6.002
CIRCUITS AND ELECTRONICS
Dependent Sources and Amplifiers
6.002 – Fall 2002: Lecture 8
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Review
Nonlinear circuits — can use the node method
Small signal trick resulted in linear response
Today
Dependent sources
Amplifiers
Reading: Chapter 7.1, 7.2
6.002 – Fall 2002: Lecture 8
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Dependent sources Seen previously Resistor Independent Current source
+ i + i
v
–
R v – I
v i= R
i=I
2-terminal 1-port devices New type of device: Dependent source iI i O
+ control port
f ( vI )
+
vI
vO
–
–
output port
2-port device E.g., Voltage Controlled Current Source Current at output port is a function of voltage at the input port 6.002 – Fall 2002: Lecture 8
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Dependent Sources: Examples
Example 1: Find V + R V –
independent current source
I = I0
V = I0R
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Dependent Sources: Examples Example 2: Find V + R V –
voltage controled current source
+ R V –
K I = f (V ) = V
iI +
f (vI ) =
K vI
iO +
vI
vO
–
–
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Dependent Sources: Examples Example 2: Find V voltage controled current source
+ R V –
K I = f (V ) = V e.g. K = 10-3 Amp·Volt R = 1kΩ
K V = IR = R V or V 2 = KR or V = KR = 10 −3 ⋅ 10 3 = 1 Volt
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Another dependent source example
RL iIN
vI + –
iD
+
+
vIN
vO
–
–
e.g.
VS + –
iD = f (vIN ) iD = f (vIN ) K 2 = (vIN − 1) for vIN ≥ 1 2 iD = 0 otherwise
Find vO as a function of vI .
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Another dependent source example VS RL iIN
vI + –
iD
+
+
vIN
vO
–
–
iD = f (vIN ) e.g.
iD = f (vIN ) K 2 = (vIN − 1) for vIN ≥ 1 2 iD = 0 otherwise
Find vO as a function of vI .
6.002 – Fall 2002: Lecture 8
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Another dependent source example VS RL vI vI
+ –
vO K 2 iD = (vIN − 1) for vIN ≥ 1 2 iD = 0 otherwise
Find vO as a function of vI .
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Another dependent source example VS RL vI vI
+ –
vO K 2 iD = (vIN − 1) for vIN ≥ 1 2 iD = 0 otherwise
KVL
− VS + iD RL + vO = 0 vO = VS − iD RL K 2 vO = VS − (vI − 1) RL 2 vO = VS
for vI ≥ 1 for vI < 1
Hold that thought
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Next, Amplifiers
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Why amplify? Signal amplification key to both analog and digital processing. Analog: AMP
IN
Input Port
OUT
Output Port
Besides the obvious advantages of being heard farther away, amplification is key to noise tolerance during communication
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Why amplify? Amplification is key to noise tolerance during communication No amplification
useful signal
1 mV
e nois
10 mV
huh?
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Try amplification e nois
AMP
not bad!
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Why amplify? Digital: Valid region 5V
5V
VIH IN VIL 0V
5V
OUT Digital System
IN
5V
VOL
OUT
V OH
VIH VIL
0V
0V
VOH
t
6.002 – Fall 2002: Lecture 8
V OL
t
0V
15
Why amplify? Digital:
Static discipline requires amplification! Minimum amplification needed: VIH VIL
6.002 – Fall 2002: Lecture 8
VOH VOL
VOH − VOL VIH − VIL
16
An amplifier is a 3-ported device, actually Power port Input port
iO
iI
+v – I
Amplifier
+ v Output – O port
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.”
POWER IN OUT
How do we build one? 6.002 – Fall 2002: Lecture 8
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Remember? VS RL vI vI
+ –
vO K 2 iD = (vIN − 1) for vIN ≥ 1 2 iD = 0 otherwise
KVL
− VS + iD RL + vO = 0 vO = VS − iD RL K 2 vO = VS − (vI − 1) RL 2 vO = VS
for vI ≥ 1 for vI < 1
Claim: This is an amplifier
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So, where’s the amplification? Let’s look at the vO versus vI curve. mA e.g. VS = 10V , K = 2 2 , RL = 5 kΩ V K 2 vO = VS − RL (vI − 1) 2 2 2 = 10 − ⋅10 −3 ⋅ 5 ⋅ 103 (vI − 1) 2 vO = 10 − 5 (vI − 1) vO VS
2
∆vO
1
∆vO >1 ∆v I 6.002 – Fall 2002: Lecture 8
∆vI
vI
amplification 19
Plot vO versus vI vO = 10 − 5 (vI − 1)
2
0.1 change in vI
Demo
vI
vO
0.0 1.0 1.5 2.0 2.1 2.2 2.3 2.4
10.00 10.00 8.75 5.00 4.00 2.80 1.50 ~ 0.00
1V change in vO
Gain!
Measure vO .
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One nit … vO
What happens here? 1
vI
Mathematically, K 2 vO = VS − RL (vI − 1) 2 So
is mathematically predicted behavior
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One nit … vO
K 2 vO = VS − RL (vI − 1) 2 What happens here? vI
1 However, from
iD =
K (vI − 1)2 2 VS
for vI ≥ 1
RL vO
VCCS
iD
For vO>0, VCCS consumes power: vO iD For vO<0, VCCS must supply power! 6.002 – Fall 2002: Lecture 8
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If VCCS is a device that can source power, then the mathematically predicted behavior will be observed —
vO
K 2 i.e. vO = VS − RL (vI − 1) 2 vI
where vO goes -ve
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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.
K 2 iD = (vI − 1) 2 will no longer be valid when vO ≤ 0 . e.g. iD saturates (stops increasing) and we observe: Commonly
vO
1
6.002 – Fall 2002: Lecture 8
vI
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