Transistors

1
Transistors

• They are unidirectional current carrying devices
  like diodes 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
• Diodes and Transistors are the basic building
  blocks of the multibillion dollar semiconductor
  industries

2
NPN Bipolar Junction Transistor

• One N-P (Base Collector) diode one P-N (Base
  Emitter) diode

3
PNP Bipolar Junction Transistor

• One P-N (Base Collector) diode one N-P (Base
  Emitter) diode

4
Analogy with Transistor Fluid-jet operated Valve
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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

8
Analogy with Transistor in Active Region
Fluid-jet operated Valve
In active region this stopper does not really
have noticeable effect on the flow rate
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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

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BJT in Active Region (3)

• Since iB (1-?) iE , the previous equation can
  be rewritten as

• Normally the above equation is never used to
  calculate 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,

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
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)
By Applying KVL to the collector emitter circuit
By using this equation along with the iC / vCE
characteristics of the base collector junction,
iC 4 mA, VCE 6V
Transistor power dissipation VCEIC 24 mW
We can also solve the problem without using the
characteristics if ? and VBE values are known
13
Deriving BJT Operating points in Active Region
An Example (3)
Output Characteristics
Input Characteristics
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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

15
Analogy with Transistor Cutoff Fluid-jet operated
Valve
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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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Analogy with Transistor in Saturation Region
Fluid-jet operated Valve(1)
This stopper is almost closed thus valve
position does not have much influence on the flow
rate
19
Analogy with Transistor Saturation Fluid-jet
operated Valve (2)
The valve is wide open changing valve position
a little bit does not have much influence on
the flow rate.
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BJT in Saturation Region Example 1
In the CE Transistor circuit shown earlier VBB
5V, RBB 107.5 k?, RCC 100 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 / IB 1mA/40 ?A 25? 100
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BJT in Saturation Region Example 1 (2)
Input Characteristics
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BJT in Saturation Region Example 2
In the CE Transistor circuit shown earlier VBB
5V, RBB 50 k?, RCC 1 k?, VCC 10V. Find
IB,IC,VCE,? and the transistor power dissipation
using the characteristics as shown below
Here IB is 80 ?A from the input characteristics
IC can be found to be only about 7.9 mA from
the output characteristics and VCE ? 0.5V(? VBC
? 0.2V or base collector junction is forward
biased (how?))
? IC / IB 7.9 mA/80 ?A 98.75 ? 100
Transistor power dissipation VCEIC ? 4 mW
Note In this case the BJT is not in very hard
saturation
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BJT in Saturation Region Example 2 (2)
10 mA
Input 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
Then IC ?IB 40040 ?A 16000 ?A and VCE
VCC-RCC IC 10- 0.0161000 -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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Analysis of 4 Resistor Bias Circuit
?
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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 LP51 Since IE IB IC IB ?IB
(1 ?)IB LP52 Replacing IE by (1 ?)IB in
LP51, we get
LP53
If we design (1 ?)RE ?? RB (say (1 ?)RE ??
100RB)
Then
LP54
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Analysis of 4 Resistor Bias Circuit (3)
LP55
And
(for large ?)
Hence IC and IE become independent of ?!
Thus we can setup a Q-point independent of ?
which tends to vary widely even within
transistors of identical part number (For
example, ? 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
k?, R2 1 k?, RC 3.3 k?, RE 1 k? and VCC
30V. Assume VBE 0.7V. Compute VCC and IC for ?
i)100 and ii) 300
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4 Resistor Bias Circuit Example (1)
i) ? 100
ICQ ?IB 3.09 mA IEQ (1 ?)IB 3.12 mA
VCEQ VCC-ICRC-IERE 30-3.093.3-3.12116.68V
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4 Resistor Bias Circuit Example (2)
i) ? 300
ICQ 300IB 3.13 mA IEQ (1 ?)IB 3.14 mA
VCEQ VCC-ICRC-IERE 30-3.133.3-3.14116.53V
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4 Resistor Bias Circuit Example (3)
 
