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Page 1
• Energy gap
???? G/si
=1.21- 3.6 × 10
-4
.T ev
???? G/Ge
=0.785- 2.23 × 10
-4
.T ev
? Energy gap depending on temperature
• E
F
= E
C
- KT ln?
???? ???? ???? ???? ? = E
v
+ KT ln ?
???? ???? ???? ???? ?
• No. of electrons n = N
c
e
-(E
c
-E
f
)/RT
(KT in ev)
• No. of holes p = N
v
e
-(E
f
-E
v
)/RT
• Mass action law n
p
= n
i
2
= N
c
N
v
e
-EG/KT
• Drift velocity ???? d
= µE (for si ???? d
= 10
7
cm/sec)
• Hall voltage ???? H
=
B.I
w
e
. Hall coefficient R
H
= 1/? . ? ? charge density = qN
0
= ne …
• Conductivity s = ?µ ; µ = sR
H
.
• Max value of electric field @ junction E
0
= -
q
?
si
N
d
. n
n0
= -
q
?
si
N
A
. n
p0
.
• Charge storage @ junction Q
+
= - Q
-
= qA x
n0
N
D
= qA x
p0
N
A
• Diffusion current densities J
p
= - q D
p
dp
dx
J
n
= - q D
n
dn
dx
• Drift current Densities = q(p µ
p
+ nµ
n
)E
• µ
p
, µ
n
decrease with increasing doping concentration .
•
D
n
µ
n
=
D
p
µ
p
= KT/q ˜ 25 mv @ 300 K
• Carrier concentration in N-type silicon n
n0
= N
D
; p
n0
= n
i
2
/ N
D
• Carrier concentration in P-type silicon p
p0
= N
A
; n
p0
= n
i
2
/ N
A
• Junction built in voltage V
0
= V
T
ln ?
???? ???? ???? ???? ???? ???? 2
?
• Width of Depletion region W
dep
= x
p
+ x
n
= ?
2e
s
q
?
1
N
A
+
1
N
D
?(V
0
+ V
R
)
* ?
2???? ???????? ???? = 12.93???? ???????????? ???????? ?
•
x
n
x
p
=
N
A
N
D
• Charge stored in depletion region q
J
=
q.N
A
N
D
N
A
+N
D
. A . W
dep
• Depletion capacitance C
j
=
e
s
A
W
dep
; C
j0
=
e
s
A
W
dep
/ V
R
=0
C
j
= C
j0
/? 1 +
V
R
V
0
?
m
C
j
= 2C
j0
(for forward Bias)
• Forward current I = I
p
+ I
n
; I
p
= Aq n
i
2
D
p
L
p
N
D
? ???? ???? /???? ???? - 1?
I
n
= Aq n
i
2
D
n
L
n
N
A
? ???? ???? /???? ???? - 1?
• Saturation Current I
s
= Aq n
i
2
?
D
p
L
p
N
D
+
D
n
L
n
N
A
?
• Minority carrier life time t
p
= L
p
2
/ D
p
; t
n
= L
n
2
/ D
n
Analog Circuits (Formula Notes/Short Notes)
Page 2
• Energy gap
???? G/si
=1.21- 3.6 × 10
-4
.T ev
???? G/Ge
=0.785- 2.23 × 10
-4
.T ev
? Energy gap depending on temperature
• E
F
= E
C
- KT ln?
???? ???? ???? ???? ? = E
v
+ KT ln ?
???? ???? ???? ???? ?
• No. of electrons n = N
c
e
-(E
c
-E
f
)/RT
(KT in ev)
• No. of holes p = N
v
e
-(E
f
-E
v
)/RT
• Mass action law n
p
= n
i
2
= N
c
N
v
e
-EG/KT
• Drift velocity ???? d
= µE (for si ???? d
= 10
7
cm/sec)
• Hall voltage ???? H
=
B.I
w
e
. Hall coefficient R
H
= 1/? . ? ? charge density = qN
0
= ne …
• Conductivity s = ?µ ; µ = sR
H
.
• Max value of electric field @ junction E
0
= -
q
?
si
N
d
. n
n0
= -
q
?
si
N
A
. n
p0
.
