Transcript BJT DC.ppsx

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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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NPN BJT Current flow
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BJT  and 
•From the previous figure iE = iB + iC
•Define  = iC / iE
•Define  = iC / iB
•Then  = iC / (iE –iC) =  /(1- )
•Then iC =  iE ; iB = (1-) iE
•Typically   100 for small signal BJTs (BJTs that
handle low power) operating in active region (region
where BJTs work as amplifiers)
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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 =  iB)
• Since the base emitter junction is forward biased, from Shockley
equation
  VBE  
  1
iC  I CS exp 
  VT  
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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 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 exp BE
V
 T



S 



 v 

1 CE 

 V 


A 



FO


F  
1

vCE 
V



A 
iB 

v
exp BE
V
 T




FO 

IS





Chap 5 - 9 9
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,
VBB  VBE
iB 
R BB
or by drawing load line on the base –emitter side
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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
k, RCC = 1 k, VCC = 10V. Find IB,IC,VCE, and the transistor
power dissipation using the characteristics as shown below
By Applying KVL to the base emitter circuit
iB
100 A
VBB  VBE
IB 
R BB
0
5V vBE
By using this equation along with the
iB / vBE characteristics of the base
emitter junction, IB = 40 A
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Deriving BJT Operating points in
Active Region –An Example (2)
iC
10 mA
By Applying KVL to the collector emitter circuit
VCC  VCE
100 A
IC 
R CC
80 A
60 A
40 A
20 A
0
20V vCE
By using this equation along with the iC /
vCE characteristics of the base collector
junction, iC = 4 mA, VCE = 6V
I C 4mA
 
 100
I B 40A
Transistor power dissipation = VCEIC = 24 mW
We can also solve the problem without using the characteristics
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if  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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BJT in Saturation Region
•Under this condition iC / iB   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 
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BJT in Saturation Region – Example 1
In the CE Transistor circuit shown earlier VBB= 5V, RBB= 107.5
k, RCC = 10 k, VCC = 10V. Find IB,IC,VCE, and the transistor
power dissipation using the characteristics as shown below
Here even though IB is still 40 A; 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
10 mA
100 A
80 A
60 A
40 A
20 A
0
 = IC / IB = 1mA/40 A = 25 100
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20V vCE
BJT in Saturation Region – Example 2
In the CE Transistor circuit shown earlier VBB= 5V, RBB= 43 k,
RCC = 1 k, VCC = 10V. Find IB,IC,VCE, and the transistor power
dissipation using the characteristics as shown below
Here IB is 100 A 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?))
 = IC / IB = 9.5 mA/100 A = 95  100
Transistor power dissipation = VCEIC  4.7 mW
Note: In this case the BJT is not in very hard saturation
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BJT in Saturation Region – Example 2
(2)
iC
iB
100 A
10 mA
100 A
80 A
60 A
40 A
20 A
0
0
5V vBE
Input Characteristics
20V vCE
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 k, RCC = 1 k, VCC = 10V,  = 400. Find IB,IC,VCE,
and the transistor power dissipation using the characteristics as
shown below
By Applying KVL to the base emitter circuit
VBB  VBE
IB 
 40A
R BB
Then IC = IB= 400*40 A = 16000 A
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  = 9.8 mA / 40 A = 245
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BJT Operating Regions at a Glance (1)
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BJT Operating Regions at a Glance (2)
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BJT Large-signal (DC) model
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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 , 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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