Circuit and analog electronics ch6

Circuits and Analog Electronics

Ch6 Basic BJT Amplifiers Circuits
6.1 Bipolar junction transistors (BJTs)
6.2 Single-Stage BJT Amplifiers
6.3 Frequency Response
6.4 Power Amplifiers

References: Floyd-Ch-3, 5, 6; Gao-Ch7;
References


Key Words:
Words
Construction of BJT
BJT in Active Mode
BJT DC Model and DC Analysis
C-E Circuits I-V Characteristics
DC Load Line and Quiescent Operation Point
BJT AC Small-Signal Model

This lecture will spend some time on understanding
how the bipolar junction transistor (BJT) works based
on what we have known about PN junctions. One way
to look at a BJT transistor is two back-to-back diodes,
but it has very different characteristics.
Once we understand how the BJT device operates, we
will take a look at the different circuits (amplifiers)
which we can build.

Construction of Bipolar junction transistors

Emitter-base
junction

Base region
(very narrow)

Emitter region

Collector

Collector region

Emitter
Base

Collector-base
junction

Construction of Bipolar junction transistors

NPN BJT shown
• 3 terminals: emitter, base, and collector
• 2 junctions: emitter-base junction (EBJ) and collector-base
junction (CBJ)
– These junctions have capacitance (high-frequency
model)
• BJTs are not symmetric devices
– doping and physical dimensions are different for emitter
and collector

Standard bipolar junction transistor symbols

Depending on the biasing across each of the junctions, different
modes of operation are obtained – cutoff, active and saturation

BJT in Active Mode

Two external voltage sources set the bias conditions for active
mode
– EBJ is forward biased and CBJ is reverse biased

BJT in Active Mode

IE ＝ IEN ＋ IEP≈ IEN

Forward bias of EBJ injects electrons from emitter into base
(small number of holes injected from base into emitter)

BJT in Active Mode

IB ＝ IBN ＋ IEP

• Most electrons shoot through the base into the collector
across the reverse bias junction
• Some electrons recombine with majority carrier in (P-type)
base region

BJT in Active Mode

IC = ICN + ICBO

Electrons that diffuse across the base to the CBJ junction are
swept across the CBJ depletion region to the collector.

BJT in Active Mode


IC = ICN + ICBO

IE = IB + IC
Let ICN ＝
IB ＝ IBN ＋
IEP

IC
α≈
---common-base current gain
IE

IE
IC (1 － ) = αIB +
ICBO

BJT in Active Mode

α≈

IB ＝ IBN ＋
IEP
IC (1 － )=

IC
IE

αIB+ICBO
α
β=
Let

1−α

Beta:

β≈

IC
IB

IC=ICN+ICBO

IE=IB+IC

I C = β I B + (1 + β ) I CBO

---common-emitter current gain

 I E = I C + I B ≈ (1 + β ) I B

 I C = β I B + I CEO = β I B
 I = αI
E
 C

BJT Equivalent Circuits
°
＋

iB→

←iC

vBE
－

IB→

←IC

•
i E↓

－

°

＋

＋

V CE

VBE＝Von

βIB
－

•

vCE

°

BJT DC model

°

β iB

°
＋

•

•
IE↓

°

－

•Use a simple constant-VBE
model
– Assume VBE = 0.7V

BJT DC Analysis
• Make sure the BJT current equations and
region of operation match
VBE > 0,
VBC < 0, → VE < VB <VC
• Utilize the relationships (β and α) between
collector, base and emitter currents to solve
for all currents

 I E = I C + I B = (1 + β ) I B

I C = β I B
 I = αI
E
 C

Base-emitter Characteristic(Input characteristic)

i B = f (v BE )

vCE = C

Collector characteristic (output characteristic)

iC = f (VCE )

iB =C

iB = 40 μA

Collector characteristic (output characteristic)

iC = f (VCE )

iB =C

Collector characteristic
Saturation
Saturation occurs when the
supply voltage, VCC, is
across the total resistance
of the collector circuit, RC.
IC(sat) = VCC/RC
Vsat
Once the base current is high enough to produce saturation, further increases in
base current have no effect on the collector current and the relationship IC = βIB is
no longer valid. When VCE reaches its saturation value, VCE(sat), the base-collector
junction becomes forward-biased.


When IB = 0, the transistor is in
cutoff and there is essentially no
collector current except for a
very tiny amount of collector
leakage current, ICEO, which can
usually be neglected. IC ≈ 0.

