Lecture #18 Summary of ideal devices,
amplifier examples
Reminder:
MIDTERM coming up one week from
today (Monday October 18th)
This week: Review and examples
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EE 42 fall 2004 lecture 18
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Midterm
• Monday, October 18,
• In class
• One page, one side of notes
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Topics
Today:
• Summary of ideal circuits
• Amplifier examples
– Comparator
– Op-Amp
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Ideal devices
Wire:
Current in =current out
No voltage differences
Resistor
V  IR
dV
I  C
dt
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Ideal devices 2
Inductor:
dI
V L
dt
Ideal diode:
Reversed bias  no current, open circuit
Forward bias  no voltage drop, just like a wire
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Ideal devices 3
+
+
Transformer
V1
V2
-
-
V1 N1

V2 N 2
N1 and N2 are the number of turns of wire
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Ideal devices 4
V (t )
I (t )
~
Voltage source:
Voltage given, current can be anything
Note: the voltage could be given as
A function of time
Current source
Current given, voltage can be anything
Note: the current could be given as
A function of time
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Ideal devices 5
+
V1
-
+
~
V2
-
+
V1
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I (t )
Dependent Voltage source:
Voltage given as a multiple of another
Voltage or a current, current can be anything
V2  KV1
Dependent Current source
Current given as a multiple of a different
current or voltage, voltage can be anything
I 2  GV1
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NPN transistor
Base
I in (t )
Collector
 I in (t )
The Base-Collector junction must be
reverse biased.
The model of the diode can be ideal,
or better with a 0.7 volt turn on
Emitter
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Beta could be anywhere from 40 to
400
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NMOS transistor
Drain
Gate
If VGS is higher than the threshold
VTH, then the switch is closed.
D
Source
G
S
The resistance from the source to the drain depends on the how much
greater than threshold VGS is, and how wide the transistor is.
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PMOS transistor
Source
If VGS is lower than the threshold VTH,
then the switch is closed.
S
G
Gate
Drain
D
The resistance from the source to the drain depends on the how much
below threshold (more negative) VGS is.
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Amplifier
VIN
+

+

V0
• V0=AVIN
• But V0 cannot rise above some physical voltage related to
the positive power supply VCC (“ upper rail”)
V0 < V+RAIL
• And V0 cannot go below most negative power supply, VEE
i.e., limited by lower “rail”
V0 > V-RAIL
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Amplifier
V+rail
VIN
+

• V0=AVIN
+

V0
V+rail
• Output is referenced to “signal ground”
• V0 cannot rise above some physical voltage related to the
positive power supply VCC (“ upper rail”)
V0 < V+RAIL
• V0 cannot go below most negative power supply, VEE i.e.,
limited by lower “rail”
V0 > V-RAIL
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OP-AMPS AND COMPARATORS
A very high-gain differential amplifier can function either in extremely
linear fashion as an operational amplifier (by using negative feedback)
or as a very nonlinear device – a comparator. Let’s see how!
Differential Amplifier
V+
V
+
A

“Differential”
V0  A( V  V )
Circuit Model in linear region
V0
Ri
+

V1
+

AV1
+

 V0 depends only on difference (V+  V-)
But if A ~ , is the
output infinite?
The output cannot be larger than the supply voltages. It will limit or
“clip” if we attempt to go too far.
“Very high gain”  A  
We call the limits of the output the “rails”.
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V0
WHAT ARE I-V CHARACTERISTICS OF AN ACTUAL
HIGH-GAIN DIFFERENTIAL AMPLIFIER ?
VIN
+

