Mba admssion in india

Admission in India 2015
By:
admission.edhole.com

2
Transistors
•They are unidirectional current carrying devices with capability to
control the current flowing through them
• The switch current can be controlled by either current or voltage
• Bipolar Junction Transistors (BJT) control current by current
• Field Effect Transistors (FET) control current by voltage
•They can be used either as switches or as amplifiers

3
NPN Bipolar Junction Transistor
•One N-P (Base Collector) diode one P-N (Base Emitter) diode

4
PNP Bipolar Junction Transistor
•One P-N (Base Collector) diode one N-P (Base Emitter) diode

5
NPN BJT Current flow

6
BJT α and β
•From the previous figure iE = iB + iC
•Define α = iC / iE
•Define β = iC / iB
•Then β = iC / (iE –iC) = α /(1- α)
•Then iC = α iE ; iB = (1-α) iE
•Typically β ≈ 100 for small signal BJTs (BJTs that
handle low power) operating in active region (region
where BJTs work as amplifiers)admission.edhole.com

7
BJT in Active Region
Common Emitter(CE) Connection
• Called CE because emitter is common to both VBB and VCC

8
BJT in Active Region (2)
•Base Emitter junction is forward biased
•Base Collector junction is reverse biased
•For a particular iB, iC is independent of RCC
⇒transistor is acting as current controlled current source (iC is
controlled by iB, and iC = β iB)
• Since the base emitter junction is forward biased, from Shockley
equation






−





= 1exp
T
BE
CSC
V
V
Ii

9
Early Effect and Early Voltage
• As reverse-bias across collector-base junction increases, width of the
collector-base depletion layer increases and width of the base decreases
(base-width modulation).
• In a practical BJT, output characteristics have a positive slope in forward-
active region; collector current is not independent of vCE.
• Early effect: When output characteristics are extrapolated back to point of
zero iC, curves intersect (approximately) at a common point vCE = -VA which
lies between 15 V and 150 V. (VA is named the Early voltage)
• Simplified equations (including Early effect):
iC
=IS
exp
vBE
VT


















1+
vCE
VA












βF
=βFO
1+
vCE
VA












iB
=
IS
βFO
exp
vBE
VT


















Chap 5 - 9admission.edhole.com

10
BJT in Active Region (3)
•Normally the above equation is never used to calculate iC, iB
Since for all small signal transistors vBE ≈ 0.7. It is only useful
for deriving the small signal characteristics of the BJT.
•For example, for the CE connection, iB can be simply
calculated as,
BB
BEBB
B
R
VV
i
−
=
or by drawing load line on the base –emitter side

11
Deriving BJT Operating points in
Active Region –An Example
In the CE Transistor circuit shown earlier VBB= 5V, RBB= 107.5
kΩ, RCC = 1 kΩ, VCC = 10V. Find IB,IC,VCE,β and the transistor
power dissipation using the characteristics as shown below
BB
BEBB
B
R
VV
I
−
=
By Applying KVL to the base emitter circuit
By using this equation along with the
iB / vBE characteristics of the base
emitter junction, IB = 40 µA
iB
100 µA
0
5V vBE

12
Deriving BJT Operating points in
Active Region –An Example (2)
By using this equation along with the iC /
vCE characteristics of the base collector
junction, iC = 4 mA, VCE = 6V
By Applying KVL to the collector emitter circuit
CC
CECC
C
R
VV
I
−
=
100
A40
mA4
I
I
B
C
=
µ
==β
Transistor power dissipation = VCEIC = 24 mW
We can also solve the problem without using the characteristics
if β and VBE values are known
iC
10 mA
0
20V vCE
100 µA
80 µA
60 µA
40 µA
20 µA

13
BJT in Cutoff Region
•Under this condition iB= 0
•As a result iC becomes negligibly small
•Both base-emitter as well base-collector junctions may be reverse
biased
•Under this condition the BJT can be treated as an off switch

14
BJT in Saturation Region
•Under this condition iC / iB < β in active region
•Both base emitter as well as base collector junctions are forward
biased
•VCE ≈ 0.2 V
•Under this condition the BJT can be treated as an on switch

15
•A BJT can enter saturation in the following ways (refer to
the CE circuit)
•For a particular value of iB,if we keep on increasing RCC
•For a particular value of RCC,if we keep on increasing iB
•For a particular value of iB,if we replace the transistor
with one with higher β
BJT in Saturation Region (2)

16
In the CE Transistor circuit shown earlier VBB= 5V, RBB= 107.5
kΩ, RCC = 10 kΩ, VCC = 10V. Find IB,IC,VCE,β and the transistor
power dissipation using the characteristics as shown below
BJT in Saturation Region – Example 1
Here even though IB is still 40 µA; from the output characteristics,
IC can be found to be only about 1mA and VCE ≈ 0.2V(⇒ VBC ≈ 0.5V
or base collector junction is forward biased (how?))
β = IC / IB = 1mA/40 µA = 25< 100
iC
10 mA
0
20V vCE
100 µA
80 µA
60 µA
40 µA
20 µA

