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Amplifiers : Basics and DC Analysis
ESc201 : Introduc=on to Electronics L14
rhegde Dept. of Electrical Engineering
IIT Kanpur
1
Objective
1. Learn ideal Transistor characteristics required for Voltage Amplification
2. Learn to build amplifiers using elements which have non-ideal characteristics.
2
Voltage Amplification
V0Vi+ +
- -
V0Vi+ +
- -
I0I1
3-terminal unilateral linear device
1>×=
GVGV io
3
V0Vi+ +
- -
I0I1
V0
+ +
- -
Vi
I1
Ri
I0
gmvi rO
iInput resistance R = i iV / I
0Output conductance: 1
i
oo o
o V=
Ig = / r =V
00
= om
i V
IgV =
Trans conductance
(Ideally large)
(Ideally small)
(Ideally large)
4
Voltage Amplifier
V0
VS RL
RS
gmvi
RS
RL
v0
rORi+vi-
vS
V0
+ +
- -
Vi
I1
Ri
I0
gmvi rO
5
V m oA g r≤ ×
o iLV m o
s o L i S
V RRA g rV r R R R
= = − × ×+ +
gmvi
RS
RL
v0
rORi+vi-
vS
ii S
i S
RV VR R
=+ o m i o LV g V r R= − ×
1m og r× >Necessary Condition for Voltage Amplifications 6
Trans-conductance >> Output Conductance
Transistor
V0Vi+ +
- -
I0I1
om gg >>1>>om rgVoltage Amplification
V0
+ +
- -
Vi
I1
Ri
I0
gmvi rO
00
= om
i V
IgV = 0
i
oo
o V=
IgV
=
7
Transistor Trans-resistor
+ Vo -
IO + VIN -
Current IO is much more sensitive to VIN than VO
• Can be used for voltage amplification • Can be used as a switch • Implement logic • …
8
V0Vi+ +
- -
I0I1
V0
+ +
- -
Vi
I1
Ri
I0
gmvi rO
o iLV m o
S o L i S
V RRA g rV r R R R
= = − × ×+ +
gmvi
RS
RL
v0
rORi+vi-
vS
9
o iLV m o
s o L i S
V RRA g rV r R R R
= = − × ×+ +
In the ideal case ro is infinite
We would ideally like input resistance Ri to be infinite as well !
V m LA g R= −
gmvi
RS
RL
v0
rORi+vi-
vS
o iV m L
s i S
V RA g RV R R
= = − ×+
10
An ideal 3-terminal device for Voltage Amplification
V0Vi+ +
- -
I0I1
V0
+ +
- -
Vi
I0
gmvi
I0Vi3Vi2Vi1V0
Vi
I0
Ideal Transistor Characteristics 11
Ideal Transistor (IT)
RL
v0
vs
IT
Making a voltage amplifier with an ideal transistor is straightforward
V0
+ +
- -
Vi
I0
gmvi
oV m L
s
vA g Rv
= = −
gmvi RL
v0
vS+vi-
In practice there is no element which has the characteristics of ideal transistor ! 12
I0Vi3Vi2Vi1V0
Vi
I0
Ideal transistor
Device X
Vi3Vi2Vi1V0
I0I0
Vα
Vi
13
I0Vi3Vi2Vi1V0
Vi
I0
Ideal transistor
Device X Vi3Vi2Vi1V0
I0I0
Vα
Vi
Device Y
Vi3Vi2Vi1
I0
V0Vsat
I0
Vα
Vi
How do we use elements such as X, Y etc to make amplifiers? 14
Bipolar Junction Transistor: BJT
C F BI Iβ=
15
16
MOSFET Metal Oxide Semiconductor Field Effect Transistor
Device X
Vi3Vi2Vi1V0
I0I0
Vα
Vi
αα
αVVVVgVVI
iim
io>−×=
≤=
for )(for 0
I0Vi3Vi2Vi1V0
Vi
I0
Ideal Characteristics
V0Vi+ +
- -
I0I1X
17
Simple amplifier will not work !
Vi
IO
1V tSinVvKRgVV
sL
mω
α5.0;101.0;1 1
==
Ω== −
RL
v0X+
vi-vs
How do we use device X to make an amplifier?
I0
Vα
Vi
Vi3Vi2Vi1V0
I0
0 0O OI V= ⇒ =No Amplification 18
Simple amplifier will not work !
ViVα
IO
RL
v0X+
vi-vs
How do we use device X to make an amplifier?
RL
v0X+
vi-
vs
vB
BV Vα>
IO
VαViVB
o o LV I R= −
tSinVvKRgVV
sL
mω
α5.0;101.0;1 1
==
Ω== −
19
When only a part of device characteristics is suitable for amplification, then we need to push the device into that region by applying suitable bias voltages. This process is called BIASING
RL
v0X+
vi-
vs
vB
ViVα
IO
IO
VαViVB
20
How should one choose the bias voltage VB ?
I0
1.0V 1.2V
2mA
Q-point
Vin
tSinVvs ω5.0=RL
v0X+
vi-
vs
VB
Quiescent point or Bias point
1.2V
21
RL
v0X+
vi-
vs
1.2V
Vin
IO 0.5 ( )sv VSin tω=
I0
1.0V 1.2V
2mA
Q-point
Vin
11 ; 0.01mV V gα−= = Ω
IO
o o LV I R= −
22 αα
αVVVVgVVI
iim
io>−×=
≤=
for )(for 0
clipped
Output voltage is distorted !
IO
O O LV I R= −
Need to choose a proper value of biasing Voltage
RL=1kΩ
23
RL
v0X+
vi-
vs
1.8V
IO
IO vs. time
0.5 ( )sv VSin tω=
I0
Vin
Q-point
1.0V 1.8V
8mA
11 ; 0.01mV V gα−= = Ω
VIN vs time
VO vs. time
Unnecessary Power Dissipation
24
RL=1kΩ
IO vs. time
I0
Vin
Q-point
1.0V 1.5V
5mA
11 ; 0.01mV V gα−= = Ω
RL
v0X+
vi-
vs
1.5V
IO
VO vs. time
Optimum Biasing ?
0.5 ( )sv VSin tω=
Vin vs. time
25
X+vi-
vs
1.5V
RL
v0IO
How do we get rid of unwanted dc voltage at the output ?
RL
v0X+
vi-
vs
1.5V
IO
0.5 ( )sv VSin tω=
I0
Vin
Q-point
1.0V 1.5V
5mA
11 ; 0.01mV V gα−= = Ω
RC
26
X+vi-
RLRC
vs
vo
X+vi-
vs
1.5V
v0
RLRC
IO
X+vi-1.5V
RC
IO ac (signal)
Capacitor is chosen large enough so that at the signal frequency 1/jωC ~0.
dc
I0
Vin
Q-point
1.0V 1.5V
5mA
11 ; 0.01mV V gα−= = Ω
27
X+vi-
v0
RLRC
vs
( ) for o m i iI g V V V Vα α= × − >
I0
Vin
Q-point
1.0V 1.5V
5mA
11 ; 0.01mV V gα−= = Ω
RL
v0
vS+vi-
RCgm(vi-Vα)
28
