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