Cutoff

In cutoff both the base-emitter
and the base-collector junctions
are reverse-biased.


linearity

Δ

Discussion of an amplification effect

vi = Ri × B
i
vo = RL ×C
i
∆ vCE
∆ vBE
Ri =
<< RL =
∆ iB
∆ iC
With iB << iC

vi < < vo

E.g. for common-base configuration transistor:

vo
Av =
= 50 ~ 300
vi

DC Load Line and Quiescent Operation Point

Q-point
VCC

ICQ

.

Q

VCEQ
DC load line

Base-emitter loop: I B =
Collector-emitter loop:

VCC − VBE VCC
≈
= 40( µA)
Rb
Rb

vCE = VCC − iC RC = 10 − iC × 4k

BJT AC Small-Signal Model
°
＋

iB→

←iC

vBE
－

β iB
•

•
i E↓

°

°
＋

°
＋

vCE

vbe

－

－

ib→

rbe

←ic

β ib

°
＋
vce
－

•
ie↓

°
rbe ≈ 300Ω + (1 + β )

26( mV )
I E ( mA)

• We can create an equivalent circuit to model the transistor for small signals
– Note that this only applies for small signals (vbe < VT)
• We can represent the small-signal model for the transistor as a voltage controlled
current source (
) or a current-controlled current source (ic = βib).
• For small enough signals, approximate exponential curve with a linear line.


BJT fundamentals:
VBE = 0.7V

I E = IC + I B = ( 1 + β ) I B ≅ IC
IC = β I B


Key Words:
Words
Common-Emitter Amplifier
Graphical Analysis
Small-Signal Models Analysis
Common-Collector Amplifier
Common-Base Amplifier

C-E Amplifiers
To operate as an amplifier, the BJT must be biased to operate in active
mode and then superimpose a small voltage signal vbe to the base.
DC + small signal

coupling capacitor
(only
passes
ac
signals)
∆
RC
vi C2 ∆ vBE → ∆ iB → ∆iC =iB → ∆ic → ∆vCE C1 vo
→
 β
→

vi → ∆iB

∆iB → ∆iC

∆iC → vO

C-E Amplifiers
+


Vi

Vi


Vi

C-E Amplifiers

Apply a small signal
input voltage and see ib

i B = I B + ib
vBE=vi+VBE

C-E Amplifiers

See how ib translates into vce.

• vi = 0 → IB 、 IC 、 VCE

vi ≠ 0

iC=ic+IC

i B = I B + ib



iC = I C + iC 
vCE = VCE + vce 


• VoM > > ViM

f ( o ) = f (i )

• vo out of phase with vi

vCE=vce+VCE

C-E Amplifiers

Considering VC (all the capacitors are replaced
by open circuits)

Considering Vi (all the capacitors are replaced
by short circuits)

C-E Amplifiers
Considering VC (all the capacitors are replaced
by open circuits)

Considering Vi (all the capacitors are replaced
by short circuits)

Graphical Analysis
• Can be useful to understand the operation of BJT
circuits.
• First, establish DC conditions by finding IB (or VBE)
• Second, figure out the DC operating point for IC

VCC

Can get a feel for whether the BJT will stay in active region of operation
– What happens if RC is larger or smaller?

Graphical Analysis

'
vce = −ic ( RC // RL ) = −ic RL

V′CC

Graphical Analysis
Q-point is centered on the ac load line:

V′CC

Graphical Analysis
Q-point closer to cutoff:

V′CC

Clipped at cutoff
(cutoff distortion)

Graphical Analysis
Q-point closer to saturation:

V′CC

Clipped at cutoff
(saturation distortion)

Graphical Analysis


Steps for using small-signal models
1. Determine the DC operating point of the BJT
－ in particular, the collector current
2. Calculate small-signal model parameters: rbe
3. Eliminate DC sources
– replace voltage sources with short circuits and
current sources with open circuits
4. Replace BJT with equivalent small-signal models
5. Analysis

Example 1

VC = ( I B + I C ) R + I B R b + VBE + I E R e

VC − VBE
→ IB =
Rb + (1 + β )( R + Re )

IC ≈ βIB,
IE = IC + IB = (1+β)IB
VCE = VC − I C RC − I E ( R + Re )

Example 2

VB =

Rb 2
VCC
Rb1 + Rb 2

IC ≈ I E =

vs

IB =

VB − VBE
≈V B/ Re
Re

IC
β

VCE = VCC − I C ( R C + R e )