17
In the CE Transistor circuit shown earlier VBB= 5V, RBB= 43 kΩ,
RCC = 1 kΩ, VCC = 10V. Find IB,IC,VCE,β and the transistor power
dissipation using the characteristics as shown below
Here IB is 100 µA from the input characteristics; IC can be found to be
only about 9.5 mA from the output characteristics and VCE ≈ 0.5V(⇒
VBC ≈ 0.2V or base collector junction is forward biased (how?))
β = IC / IB = 9.5 mA/100 µA = 95 < 100
Note: In this case the BJT is not in very hard saturation
Transistor power dissipation = VCEIC ≈ 4.7 mW

18
10 mA
Output Characteristics
iC
0
20V vCE
100 µA
80 µA
60 µA
40 µA
20 µA
iB
100 µA
0
5V vBE
Input Characteristics
(2)

19
In the CE Transistor circuit shown earlier VBB= 5V, VBE = 0.7V
RBB= 107.5 kΩ, RCC = 1 kΩ, VCC = 10V, β = 400. Find IB,IC,VCE,
and the transistor power dissipation using the characteristics as
shown below
A40
R
VV
I
BB
BEBB
B µ=
−
=
By Applying KVL to the base emitter circuit
Then IC = βIB= 400*40 µA = 16000 µA
and VCE = VCC-RCC* IC =10- 0.016*1000 = -6V(?)
But VCE cannot become negative (since current can flow only
from collector to emitter).
Hence the transistor is in saturation

20
BJT in Saturation Region – Example 3(2)
Hence VCE ≈ 0.2V
∴IC = (10 –0.2) /1 = 9.8 mA
Hence the operating β = 9.8 mA / 40 µA = 245

21
BJT Operating Regions at a Glance (1)

22
BJT Operating Regions at a Glance (2)

23
BJT Large-signal (DC) model

24
BJT ‘Q’ Point (Bias Point)
•Q point means Quiescent or Operating point
• Very important for amplifiers because wrong ‘Q’ point
selection increases amplifier distortion
•Need to have a stable ‘Q’ point, meaning the the operating
point should not be sensitive to variation to temperature or
BJT β, which can vary widely

25
Four Resistor bias Circuit for Stable ‘Q’
Point
≡
By far best circuit for providing stable bias point

26
Analysis of 4 Resistor Bias Circuit
≡
21
2
THB
RR
RVcc
VV
+
==
21
21
THB
RR
RR
RR
+
==

27
Applying KVL to the base-emitter circuit of the Thevenized
Equivalent form
VB - IB RB -VBE - IE RE = 0 (1)
Since IE = IB + IC = IB + βIB= (1+ β)IB (2)
Replacing IE by (1+ β)IB in (1), we get
Analysis of 4 Resistor Bias Circuit (2)
EB
BEB
B
R)1(R
VV
I
β++
−
= (3)
If we design (1+ β)RE >> RB (say (1+ β)RE >> 100RB)
Then
E
BEB
B
R)1(
VV
I
β+
−
≈ (4)

28
Analysis of 4 Resistor Bias Circuit (3)
E
BEB
EC
R
VV
II
−
≈= (5)And (for large β)
Hence IC and IE become independent of β!
Thus we can setup a Q-point independent of β which tends to
vary widely even within transistors of identical part number
(For example, β of 2N2222A, a NPN BJT can vary between
75 and 325 for IC = 1 mA and VCE = 10V)

29
4 Resistor Bias Circuit -Example
A 2N2222A is connected as shown
with R1 = 6.8 kΩ,R2 = 1 kΩ,RC = 3.3 kΩ,
RE = 1 kΩ and VCC = 30V. Assume
VBE = 0.7V.
Compute VCC and IC for β = i)100
and ii) 300

30
4 Resistor Bias Circuit –Example (1)
i) β = 100
V85.3
18.6
1*30
RR
RVcc
VV
21
2
THB =
+
=
+
==
Ω=
+
=
+
== k872.0
18.6
1*8.6
RR
RR
RR
21
21
THB
A92.30
1*101872.0
7.085.3
R)1(R
VV
I
EB
BEB
B µ=
+
−
=
β++
−
=
ICQ = βIB = 3.09 mA
IEQ = (1+ β)IB = 3.12 mA
VCEQ = VCC-ICRC-IERE = 30-3.09*3.3-3.12*1=16.68V

