2b BJT fundamentals

BQ ICQ VCEQ  To determine the Q-point on dc load line, we must know: -- the biasing base current, IB from the biasing or input circuit; -- the dc load line of the transistor circ

DC and AC Load Line

Analysis

1

Transistor Amplifiers

 Although in modern electronics, amplifiers are more

conveniently implemented using integrated circuits (IC) like ope rational amplifiers , understanding of transistor amplifiers is usef

ul as

 they serve as the building blocks of integrated-circuit amplifiers

 they are used in high frequency applications e.g in radio

communications circuits, where discrete transistor amplifiers are stil

l popularly in use

 The analysis of BJT as amplifier is studied They are divided

into

 dc biasing in amplifier circuits,

 ac small-signal analysis by modeling the ac amplifier as a two-port

network

 ac large-signal analysis which uses ac load line to determine the

maximum unclipped

DC bias provides for proper operation of an amplifier If

an amplifier is not biased correctly, it can go into saturation or cutoff when an input signal is applied.

Linear operation:

Output voltage limited by cutoff:

Output voltage limited by saturation:

The positive portion of output signal has been clipped due to transistor’s cutoff.

The negative portion of output signal has been clipped due to transistor’s saturation.

 Input loop consists of V BB ,

DC bias circuit of CE amplifier

The purpose of the dc biasing circuit is to set up

the initial dc values of I B , I C , and V CE

 There are three important observations to be made from

the dc bias circuit:

a) When conducting, the base-emitter junction acts as a

forward biased diode with forward current I B Therefore, V BE = 0.7 V

b) By applying KCL at the emitter terminal (E),

I E = I B + I C

c) The collector current is represented by a dependent

current source Because the amount of collector curren

t, I C depends on the base current, I B This relationship is : I C = β DC I B

It is the current gain feature that enables BJTs to be used for amplifying signals.

Biasing Equation:

V BB = R B I B + V BE + R E I E

DC biasing circuit

 Since the transistor is operating in the active region, the emitter

current can be expressed by:

Or I B = (V BB –V BE ) / (R B + β DC R E ) This is the biasing base current.

 I C = β DC I B - Transistor operates in the active region

 Applying KVL around the output circuit we obtain the dc load line

equation:

V CC = R C I C + V CE + R E I E For β DC  50, assume I E = I C,

V CE = V CC – (R C + R E ) I C This is the transistor collector-emitter voltage.

 The operating point (Q point) of the transistor is at (V CE , I C ).

DC biasing circuit (cont’d)

Q-point

 IC and VCE represents the operating-point of the transistor on the

output characteristic It is also known as the Quiescent Point (Q-point) or the bias-point of the transistor, Q-point (VCEQ, ICQ )

 The following are the same:

• Biasing point

• Quiescent point

• Operating point (OP)

• DC point

 Occasionally, a subscript Q is added to the

current or the voltage variables so to indicate

the Q-point values

 The Q-point (quiescent-point) specifies the dc output current IC and the dc output voltage VCE when no ac signal is superimposed at the i nput of the amplifier.

 Determined by using transistor output characteristic and DC load line

The silicon transistor shown

below (β DC = 200) is used in an

amplifier circuit having a base bi

as resistor R B = 10kΩ, a base volt

age source V BB = 10V, a collector

resistor R C = 100Ω, an emitter re

sistor R E = 680Ω and a d.c power

supply V CC = 20V Determine the

operating point (V CEQ , I CQ )of the t

VCC = RCIC + VCE + REIE20V = 100ΩIC + VCE + 680ΩIE

Since β DC is large, I E = I C

VCE= 20V – 100xIC – 680IE

VCE= 20V –100x12.74mA – 680 x 12.74mA = 10.06V

The operating point of the transistor is at (V CEQ = 10.06V, I CQ = 12.74mA).

Potentials at base, emitter and collector

Exercise 1

IC

VCC20V

RB 52KΩ

VCE

RE2.2KΩ

Find IB, IC, VCE and determine the

operating region of this transistor

and indicate Q-point.