There are three basic configurations for single-stage
BJT amplifiers:
– Common-Emitter
– Common-Base
– Common-Collector

VE < VB < VC

VE < VB < VC

VE < VB < VC

VCC = I B Rb + VBE + I E Re = I B Rb + VBE + (1 + β ) I B Re
IB =

VCC − VBE
VCC
≈
Rb + (1 + β ) Re Rb + (1 + β ) Re

I C = βI B
VCC = VCE + I E Re ≈ VCE + I C Re

VCE ≈ VCC − I C Re

&
&
Note : Vo is slightly less than Vi due to the voltage drop introduced by VBE
AV ≅1

The last basic configuration is to tie the collector to a fixed voltage, drive
an input signal into the base and observe the output at the emitter.


Let’s find Av ，
Ai ：
⋅

⋅

Vo = I e ( Re // RL ) = I b (1 + β )( Re // RL )
⋅

⋅

Vi = I b [rbe + (1 + β )( Re // RL )] = I b rbe + I e ( Re // RL )
⋅

(1 + β )( Re // RL )
β ( Re // RL )
∴ AV = ⋅ =
≈
<1
Vi rbe + (1 + β )( Re // RL ) rbe + (1 + β )( Re // RL )
VO


Let’s find Av ，
A e (R
I o RL = iI： e // RL ) = (1 + β ) I b ( Re // RL )
(1 + β )( Re // RL )
Io = Ib
RL
I b (rbe + (1 + β )( Re // RL )) = ( I i − I b ) Rb
rbe + (1 + β )( Re // RL ) + Rb
(1 + β )( Re // RL ) + Rb
Ii = Ib
≈ Ib
Rb
Rb

Ai =

(1 + β )( Re // RL )
Rb
(1 + β )( Re // RL )
×
≈
RL
(1 + β )( Re // RL ) + Rb
RL

Ai ≈

(1 + β )( Re // RL )
>>1
RL

&
Ii
&
Io

(1 + β )( Re // RL ) << Rb

Let’s find Ri ：

v i = ib rbe + i e ( Re // R L ) = ib [ rbe + (1 + β )( Re // R L )]
Ri′ =

vi
= rbe + (1 + β )( Re // R L )
ib

Ri = Ri′ // Rb = [rbe + (1 + β )( Re // RL )] // Rb ≈ Rb // β ( Re // RL )

Let’s find Ro ：

I Re = I + I e

I = I Re − I e

I = I Re − I e = I Re − ( 1 + β ) I b

I = I Re − I b − β I b
v
v
=
+ (1 + β )
Re
rbe + Rs // Rb

Ie
I Re

I

v
1
Ro = =
1
i 1 +
Re (rbe + Rs // Rb ) (1 + β )

(rbe + Rs // Rb )
= Re //
1+ β


Ri = [rbe + (1 + β )( Re // RL )] // Rb

(rbe + Rs // Rb )
Ro = Re //
1+ β


⋅

⋅

β ( Re // RL )
AV = ⋅ ≈
≈1
rbe + (1 + β )( Re // RL )
Vi
VO

Ai ≈

(1 + β )( Re // RL )
>>1
RL

Ri = [rbe + (1 + β )( Re // RL )] // Rb
(rbe + Rs // Rb )
Ro = Re //
1+ β

C-C amp characteristics:
• Voltage gain is less than unity, but close (to unity) since β is large and rbe is small.
• Also called an emitter follower since the emitter follows the input signal.
• Input resistance is higher, output resistance is lower.
- Used for connecting a source with a large Rs to a load with low
resistance.

Ground the base and drive the input signal into the emitter

VB − VBE = I E R e

VB =

VCC
R b2
R b1 + R b 2

VCE = VCC − I C RC − I E Re ≈ VCC − I C ( RC + Re )

IC ≈ I E =

IB =

IC
β

VB − V BE V B
≈
Re
Re


Ro

Ri

→ Av =

−ic ( Rc // RL ) β ( Rc // RL )
=
− ib rbe
rbe

rbe
// Re
Ri=
(1 + β )

R o≈R C

R
−β C
β RC ( R + R )
&
( RC + RL )
I
C
L
& = Io =
≈
≅ E ≈1
Ai
(1 + β )
IC
&
I i − rbe r
be

(1 + β )

// Re

β
≈ 1( sinceI E ≅ I C )
For RL<<RC, Ai ≈
(1 + β )


