# Transistors NPN Bipolar Junction Transistor

advertisement ```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
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6
BJT in Saturation Region
•Under this condition iC / iB &lt; 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
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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
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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&lt; 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 &lt; 100
Transistor power dissipation = VCEIC ≈ 4.7 mW
Note: In this case the BJT is not in very hard saturation
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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
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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
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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
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BJT Operating Regions at a Glance (1)
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10
BJT Operating Regions at a Glance (2)
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BJT Large-signal (DC) model
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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
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Four Resistor bias Circuit for Stable ‘Q’
Point
≡
By far best circuit for providing stable bias point
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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
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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 &gt;&gt; RB (say (1+ β)RE &gt;&gt; 100RB)
Then
IB ≈
VB − VBE
(1 + β)R E
(4)
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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)
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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
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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
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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
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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.
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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 &times;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)
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16
Four-Resistor Bias Network for BJT
(cont.)
• All calculated currents &gt; 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.
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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 &lt;&lt; (V EQ − V BE )
• This implies that IB &lt;&lt; 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 &lt; 17 % of total
quiescent power consumed by circuit and I2 &gt;&gt; IB for β &gt; 50.
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17
Four-Resistor Bias Network for BJT:
Design Guidelines
V CC
V
≤ V EQ ≤ CC
4
2
• Choose Th&eacute;venin 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
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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
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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)
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