Solution

Step 1: Determine IB Apply KVL around the input circuit:

VBB = RBIB + VBE + βDC IB RE5V = 52kΩIB+0.7V+100 x 2.2KΩ IB

IB=(5V–0.7V)/(52kΩ+100 x 2.2KΩ) = 15.8 A

Step 2: Determine the collector current IC.

IC= βDC IB= 100 x 15.8A= 1.58 mA

Step 3: Determine VCE Applying KVL around the output circuit

VCC = RCIC + VCE + REIE20V = 4.7KΩIC + VCE + 2.2KΩIE

Since βDC is large, IE = IC

VCE=20V – 4.7KΩIC – 2.2KΩIC

VCE=20V–4.7KΩ x 1.58mA–2.2KΩ x1.58mA = 9.098V

Because VBE=0.7V>0 BE forward biased ∴VBC=VBE–VCE=0.7–9.098=–8.398V<0

-BC reverse biased ∴this transistor is in active region

The Q-point of the transistor

is at:

(V CEQ = 9.098V, I CQ = 1.58mA).

DC=100

RB 47KΩ

IB

VBB

4V

RC2.2KΩ

RE1.8KΩ

CB

E

Solution:

Step 1: find I B

IB = (VBB – VBE) / (RB + βDC RE) = (4–0.7)/(47KΩ+1001.8KΩ) =14.54 A

DC operating point (Q-point)

 A transistor must first be dc biased before it can be operated as an ac

signal amplifier

 A transistor, like a diode, allows current to flow only in one direction

In order to reproduce and amplify a fluctuating input current signal, th

e transistor must first be input with a dc base current such that the flu ctuating signal can be imposed correctly onto the base current.

DC operating point (Q-point) (cont’d)

signal at the base terminal is amplified, and its wave-shape i

s accurately reproduced in the collector current.

 The output signal i C has a larger amplitude than the input

signal i B.

and voltage conditions are established.

 The dc collector current I C and the dc collector-to-emitter

voltage V CE are used to specify this conditions

B

DC load line

 The dc load line is a graph that represents all the possible

combinations of I C and V CE for a given amplifier.

 Q-point on the dc load line that indicates the values of V CE and I C

for an amplifier at rest, (Quiescent means at rest).

 A questcent amplifier is one that has no ac signal applied and

therefore has constant dc values of I C and V CE

( BQ)

ICQ

VCEQ

 To determine the Q-point

on dc load line, we must

know:

the biasing base current,

IB from the biasing or input

circuit;

the dc load line of the

transistor circuit.

 A straight line intersecting the

vertical axis at approximately

I C(sat) and the horizontal axis at

V CE (off)

 I C(sat) occurs when transistor

operating in saturation region

 V CE(off) occurs when transistor

operating in cut-off region

CC

C R

V I

0

)

C off CC C C I

V

• If the dc load line has a collector saturation current

I C(sat) = 8mA and a cut-off collector-emitter voltage

V CE(cut-off) = 15V The biasing base current I B was 40 A Draw

dc load line on output characteristic We can determine the Q-point is at I CQ = 4.12mA and V CEQ = 7.23V

12345678

DC load line equation  The KVL around the output loop

equation is given by the following expression:

 This equation can be arranged into

the following expression.

I C = (V CC - V CE ) / (R C + R E )

- dc load Line equation of the amplifier

 This is a linear equation drawn

onto the output characteristic curve of the BJT This line is known

as the dc load line of the amplifier circuit.

IE

Draw dc load line on the output characteristic of a BJT

 From the dc load line equation: I C = (V CC – V CE ) / (R C + R E )

 As this is a linear equation, we need to identify only two points on the

line; and by joining these two points, the dc load line is then construct

ed on the output characteristic of the BJT.

 Point 1: Let I C = 0, then V CE(off) = V CC

This is the horizontal axis intersect point and is known as the Cut-off

Point.

 Point 2: Let V CE = 0, then I C(sat) = V CC / (R C + R E )

This is the vertical axis intersect point and is known as the Saturation

IE

IC

VCE

DC Load Line

Saturation point

Cut-off point

IC(sat)

VCE(off)

Effect of changes in RC or RE on dc load line

 The gradient of the dc load line is equal to 1/(R C + R E ).

