6. In what mode (active, cutoff and saturation) do …1 Key educational goals: Develop the basic...

1

Key educational goals:

Develop the basic principle of operation of a BJT. Classify the different

types of BJTs and analyze their applications.

Reading/Preparatory activities for class

i)Textbook: Chapter 13.

ii) Power-point file:BJT

Questions to guide your reading and to think about ahead of time.

1. Why is the rationale for different symbols for NPN and PNP BJT?

2. What are the three different modes of operation of a BJT?

3. What are the three reasons for a BJT to go into saturation?

4. Why should the bias point of a BJT be invariant to temperature, etc.changes?

5. What is the difference between a large signal BJT model and a small signal BJT model?

6. In what mode (active, cutoff and saturation) do amplifiers operate?

7. What type of an amplifier is a BJT based voltage regulator and why is it

better than a zener diode based voltage regulator?

8. Which type of BJT is more convenient to use for current source application and why?

2

Introduction

Chapter 1: Introduction and Chapter 2:Resistive circuits

The main concepts for the module

1. Analyze how a BJT conducts in active region even though the base collector junction is

reverse biased.

2. Evaluate physically how a BJT can get into saturation following the fluid-jet controlled

valve models.

3. Develop the concept of a load line to determine an operating point in a BJT circuit.

4. Analyze how the transistor variation can drive a BJT from active to saturation region

and vice-versa.

5.Evaluate the position of the bias point for the best possible amplifier operation.

6. Analyze the need for a common emitter amplifier to be different from emitter follower

amplifier?

5. Identify an application area where BJTs are made to work in saturation region.

3

Summary

The knowledge gained from this module will be useful for designing amplifiers for increasing

transducer signals, voltage regulators for supplying regulated dc voltage and simple logic circuits

(known as TTL or transistor transistor logic ).

For next time

We will next look into the Metal Oxide Junction Field effect transistor (MOSFET) which works like

a voltage controlled current source.

Sample test/exam questions/problems to help you study:

1. Is Q1 in the next slide (Fig. 1) in saturation or active region? The manufacturer suggest

an active region of 200. See flowchart on slide 5 to solve.

2. In Fig.2 (next slide) if V1 = 15V, R1 = 9.1 kΩ, R2 = 3.9 k, RE = 100Ω , = 75 for Q2; calculate

IB and the upper limit of RC for the above current source to work properly as a current source.

4

Fig.1

Fig.2

BJT 5

Flowchart to determine BJT’s region of operation

2. Use suggested

to

compute IC = IB

2. Compute

VCE = VCC - IC RC

Is VCE > 0.2?

3. BJT is in active or

quasi-saturation

region.

Keep as

suggested and use

IC

as in 2.

4. BJT is in

saturation.

Choose

VCE = 0.2 and

compute new

IC = (VCC - VCE)/RC

No

1. Find IB = (VCC -

VBE)/RB

5. Choose

new

= IC / IB

Yes

VCC RB

RC

+-VBE

IC

IB

VCE

+

-

BJT 6

Transistors

•Different types of transistors and their descriptions

•Classification of BJTs (NPN and PNP)

•Operational characteristics of a NPN BJT and analogy with a controlled valve

•BJT and

•Different operating regions of a BJT: active, cut-off and saturation with

examples

•Large signal (DC) model of BJT and setting up bias (Q) points

•Analysis of 4 resistor bias circuit

•Small signal model of a BJT

•Analysis of CE and Emitter follower amplifiers with examples

•Voltage regulator analysis with example

•PNP BJT current source with example

BJT 7

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

BJT 8

NPN Bipolar Junction Transistor•One N-P (Base Collector) diode one P-N (Base Emitter) diode

BJT 9

PNP Bipolar Junction Transistor

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

BJT 10

Analogy with Transistor :Fluid-jet

operated Valve

BJT 11

NPN BJT Current flow

BJT 12

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)

BJT 13

BJT in Active Region

Common Emitter(CE) Connection

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

BJT 14

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

BJT 15

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

1

V

VexpIi

T

BEESE

BJT 16

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 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

BJT 17

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

BJT 18


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 characteristics

if and VBE values are known

BJT 19

iB

100 A

0

5V vBE

iC

10 mA

0

20V vCE

100 A

80 A

60 A

40 A

20 A


Active Region –An Example (3)

Output CharacteristicsInput Characteristics

BJT 20

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

BJT 21

Analogy with Transistor Cutoff

Fluid-jet operated Valve

BJT 22

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

BJT 23

•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)

BJT 24

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

BJT 25

Analogy with Transistor Saturation

Fluid-jet operated Valve (2)

The valve is wide open; thus changing valve position a little

does not have much influence on the flow rate.

