Dependent Sources and Amplifiers · 2013. 5. 22. · 6.002 – Fall 2002: Lecture 8 3 Dependent...
Transcript of Dependent Sources and Amplifiers · 2013. 5. 22. · 6.002 – Fall 2002: Lecture 8 3 Dependent...
![Page 1: Dependent Sources and Amplifiers · 2013. 5. 22. · 6.002 – Fall 2002: Lecture 8 3 Dependent sources + v – i R R v Resistor i = 2-terminal 1-port devices + v – i I i =I Independent](https://reader033.fdocuments.net/reader033/viewer/2022051900/5fef7e43ae34407b900eed04/html5/thumbnails/1.jpg)
6.002 – Fall 2002: Lecture 8 1
6.002 CIRCUITS ANDELECTRONICS
Dependent Sourcesand Amplifiers
![Page 2: Dependent Sources and Amplifiers · 2013. 5. 22. · 6.002 – Fall 2002: Lecture 8 3 Dependent sources + v – i R R v Resistor i = 2-terminal 1-port devices + v – i I i =I Independent](https://reader033.fdocuments.net/reader033/viewer/2022051900/5fef7e43ae34407b900eed04/html5/thumbnails/2.jpg)
6.002 – Fall 2002: Lecture 8 2
Nonlinear circuits — can use thenode methodSmall signal trick resulted in linearresponse
TodayDependent sources
Reading: Chapter 7.1, 7.2
Review
Amplifiers
![Page 3: Dependent Sources and Amplifiers · 2013. 5. 22. · 6.002 – Fall 2002: Lecture 8 3 Dependent sources + v – i R R v Resistor i = 2-terminal 1-port devices + v – i I i =I Independent](https://reader033.fdocuments.net/reader033/viewer/2022051900/5fef7e43ae34407b900eed04/html5/thumbnails/3.jpg)
6.002 – Fall 2002: Lecture 8 3
Dependent sources
+ –v
Ri Rvi =Resistor
2-terminal 1-port devices
+ –v
i Ii =IIndependentCurrent source
Seen previously
control port
outputport
Ii
Iv
Oi
Ov
+
–
+
–
New type of device: Dependent source
2-port device
E.g., Voltage Controlled Current SourceCurrent at output port is a function of voltage at the input port
)v(f I
![Page 4: Dependent Sources and Amplifiers · 2013. 5. 22. · 6.002 – Fall 2002: Lecture 8 3 Dependent sources + v – i R R v Resistor i = 2-terminal 1-port devices + v – i I i =I Independent](https://reader033.fdocuments.net/reader033/viewer/2022051900/5fef7e43ae34407b900eed04/html5/thumbnails/4.jpg)
6.002 – Fall 2002: Lecture 8 4
Dependent Sources: Examples
independentcurrentsource
Example 1: Find V
0II =
+–VR
RIV 0=
![Page 5: Dependent Sources and Amplifiers · 2013. 5. 22. · 6.002 – Fall 2002: Lecture 8 3 Dependent sources + v – i R R v Resistor i = 2-terminal 1-port devices + v – i I i =I Independent](https://reader033.fdocuments.net/reader033/viewer/2022051900/5fef7e43ae34407b900eed04/html5/thumbnails/5.jpg)
6.002 – Fall 2002: Lecture 8 5
voltagecontroledcurrentsource
Example 2: Find V
( )VKVfI ==
+–VR
Ii
Iv
Oi
Ov
+
–
+
–
+–VR
( )I
I vKvf =
Dependent Sources: Examples
![Page 6: Dependent Sources and Amplifiers · 2013. 5. 22. · 6.002 – Fall 2002: Lecture 8 3 Dependent sources + v – i R R v Resistor i = 2-terminal 1-port devices + v – i I i =I Independent](https://reader033.fdocuments.net/reader033/viewer/2022051900/5fef7e43ae34407b900eed04/html5/thumbnails/6.jpg)
6.002 – Fall 2002: Lecture 8 6
voltagecontroledcurrentsource
RVKIRV ==
KRV =2
KRV =33 1010 ⋅= −
Volt1=
oror
Example 2: Find V
( )VKVfI ==
+–VR
e.g. K = 10-3 Amp·VoltR = 1kΩ
Dependent Sources: Examples
![Page 7: Dependent Sources and Amplifiers · 2013. 5. 22. · 6.002 – Fall 2002: Lecture 8 3 Dependent sources + v – i R R v Resistor i = 2-terminal 1-port devices + v – i I i =I Independent](https://reader033.fdocuments.net/reader033/viewer/2022051900/5fef7e43ae34407b900eed04/html5/thumbnails/7.jpg)
6.002 – Fall 2002: Lecture 8 7
Another dependent source example
Iv +–
( )IND vfi =
LR +–SV
e.g. ( )IND vfi =
( )2IN 1v2K
−= for vIN ≥ 1
INi
INv
Di
Ov+
–
+
–
otherwise0iD =
Find vO as a function of vI .