31
i) β = 300
V85.3
18.6
1*30
RR
RVcc
VV
21
2
THB =
+
=
+
==
Ω=
+
=
+
== k872.0
18.6
1*8.6
RR
RR
RR
21
21
THB
A43.10
1*301872.0
7.085.3
R)1(R
VV
I
EB
BEB
B µ=
+
−
=
β++
−
=
ICQ = 300IB = 3.13 mA
IEQ = (1+ β)IB = 3.14 mA
VCEQ = VCC-ICRC-IERE = 30-3.13*3.3-3.14*1=16.53V

32
The above table shows that even with wide variation
of β the bias points are very stable.
β = 100 β = 300
%
Change
VCEQ 16.68 V 16.53 V 0.9 %
ICQ 3.09 mA 3.13 mA 1.29 %

33
Four-Resistor Bias Network for BJT
VEQ
=VCC
R1
R1
+R2
REQ
=R1
R2
=
R1
R2
R1
+R2
VEQ
=REQ
IB
+VBE
+RE
IE
4=12,000IB
+0.7+16,000(βF
+1)IB
∴IB
=
VEQ
−VBE
REQ
+(βF
+1)RE
=
4V-0.7V
1.23×106Ω
=2.68µA
IC
=βF
IB
=201µA
IE
=(βF
+1)IB
=204 µA
VCE
=VCC
−RC
IC
−RE
IE
VCE
=VCC
− RC
+
RF
αF










IC
=4.32 V
F. A. region correct - Q-point is (201 µA, 4.32 V)
βF=75

34
Four-Resistor Bias Network for BJT
(cont.)
• All calculated currents > 0, VBC = VBE - VCE = 0.7 - 4.32
= - 3.62 V
• Hence, base-collector junction is reverse-biased,
and assumption of forward-active region operation
is correct.
• Load-line for the circuit is:
VCE
=VCC
−RC
+
RF
αF










IC
=12−38,200IC
The two points needed to plot the load
line are (0, 12 V) and (314 µA, 0).
Resulting load line is plotted on
common-emitter output characteristics.
IB = 2.7 µA, intersection of
corresponding characteristic with load
line gives Q-point.

35
Four-Resistor Bias Network for BJT:
Design Objectives
• We know that
• This implies that IB << I2, so that I1 = I2. So base current doesn’t disturb
voltage divider action. Thus, Q-point is independent of base current as
well as current gain.
• Also, VEQ is designed to be large enough that small variations in the
assumed value of VBE won’t affect IE.
• Current in base voltage divider network is limited by choosing I2 ≤ IC/5.
This ensures that power dissipation in bias resistors is < 17 % of total
quiescent power consumed by circuit and I2 >> IB for β > 50.
IE
=
VEQ
−VBE
−REQ
IB
RE
≅
VEQ
−VBE
RE
for REQ
IB
<<(VEQ
−VBE
)

36
Design Guidelines
• Choose Thévenin equivalent base voltage
• Select R1 to set I1 = 9IB.
• Select R2 to set I2 = 10IB.
• RE is determined by VEQ and desired IC.
• RC is determined by desired VCE.
VCC
4
≤VEQ
≤
VCC
2
R1
=
VEQ
9IB
R2
=
VCC
−VEQ
10IB
RE
≅
VEQ
−VBE
IC
RC
≅
VCC
−VCE
IC
−RE

37
Example
• Problem: Design 4-resistor bias circuit with given parameters.
• Given data: IC = 750 µA, βF = 100, VCC = 15 V, VCE = 5 V
• Assumptions: Forward-active operation region, VBE = 0.7 V
• Analysis: Divide (VCC - VCE) equally between RE and RC. Thus, VE = 5 V
and VC = 10 V
RC
=
VCC
−VC
IC
=6.67 kΩ
RE
=
VE
IE
=6.60 kΩ
VB
=VE
+VBE
=5.7 V
IB
=
IC
βF
=7.5 µA
I2
=10IB
= 75.0 µA
I1
=9IB
=67.5 µA
R1
=
VB
9IB
=84.4 kΩ
R2
=
VCC
−VB
10IB
=124 kΩ

38
Two-Resistor Bias Network for BJT:
Example
• Problem: Find Q-point for pnp transistor in 2-resistor bias circuit with
given parameters.
• Given data: βF = 50, VCC = 9 V
• Assumptions: Forward-active operation region, VEB = 0.7 V
• Analysis: 9=VEB
+18,000IB
+1000(IC
+IB
)
∴9=VEB
+18,000IB
+1000(51)IB
∴IB
=
9V−0.7V
69,000Ω
=120 µA
IC
=50IB
=6.01mA
VEC
=9−1000(IC
+IB
)=2.88 V
VBC
=2.18 V
Forward-active region operation is
correct Q-point is : (6.01 mA, 2.88 V)

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