 
The above table shows that even with wide
variation of ? the bias points are very stable.
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Common Emitter Amplifier
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How does the CE Amplifier appears to
DC Source?
Why? Since all the capacitors appears as open to
dc source
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How does the CE Amplifier appears to
AC Source?
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Answer

• C1, C2, CE are chosen such that 1/?C1, 1/?C2,
  1/?CE ? 0, that is all capacitors appear as
  short to the AC source
• The DC source appears as a short to the AC
  source
• The base-emitter junction appears a resistance r?
  to the small signal variation around Q point
• The BJT model appears as shown below

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Computing r?
The Shockley equation for a base emitter junction
working at a stable Q point is
LP56
Then
LP57
describes the small variation in IEQ due to the
small variation in VBEQ
Since in a forward biased B-E junction
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Computing r? (2)
LP58
IEQ
LP59
?
LP510
Define
LP511
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Computing r? (3)
Since VT 26 mV at 3000 K
at 3000 K
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Compute AV, Zin etc. for Common Emitter Amplifier
using Greenboard
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Common Emitter Amplifier
30 V
6.8 kW
3.3 kW
2.7 kW
100W
1 kW
1 kW
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Common Emitter Amplifier Performance
(Note the phase inversion)
Input
Output
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Emitter Follower
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Compute AV, Zin etc. for Emitter Follower using
Greenboard
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Practical Multistage Amplifier
Emitter Follower (For high input impedance
so that the source does not get loaded)
Common Emitter Amplifier (For voltage gain)
Emitter Follower (For low output impedance
so that load cannot affect amplifier gain)
Load
VS
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Voltage Regulators(Derived from Emitter Followers)
(a)
(b)
The similarity between Emitter follower and
voltage regulator can be readily appreciated by
redrawing (a) as (b) without the capacitor
(mainly used for improving transient
performance)
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How does the Regulator work

• Note that Vo VZ VBE
• As Vo tries to change, VBE is disturbed since VZ
  is constant
• This causes IB the base current of Q1 to change
• A change of IB causes VCE to adjust in such a way
  so as to
• compensate for the change in Vo

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Voltage Regulator Example

• If V1 20V, RL 15 ?, RS 680 ?, Vz 10V, ?
  80 calculate
• Iz and power dissipation in Q1
• Vo VZ VBE 10-0.7 9.3V
• ?IL Vo/ RL 9.3/15 0.62A Emitter current
• ?IB IL/(1 ?) 0.62/81 7.65 mA
• Now Is (V1- Vz)/ Rs 10/680 14.71 mA
• Iz Is- IB (14.71-7.65) mA 7.06 mA

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Voltage Regulator Example(2)

• Now IE IL 0.62 A
• IC IE IB ? IE 0.62 A
• Now VCE V1 Vo 20 9.3 10. 7V
• PD Q1 dissipation VCEIC 10.7 0.62 6.63W
• Hence Q1 will definitely need a heatsink.

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Biasing PNP BJT
?
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PNP BJT Current Source
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PNP BJT Current Source
?
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Analyzing PNP BJT Current Source
Applying KVL to the base-emitter circuit of the
Thevenized Equivalent form V1 - IB RB -VEB - IE
RE-VB 0
By replacing IB IE/(1?)
?
?
Thus IC is independent of RC
59
Analyzing PNP BJT Current Source (2)
The circuit works only in the active region Thus
the current source works until IERE ICRC ? V1
VEC (sat)
or
If RC is above this value the transistor will be
driven into saturation
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Example of PNP BJT Current Source (2)
If V1 15V, R1 9.1 k?, R2 3.9 k?, RE 100
?, ? 75 calculate IB and the upper limit of RC
for the above current source to work properly as
a current source.
k?
?
?
61
Example of PNP BJT Current Source (3)
IC ?IB 71.25 mA
?