![Page 8: Dependent Sources and Amplifiers · 2013. 5. 22. · 6.002 – Fall 2002: Lecture 8 3 Dependent sources + v – i R R v Resistor i = 2-terminal 1-port devices + v – i I i =I Independent](https://reader033.fdocuments.net/reader033/viewer/2022051900/5fef7e43ae34407b900eed04/html5/thumbnails/8.jpg)
6.002 – Fall 2002: Lecture 8 8
Another dependent source example
Iv +–
( )IND vfi =
LR
SV
e.g. ( )IND vfi =
( )2IN 1v2K
−= for vIN ≥ 1
INi
INv
Di
Ov+
–
+
–
otherwise0iD =
Find vO as a function of vI .
![Page 9: Dependent Sources and Amplifiers · 2013. 5. 22. · 6.002 – Fall 2002: Lecture 8 3 Dependent sources + v – i R R v Resistor i = 2-terminal 1-port devices + v – i I i =I Independent](https://reader033.fdocuments.net/reader033/viewer/2022051900/5fef7e43ae34407b900eed04/html5/thumbnails/9.jpg)
6.002 – Fall 2002: Lecture 8 9
Another dependent source example
Find vO as a function of vI .
Iv +–
Iv
SV
OvLR
( )2IND 1v2Ki −= for vIN ≥ 1
otherwise0iD =
![Page 10: Dependent Sources and Amplifiers · 2013. 5. 22. · 6.002 – Fall 2002: Lecture 8 3 Dependent sources + v – i R R v Resistor i = 2-terminal 1-port devices + v – i I i =I Independent](https://reader033.fdocuments.net/reader033/viewer/2022051900/5fef7e43ae34407b900eed04/html5/thumbnails/10.jpg)
6.002 – Fall 2002: Lecture 8 10
Another dependent source example
0=++− OLDS vRiVKVL
LDSO RiVv −=
( ) LISO RvKVv 212
−−= for vI ≥ 1
SO Vv = for vI < 1
Iv +–
Iv
SV
OvLR
( )2IND 1v2Ki −= for vIN ≥ 1
otherwise0iD =
Hold that thought
![Page 11: Dependent Sources and Amplifiers · 2013. 5. 22. · 6.002 – Fall 2002: Lecture 8 3 Dependent sources + v – i R R v Resistor i = 2-terminal 1-port devices + v – i I i =I Independent](https://reader033.fdocuments.net/reader033/viewer/2022051900/5fef7e43ae34407b900eed04/html5/thumbnails/11.jpg)
6.002 – Fall 2002: Lecture 8 11
Next, Amplifiers
![Page 12: Dependent Sources and Amplifiers · 2013. 5. 22. · 6.002 – Fall 2002: Lecture 8 3 Dependent sources + v – i R R v Resistor i = 2-terminal 1-port devices + v – i I i =I Independent](https://reader033.fdocuments.net/reader033/viewer/2022051900/5fef7e43ae34407b900eed04/html5/thumbnails/12.jpg)
6.002 – Fall 2002: Lecture 8 12
Why amplify?Signal amplification key to both analogand digital processing.