• Charge storage @ junction Q
+
= - Q
-
= qA x
n0
N
D
= qA x
p0
N
A
• Diffusion current densities J
p
= - q D
p
dp
dx
J
n
= - q D
n
dn
dx
• Drift current Densities = q(p µ
p
+ nµ
n
)E
• µ
p
, µ
n
decrease with increasing doping concentration .
•
D
n
µ
n
=
D
p
µ
p
= KT/q ˜ 25 mv @ 300 K
• Carrier concentration in N-type silicon n
n0
= N
D
; p
n0
= n
i
2
/ N
D
• Carrier concentration in P-type silicon p
p0
= N
A
; n
p0
= n
i
2
/ N
A
• Junction built in voltage V
0
= V
T
ln ?
???? ???? ???? ???? ???? ???? 2
?
• Width of Depletion region W
dep
= x
p
+ x
n
= ?
2e
s
q
?
1
N
A
+
1
N
D
?(V
0
+ V
R
)
* ?
2???? ???????? ???? = 12.93???? ???????????? ???????? ?
•
x
n
x
p
=
N
A
N
D
• Charge stored in depletion region q
J
=
q.N
A
N
D
N
A
+N
D
. A . W
dep
• Depletion capacitance C
j
=
e
s
A
W
dep
; C
j0
=
e
s
A
W
dep
/ V
R
=0
C
j
= C
j0
/? 1 +
V
R
V
0
?
m
C
j
= 2C
j0
(for forward Bias)
• Forward current I = I
p
+ I
n
; I
p
= Aq n
i
2
D
p
L
p
N
D
? ???? ???? /???? ???? - 1?
I
n
= Aq n
i
2
D
n
L
n
N
A
? ???? ???? /???? ???? - 1?
• Saturation Current I
s
= Aq n
i
2
?
D
p
L
p
N
D
+
D
n
L
n
N
A
?
• Minority carrier life time t
p
= L
p
2
/ D
p
; t
n
= L
n
2
/ D
n
Analog Circuits (Formula Notes/Short Notes)
• Minority carrier charge storage Q
p
= t
p
I
p
, Q
n
= t
p
I
n
Q = Q
p
+ Q
n
= t
T
I t
T
= mean transist time
• Diffusion capacitance C
d
= ?
???? ???? ???????? ???? ? I = t.g ? C
d
? I.
t? carrier life time , g = conductance = I / ???????? ????
• I
02
= 2
(T
2
-T
1
)/10
I
01
• Junction Barrier Voltage V
j
= V
B
= V
r
(open condition)
= V
r
- V (forward Bias)
= V
r
+ V (Reverse Bias)
• Probability of filled states above ‘E’ f(E) =
1
1+e
(E-E
f
)/KT
• Drift velocity of e
-
???? d
= 10
7
cm/sec
• Poisson equation
d
2
V
dx
2
=
-?
v
?
=
-nq
?
?
dv
dx
= E =
-nqx
?
Transistor :-
• I
E
= I
DE
+ I
nE
• I
C
= I
Co
– a I
E
? Active region
• I
C
= – a I
E
+ I
Co
(1- e
V
C
/V
T
)
Common Emitter :-
• I
C
= (1+ ß) I
Co
+ ßI
B
ß =
a
1-a
• I
CEO
=
I
Co
1-a
? Collector current when base open
• I
CBO
? Collector current when I
E
= 0 I
CBO
> I
Co
.
• V
BE,sat
or V
BC,sat
? - 2.5 mv /
0
C ; V
CE,sat
?
V
BE,sat
10
= - 0.25 mv /
0
C
• Large signal Current gain ß =
I
C
- I
CBo
I
B
+ I
CBo
• D.C current gain ß
dc
=
I
C
I
B
= h
FE
• (ß
dc
= h
FE
) ˜ ß when I
B
> I
CBo
• Small signal current gain ß
'
=
?I
C
?I
R
?