 Changes in R C or R E or both resistors (R C + R E ) will affect the

gradient of the dc load line.

 However, the horizontal axis intersect point will remain the

pivot at V CE = V CC while the dc load line tilts towards or away fro

m the origin of the co-ordinate due to changes in R C or R E .

E C

CC CE

E C

C

R R

V V

R R

1 I

redu

ces

Whe

n (RC + R

E

)

incr

eases

Origin (0,0) Pivot Point

E C

CC C(sat)

R R

V I





dc load line equation:

 When (RC + RE) increases, dc load line becomes less steeper.

 When (RC + RE) decreases, dc load line becomes more steeper.

Effect of changes in VCC on dc load line

The gradient = 1/(R C + R E ) of the dc load line will remain unchanged while the supply voltage V CC under go changes.

The dc load line moves parallel away or towards the origin

of the co-ordinate when V CC increases or reduces

E C

CC C(sat)

R R

V I



CC CE

E C

C

R R

V V

R R

1 I

DC Biasing + AC signal

• When an ac signal is applied to the

base of the transistor, IC and VCE will b

oth vary around their Q-point values

• When the Q-point is centered, IC and

VCE can both make the maximum

possible transitions above and below

their initial dc values

• When the Q-point is above the center

on the load line, the input signal may

cause the transistor to saturate Whe

n this happens, a part of the output si

gnal will be clipped off

• When the Q-point is below midpoint

on the load line, the input signal may

cause the transistor to cutoff This ca

n also cause a portion of the output si

gnal to be clipped

Mid-point bias

 It is desirable to have the Q-point centered on the load line

Why?

 When a circuit is designed to have a centered Q-point, the amplifier is

said to be midpoint biased

 Midpoint biasing allows optimum ac operation of the amplifier

 It is often desirable to start the design of the amplifier by biasing the BJT bias near the midpoint of its dc load line

 At midpoint bias,

IC= IC(sat) /2, and VCE = VCC/2

or IC = 0.5 IC(sat) , and VCE = 0.5 VCC

DC Biasing + AC signal

Max v ce without

distortion

Mid-point bias (cont’d)

Note that when i c is positive, v ce

DC equivalent circuit

 To transform the amplifier circuit to its dc equivalent circuit, the

following procedures should be followed

1 Reduce all ac sources to ZERO.

2 Remove all capacitors from the circuit.

3 Replace all inductors or coils with a wire (or short-circuit).

4 Redraw the amplifier circuit

 DC equivalent circuit is used to determine the dc biasing currents I B &

I C and the dc biasing voltage V CE of the amplifier circuit.

V BB

V out

DC equivalent circuit exercise 1

Draw dc equivalent circuit of the

V BB

Equivalent to

RE

DC equivalent circuit exercise 2

Draw dc equivalent circuit of the

Analysis for Voltage Divider bias

 A voltage-divider biased BJT amplifier is shown below.

 It is the most commonly used biasing circuit because it can have

voltage gain, current gain or power gain.

 Advantages of voltage divider bias

 It has a stable Q point.

 Single dc power supply V CC .

 Disadvantage of voltage divider bias method

 The biasing circuit is more complicated

 Step 1: Convert dc equivalent circuit to Thevenin’s equivalent biasing circuit.

Analysis for Voltage Divider bias (cont’d)

RE

VTH

RTH

RC+VCC

Q A

B

IC

IE

IB

 The dc biasing circuit formed by VCC, R1 & R2 across

point A & B is first converted to its Thevenin’s

Equivalent.

 The component values of the Thevenin’s Equivalent

Circuit are given as follows:

2 1

CC 2 TH

R R

V R V





2 1

2 1 TH

R R

R R R

 Step 2: Determine the base current IB from Thevenin’s equivalent biasing circuit.

Apply KVL on the input loop,

Q A

 Step 3: Determine I C

Since the transistor must operate in the active

region, the collector current I C can be determin

ed as:

 Step 4: Determine VCE.

Express the KVL for the circuit that the

collector current I C passes through:

Q A

 Potential at Base, Collector and Emitter:

Thevenin’s equivalent biasing circuit

Example ( *** )

Determine the Q-point of the

transistor with a β DC = 100 of the

voltage-divider biased amplifier

Draw dc load line and indicate

Q-point on dc load line.