Analog:
Besides the obvious advantages of beingheard farther away, amplification is keyto noise tolerance during communication
AMPIN OUT
InputPort
OutputPort
![Page 13: Dependent Sources and Amplifiers · 2013. 5. 22. · 6.002 – Fall 2002: Lecture 8 3 Dependent sources + v – i R R v Resistor i = 2-terminal 1-port devices + v – i I i =I Independent](https://reader033.fdocuments.net/reader033/viewer/2022051900/5fef7e43ae34407b900eed04/html5/thumbnails/13.jpg)
6.002 – Fall 2002: Lecture 8 13
Why amplify?
Amplification is key to noise tolerance during communication
usefulsignal
huh?
1 mVnoise
10 mV
No amplification
![Page 14: Dependent Sources and Amplifiers · 2013. 5. 22. · 6.002 – Fall 2002: Lecture 8 3 Dependent sources + v – i R R v Resistor i = 2-terminal 1-port devices + v – i I i =I Independent](https://reader033.fdocuments.net/reader033/viewer/2022051900/5fef7e43ae34407b900eed04/html5/thumbnails/14.jpg)
6.002 – Fall 2002: Lecture 8 14
AMP
Try amplification
not bad!
noise
![Page 15: Dependent Sources and Amplifiers · 2013. 5. 22. · 6.002 – Fall 2002: Lecture 8 3 Dependent sources + v – i R R v Resistor i = 2-terminal 1-port devices + v – i I i =I Independent](https://reader033.fdocuments.net/reader033/viewer/2022051900/5fef7e43ae34407b900eed04/html5/thumbnails/15.jpg)
6.002 – Fall 2002: Lecture 8 15
Why amplify?Digital:
IN OUT
Digital System
ILVIHV
5V
0V OLV
OHV5V
0V
t
5V
0V
ILVIHV
IN OUT
t
5V
0VOLV
OHV
Valid region
![Page 16: Dependent Sources and Amplifiers · 2013. 5. 22. · 6.002 – Fall 2002: Lecture 8 3 Dependent sources + v – i R R v Resistor i = 2-terminal 1-port devices + v – i I i =I Independent](https://reader033.fdocuments.net/reader033/viewer/2022051900/5fef7e43ae34407b900eed04/html5/thumbnails/16.jpg)
6.002 – Fall 2002: Lecture 8 16
Why amplify?Digital:
Static discipline requires amplification!Minimum amplification needed:
ILVIHV
OLV
OHV
ILIH
OLOH
VVVV
−−
![Page 17: Dependent Sources and Amplifiers · 2013. 5. 22. · 6.002 – Fall 2002: Lecture 8 3 Dependent sources + v – i R R v Resistor i = 2-terminal 1-port devices + v – i I i =I Independent](https://reader033.fdocuments.net/reader033/viewer/2022051900/5fef7e43ae34407b900eed04/html5/thumbnails/17.jpg)
6.002 – Fall 2002: Lecture 8 17
An amplifier is a 3-ported device, actually
We often don’t show the power port.
Also, for convenience we commonly observe“the common ground discipline.”In other words, all ports often share a common reference point called “ground.”
How do we build one?
POWERIN OUT
Amplifier
Power port
Inputport
Outputport
Ii
IvOi
Ov+–
+–
![Page 18: Dependent Sources and Amplifiers · 2013. 5. 22. · 6.002 – Fall 2002: Lecture 8 3 Dependent sources + v – i R R v Resistor i = 2-terminal 1-port devices + v – i I i =I Independent](https://reader033.fdocuments.net/reader033/viewer/2022051900/5fef7e43ae34407b900eed04/html5/thumbnails/18.jpg)
6.002 – Fall 2002: Lecture 8 18
Remember?