V
CE
= h
fe
=
h
FE
1-(I
CBo
+ I
B
)
?h
FE
?I
C
• Over drive factor =
ß
active
ß
forced
?under saturation
? I
C sat
= ß
forced
I
B sat
Conversion formula :-
CC ? CE
• h
ic
= h
ie
; h
rc
= 1 ; h
fc
= - (1+ h
fe
) ; h
oc
= h
oe
CB ? CE
• h
ib
=
h
ie
1+h
fe
; h
ib
=
h
ie
h
oe
1+h
fe
- h
re
; h
fb
=
-h
fe
1+h
fe
; h
ob
=
h
oe
1+h
fe
CE parameters in terms of CB can be obtained by interchanging B & E .
Page 3
• Energy gap
???? G/si
=1.21- 3.6 × 10
-4
.T ev
???? G/Ge
=0.785- 2.23 × 10
-4
.T ev
? Energy gap depending on temperature
• E
F
= E
C
- KT ln?
???? ???? ???? ???? ? = E
v
+ KT ln ?
???? ???? ???? ???? ?
• No. of electrons n = N
c
e
-(E
c
-E
f
)/RT
(KT in ev)
• No. of holes p = N
v
e
-(E
f
-E
v
)/RT
• Mass action law n
p
= n
i
2
= N
c
N
v
e
-EG/KT
• Drift velocity ???? d
= µE (for si ???? d
= 10
7
cm/sec)
• Hall voltage ???? H
=
B.I
w
e
. Hall coefficient R
H
= 1/? . ? ? charge density = qN
0
= ne …
• Conductivity s = ?µ ; µ = sR
H
.
• Max value of electric field @ junction E
0
= -
q
?
si
N
d
. n
n0
= -
q
?
si
N
A
. n
p0
.
• Charge storage @ junction Q
+
= - Q
-
= qA x
n0
N
D
= qA x
p0
N
A
• Diffusion current densities J
p
= - q D
p
dp
dx
J
n
= - q D
n
dn
dx
• Drift current Densities = q(p µ
p
+ nµ
n
)E
• µ
p
, µ
n
decrease with increasing doping concentration .
•
D
n
µ
n
=
D
p
µ
p
= KT/q ˜ 25 mv @ 300 K
• Carrier concentration in N-type silicon n
n0
= N
D
; p
n0
= n
i
2
/ N
D
• Carrier concentration in P-type silicon p
p0
= N
A
; n
p0
= n
i
2
/ N
A
• Junction built in voltage V
0
= V
T
ln ?
???? ???? ???? ???? ???? ???? 2
?
• Width of Depletion region W
dep
= x
p
+ x
n
= ?
2e
s
q
?
1
N
A
+
1
N
D
?(V
0
+ V
R
)
* ?
2???? ???????? ???? = 12.93???? ???????????? ???????? ?
•
x
n
x
p
=
N
A
N
D
• Charge stored in depletion region q
J
=
q.N
A
N
D
N
A
+N
D
. A . W
dep
• Depletion capacitance C
j
=
e
s
A
W
dep
; C
j0
=
e
s
A
W
dep
/ V
R
=0
C
j
= C
j0
/? 1 +
V
R
V
0
?
m
C
j
= 2C
j0
(for forward Bias)
• Forward current I = I
p
+ I
n
; I
p
= Aq n
i
2
D
p
L
p
N
D
? ???? ???? /???? ???? - 1?
I
n
= Aq n
i
2
D
n
L
n
N
A
? ???? ???? /???? ???? - 1?
• Saturation Current I
s
= Aq n
i
2
?
D
p
L
p
N
D
+
D
n
L
n
N
A
?
• Minority carrier life time t
p
= L
p
2
/ D
p
; t
n
= L
n
2
/ D
n
Analog Circuits (Formula Notes/Short Notes)
• Minority carrier charge storage Q
p
= t
p
I
p
, Q
n
= t
p
I
n
Q = Q
p
+ Q
n
= t
T
I t
T
= mean transist time
• Diffusion capacitance C
d
= ?
???? ???? ???????? ???? ? I = t.g ? C
d
? I.
t? carrier life time , g = conductance = I / ???????? ????
