Important Formulas: Semiconductor Electronics- Materials, Devices & Simple Circuits | Physics Class 12 - NEET PDF Download

 Page 1


SEMICONDUCTOR
Conductivity and resistivity
? P ( ? ?– m) ? ?( ?
–1
m
–1
)
Metals 10
–2
 -10
–6
10
2
 – 10
8
semiconductors 10
–5
 -10
–6
10
5
 – 10
–6
Insulators 10
11
 –10
19
10
–11
 – 10
–19
Charge concentration and current
? ?[ ?
n
 =  ?
e
] In case of intrinsic semiconductors
? P type ?
n
 >>  ?
e
? i = i
e
 + i
h
? ?
e
  ?
n
 = 
2
i
?
? Number of electrons reaching from valence bond to conduction bond.
? ? ? ? ?= 
kT 2 / Eg – 2 / 3
e T A (A is positive constant)
? ? = e ( ?
e
 m
e
 +  ?
n
  ?
n
)
for ?  hype ?
n
 = Na >>  ?
e
.
for ? – type ?
e
 = Na >>  ?
h
? Dynamic Resistance of P-N junction in forward biasing  =  
? ?
?V
Transistor
? CB amplifier
(i) ac current gain  ?
c
 = 
) i ( current collector in change Samll
) i ( current collector in change Samll
e
c
?
?
(ii) dc current gain ?
dc
 = 
) i ( current Emitter
) i ( current Collector
e
c
  value of  ?
dc
 lies
between 0.95 to 0.99
(iii) Voltage gain A
V
 =
) V ( voltage input in Change
) V ( voltage output in Change
f
0
?
?
? ?A
V
 = a
ac
 × Resistance gain
(iv) Power gain =
) P ( voltage input in Change
) P ( power output in Change
C
0
?
?
? Power gain = a
2
ac
 × Resistance gain
(v) Phase difference (between output and input) : same phase
(vi) Application : For High frequency
Page 2


SEMICONDUCTOR
Conductivity and resistivity
? P ( ? ?– m) ? ?( ?
–1
m
–1
)
Metals 10
–2
 -10
–6
10
2
 – 10
8
semiconductors 10
–5
 -10
–6
10
5
 – 10
–6
Insulators 10
11
 –10
19
10
–11
 – 10
–19
Charge concentration and current
? ?[ ?
n
 =  ?
e
] In case of intrinsic semiconductors
? P type ?
n
 >>  ?
e
? i = i
e
 + i
h
? ?
e
  ?
n
 = 
2
i
?
? Number of electrons reaching from valence bond to conduction bond.
? ? ? ? ?= 
kT 2 / Eg – 2 / 3
e T A (A is positive constant)
? ? = e ( ?
e
 m
e
 +  ?
n
  ?
n
)
for ?  hype ?
n
 = Na >>  ?
e
.
for ? – type ?
e
 = Na >>  ?
h
? Dynamic Resistance of P-N junction in forward biasing  =  
? ?
?V
Transistor
? CB amplifier
(i) ac current gain  ?
c
 = 
) i ( current collector in change Samll
) i ( current collector in change Samll
e
c
?
?
(ii) dc current gain ?
dc
 = 
) i ( current Emitter
) i ( current Collector
e
c
  value of  ?
dc
 lies
between 0.95 to 0.99
(iii) Voltage gain A
V
 =
) V ( voltage input in Change
) V ( voltage output in Change
f
0
?
?
? ?A
V
 = a
ac
 × Resistance gain
(iv) Power gain =
) P ( voltage input in Change
) P ( power output in Change
C
0
?
?
? Power gain = a
2
ac
 × Resistance gain
(v) Phase difference (between output and input) : same phase
(vi) Application : For High frequency
CE Amplifier
(i) ac current gain ?
ac
 = 
?
?
?
?
?
?
?
?
?
?
b
c
i
i
 V
CE
 = constant
(ii) dc current gain ?
dc
 =
b
c
i
i
(iii) Voltage gain : A
V
 =  
i
0
V
V
?
?
=  ?
ac
 × Resistance gain
(iv) Power gain = 
i
0
P
P
?
?
 =  ?
2
ac × Resistance
(v) Transconductance (g
m
) :  The ratio of the change in collector in
collector current to the change in emitter base voltage is called trans
conductance i.e. g
m
 = 
EB
c
V
i
?
?
 . Also g
m
 =  
L
V
R
A
 R
L
 = Load resistance.
? Relation between  ? and ?  :  
?
?
? ?
– 1
 or ?  = 
? ?
?
1
(v) Transconductance (gm) :  The ratio of the change in collector in collec-
tor current to the change in emitter base voltage is called trans conductance
i.e. gm =  . Also gm =  RL = Load resistance.