Electronic Devices and Circuits. Graduate Aptitude Test in Engineering.
Electronics and Communication Engineering. GATE ...
GA GATE Graduate Aptitude Test in Engineering
Electronics onics and Communication Communica Engineering
Electronic Electr onic Devices Devices and Cir Circuits
ELECTRO E ONIC DEVIC CE AND CIIRCUITS-T THEORY Introducttion: *Semicon nductor It’s a materrial whose con nductivity is in between m metal and insu ulator. Germaanium (Ge) and ssilicon (Si) are e most comm monly used semiconductor in VLSI techn nology Orbital strructure of Si 2, 8, 4 so o there are 4 eelectrons in o outermost band and readyy to form bonding. For Ge orbital structure will be 2, 8,,18,4 and thu us electrons are there to fo orm bonding. erature more and more eleectrons get frree. So in other word mateerial As we increase tempe Become m more and more conductivee; but R ∝
1 coonductivity
Hence as temperature increases { TT, }, Resistance decreases ((R, } on Applicatio All waferss on which cirrcuits are in faabricated aree made up of ssilicon anotheer reason of u using silicon is that it is aabundant in n nature si
Si
Si
Si
Si
mation of siliccon Bond form Copyrigh hts @ BodhBridge Educational Services Pvt. Ltd, Chennai
DEVIICE MODE ELLING - QUESTION Q NS ng 100 μ m area (cross-ssectional) dop ped with 10 /cm3 phosp phorous find 1) A Si-baar 0.1cm lon 2
1 17
current?((neglect temp perature effeect) μ n = 700 0cm 2 / vs ,q= =1.6x 10 (a)1.5m mA
(b)1.3m mA
(c)1..12mA
−19
C ; applied volttage is 10V Resistivity
(dd)1 μ A
m / v − s andd due to latticce 2cm / v − s .what is 2) If mob bility due to impurity scatttering is 1cm resultant scattering in n material. Mobilitty 2
(a) 2cm / v − s 2
(b)) 1cm / v − s 2
2
5cm / v − s (c) 0.5 2
(d) 0.67ccm / v − s 2
3) What is i mobility att point x in fiigure below.
(a) 10 cm / v − s 2
cm / v − s (b)100 ( 2
Mobilitty
(c))1000 cm / v − s 2
(d)10000 cm / v − s 2
16
3
c and acceeptor 4) A hypoothetical sem miconductor having h donorr concentratiion( N d ) equ ual to 10 / cm concentraation ( N a ) 2 ×10 / cm .what is new w electron and d hole concen ntration. 16
3
Intrinsic and EExtrinsic Silico on
04 / cm3 (a) n0 = 10
p0 = 1016 / cm m3
( p0 = 1016 / cm3 (b)
(c) p0 = 1016 / cm3
n0 = 1016 / cm m3
( p0 = 104 / cm3 (d)
n0 = 104 / cm3 n0 = 104 / cm3
Copyrigh hts @ BodhBridge Educational Services Pvt. Ltd, Chennai
DEV VICE MODE ELLING – SOLUTIONS S 1) SOLU UTION:
(c)
σ = nqμn (conductivitty) = 10 ×1.6 ×10 17
R=
σL A
−19
× 700 = 11.2
ρ=
1
σ
I=
= 8928.57 7
= 0.08928 = resistiivity
V = 1.122mA R
2) SOLU UTION: (d)
1
μ
=
1
μ1
+
1
μ2
+ ..... +
Here n=2
1
μ
=
1
μn 1
μ1
1
1 1 = + = 1.5 μ 1 2
+
1
μ2
μ = 0.67cm m2 / v − s
3) SOLU UTION: (c) ν d = μ E
106 = μ = 1000cm2 / v − s 3 10 4) SOLU UTION: (b)
Na > Nd n0 = N a − N d = 1016 / cm3 n0 p0 = ni 2
n0 = 104 / cm3
Copyrigh hts @ BodhBridge Educational Services Pvt. Ltd, Chennai