+

V0
• Circuit model gives the essential linear part
• The gain may be 100 to 100,000 or more
• But V0 cannot rise above some physical voltage V0 < V+RAIL
• And V0 cannot go below the lower “rail”
V0 > V-RAIL
• CMOS based amplifiers can often go all the way to their
power supplies, perhaps ± 5 volts
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I-V Characteristics of a real high-gain amplifier
Example: Amplifier with gain of 105, with max V0 of 3V and min V0 of 3V.
(a)
V-V near V0 (V)
origin
(b)
V-V over wider V0 (V)
range
3
0.2
0.1
3 2 1
.2
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upper “rail”
2
1
1
2
3
VIN(V)
lower “rail”
30 20 10
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1
2
3
10 20 30
VIN(V)
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I-V CHARACTERISTICS OF AN ACTUAL HIGH-GAIN
DIFFERENTIAL AMPLIFIER (cont.)
Example: Amplifier with gain of 105, with upper rail of 3V and lower rail
of 3V. We plot the V0 vs VIN characteristics on two different scales
(b)
V0 (V)
V-V over
wide range 3
upper
“rail”
(c)
Same V0 vs VIN over even
wider range
V0 (V)
3
2
1
2
1
lower
“rail”
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30 20 10
1
2
3
10 20 30
VIN(V)
3 2 1
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1
2
3
1
2
3
VIN(V)
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I-V CHARACTERISTICS OF AN ACTUAL HIGH-GAIN
DIFFERENTIAL AMPLIFIER (cont.)
Now plot same thing but with equal horizontal and vertical scales
(volts versus volts)
(c)
V-V with equal
Note:
X and Y axes
• (a) displays linear amplifier
V0 (V)
behavior
3
2
1
3 2 1
1
2
3
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1
2
3
VIN(V)
•
(b) shows limit of linear region –
(|VIN| < 30 V)
•
(c) shows comparator function (1 bit
A/D converter centered at VIN = 0)
where lower rail = logic “0” and
upper rail = logic “1”
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EXAMPLE OF A HIGH-GAIN DIFFERENTIAL AMPLIFIER
OPERATING IN COMPARATOR (A/D) MODE
Simple comparator with threshold at 1V. Design lower rail at 0V and
upper rail at 2V (logic “1”). A = large (e.g. 102 to105 )
VIN
+

V0
2
1
V0
1V +

If VIN > 1.010 V,
V0 = 2V = Logic “1”
0
1
2
VIN
If VIN < 0.99 V,
V0 = 0V = Logic “0”
NOTE: The actual diagram of a comparator would not show an
amplifier with “offset” power supply as above. It would be a simple
triangle, perhaps with the threshold level (here 1V) specified.
VIN
V0
Comparator
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ONE-BIT A/D CONVERSION
REQUIRED IN DIGITAL SYSTEMS


pulses in
transmission
pulses out
comparator regenerated
pulses
As we saw, we set comparator threshold at a suitable value
(e.g., halfway between the logic levels) and comparator
output goes to +rail if VIN > VTHRESHOLD and to rail if VIN <
VTHRESHOLD.
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OP-AMPS AND COMPARATORS
A very high-gain differential amplifier can function in extremely linear
fashion as an operational amplifier by using negative feedback.
R1
R2
R1
R2

+
VIN
EXAMPLE
V0
Ri
VIN
+
V1
+
AV1

+

V0
Circuit Model
Negative feedback
 Stabilizes the output
We can show that that for A   (and Ri  0 for simplicity)
V0  VIN 
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R1  R 2
R1
Stable, finite, and independent of
the properties of the OP AMP !
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OP-AMPS – “TAMING” THE WILD HIGH-GAIN AMPLIFIER
KEY CONCEPT: Negative feedback
Example:
R1
1K
VIN
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Circuit (assume R IN  )
R1
R2
R2
9K
+
1K
V0
VIN
EE 42 fall 2004 lecture 18
()
(+)
9K
V0
+ A  ( V  V )

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OP-AMP very high gain →predictable results
Analysis:
V(  ) 
R1
 A( V  V )
R1  R 2
V  VIN
Lets solve for Vthen find VO from
VO = A (V+ - V-)


AR 1

 V0  A(V  V )  AVIN 1 
 (A  1)R 1  R 2 
A(R 1  R 2 )
R1  R 2
V0  VIN
 VIN
(A  1)R 1  R 2
R1
V0  10VIN
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
 R1 
R1A 
V(  ) 1 
  AVIN 

 R1  R 2 
 R1  R 2 
AR1
V(  ) 
VIN
( A  1) R1  R 2
if A  
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OP-AMPS – Another Basic Circuit
Now lets look at the Inverting Amplifier
Example:
R1
VIN
1K
When the input is
not so large that
the output is hitting
the rails, we have
a circuit model:
R2
9K
+
VIN
V0
R1
1K
R2
()
(+)
9K
V0
+
 A  ( V  V )
Assume RIN  
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Inverting amplifier analysis
V(  )  0 so VO  -AV(  )
V(  )
Analysis:
R1
 VIN 
 (-AV(  )  VIN )
R1  R 2
Solve for V- then find VO from
VO = - AV-

 R2 
R 1A 
  VIN 

V(  ) 1 
 R1  R 2 
 R1  R 2 
R2
V(  ) 
VIN
(A  1)R 1  R 2
taking A  
R2
V0  -AVIN  -VIN
 -9VIN
R1
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