1

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

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

NPN BJT Current flow

5

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)

6

BJT in Active Region

Common Emitter(CE) Connection

• Called CE because emitter is common to both VBB and VCC

7

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

1exp

T

BECSC V

VIi

8

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):

iCI

Sexp

vBE

VT

1v

CE

VA

F

FO1

vCE

VA

iB

IS

FO

expv

BE

VT

Chap 5 - 8

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,

BB

BEBBB R

VVi

or by drawing load line on the base –emitter side

10

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

iB

100 A

0

5V vBE

11

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

iC

10 mA

0

20V vCE

100 A

80 A

60 A

40 A

20 A

12

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

13

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

14

•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)

15

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

BJT in Saturation Region – Example 1

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

iC

10 mA

0

20V vCE

100 A

80 A

60 A

40 A

20 A

16

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

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

Transistor power dissipation = VCEIC 4.7 mW

17

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)

18

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

19

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

20

BJT Operating Regions at a Glance (1)

21

BJT Operating Regions at a Glance (2)

22

BJT Large-signal (DC) model

23

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

24

Four Resistor bias Circuit for Stable ‘Q’ Point

By far best circuit for providing stable bias point

25

Analysis of 4 Resistor Bias Circuit

21

2THB RR

RVccVV

21

21THB RR

RRRR

26

Applying KVL to the base-emitter circuit of the Thevenized Equivalent formVB - IB RB -VBE - IE RE = 0 (1)

Since IE = IB + IC = IB + IB= (1+ )IB (2)

Replacing IE by (1+ )IB in (1), we get

Analysis of 4 Resistor Bias Circuit (2)

EB

BEBB R)1(R

VVI

(3)

If we design (1+ )RE RB (say (1+ )RE 100RB)

ThenE

BEBB R)1(

VVI

(4)

27

Analysis of 4 Resistor Bias Circuit (3)

E

BEBEC R

VVII

(5)And (for large )

Hence IC and IE become independent of !

Thus we can setup a Q-point independent of which tends tovary widely even within transistors of identical part number (For example, of 2N2222A, a NPN BJT can vary between75 and 325 for IC = 1 mA and VCE = 10V)

28

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. AssumeVBE = 0.7V. Compute VCC and IC for = i)100and ii) 300

29

4 Resistor Bias Circuit –Example (1)i) = 100

V85.318.6

1*30

RR

RVccVV

21

2THB

k872.018.6

1*8.6

RR

RRRR

21

21THB

A92.301*101872.0

7.085.3

R)1(R

VVI

EB

BEBB

ICQ = IB = 3.09 mAIEQ = (1+ )IB = 3.12 mA

VCEQ = VCC-ICRC-IERE = 30-3.09*3.3-3.12*1=16.68V

30

i) = 300

V85.318.6

1*30

RR

RVccVV

21

2THB

k872.018.6

1*8.6

RR

RRRR

21

21THB

A43.101*301872.0

7.085.3

R)1(R

VVI

EB

BEBB

ICQ = 300IB = 3.13 mAIEQ = (1+ )IB = 3.14 mA

VCEQ = VCC-ICRC-IERE = 30-3.13*3.3-3.14*1=16.53V

4 Resistor Bias Circuit –Example (2)

31

 

4 Resistor Bias Circuit –Example (3)

The above table shows that even with wide variation of the bias points are very stable.

 

= 100  = 300%

Change 

VCEQ 16.68 V 

16.53 V 

0.9 %

ICQ 

3.09 mA 

3.13 mA 

1.29 %

 

32

Four-Resistor Bias Network for BJT

VEQ

VCC

R1

R1R

2

REQ

R1

R2

R1R

2

R1R

2

VEQ

REQ

IBV

BER

EI

E

412,000IB0.716,000(

F1)I

B

IB

VEQ

VBE

REQ

(F1)R

E

4V-0.7V

1.231062.68 A

IC

FI

B201 A

IE(

F1)I

B204 A

VCE

VCC

RCI

C R

EI

E

VCE

VCC

RC

RF

F

I

C4.32 V

F. A. region correct - Q-point is (201 A, 4.32 V)

F 75

33

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

RF

F

I

C12 38,200I

C

The two points needed to plot the load line are (0, 12 V) and (314 A, 0). Resulting load line is plotted on common-emitter output characteristics.

IB = 2.7 A, intersection of corresponding characteristic with load line gives Q-point.

34

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 > 50.

IE

VEQ

VBE

REQ

IB

RE

V

EQ V

BE

RE

for REQ

IB(V

EQ V

BE)

35

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

VCC2

R1

VEQ

9IB

R2

VCC

VEQ

10IB

RE

VEQ

VBE

IC

RC

VCC

VCE

IC

RE

36

Four-Resistor Bias Network for BJT: Example

• Problem: Design 4-resistor bias circuit with given parameters.• Given data: IC = 750 A, F = 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 k

RE

VE

IE

6.60 k

VBV

EV

BE5.7 V

IBI

C

F

7.5 A

I210I

B75.0 A

I19I

B67.5 A

R1

VB

9IB

84.4 k

R2

VCC

VB

10IB

124 k

37

Two-Resistor Bias Network for BJT: Example

• Problem: Find Q-point for pnp transistor in 2-resistor bias circuit with given parameters.• Given data: F = 50, VCC = 9 V• Assumptions: Forward-active operation region, VEB = 0.7 V• Analysis:

9VEB

18,000IB1000(I

CI

B)

9VEB

18,000IB1000(51)I

B

IB9V 0.7V

69,000120 A

IC50I

B6.01 mA

VEC

9 1000(ICI

B)2.88 V

VBC

2.18 VForward-active region operation is correct Q-point is : (6.01 mA, 2.88 V)