• I
02
= 2
(T
2
-T
1
)/10
I
01
• Junction Barrier Voltage V
j
= V
B
= V
r
(open condition)
= V
r
- V (forward Bias)
= V
r
+ V (Reverse Bias)
• Probability of filled states above ‘E’ f(E) =
1
1+e
(E-E
f
)/KT
• Drift velocity of e
-
???? d
= 10
7
cm/sec
• Poisson equation
d
2
V
dx
2
=
-?
v
?
=
-nq
?
?
dv
dx
= E =
-nqx
?
Transistor :-
• I
E
= I
DE
+ I
nE
• I
C
= I
Co
– a I
E
? Active region
• I
C
= – a I
E
+ I
Co
(1- e
V
C
/V
T
)
Common Emitter :-
• I
C
= (1+ ß) I
Co
+ ßI
B
ß =
a
1-a
• I
CEO
=
I
Co
1-a
? Collector current when base open
• I
CBO
? Collector current when I
E
= 0 I
CBO
> I
Co
.
• V
BE,sat
or V
BC,sat
? - 2.5 mv /
0
C ; V
CE,sat
?
V
BE,sat
10
= - 0.25 mv /
0
C
• Large signal Current gain ß =
I
C
- I
CBo
I
B
+ I
CBo
• D.C current gain ß
dc
=
I
C
I
B
= h
FE
• (ß
dc
= h
FE
) ˜ ß when I
B
> I
CBo
• Small signal current gain ß
'
=
?I
C
?I
R
?
V
CE
= h
fe
=
h
FE
1-(I
CBo
+ I
B
)
?h
FE
?I
C
• Over drive factor =
ß
active
ß
forced
?under saturation
? I
C sat
= ß
forced
I
B sat
Conversion formula :-
CC ? CE
• h
ic
= h
ie
; h
rc
= 1 ; h
fc
= - (1+ h
fe
) ; h
oc
= h
oe
CB ? CE
• h
ib
=
h
ie
1+h
fe
; h
ib
=
h
ie
h
oe
1+h
fe
- h
re
; h
fb
=
-h
fe
1+h
fe
; h
ob
=
h
oe
1+h
fe
CE parameters in terms of CB can be obtained by interchanging B & E .
• A
I
=
-h
f
1+h
0
Z
L
Z
i
= h
i
+ h
r
A
I
Z
L
A
vs
=
A
v
.Z
i
Z
i
+R
s
Z
i
+R
s
=
A
I
.Z
L
=
A
I
s
.Z
L
R
s
A
V
=
A
I
Z
L
Z
i
Y
0
= h
o
-
h
f
h
r
h
i
+ R
s
A
Is
=
A
v
.R
s
Z
i
+R
s
=
A
vs
.R
s
Z
L
Choice of Transistor Configuration :-
• For intermediate stages CC can’t be used as A
V
< 1
• CE can be used as intermediate stage
• CC can be used as o/p stage as it has low o/p impedance
• CC/CB can be used as i/p stage because of i/p considerations.
Stability & Biasing :- ( Should be as min as possible)
• For S =
?I
C
?I
Co
?
V
B0,ß
S
'
=
?I
C
?V
BE
?
I
C0,ß
S
''
=
?I
C
?ß
?
V
BE,I
Co
?I
C
= S. ?I
Co
+ S
'
?V
BE
+ S
''
?ß
• For fixed bias S =
1+ß
1-ß
dI
B
dI
C
= 1 + ß
• Collector to Base bias S =
1+ß
1+ß
R
C
R
C
+R
B
0 < s < 1+ ß =
1+ß
1+ß?
R
C
+ R
E
R
C
+ R
E
+ R
B
?
