Diodes Two - Engineering Electrical Circuits - Lecture Slides, Slides of Electrical Circuit Analysis

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• Diodes-

Learning Goals

• Understand the Basic Physics of

Semiconductor PN Junctions which form most

Diode Devices

• Sketch the IV Characteristics of Typical PN

Junction Diodes

• Use the Graphical LOAD-LINE method to

determine the “Operating Point” of Nonlinear

(includes Diodes) Circuits

Diode Models

• LoadLine Analysis works

well when the ckt

connected to a SINGLE

Diode can be

“Thevenized”

• However, for NONLinear

ckts, such as those

containing multiple

diodes, construction of

the LOAD-Curve Eqn

may be difficult, or even

impossible.

• Many such ckts can be

analyzed by Idealizing

the diode

Diode Models

• Consider an Electrical Diode →

 We can MODEL the V-I

Behavior of this Device in

Several ways

V

I

REAL Behavior

IDEAL Model

OFFSET Model

LINEAR Model

Ideal Model (Ideal Rectifier)

3. Check to see if guesses for i-flow, ∆V, and

BIAS-State are consistent with the Ideal-

Diode Model

4. If i-flow, ∆V, and bias-V are consistent with

the ideal model, then We’re DONE.

• If we arrive at even a SINGLE Inconsistency, then

START OVER at step-

Diode OFF^ Diode ON

Example  Ideal Diode

• Find For Ckt Below find:
  • Use the

Ideal

Diode

Model

I (^) d 1 & V o

↓ Id 1

Id 2 ↓

A

Example  Ideal Diode

• Using I (^) D2 = 1 mA
  • Thus
  • Now must Check that

both Diodes are indeed

conducting

• From the analysis
• Thus the current thru

both Diodes is positive

which is consistent with

the assumption 

↓ Id 1

Id 2 ↓

A

( [ ]) 1 mA

1. 9 k

0 10 V

Id =

Id 1 = + 0. 0101 mA

Id 1 = + 10 μA Id 2 = + 1 mA

Example  Ideal Diode

• Since both Diodes conduct the Top of Vo is connected to GND thru D2 & D - Another way to think about this is that since VD2 = 0 and V (^) D1 = 0 (by Short Assumption) Find Vo = GND+V (^) D2+VD1 = GND + 0 + 0 = 0 - Thus the Answer

↓ Id 1

Id 2 ↓

A

Id 1 = + 10 μA Vo = 0 V

Example  Ideal Diode

• Again Assume BOTH Diodes are ON, or Conducting - As Before VD1 = V (^) D2 = 0 - Again V (^) B shorted to GND thru D - Then Find by Ohm - Now use KCL at Node-B (in = out)

( )

1. 01 mA
2. 9 k

10 0 V

Id =

( [ ]) Ω

10 k

0 10 V

Id 1 Id 2

↓ Id 1

Id 2 ↓

B

Example  Ideal Diode

• Using I (^) D2 = 1.01 mA
  • Thus
  • Now must Check that

both Diodes are indeed

conducting

• From the analysis
• We find and

INCONSISTENCY and

our Assumption is

WRONG 

↓ Id 1

Id 2 ↓

B

( [ ])

1. 01 mA 10 k

0 10 V

Id =

Id 1 = − 0. 01 mA

Id 1 = − 10 μA Id 2 = + 1. 01 mA

Example  Ideal Diode

• Must Check that D1 is REVERSE Biased as it is assumed OFF - By KVL & Ohm - Thus D1 is INDEED Reverse-Biased, Thus the Ckt operation is Consistent with our Assumption 

↓ Id 1

Id 2 ↓

B

1. 05 V 50 mV

10 V 10 kΩ 1. 005 mA

= + = +

= − + ⋅

B

B V

V

Example  Ideal Diode

• Calculate Vo by noting

that:

• D2 is ON → VD2 = 0

D1 is OFF → Current

can only flow thru D

• In this case Vo = VB
• By the Previous

Calculation, Find

↓ Id 1

Id 2 ↓

B

Id 1 = 0 A Vo = + 50 mV

Point Slope Line Eqn

• When constructing

multipiece-wise linear

models, the Point-Slope

Equation is extremely

Useful

• Where
  • (x 1 , y 1 ) & (x 2 , y 2 ) are KNOWN Points - Example: Find Eqn for

line-segment:

(^42 4 6 8 10 12 14 16 18 )

6

8

10

12

14

16

18

x

y

( y − y 1 ) = m ( x − x 1 )

2 1

2 1 x x

y y x

y m −

(3,17)

(19,5)

4

3

16

12 19 3

5 17

= −

= − −

−

∆

∆

m

x

y m

Point Slope Line Eqn

• Using the 2nd^ Point
• Can easily convert to y =

mx+b

• Multiply by m, move −

to other side of =

(^42 4 6 8 10 12 14 16 18 )

6

8

10

12

14

16

18

x

y

( ) ( 19 ) 4

y − 5 = − x −

(3,17)

(19,5)

( ) ( )

y x

y x

y x

y x

y x