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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

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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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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 thathandle 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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Analogy with Transistor in Active Region: Fluid-jet operated Valve

In active region this stopper does not really havenoticeable 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 Shockleyequation

1

V

VexpIi

T

BEESE

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BJT in Active Region (3)

1

V

VexpI)-(1 i

T

BEESB

•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 usefulfor 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 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

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 A

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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

Transistor power dissipation = VCEIC = 24 mW

We can also solve the problem without using the characteristicsif and VBE values are known

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iB

100 A

0

5V vBE

iC

10 mA

0

20V vCE

100 A

80 A

60 A

40 A

20 A

Deriving BJT Operating points in Active Region –An Example (3)

Output CharacteristicsInput 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

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Analogy with Transistor CutoffFluid-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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•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

BJT in Saturation Region (2)

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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

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Analogy with Transistor SaturationFluid-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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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

BJT in Saturation Region – Example 1

Here even though IB is still 40 A; from the output characteristicsIC 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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iB

100 A

0

5V vBE

Input Characteristics

BJT in Saturation Region – Example 1 (2)

iC

10 mA

0

20V vCE

100 A

80 A

60 A

40 A

20 A

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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

Note: In this case the BJT is not in very hard saturation

Transistor power dissipation = VCEIC 4 mW

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10 mA

Output Characteristics

iC

0

20V vCE

100 A

80 A

60 A

40 A

20 A

iB

100 A

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 k, RCC = 1 k, VCC = 10V, = 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

By Applying KVL to the base emitter circuit

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