Post on 06-Jul-2018
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Analog Electronics
Section 1
Prof. Dr. Murat Aşkar D-211
E-mail: askar@eee.metu.edu.tr
Web: www.eee.metu.edu.tr/~askar
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Analog and Digital Signals
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Analog Signals are links to the real world
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Transreceiver
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Bio-eng: Neuronal implants
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Linear Amplifiers (1)
Multiply amplitude of a signal by a constant scalar quantity
xo(t) = A xi(t)
Non-scalar or
non-uniformamplification
is called
distortion
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Linear Amplifiers (2)
• Symbol for a single-ended
input linear voltage amplifier.
• Ideally provides linear
voltage gain regardless of
the amplitude of the input
signal
• Real amplifiers have power
supplies that limit theamplitude of the output
• If input is too large, output
clamps
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Transfer Characteristics (1)
Plot of amplifier output versus amplifier input
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Transfer Characteristics (2)
• Gain = Slope
• To operate amplifier in its linear region, the input
must be kept small enough
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Real Transfer Characteristics
• Each circle represents
a different DC
component for the
input and output
signals – called an
operating point
• Location of operating
point has an effect on
• input signal range
• amplifier gain magnitude
• amount of distortion
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Operating Point (1)
Voltage gain, output DC voltage, allowable input
magnitude range are affected
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Operating Point (2)
Derivative of
transfer
characteristicgives measure
of amplifier gain
linearity (and
distortion)
Input and output signal amplitude ranges are
maximized when operating point is near middle of
linear region
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Signal Convention (1)
• DC magnitudes in uppercase symbol and
subscript
– Example: ID, VD
• Ac signal quantities in lowercase symbol &
subscript
– Example: id(t), vd(t)
• Total DC + ac signal quantities in lowercase
symbol, uppercase subscript – Example: iD(t), vD (t)
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Signal Convention (2)In general
vD(t) = VD + vd(t)
iC(t) = IC + ic(t)
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Superposition
• If the amplifier is linear, superposition can be
applied. Each component can be determined
seperately.• Determine the DC magnitude of the output using
DC model , i.e capacitors are open circuited,
inductors are short circuited. ( Example: IY, VX)• Determine the ac signal component of the output
using the ac model , i.e., DC sources are killed
and capacitors are short circuited, inductors are
open circuited. (Example: vy(t) = Av vi(t))
• Total output is then: vY(t) = V
Y + v
y(t)
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Amplifier Classification
CurrentCurrentCurrent
TransresistanceVoltageCurrent
TransconductanceCurrentVoltage
VoltageVoltageVoltage
TypeOutputInput
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+
vi-
+
vo-
Ideal Voltage Amplifier
Avi
+
R oAvi
+
vi-
+
vo-
R i
+
Real Voltage Amplifier
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Loaded Ideal Voltage Amplifier
Avi
+
vi-
+
vo-
+
R LR s
vs
+
Av
v
s
o=
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Loaded Voltage Amplifier
R oAvi
+
vi-
+
vo-
R i
+
R LR s
vs
+
Lo
L
is
i
s
o
R R
R
R R
R A
v
v
++=
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Ideal Current Amplifier
Real Current Amplifier
Aii
ii io
Aiiii ioR i R o
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Loaded Ideal Current Amplifier
Ai
i
s
o=
Aii R L
ii
R sis
io
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Loaded Current Amplifier
Aii R L
ii
R Sis
io
R i R o
Lo
o
is
s
s
o
R R
R
R R
R A
i
i
++=
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To determine Input Resistance R i
R oAvi
+
vi-
+
vo-
R i
+
• Apply input voltage vi
(or input current ii)• Determine the input
current ii (or input
voltage vi)Then
R i
= vi
/ iiAii
ii io
R i R o
ii
+
vi-
Voltage Amplifier
Current Amplifier
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To determine Output Resistance R o
R oAvi
+
vi-
+
vx-
R i
+
• Kill the input signal
(set vi or ii to zero)• Apply a test voltage
vx to the output node
• Determine the currentix that the source
delivers to the circuit
Then
R o = vx / ix
Aii
ii ix
R i R o
ii
+
vi-
Voltage Amplifier
Current Amplifier
+
vx-
ix