bruce mayer, pe registered electrical & mechanical engineer bmayer@chabotcollege

[email protected] • ENGR-44_Lec-02-8_Dep-Sources.ppt1

Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis

Bruce Mayer, PERegistered Electrical & Mechanical Engineer

[email protected]

Engineering 43

Chp 2.8Chp 2.8DependentDependent

SourcesSources



Ckts With Dependent SourcesCkts With Dependent Sources A Convention About Dependent Sources

• Unless Otherwise Specified The Current And Voltage Variables Are Assumed In SI Units Of Amps And Volts

XDIV

CONTROLLINGVARIABLE

DEPENDENTVARIABLE

• For This Example The Multiplier, , Must Have Units Of V/A or Ω



Other Dependent SourcesOther Dependent Sources

Three More Types of Dependent Sources• VD = VX ( a Unitless Scalar)

• ID = VX ( in Units of A/V, or Siemens)

• ID = IX ( a Unitless Scalar)



Analysis StrategyAnalysis Strategy General Strategy

• Treat Dependent Sources As Regular Sources And Add One More Equation For The Controlling Variable

Plan for Ckt at Rt• Single Loop Circuit

• Use KVL To Determine Current– KVL:

AV

KVL

05312 11 IkVIkV A

• 1-Equation, TWO-Unknowns: I1 and VA

– Controlling Variable Provides the Extra Equation



Analysis Example cont.Analysis Example cont. The Dependent Source

Control Equation

11 k22000 IIVA

mAk

VIVIk

IkIkIkV

26

12126

052312

11

111

VmAkIkVO 10255 1

Substitute VA Relation Into KVL

Use Ohm’s Law to Find VO



ExampleExample Find VO a

Strategy• If VS Is Known, Then VO Can Be Determined

Using Voltage Divider

• To Find VS We Note That We Have A Single Node-Pair– Mark Node-A



Example cont.1Example cont.1 Apply KCL (out POS) & Ohm at Node-A

kVI SO 3

This Single Eqn Contains TWO Unknowns• Use The Equation For The Dependent Source

Controlling Variable To Eliminate IO In Favor Of VS

• Sub For IO (Be Sure Maintain Consistent Units)



Example cont.2Example cont.2• Solving for VS (mult. Both sides by LCD of 6 kΩ)

VVVVk

V

k

V

k

V

kk SS

SSS 1256003

4

361

106

VVkk

kVV SO 8

3

212

42

4

• Now Apply The Voltage Divider Equation to Find VO



ExampleExample Find VO

The Plan• A One Loop Problem

• Find The Current

• Then Use Ohm’s Law

The Dependent Source Is Just One More Voltage Source• Apply KVL

KVL TOTHIS LOOP

012312 IkVIkV O



Example cont.Example cont. Use the Dependent Source Control Equation

to Eliminate VO

Solving for the Current, I

Finally, Solve for VO Using Ohm’s Law

mA2

121123

I

VIkIkIk



FET Amp ExampleFET Amp Example Find Voltage Gain or Amplification

tvtv

Ai

ov

KVL

KCL

The Plan• One Loop On The Left → KVL

• One Node-pair On Right → KCL

KVL



FET Amp Example cont.1FET Amp Example cont.1 Ohm’s Law

0)(

)( L

Ogm R

tvtvg

LoL Rtvti

Determined by a Voltage Divider

Load Resistor & vo

Apply KCL At Large Node

Then vo

Li

And from Above have vg as a function of vi



FET Amp Example cont.2FET Amp Example cont.2 Finally the Gain

Note That the Common-Source FET is an INVERTING Amplifier



WhiteBoard WorkWhiteBoard Work

Let’s Work Problem This Problem

8

4

4 3 I S

I o

6 V

If the OutPut Current Io is 2A

Then Find the Supply Current IS

bruce mayer, pe registered electrical & mechanical engineer bmayer@chabotcollege

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