6.002x CIRCUITS AND ELECTRONICS · 5 Dependent sources A new element for our toolchest; can be...
Transcript of 6.002x CIRCUITS AND ELECTRONICS · 5 Dependent sources A new element for our toolchest; can be...
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6.002x CIRCUITS AND ELECTRONICS
Dependent Sources Amplifiers
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n Nonlinear circuits — can use the
node method
n Small signal trick resulted in linear response
Today
n Dependent sources
Reading: Chapter 7.1, 7.2
Review
n Amplifiers
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Dependent Sources
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4 2-terminal 1-port devices
Independent Current source
Resistor
Elements we have seen previously
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Dependent sources A new element for our toolchest; can be linear or nonlinear
E.g., Voltage Controlled Current Source Current at output port is a function of voltage at the input port
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Dependent source in a circuit
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First, an Example with an independent source Example 1: Find V
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voltage controled current source
Example 2: Find V
Dependent Sources: Example
Note that above is abstracted view of:
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voltage controled current source
Example 2: Find V
Dependent Sources: Examples
( )VKVfI ==
+
– V e.g. K = 10 -3 Amp·Volt, R = 1kΩ
R
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Other Types of Dependent Sources
Similarly, CCVS, VCVS
iI iO
control port
output port
+
–
+
– vO vI
control port
output port
+
–
+
–
iI iO
vO vI
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Another dependent source example
Di+
–
+
–
vO
iIN
vIN
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Simplify our Drawing
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Solve
Hold that thought
+ –
( )2IND 1v2Ki −= for vIN ≥ 1
otherwise 0iD =
VS
vI
vO
RL
vIN
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Plot vO versus vI curve for our dependent source example
0
vO
vI
+ –
VS
vI
vO
RL
vIN
( ) 1for 12
2 VvRvKVv ILISO ≥−−=
VvVv ISO 1for <=
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One way - leave all dependent sources in (note, dependent sources must be linear!) - solve for one independent source at a time - [section 3.5.1 of the text]
Superposition with Dependent Sources
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Next, Amplifiers
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Why amplify? Signal amplification key to both analog and digital processing.
Analog:
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
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Try amplification
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Why amplify? Amplification if fundamental to the digital domain as well
IN OUT
Digital System
5 V
0 V
5 V
0 V
5 V
0 V
IN OUT 5 V
0 V
VOH
VOL
VIH
VIL
t t
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Why amplify? Digital domain:
Static discipline requires amplification! Minimum amplification needed:
VOH
VOL VIL
VIH
IN OUT
Digital System
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An amplifier is a 3-ported device, actually
How do we build one?
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.”
IN OUT Amplifier
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You already have! Remember…
Claim: This is an amplifier
Node method:
LDSO RiVv −=
( ) LISO RvKVv 212
−−= for vI ≥ 1
for vI < 1
0=+−
DL
SO iRVv
( )2IND 1v2Ki −= for vIN ≥ 1
otherwise
+ –
VS
RL vO
vI
vI
iD = 0
vO = VS
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So, where’s the amplification? vO versus vI curve
1
VS
vO
vI
Ω=== kRVmAKVV LS 5,2,10 2e.g.
( )212
−−= ILSO vRKVv
( )21510 −−= IO vv
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Plot vO versus vI ( )2IO 1v510v −−=
vI vO
Measure vO .
Demo
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One nit … Mathematically,
( )212
−−= ILSO vRKVv
So this is mathematically predicted behavior in our amplifier built with an abstract dependent source
But, what happens Here with a practical device?
1 0
vO
vI
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One nit …
1
( )212
−−= ILSO vRKVv
What happens here?
vO
vI
However, looking at the circuit
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If our VCCS is a device that can source (or supply) power then the mathematically predicted behavior will be observed
( )212
−−= ILSO vRKVvi.e.
where vO goes -ve
vO
vI
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However, if our VCCS is a passive device i.e., it does not provide power gain
( )212
−= ID vKiCommonly will no longer be valid when vO ≤ 0 .
So, something must give!
Turns out, our model breaks down.
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e.g. iD saturates (stops increasing) and we observe:
vO
vI
We will look at a practical device shortly…
Demo
Then it cannot source power, so vO cannot go -ve.