6.002 – Fall 2002: Lecture 8 1
6.002 CIRCUITS ANDELECTRONICS
Dependent Sourcesand Amplifiers
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
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
6.002 – Fall 2002: Lecture 8 4
Dependent Sources: Examples
independentcurrentsource
Example 1: Find V
0II =
+–VR
RIV 0=
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
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
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 .
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 .
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 =
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
6.002 – Fall 2002: Lecture 8 11
Next, Amplifiers
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
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
6.002 – Fall 2002: Lecture 8 14
AMP
Try amplification
not bad!
noise
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
6.002 – Fall 2002: Lecture 8 16
Why amplify?Digital:
Static discipline requires amplification!Minimum amplification needed:
ILVIHV
OLV
OHV
ILIH
OLOH
VVVV
−−
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+–
+–
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 =
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
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
6.002 – Fall 2002: Lecture 8 21
One nit …
1Iv
Ov
Mathematically,( )21
2−−= ILSO vRKVv
Whathappenshere?
So is mathematically predicted behavior
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?
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
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