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
1
NPN Bipolar Junction Transistor
•One N-P (Base Collector) diode one P-N (Base Emitter) diode
2
1
PNP Bipolar Junction Transistor
•One P-N (Base Collector) diode one N-P (Base Emitter) diode
3
NPN BJT Current flow
4
2
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)
5
BJT in Active Region
Common Emitter(CE) Connection
• Called CE because emitter is common to both VBB and VCC
6
3
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
⎡ ⎛V
iC = I CS ⎢exp⎜⎜ BE
⎣ ⎝ VT
⎞ ⎤
⎟⎟ − 1⎥
⎠ ⎦
7
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 forwardactive 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):
⎡
⎛v
⎢
iC = I S ⎢exp ⎜ BE
⎢
⎜V
⎢⎣
⎝ T
⎞⎥⎤⎡⎢ v
⎟⎟⎥⎥⎢1+ CE
⎢
⎠⎥⎦⎢⎣ V A
⎤
⎥
⎥
⎥
⎥⎦
⎡
⎢
β F = β FO ⎢1+
⎢
⎢⎣
⎤
⎥
⎥
⎥
A ⎥⎦
v CE
V
iB =
⎡
⎢
⎢
⎢
FO ⎢⎣
IS
β
⎛v
exp ⎜ BE
⎜V
⎝ T
⎞⎥⎤
⎟⎟⎥⎥
⎠⎥⎦
Chap 5 - 8 8
4
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,
V − VBE
i B = BB
R BB
or by drawing load line on the base –emitter side
9
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
By Applying KVL to the base emitter circuit
iB
100 μA
IB =
0
5V vBE
VBB − VBE
R BB
By using this equation along with the
iB / vBE characteristics of the base
emitter junction, IB = 40 μA
10
5
Deriving BJT Operating points in
Active Region –An Example (2)
iC
10 mA
By Applying KVL to the collector emitter circuit
V − VCE
100 μA
I C = CC
R CC
80 μA
60 μA
40 μA
20 μA
By using this equation along with the iC /
vCE characteristics of the base collector
junction, iC = 4 mA, VCE = 6V
0
20V vCE
β=
I C 4mA
=
= 100
I B 40μA
Transistor power dissipation = VCEIC = 24 mW
We can also solve the problem without using the characteristics
11
if β and VBE values are known
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
12
6
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
13
BJT in Saturation Region (2)
•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
14
7
BJT in Saturation Region – Example 1
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
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?))
iC
100 μA
10 mA
80 μA
60 μA
40 μA
20 μA
β = IC / IB = 1mA/40 μA = 25< 100
0
15
20V vCE
BJT in Saturation Region – Example 2
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
Transistor power dissipation = VCEIC ≈ 4.7 mW
Note: In this case the BJT is not in very hard saturation
16
8
BJT in Saturation Region – Example 2
(2)
iC
iB
100 μA
10 mA
100 μA
80 μA
60 μA
40 μA
20 μA
0
0
20V vCE
5V vBE
Input Characteristics
Output Characteristics
17
BJT in Saturation Region – Example 3
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
By Applying KVL to the base emitter circuit
IB =
VBB − VBE
= 40μA
R BB
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
18
9
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
19
BJT Operating Regions at a Glance (1)
20
10
BJT Operating Regions at a Glance (2)
21
BJT Large-signal (DC) model
22
11
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
23
Four Resistor bias Circuit for Stable ‘Q’
Point
≡
By far best circuit for providing stable bias point
24
12
Analysis of 4 Resistor Bias Circuit
≡
VB = VTH =
Vcc R 2
R1 + R 2
R B = R TH =
R1 R 2
R1 + R 2
25
Analysis of 4 Resistor Bias Circuit (2)
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
IB =
VB − VBE
R B + (1 + β)R E
(3)
If we design (1+ β)RE >> RB (say (1+ β)RE >> 100RB)
Then
IB ≈
VB − VBE
(1 + β)R E
(4)
26
13
Analysis of 4 Resistor Bias Circuit (3)
And I C = I E ≈
VB − VBE
(for large β) (5)
RE
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)
27
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
28
14
4 Resistor Bias Circuit –Example (1)
i) β = 100
Vcc R 2 30 *1
=
= 3.85V
R 1 + R 2 6.8 + 1
VB = VTH =
R B = R TH =
IB =
R1 R 2
6.8 *1
=
= 0.872kΩ
R 1 + R 2 6.8 + 1
VB − VBE
3.85 − 0.7
=
= 30.92μA
R B + (1 + β)R E 0.872 + 101*1
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
29
4 Resistor Bias Circuit –Example (2)
i) β = 300
VB = VTH =
Vcc R 2 30 *1
=
= 3.85V
R 1 + R 2 6.8 + 1
R1 R 2
6.8 *1
=
= 0.872kΩ
R 1 + R 2 6.8 + 1
VB − VBE
3.85 − 0.7
IB =
=
= 10.43μA
R B + (1 + β)R E 0.872 + 301*1
R B = R TH =
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
30
15
4 Resistor Bias Circuit –Example (3)
β = 100
β = 300
%
Change
VCEQ
16.68 V
16.53 V
0.9 %
ICQ
3.09 mA
3.13 mA
1.29 %
The above table shows that even with wide variation
of β the bias points are very stable.
