10/11/2004ee 42 fall 2004 lecture 181 lecture #18 summary of ideal devices, amplifier examples...
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10/11/2004 EE 42 fall 2004 lecture 18 1
Lecture #18 Summary of ideal devices, amplifier examples
Reminder:
MIDTERM coming up one week from today (Monday October 18th)
This week: Review and examples
10/11/2004 EE 42 fall 2004 lecture 18 2
Midterm
• Monday, October 18,
• In class
• One page, one side of notes
10/11/2004 EE 42 fall 2004 lecture 18 3
Topics
Today:
• Summary of ideal circuits
• Amplifier examples– Comparator– Op-Amp
10/11/2004 EE 42 fall 2004 lecture 18 4
Ideal devices
Wire:Current in =current outNo voltage differences
Resistor
IRV
dt
dVCI
10/11/2004 EE 42 fall 2004 lecture 18 5
Ideal devices 2
Inductor:
dt
dILV
Ideal diode:Reversed bias no current, open circuitForward bias no voltage drop, just like a wire
10/11/2004 EE 42 fall 2004 lecture 18 6
Ideal devices 3
Transformer
2
1
2
1
N
N
V
V
+
V1
-
+
V2
-
N1 and N2 are the number of turns of wire
10/11/2004 EE 42 fall 2004 lecture 18 7
Ideal devices 4
~Voltage source:Voltage given, current can be anything
Note: the voltage could be given as A function of time
)(tV
Current sourceCurrent given, voltage can be anything
Note: the current could be given as A function of time
)(tI
10/11/2004 EE 42 fall 2004 lecture 18 8
Ideal devices 5
~Dependent Voltage source:Voltage given as a multiple of another Voltage or a current, current can be anything
12 KVV
Dependent Current sourceCurrent given as a multiple of a differentcurrent or voltage, voltage can be anything)(tI
+
V1
-
+
V2
-
+
V1
- 12 GVI
10/11/2004 EE 42 fall 2004 lecture 18 9
NPN transistor
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
Beta could be anywhere from 40 to 400
)( in tI
)(in tI
Emitter
Base
Collector
10/11/2004 EE 42 fall 2004 lecture 18 10
NMOS transistor
If VGS is higher than the threshold VTH, then the switch is closed.
Source
GateDrain
D
G
SThe resistance from the source to the drain depends on the how much greater than threshold VGS is, and how wide the transistor is.
10/11/2004 EE 42 fall 2004 lecture 18 11
PMOS transistor
If VGS is lower than the threshold VTH, then the switch is closed.
Source
Gate Drain
S
G
D
The resistance from the source to the drain depends on the how much below threshold (more negative) VGS is.
10/11/2004 EE 42 fall 2004 lecture 18 12
Amplifier
+
V0
+
VIN
• 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
10/11/2004 EE 42 fall 2004 lecture 18 13
Amplifier
+
V0
+
VIN
• V0=AVIN
• 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
V+rail
V+rail
10/11/2004 EE 42 fall 2004 lecture 18 14
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!
+
+
V0AV1
+
V1
Ri
Circuit Model in linear region)VV(AV0
+
AV+
V
V0
Differential Amplifier
“Differential” V0 depends only on difference (V+ V-)
“Very high gain” A 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.
We call the limits of the output the “rails”.
10/11/2004 EE 42 fall 2004 lecture 18 15
WHAT ARE I-V CHARACTERISTICS OF AN ACTUAL HIGH-GAIN DIFFERENTIAL AMPLIFIER ?
+
V0
+
VIN
• 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
10/11/2004 EE 42 fall 2004 lecture 18 16
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.
VIN(V)1 2 3
V0 (V)
0.1
0.2
3 2 1
.2
(a)V-V near origin
3
(b)V-V over wider range
VIN(V)10 20 30
V0 (V)
1
30 20 10
21
23
upper “rail”
lower “rail”
10/11/2004 EE 42 fall 2004 lecture 18 17
I-V CHARACTERISTICS OF AN ACTUAL HIGH-GAIN DIFFERENTIAL AMPLIFIER (cont.)
VIN(V)1 2 3
V0 (V)
1
2
3 2 1
23
1
3
(c)Same V0 vs VIN over even wider range
3
(b)V-V over wide range
VIN(V)10 20 30
V0 (V)
1
30 20 10
21
23
upper “rail”
lower “rail”
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
10/11/2004 EE 42 fall 2004 lecture 18 18
I-V CHARACTERISTICS OF AN ACTUAL HIGH-GAIN DIFFERENTIAL AMPLIFIER (cont.)
VIN(V)1 2 3
V0 (V)
1
2
3 2 1
23
1
3
(c)V-V with equal X and Y axes
Note:
• (a) displays linear amplifier behavior
• (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”
Now plot same thing but with equal horizontal and vertical scales (volts versus volts)
10/11/2004 EE 42 fall 2004 lecture 18 19
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 )
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.
If VIN > 1.010 V,V0 = 2V = Logic “1”
If VIN < 0.99 V,V0 = 0V = Logic “0”
V0
VIN1 20
1
2+ V0
VIN
+1V
V0VIN
Comparator
10/11/2004 EE 42 fall 2004 lecture 18 20
ONE-BIT A/D CONVERSION REQUIRED IN DIGITAL SYSTEMS
pulses in transmission
comparator regenerated
pulsespulses out
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.
10/11/2004 EE 42 fall 2004 lecture 18 21
OP-AMPS AND COMPARATORS
A very high-gain differential amplifier can function in extremely linear fashion as an operational amplifier by using negative feedback.
Negative feedback Stabilizes the output
R2R1
+ V0VIN
EXAMPLE
We can show that that for A (and Ri 0 for simplicity)
1
21IN0 R
RRVV
+
+
V0AV1
-
+V1
Ri
R2
Circuit Model
R1
VIN
Stable, finite, and independent of the properties of the OP AMP !
10/11/2004 EE 42 fall 2004 lecture 18 22
OP-AMPS – “TAMING” THE WILD HIGH-GAIN AMPLIFIER
KEY CONCEPT: Negative feedback
Circuit (assume )R IN
V0
(+)()1K
VIN
9K
R2R1
+ )VV(A
V0-+
1K
VIN
9K
R2R1
Example:
10/11/2004 EE 42 fall 2004 lecture 18 23
OP-AMP very high gain →predictable results
Analysis:
IN
21
1)(
VV
)VV(ARR
RV
IN21
1)(
21
1IN
21
1)(
VRR)1A(
ARV
RRR
AVRRAR
1V
A if 10VV
R
RRV
R1)R(A
)RA(RVV
R1)R(A
AR1AV)VA(VV
IN0
1
21IN
21
21IN0
21
1IN0Lets solve for V-
then find VO from
VO = A (V+ - V-)
10/11/2004 EE 42 fall 2004 lecture 18 24
OP-AMPS – Another Basic Circuit
Now lets look at the Inverting Amplifier
INR Assume
V0
(+)()1K
VIN
9K
R2R1
+ )VV(A
V0-+
1KVIN
9K
R2R1
Example:
When the input is not so large that the output is hitting the rails, we have a circuit model:
10/11/2004 EE 42 fall 2004 lecture 18 25
Inverting amplifier analysis
Analysis:
)V(-AVRR
RVV
-AVV so 0V
IN)(21
1IN)(
)(O)(
IN21
2)(
21
2IN
21
1)(
VR1)R(A
RV
RR
RV
RR
AR1V
-9VR
R-V-AVV IN
1
2ININ0
Solve for V- then find VO from
VO = - AV-
A taking
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