β RC ( R + R )
&
I
RC
&
C
L
Ai = o ≈
≈
&
(1 + β )
RC + RL
Ii

For RL<<RC, Ai ≈

β
≈1
(1 + β)

β ( Rc // RL )
rbe
rbe
r
// Re ≈ be
Ri=
(1 + β )
(1 + β )
Av =

R o≈R C

CB amp characteristics:
• current gain has little dependence on β
• is non-inverting
• most commonly used as a unity-gain current amplifier or current buffer and not
as a voltage amplifier: accepts an input signal current with low input resistance
and delivers a nearly equal current with high output impedance
• most significant advantage is its excellent frequency response


Summary for three types of diodes:
C-C
Input
Output
Functions

C-E

C-B

IB

IB

IB

IE

IC

IC

Zout < Zin Zout > Zin Zout > Zin
Vout ≈ Vin Vout > Vin Vout > Vin


Key Words:
Words
Basic Concepts
High-Frequency BJT Model
Frequency Response of the CE Amplifier

Basic Concepts

1.0V

0.5V

0V

-0.5V

-1.0V
0.5ms
V(1)

V(2)

1.0ms

1.5ms

2.0ms

2.5ms
Time

3.0ms

3.5ms

4.0ms

Basic Concepts
800mV
600mV
400mV
200mV
0V
0Hz
V(2)

2KHz
V(1)

4KHz

6KHz

8KHz

10KHz
Frequency

12KHz

14KHz

16KHz

18KHz

1. 0V

0. 5V

0V
1 0 Hz

1 0 0 Hz
V( 2 )

1 . 0 KHz

1 0 KHz

1 0 0 KHz

1 . 0 MHz

20KHz

Basic Concepts

&
Av = Av ( f )∠ϕ ( f )

Lower cut off frequency

or

&
A = Av (ω )∠ϕ (ω )

Upper cut off frequency

The drops of voltage gain (output/input) is mainly due to:
1 、 Increasing reactance ofC s , Cc , Ce (at low f)
2 、 Parasitic capacitive elements of the network (at high f)
3 、 Dissappearance of changing current (for transformer coupled amp.)

High-Frequency BJT Model
In BJTs, the PN junctions (EBJ and CBJ) also have capacitances associated
with them
Cµ

rbe

Cπ

Cµ

Cπ

rbe

C'π

C'µ


rbe

C'π

vs

There are three capacitors in the circuit.
At the mid frequency band, these are considered to be short
circuits and internal capacitors C'µ and
are considered to be open
C'π,
circuits.

C'µ

At low frequencies, C1, C2 are an
open circuit and the gain is zero.
Thus C1 has a high pass effect on the
gain, i.e. it affects the lower cutoff
frequency of the amplifier.

vs

τ 1 = C1 ( Rs + Rb1 // Rb 2 // rbe )

f L1
τ2 is the time constant for C2.

τ 2 >> τ 1

1
=
2πτ 1

---is neglected


τ 1 = C1 ( Rs + Rb1 // Rb 2 // rbe )
τ 2 >> τ 1

---is neglected

Capacitor Ce is an open circuit. The
pole time constant is given by the
resistance multiplied by Ce.

vs

′
f L ≈ 1.1 f L1 + f L 2 +  f Le
2

2

2

 ( R // Rs + rbe )

τe =  b
// Re Ce


1+ β


1
f Le =
2πτ e

At high frequencies, C1, C2 Ce are all
short circuit.
The frequency that dominates is the
lowest pole frequency.
vs

The time constant is neglected for C'µ
'
′
( RL << 1 jωCµ )

rbe

C'π

C'µ

′
τ Cπ′ = ( Rb // Rs // rbe )Cπ

1
fH =
2πτ Cπ′

In summary:the lower cut off frequency is determined by network capacitence.
e.g.C1 ( C2 , Ce ) The higher cut off frequency is determined by the parasitic
′
ferquency of the BJT. e.g. Cπ


rbe

C'π

j

⋅

A v = Avm ⋅
vs

f
→ ∞,
fL

C'µ

(1 + j

f
fL

f
f
)(1 + j
)
fL
fH

⋅
f
For f L << f << f H ,
→ 0 ⇒ Av = Avm — mid - frequency
fH
f
j
⋅
f
fL
For f < f L ( f << f H ),
→ 0, ⇒ Av = Avm
— low - frequency
f
fH
1+ j
fL
⋅
fL
1
For f > f H ( f >> f L ) → 0, ⇒ Av = Avm
— High − frequency
f
f
1+ j
fH


rbe
vs

j

⋅

A v = Avm ⋅

fL =

(1 + j

ωL
1
=
2π 2πτ L

f
fL

f
f
)(1 + j
)
fL
fH

fH =

ωH
1
=
2π 2πτ H

C'π

C'µ


decade

0

decade


Key Words:
Words
Power Calculation
Class-A, B, AB Amplifiers
Complementary Symmetry(Push-Pull) Amplifier
Biasing the Push-Pull Amplifier (OCL)
Single-Supply Push-Pull Amplifier (OTL)