BJT 26

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 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

BJT 27

iB

100 A

0

5V vBE

Input Characteristics


(2)

iC

10 mA

0

20V vCE

100 A

80 A

60 A

40 A

20 A

BJT 28


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

BJT 29

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


(2)

BJT 30

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


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

BJT 31

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

BJT 32

BJT Operating Regions at a Glance (1)

BJT 33

BJT Operating Regions at a Glance (2)

BJT 34

BJT Large-signal (DC) model

BJT 35

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

BJT 36

Four Resistor bias Circuit for Stable ‘Q’

Point

By far best circuit for providing stable bias point

BJT 37

Analysis of 4 Resistor Bias Circuit

21

2THB

RR

RVccVV

21

21THB

RR

RRRR

BJT 38

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

Analysis of 4 Resistor Bias Circuit (2)

EB

BEBB

R)1(R

VVI

LP53

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

ThenE

BEBB

R)1(

VVI

LP54

BJT 39

Analysis of 4 Resistor Bias Circuit (3)

E

BEBEC

R

VVII

LP55And (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)

BJT 40

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

BJT 41

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 mA

IEQ = (1+ )IB = 3.12 mA

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

BJT 42

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 mA

IEQ = (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)

BJT 43

= 100 = 300

%

Change

VCEQ 16.68 V 16.53 V 0.9 %

ICQ 3.09 mA 3.13 mA 1.29 %

4 Resistor Bias Circuit –Example (3)

The above table shows that even with wide variation

of the bias points are very stable.

BJT 44

iC

7 mA

0

30V vCE

75 A

60 A

45 A

30 A

15 A

iC

7 mA

0

30V vCE

25 A

20 A

15 A

10 A

5 A

i) = 100 i) = 300

In both cases BJT is in Active Region. The BJT model on

the left is applicable for this dc bias circuit. However this

model is NOT APPLICABLE for the small signal ac

equivalent circuit drawn in next slide.

BJT 45

Common Emitter Amplifier

BJT 46

How does the CE Amplifier appears to

DC Source?

Why? Since all the capacitors appears as open to dc source

BJT 47

How does the CE Amplifier appears to

AC Source?

BJT 48

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

BJT 49

Computing r

1exp

T

BEQ

ESEQV

VIi

The Shockley equation for a base emitter junction working at a

stable ‘Q’ point is

Then

Since in a forward biased B-E junction

T

BEQ

T

ES

BEQ

EQ

V

Vexp

V

I

dV

dI

1V

Vexp

T

BEQ

LP56

LP57

BJT 50

Computing r (2)

1

V

Vexp

V

I

dV

dI

T

BEQ

T

ES

BEQ

EQ

LP58

= IEQ

T

EQ

BEQ

EQ

V

I

dV

dI LP59

BJT

51

Computing r (3)

Since VT = 26 mV at 3000 K

EQI

026.0r

at 3000 K

which describes the small variation in IBQ due to the small variation in

VBEQ. This is basically the reciprocal of the gradient of the IB versus

VBE charactersistics at the quiescent point as shown below.

BQdI

BEQdV

bi

bevr Define LP510

EQI

TV

EQdI

BEQdV

EQdI

BEQdV

BQI

BEQVr

BEQV

BJT 52

Compute AV, Zin etc. for

Common Emitter Amplifier using

“Greenboard”

BJT 53


Find Av, Av', Avo, Ai, Zin, G, Zo, vo' of the CE

Amplifier shown in the next slide. Temperature

=3000 K. VT = 26 mV. IEQ ≈ ICQ =3.13 mA,VT

=26 mV, VBEQ = 0.7 V.

BJT 54

+30 V

RE=

1 k

R1=6.8 k

R2=1 k

RC=3.3 k

RL

=2.7 kRs =100

0

VS

1kHz;

20mVpp


C1 =50F

C2 =50F

CE =50F

BJT 55

+3.13 V

+30 V

1 k

6.8 k

1 k

3.3 k

2.7 k100

0

VS

1kHz;

20mVpp


+3.85 V

17.32mVpp

+16.53 V

2.96Vpp

0

Vo

2.96Vpp

BJT 56

Vo

0 1 2 3 mS-1.6

0

1.6

V

10

0 1 2 3 mS-10

0mV

Vs

Input

Output

Common Emitter Amplifier Performance(Note the phase inversion)

BJT 57

Emitter Follower

BJT 58

Compute AV, Zin etc. for

Emitter Follower using “Greenboard”

BJT 59

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)

LoadVS

Practical Multistage Amplifier

BJT 60

Voltage Regulators(Derived from

Emitter Followers)

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)

(a) (b)

BJT 61

•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

How does the Regulator work

BJT 62

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

BJT 63

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.

BJT 64

Biasing PNP BJT

BJT 65

PNP BJT Current Source

BJT 66

PNP BJT Current Source

21

2THB

RR

R1VVV

21

21THB

RR

RRRR

BJT 67

Applying KVL to the base-emitter circuit of the Thevenized

Equivalent form

V1 - IB RB -VEB - IE RE-VB = 0

Analyzing PNP BJT Current Source

)1/(RR

VV1VI

BE

EBBE

By replacing IB= IE/(1+)

)1(RR

)VV1V(I

)1(II

EB

EBBEBC

Thus IC is independent of RC

BJT 68

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)

C

ECEEC

I

)sat(VRI1VR

or

If RC is above this value the transistor will be driven

into saturation

BJT 69

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.

73.29.31.9

9.3*1.9

RR

RRR

21

21B

k

V5.49.31.9

9.3*1VVB

mA2.72)1/(RR

VV1VI

BE

EBBE

mA95.0)1(

II E

B

BJT 70

Example of PNP BJT Current Source (3)

10125.71

2.01.0*2.7215

I

)sat(VRI1VR

C

ECEEC

IC =IB = 71.25 mA

6. In what mode (active, cutoff and saturation) do …1 Key educational goals: Develop the basic...

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Transcript of 6. In what mode (active, cutoff and saturation) do …1 Key educational goals: Develop the basic...