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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

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.