31
Four-Resistor Bias Network for BJT
V EQ = V CC
β F = 75
R1
R EQ = R1 R 2 =
R1 R 2
R1 + R 2
R1 + R 2
V EQ = R EQ I B + V BE + R E I E
4 = 12 ,000 I B + 0 .7 + 16 ,000 (β F + 1)I B
V EQ − V BE
4 V - 0.7V
∴ IB =
=
= 2 .68 μ A
R EQ + (β F + 1)R E 1 .23 ×10 6 Ω
I C = β F I B = 201 μ A
IE =(βF +1)IB = 204 μA
VCE =VCC − RC IC − RE I E
⎛
⎜
VCE =VCC − ⎜ RC +
⎜⎜
⎝
⎞
RF ⎟
⎟
α F ⎟⎟⎠
IC = 4.32 V
F. A. region correct - Q-point is (201 μA, 4.32 V)
32
16
Four-Resistor Bias Network for BJT
(cont.)
• All calculated currents > 0, VBC = VBE - VCE = 0.7 4.32 = - 3.62 V
R
• Hence, base-collector junction
V = V −isRreverse-biased,
+
I = 12 − 38 ,200 I
α
and assumption of forward-active region operation
The two points needed to plot the load
is correct.
line are (0, 12 V) and (314 μA, 0).
• Load-line for the circuitResulting
is: load line is plotted on
CE
CC
⎛
⎜
⎜
⎜⎜
⎝
C
⎞
⎟
F ⎟
⎟⎟ C
F ⎠
C
common-emitter output characteristics.
IB = 2.7 μA, intersection of
corresponding characteristic with load
line gives Q-point.
33
Four-Resistor Bias Network for BJT:
Design Objectives
• We know that
IE =
V EQ − V BE − R EQ I B
RE
≅
V EQ − V BE
RE
for
R EQ I B << (V EQ − V BE )
• 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.
34
17
Four-Resistor Bias Network for BJT:
Design Guidelines
V CC
V
≤ V EQ ≤ CC
4
2
• Choose Thévenin equivalent base voltage
• Select R1 to set I1 = 9IB.
• Select R2 to set I2 = 10IB.
R1 =
V EQ
9IB
R2 =
V CC − V EQ
10 I B
• RE is determined by VEQ and desired IC.
• RC is determined by desired VCE.
RE ≅
RC ≅
V EQ − V BE
V CC − V CE
IC
IC
− RE
35
Four-Resistor Bias Network for BJT:
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 =
RE =
V CC − V C
IC
VE
= 6 .67 k Ω
= 6 .60 k Ω
IE
V B = V E + V BE = 5 .7 V
I
I B = C = 7 .5 μ A
βF
I2 =10IB = 75.0 μA
I1 = 9IB = 67.5 μA
R1 =
VB
= 84.4 kΩ
9I B
V −V
R2 = CC B =124 kΩ
10I B
36
18
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 = V EB + 18 ,000 I B + 1000 (I C + I B )
∴ 9 = V EB + 18 ,000 I B + 1000 (51 )I B
9V − 0.7V
= 120 μ A
69 ,000 Ω
I C = 50 I B = 6.01 mA
∴ IB =
V EC = 9 − 1000 (I C + I B ) = 2.88 V
V BC = 2.18 V
Forward-active region operation is
correct Q-point is : (6.01 mA, 2.88 V)
37
19