0=++− OLDS vRiVKVL
LDSO RiVv −=
( ) LISO RvKVv 212
−−= for vI ≥ 1
SO Vv = for vI < 1
Claim: This is an amplifier
Iv +–
Iv
SV
OvLR
( )2IND 1v2Ki −= for vIN ≥ 1
otherwise0iD =
![Page 19: Dependent Sources and Amplifiers · 2013. 5. 22. · 6.002 – Fall 2002: Lecture 8 3 Dependent sources + v – i R R v Resistor i = 2-terminal 1-port devices + v – i I i =I Independent](https://reader033.fdocuments.net/reader033/viewer/2022051900/5fef7e43ae34407b900eed04/html5/thumbnails/19.jpg)
6.002 – Fall 2002: Lecture 8 19
So, where’s the amplification?Let’s look at the vO versus vI curve.
amplification1>∆∆
I
O
vv
Ω=== k5R,VmA2K,V10V L2Se.g.
Ov∆
Iv∆
( )212
−−= ILSO vRKVv
( )21510 −−= IO vv
( )233 1105102210 −⋅⋅⋅−= −
Iv
1 Iv
SVOv
![Page 20: Dependent Sources and Amplifiers · 2013. 5. 22. · 6.002 – Fall 2002: Lecture 8 3 Dependent sources + v – i R R v Resistor i = 2-terminal 1-port devices + v – i I i =I Independent](https://reader033.fdocuments.net/reader033/viewer/2022051900/5fef7e43ae34407b900eed04/html5/thumbnails/20.jpg)
6.002 – Fall 2002: Lecture 8 20
Plot vO versus vI
( )2IO 1v510v −−=
10.001.0
~ 0.002.41.502.32.802.24.002.15.002.08.751.5
10.000.0vOvI
0.1 changein vI
1V changein vO
Gain!
Measure vO .Demo
![Page 21: Dependent Sources and Amplifiers · 2013. 5. 22. · 6.002 – Fall 2002: Lecture 8 3 Dependent sources + v – i R R v Resistor i = 2-terminal 1-port devices + v – i I i =I Independent](https://reader033.fdocuments.net/reader033/viewer/2022051900/5fef7e43ae34407b900eed04/html5/thumbnails/21.jpg)
6.002 – Fall 2002: Lecture 8 21
One nit …
1Iv
Ov
Mathematically,( )21
2−−= ILSO vRKVv
Whathappenshere?
So is mathematically predicted behavior
![Page 22: Dependent Sources and Amplifiers · 2013. 5. 22. · 6.002 – Fall 2002: Lecture 8 3 Dependent sources + v – i R R v Resistor i = 2-terminal 1-port devices + v – i I i =I Independent](https://reader033.fdocuments.net/reader033/viewer/2022051900/5fef7e43ae34407b900eed04/html5/thumbnails/22.jpg)
6.002 – Fall 2002: Lecture 8 22
One nit …
Di
SV
OvLR
VCCS
1Iv
Ov
For vO>0, VCCS consumes power: vO iDFor vO<0, VCCS must supply power!
( )212
−−= ILSO vRKVv
( )212
−= ID vKi for vI ≥ 1However, from
Whathappenshere?
![Page 23: Dependent Sources and Amplifiers · 2013. 5. 22. · 6.002 – Fall 2002: Lecture 8 3 Dependent sources + v – i R R v Resistor i = 2-terminal 1-port devices + v – i I i =I Independent](https://reader033.fdocuments.net/reader033/viewer/2022051900/5fef7e43ae34407b900eed04/html5/thumbnails/23.jpg)
6.002 – Fall 2002: Lecture 8 23
If VCCS is a device that can source power, then the mathematically predicted behavior will be observed —
( )212
−−= ILSO vRKVvi.e.
where vO goes -veIv
Ov
![Page 24: Dependent Sources and Amplifiers · 2013. 5. 22. · 6.002 – Fall 2002: Lecture 8 3 Dependent sources + v – i R R v Resistor i = 2-terminal 1-port devices + v – i I i =I Independent](https://reader033.fdocuments.net/reader033/viewer/2022051900/5fef7e43ae34407b900eed04/html5/thumbnails/24.jpg)
6.002 – Fall 2002: Lecture 8 24
If VCCS is a passive device,then it cannot source power, so vO cannot go -ve.So, something must give!Turns out, our model breaks down.
( )212
−= ID vKiCommonly
will no longer be valid when vO ≤ 0 .e.g. iD saturates (stops increasing)and we observe:
Iv
Ov
1