Post on 12-Jul-2015
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
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NPN Bipolar Junction Transistor•One N-P (Base Collector) diode one P-N (Base Emitter) diode
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PNP Bipolar Junction Transistor•One P-N (Base Collector) diode one N-P (Base Emitter) diode
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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 thathandle low power) operating in active region (region where BJTs work as amplifiers)admission.edhole.com
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BJT in Active Region
Common Emitter(CE) Connection
• Called CE because emitter is common to both VBB and VCC
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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 Shockleyequation
−
= 1exp
T
BECSC V
VIi
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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=I
Sexp
vBE
VT
1+vCE
VA
βF=β
FO1+
vCE
VA
iB=
IS
βFO
expvBE
VT
Chap 5 - 9admission.edhole.com
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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
BEBBB R
VVi
−=
or by drawing load line on the base –emitter sideadmission.edhole.com
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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
BEBBB R
VVI
−=
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 vBEadmission.edhole.com
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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
CECCC R
VVI
−=
100A40
mA4
I
I
B
C =µ
==β
Transistor power dissipation = VCEIC = 24 mW
We can also solve the problem without using the characteristicsif β and VBE values are known
iC
10 mA
0
20V vCE
100 µA
80 µA
60 µA
40 µA
20 µA
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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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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
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•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)
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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
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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
Note: In this case the BJT is not in very hard saturation
Transistor power dissipation = VCEIC ≈ 4.7 mW
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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
BJT in Saturation Region – Example 2 (2)
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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
BJT in Saturation Region – Example 3
A40R
VVI
BB
BEBBB 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
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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 ‘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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Analysis of 4 Resistor Bias Circuit
≡
21
2THB RR
RVccVV
+==
21
21THB RR
RRRR
+==
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Applying KVL to the base-emitter circuit of the Thevenized Equivalent formVB - 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
BEBB R)1(R
VVI
β++−= (3)
If we design (1+ β)RE >> RB (say (1+ β)RE >> 100RB)
ThenE
BEBB R)1(
VVI
β+−
≈ (4)
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Analysis of 4 Resistor Bias Circuit (3)
E
BEBEC R
VVII
−≈= (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 tovary widely even within transistors of identical part number (For example, β of 2N2222A, a NPN BJT can vary between75 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. AssumeVBE = 0.7V. Compute VCC and IC for β = i)100and ii) 300
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4 Resistor Bias Circuit –Example (1)i) β = 100
V85.318.6
1*30
RR
RVccVV
21
2THB =
+=
+==
Ω=+
=+
== k872.018.6
1*8.6
RR
RRRR
21
21THB
A92.301*101872.0
7.085.3
R)1(R
VVI
EB
BEBB µ=
+−=
β++−=
ICQ = βIB = 3.09 mAIEQ = (1+ β)IB = 3.12 mA
VCEQ = VCC-ICRC-IERE = 30-3.09*3.3-3.12*1=16.68V
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i) β = 300
V85.318.6
1*30
RR
RVccVV
21
2THB =
+=
+==
Ω=+
=+
== k872.018.6
1*8.6
RR
RRRR
21
21THB
A43.101*301872.0
7.085.3
R)1(R
VVI
EB
BEBB µ=
+−=
β++−=
ICQ = 300IB = 3.13 mAIEQ = (1+ β)IB = 3.14 mA
VCEQ = VCC-ICRC-IERE = 30-3.13*3.3-3.14*1=16.53V
4 Resistor Bias Circuit –Example (2)
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4 Resistor Bias Circuit –Example (3)
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 %
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Four-Resistor Bias Network for BJT
€
VEQ
=VCC
R1
R1+R
2
REQ
=R1R
2=
R1R
2
R1+R
2
VEQ
=REQ
IB+V
BE+R
EIE
4=12,000IB+0.7+16,000(β
F+1)I
B
∴IB=
VEQ
−VBE
REQ
+(βF+1)R
E
= 4V-0.7V
1.23×106Ω=2.68 µA
IC=β
FIB=201 µA
€
IE=(β
F+1)I
B=204 µA
VCE
=VCC
−RCIC−R
EI
E
VCE
=VCC
− RC
+R
F
αF
IC=4.32 V
F. A. region correct - Q-point is (201 µA, 4.32 V)
€
βF=75
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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+R
F
αF
IC=12−38,200I
C
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.
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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=V
EQ−V
BE−R
EQIB
RE
≅V
EQ−V
BE
RE
for REQIB<<(V
EQ−V
BE)
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Four-Resistor Bias Network for BJT: 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.
€
VCC4
≤VEQ≤VCC
2
€
R1=
VEQ
9IB
€
R2=V
CC−V
EQ
10IB
€
RE≅V
EQ−V
BE
IC
€
RC≅V
CC−V
CE
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=
VCC
−VC
IC
=6.67 kΩ
RE=
VE
IE
=6.60 kΩ
VB=V
E+V
BE=5.7 V
IB= I
C
βF
=7.5 µA
€
I2=10I
B= 75.0 µA
I1= 9I
B= 67.5 µA
R1=
VB
9IB
= 84.4 kΩ
R2=
VCC
−VB
10IB
=124 kΩ
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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(I
C+I
B)
∴9=VEB
+18,000IB+1000(51)I
B
∴IB=9V−0.7V
69,000Ω=120 µA
IC=50I
B=6.01 mA
VEC
=9−1000(IC+I
B)=2.88 V
VBC
=2.18 VForward-active region operation is correct Q-point is : (6.01 mA, 2.88 V)
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