An Analog Electronics System Block

Sensor

Energy
conversion

Voltage
Amplifiers

Power
Amplifiers

Signal
Amplifiers

Load

Energy
conversion

The output power delivered to the load RL:

PoM =

Vom I om
2

1
= Vom I om
2 2

The average power delivered by the supply:

1
PS =
T

T

∫
0

1
VCC iC (t )dt = VCC
T

T

∫

iC (t )dt = VCC I C

0

The efficiency in converting supply power to useful output power is
defined as
P
η = OM × 100%
PS

Power Calculation

The DC power by
the supply

PC = VCEQ I CQ = (VCC − I CQ RC ) I CQ
2
= PS − I CQ RC
The DC power delivered to BJT by the supply

Power Calculation
The average power dissipated as heat in the BJT:
1 T
∫ vCE iC dt
T 0
1 T
= ∫0 ( I CQ + I m cos ω t )(VCEQ − Vm cos ω t )dt
T
1
= I CQVCEQ − I mVm = PC − PCL
2

PT =

Class-A Amplifiers

Class-B Amplifiers

Class-AB Amplifiers

Complementary Symmetry Power Amplifier (Class-B)

I om

PO =

2

×

2
1 Vom
≈
2 2 RL

Vom

Assuming vo = Vom sin ω t
PT 1 =

POM

2
1 VCC
≈
2 RL

→ vCE = VCC − vo

vO
1 π
1 π
d (ω t )
∫0 vCE iC d ( ω t ) =
∫0 (VCC − vO )
2π
2π
RL

2
2
πV

1  π VCCVon
on sin ωt
P1 =
sin ωtd ωt − ∫
d ωt 
T

0
2π ∫0 RL
RL



Von 2 π 2
1  VCCVon π
=

∫0 sin ωtdωt − RL ∫0 sin ωtdωt 
2π  RL

2

Von 1 π
1  VCCVon π
=

∫0 sin ω tdω t − RL 2 ∫0 1 − cos 2ω tdω t 
2π  RL

1  VCCVon
=
( − cos ω t )

2π  RL

π
0

Von 2  1  VCCVon
Von 2  1  VCCVon Von 2 
−
×2 −
−
=

= 

2 RL  2π  RL
2 RL  RL  π
4 


VOm ≈ VCC

2
VCC 4 − π
PT 1 =
×
RL
4π

2
4 − π VCC
PT = PT 1 + PT 2 =
×
2π
RL

Note: PT represents the amount of
power dissipated by the BJT as heat

I om

2
1 Vom
≈
2 2 RL

Vom

2
1 VCC
POM ≈
2
2 RL
2
VCC 4 − π
VOm ≈ VCC
PT 1 =
×
RL
4π
2
4 − π VCC
PT = PT 1 + PT 2 =
×
2π
RL

PO =

×

2
2VCC
PE = PT + PO =
πR L
2
1 VCC
P
π
η = O = 2 RL
=
2
2 VCC 4
PE
Note that for class A: η﹦25﹪ ~ 50﹪
π RL

class B: η﹦78.5﹪
class AB: η=25﹪ ~ 78.5﹪

=78.5%


Crossover
distortion

Biasing the Push-Pull Amplifier (Class-AB) (OCL)
To overcome crossover distortion, the biasing is adjusted to just overcome the
VBE of the transistors; this results in a modified form of operation called class
AB. In class AB operation, the push-pull stages are biased into slight
conduction, even when no input signal is present.

}V

CC

}V

CC

Power Calculation is the same as class-B

Single-Supply Push-Pull Amplifier (OTL)
The circuit operation is the same as that described previously, except
the bias is set to force the output emitter voltage to be VCC/2 instead of
zero volts used with two supplies. Because the output is not biased at
zero volts, capacitive coupling for the input and output is necessary to
block the bias voltage from the source and the load resistor.

Circuit and analog electronics ch6

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