Step 1: Convert dc equivalent circuit to Thevenin’s equivalent biasing circuit

VTH = R2 / (R1 + R2) x VCC = 5.6k  /(10k  + 5.6k) x 10 = 3.59V

RTH = R1 x R2 / (R1 + R2) = (10k x5.6k)/(10k+5.6k) = 3.59 k 

Solution:

RTH 3.59k

IB

VTH3.59V

VBE

RC 1k

VCC10V

DC 100

RE

560

++

IE

_

+

DC equivalent circuit:

+VCC = 10V

RC = 1k

R1 10k

RE

560

R2 5.6k

Thevenin’s equivalent biasing circuit

RE560Ω

R25.6KΩ

R110KΩ

RC1KΩ

+VCC 10V

Q

 Step 2: Determine the base current I B from Thevenin’s equivalent circuit.

Apply KVL on the input loop,



IB

VTH 3.59V

VBE

RC = 1k

VCC = 10V

DC 100

RE

560

++

IE

_Q

+

 Step 3: Determine the collector current IC

Since the transistor must operate in the

active region, the collector current IC can be

determined as:

IC = DCIB=100x48.5A= 4.85mA

The Q-point is at (2.434V, 4.85mA)

 Step 5: plot dc load line and indicate

Saturation point

VCE(off)=10V

IC(sat)=6.41

Cutoff point2.434

RTH 3.59k

IB

VTH 3.59V

VBE

RC = 1k

VCC = 10V

DC 100

RE

560

++

IE

_Q

+

0

Voltage divider

biasing CE amplifier

A voltage divider biasing CE

amplifier is shown below Draw

dc equivalent circuit and Theven

ine’s equivalent biasing circuit

Determine Q-point and indicate i

t on dc load line.

+VCC = 12V

RC = 2.7k

R1 22k

RE 2.2K

R2 10k

Solution:

Step1: DC equivalent circuit:

RE2.2KΩ

R210KΩ

R122KΩ

RC2.7KΩ

+VCC 12V

VCC = 12V

DC 100

RE 2.2K

+

IE

+

VTH =R2 /(R1+R2) x VCC = 10/(10+22) x 12 = 3.75V

RTH =R1 x R2/(R1+R2) = (10x22)/(10+22) = 6.88 k 

RTH6.88KΩ

IB

VTH3.75V

VBE

RC2.7k

VCC = 12V

DC 100

RE 2.2K

+

IE

Step 6: draw dc load line and

indicate Q-point on dc load line

Saturation point

VCE(off)=12V

IC(sat)=2.449

Cutoff point

AC equivalent circuit

 To transform the amplifier circuit to its ac equivalent

circuit, the following procedures should be followed.

1 Reduce all dc sources to ZERO.

2 Replace all capacitors with a wire (or short-circuit).

3 Remove all inductors or coils from the circuit.

4 Redraw the amplifier circuit

V out

AC small-signal equivalent circuit

amplifier circuit, it is often useful to represent the BJT by an e quivalent circuit.

 An equivalent circuit uses various internal transistor

parameters (usually specified by the manufacturer of the BJT)

to represent the BJT’s operation.

- The Norton Equivalent of Eber Moll’s model

Norton equivalent of Eber Moll’s model

 Using the Norton equivalent form of

the Eber Moll’s BJT model convert

the ac equivalent circuit to its ac

small signal equivalent circuit.

 Analysis of this small-signal

equivalent circuit is much simpler

than the actual amplifier circuit.

DC Biasing + AC signal

• When an ac signal is applied to the

base of the transistor, IC and VCE will

both vary around their Q-point values

• When the Q-point is centered, IC and

VCE can both make the maximum

possible transitions above and below

their initial dc values

• When the Q-point is above the center

on the load line, the input signal may

cause the transistor to saturate Whe

n this happens, a part of the output si

gnal will be clipped off

• When the Q-point is below midpoint

on the load line, the input signal may

cause the transistor to cutoff This ca

n also cause a portion of the output si

gnal to be clipped