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(ELEC102)final_pasta.pdf
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(321) (16)

(a)
Name two advantages of MOSFET. (b) Draw the cross sectional diagram for an enhancement NMOSFET and describe very briefly the structure.

(16)


(a)
small size (scaled down easily)
and low power consumption.

(4) (b)
D : drain
Oxide layer
VD B : Body ID
(insulator)
(or substrate)

VG
IG
source
G : Gate
usually
Metallic film
shorted
to body IS VS
IG = 0
S : source

ID = IS
An NMOSFET consists of a metal gate insulated from a p-type semiconductor substrate (or body) by an insulating layer of silicon dioxide. On either side of the gate there are n type regions forming the drain and source.
(12)
(12)

(c) Draw the cross sectional structure and draw the n-channel conditions when the NMOS is in Triode (VDS << VGS -VT) and Saturation (VDS > VGS -VT) mode. Sketch also the ID-VDS curve in each mode. (12)
(c)
Triode (ohmic)
VD (small)

ID


n+
n+
n layer

ID

(5)

VD

saturation
electrons



.
D
P
space charge region

IDS = IDSAT
ID

VGS

IDSAT
(7)
VDS

VDSAT

(22)

Find Vo . (20) 6V
Show clearly your reasons.
1k. VoGiven the NMOS are all identical, VT = 1V
I

K = 0.25mA/V2 .

Assume 2 NMOS are in saturation (2)
At triode region , VGS VT , VDS < VGS -VT , ID =
2K(VGS CVT)VDS C KVDS2
At saturation region , VGS VT , VDS VGS -VT ,
ID = K[(VGS CVT)2 ]

ID = K (VGS .VT )2 26. Vo 1
I = 0.25m*(Vo.1) = * (8)
1k. 2 0.5(1.2Vo+ Vo2) = 6.Vo 1.2Vo+ Vo2 =12.2Vo Vo2 =11 Vo =
11 . 3.31V (8)
NMOS is saturate since
1.
VGS > VT 3.31 > 1 (4)

2.
VDS 3.31 > 3.31-1


> VGS CVT
6V
2.68mA
1k. 3.31V

I = 0.25(3.31.1)1.34mA
1.34mA 2 .
(22)

Show that V2 = 6V. Hence find V1 . Show clearly your reasons. (22)
10V
12V


2k.2k.

V1
V2 2k. 0V

Given VT = 1V K = 0.5mA/V2
10V V2 = 6V since IG = 0. (4)

I Assume NMOS is triode
2k. V1 2


10V . V1 V1
6V

I == 2*0.5m *[(6V .1V)V1. )]
2k. 2 V12 (8)
10V . V1 = 2*[5V1. ]
2 V12 .11V1+10 = 0
V1 =10V or 1V
(6)
When Vo = 10V, MOS is OFF
(2)
But MOS is not OFF (VGS > VT) . Hence Vo = 1V 10V
NMOS is triode since
1. VGS > VT 6 > 1 (2)

4.5m
2. VDS 1 > 6 -1
< VGS CVT
1V
6V
2k.

I =1(5*1.0.5) = 4.5mA
At triode region, VGS VT , VDS < VGS -VT , ID = 2K(VGS CVT)VDS C KVDS2
At saturation region, VGS VT , VDS VGS -VT , ID = K [(VGS CVT)2]
(17)

Draw the cross sectional structure of a NPN BJT transistor operated in the amplifier mode, Describe the movement of electrons, the designs in the emitter/base/collector, and explain briefly the equation IE IC / . If IC .IB, find in terms of . (17)
(a)
IE IC
EBC
e -

e

IB
VBE

VCB






1.
EB Junction is a forward bias (on) diode and BC is reverse bias (off) diode

2.
E is very heavily doped (N + for NPN). E has many electrons,

3.
B is very thin. So most electrons injected from E