• Self bias S =
1+ß
1+ß
R
E
R
E
+R
th
˜ 1+
R
th
R
e
ßR
E
> 10 R
2
• R
1
=
V
cc
R
th
V
th
; R
2
=
V
cc
R
th
V
cc
-V
th
• For thermal stability [ V
cc
- 2I
c
(R
C
+ R
E
)] [ 0.07 I
co
. S] < 1/? ; V
CE
<
V
CC
2
Hybrid –pi(p)- Model :-
g
m
= |I
C
| / V
T
r
b
'
e
= h
fe
/ g
m
r
b
'
b
= h
ie
- r
b
'
e
r
b
'
c
= r
b
'
e
/ h
re
g
ce
= h
oe
- (1+ h
fe
) g
b
'
c
Specifications of An amplifier :-
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FAQs on Analog Electronics Formulas for GATE EE Exam - Analog and Digital Electronics - Electrical Engineering (EE)
| 1. What are some important analog electronics formulas for the GATE EE exam in Electrical Engineering? |  |
Ans. Some important analog electronics formulas for the GATE EE exam are: - Voltage Divider Rule: Vout = Vin * (R2 / (R1 + R2)) - Ohm's Law: V = I * R, where V is the voltage, I is the current, and R is the resistance. - Kirchhoff's Voltage Law (KVL): The sum of voltages in a closed loop is zero. - Kirchhoff's Current Law (KCL): The sum of currents entering a node is equal to the sum of currents leaving the node. - Gain of an Amplifier: Gain = Vout / Vin = -Rf / Rin, where Rf is the feedback resistor and Rin is the input resistor.
| 2. How can I apply the voltage divider rule in analog electronics problems? | |
Ans. The voltage divider rule is used to calculate the output voltage in a resistive voltage divider circuit. To apply the voltage divider rule, follow these steps: 1. Identify the resistors in the circuit and assign them values (in ohms). 2. Determine the input voltage (Vin) and assign it a value (in volts). 3. Use the voltage divider formula: Vout = Vin * (R2 / (R1 + R2)), where R1 and R2 are the resistors in the circuit. 4. Substitute the given resistor values and input voltage into the formula to calculate the output voltage (Vout).
| 3. What is Kirchhoff's Voltage Law (KVL) and how is it used in analog electronics? | |
Ans. Kirchhoff's Voltage Law (KVL) states that the sum of voltages in a closed loop in a circuit is zero. It is used in analog electronics to analyze and solve circuit problems. KVL is based on the principle of conservation of energy, which states that the total energy in a closed system remains constant. To apply KVL, follow these steps: 1. Identify the closed loop in the circuit. 2. Assign polarities to the voltage sources and resistors in the loop. 3. Write an equation for each voltage drop in the loop, considering the polarities. 4. Sum up all the voltage drops and set the sum equal to zero. 5. Solve the resulting equation to find the unknown voltages or current in the circuit.
| 4. How can Ohm's Law be used in analog electronics? | |
Ans. Ohm's Law is a fundamental law in analog electronics that relates voltage, current, and resistance. It can be used to calculate any one of these quantities if the other two are known. Ohm's Law states that the current (I) flowing through a resistor is directly proportional to the voltage (V) across it and inversely proportional to its resistance (R). The equation for Ohm's Law is V = I * R. To use Ohm's Law, you can rearrange the equation to solve for the desired quantity. For example, if you know the voltage and resistance, you can calculate the current by dividing the voltage by the resistance.
| 5. What is the gain of an amplifier and how is it calculated in analog electronics? | |
Ans. The gain of an amplifier in analog electronics refers to the amplification factor or the ratio of the output voltage to the input voltage. It indicates the level of amplification or attenuation provided by the amplifier. The gain can be expressed as a positive or negative value, indicating amplification or inversion, respectively. In an amplifier circuit, the gain is determined by the feedback resistor (Rf) and the input resistor (Rin). The formula for calculating the gain is: Gain = Vout / Vin = -Rf / Rin. By knowing the values of the feedback resistor and input resistor, you can calculate the gain of the amplifier.
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Feb 07, 2026 Last updated
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Document Description: Analog Electronics Formulas for GATE EE Exam for Electrical Engineering (EE) 2026 is part of Analog and Digital Electronics preparation. The notes and questions for Analog Electronics Formulas for GATE EE Exam have been prepared according to the Electrical Engineering (EE) exam syllabus. Information about Analog Electronics Formulas for GATE EE Exam covers topics like and Analog Electronics Formulas for GATE EE Exam Example, for Electrical Engineering (EE) 2026 Exam. Find important definitions, questions, notes, meanings, examples, exercises and tests below for Analog Electronics Formulas for GATE EE Exam.
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