in both bias circuits with same k’n(W/L) and Vt=1.5V.
Plot iD – vGS characteristics for both types of biasing technique.
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3
USN
1 P E
PESIT Bangalore South Campus
Hosur road, 1km before Electronic City, Bengaluru -100
Department of Electronics & Communication Engineering
3
Determine Rin, Gv and Rout of CS amplifier, CG amplifier and CD amplifier. Given
VDD=VSS=10V, RG=4.7kΩ, RD=15kΩ, k’n(W/L)= 1mA/V2 and Vt=1.5V. Biasing is
done using constant current source of 0.5mA. RL = 15kΩ.
Solution
1. CS Amplifier : Rin= RG=4.7kΩ , Rout= RD=15kΩ, Gv=Av=-15V/V
2. CG Amplifier: Rin= RG=1/gm=1kΩ , Rout= RD=15kΩ, Gv= 7V/V
3. CD Amplifier: Rin= RG=4.7kΩ , Rout= RD=15kΩ, Gv=0.135V/V
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USN
1 P E
PESIT Bangalore South Campus
Hosur road, 1km before Electronic City, Bengaluru -100
Department of Electronics & Communication Engineering
4
a. Derive the analytical expression for the transfer characteristics of CS amplifier.
Solution: Circuit Diagram- 1M
Av in Cut-off -2M
Av in saturation region- 2M
Av in triode region – 2M
7+3
b. Construct VTC for the case k’n(W/L)= 1mA/V2, Vt=1 V, RL = 18kΩ and VDD =
10V.
Solution: Point X=(0,10V)
Point A =(1V,10V)
Point B = (2V,1V)
Point C= (0V,0.061V)
5
Using two transistors Q1 and Q2 having equal lengths but widths related W2/W1=5.
Given VDD=-VSS=5V, k’n(W/L)= 0.8 mA/V2, Vt=1 V and λ=0
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a. Design constant current source to obtain I=0.5mA.
Solution: R=85KΩ
b. What is the voltage at the gates of Q1 and Q2.
Solution: VG1=VG2 = -3.5V
c. What is the lowest allowed voltage at the drain of Q2 while Q2 remains in
saturation region?
Solution : VDmin = -4.5V
6
Consider a cmos inverter fabricated in a 0.25µm process for which Cox=6fF/µm2, µn
Cox = 115µA/V2 , Vtn= -Vtp=0.4V and VDD = 2.5V. The W/L ratio of QN is
0.375µm/0.25µm and Qp =1.125µm/0.25µm. The equivalent capacitence value is
6.25fF. Find tPHL, tPLH and tp.
Solution: tPHL=23.3ps, tPLH=30ps, tp=26.5ps
5
10
USN
1 P E
PESIT Bangalore South Campus
Hosur road, 1km before Electronic City, Bengaluru -100
Department of Electronics & Communication Engineering
7
For each of the circuits shown in Figure. Find the labeled node voltages.
The NMOS transistors have k’n(W/L)= 2 mA/V2, Vt=1 V
Solution: (a) V1=0.9V, V2=-1.6V
(b) V1=4.1V, V2=2.5V, V3=0.9V
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