files 3-handouts handout 2d

8/3/2019 Files 3-Handouts Handout 2d

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Diode Circuits Analysis

Graphical Method The intersection of the diode and theresistor current equations is calculated

Gives good understanding to the circuit operation Time consuming Not suitable for large circuits

Analyt ical Method Both diode and resistor current equations

are solved simultaneously or iteratively Faster than graphical method More accurate Time consuming

Not suitable for large circuits

Diode Models Piecewise-linear diode model is used to

replace the diode Least time consuming method Suitable for larger circuits

Accuracy depends on the used model

A simple diode circuit.

R

V V I

e I I

DS

R

nV

V

S D

T

D

−=

⎟⎟ ⎠ ⎞

⎜⎜⎝ ⎛ −=

EquationCurrentResistor

EquationCurrentDiode

1



Graphical MethodStep 1 : Locating the Operating Points Step 2 : KVL & KCL Constraints

Step 3 : Enforcing KVL & KCL Solution



Graphical Method (cont.)

Gives good understanding tothe circuit operation

Not suitable for large circuits

General Problem Basic Load Line Construction

Alternative Construction



Graphical Method (Example)Problem: Find Q-point

Given data: V s =10 V, R =10k Ω. Analysis:

To define the load line we use,V D = 0

V D = 5 V, I

D =0.5 mA

These points and the resultingload line are plotted.Q-point isgiven by the intersection of theload line and the diodecharacteristic:

Q-point = (0.95 mA, 0.6 V)



Analytical Method

R

V V I

e I I

DS

R

nV

V

S D

T

D

−=

⎟

⎟

⎠

⎞

⎜

⎜

⎝

⎛ −=

EquationCurrentResistor

EquationCurrentDiode

1

Example:Is=10-15 A, n=1, Vs=5 V, R=1K

Step 1:

Assume VD=VDo=0.7VID1=(Vs-VDo)/R=4.3 m.A

VD1=0.025ln(ID1 /Is)=0.7272 V

Step 2:ID2=(Vs-VD1)/R=4.2728 m.A VD2=0.025ln(ID2 /Is)=0.7271 V

Exact Value:ID=4.2729 m.A, VD=0.7271 V



Diode Models

Approximating the diode forwardcharacteristic with two straight lines.

Piecewise-linear model of the diodeforward characteristic and its equivalentcircuit representation.



Diode Models (cont.)

Development of the constant-voltage-dropmodel of the diode forward characteristics. A vertical straight line (b) is used toapproximate the fast-rising exponential.

The constant-voltage-drop model of the diode forwardcharacteristic and its equivalent circuit representation.

Also Ideal diode model may be used

which means short circuit in theforward direction and open circuitin the reverse direction.



Which Model to Use?



Analysis using Diode Models

Diode is assumed to be either on or off.

Analysis is conducted as follows:

1. Select diode model.2. Identify anode and cathode of diode and label v D and i D .3. Guess diode’s region of operation from circuit.4. Analyze circuit using diode model appropriate for

assumed operating region.5. Check results to check consistency with assumptions.

(For forward assumption check that iD > 0, for reverse

biased check that vD < VDO or zero for ideal diode model)



Since source is forcing positive currentthrough diode assume diode is on.

Q-point is(1 mA, 0V)

Since source is forcing currentbackward through diode assume diode

is off. Hence I D =0 . Loop equation is:

Q-point is (0, -10 V)

Analysis using Ideal Model for Diode: Example

0

m1k 10

V)010(

≥

=Ω

−=

D

D

I

A I

∵our assumption is right.

V10

010104

−=∴

=++

D

D D

V

I V

our assumption is right.

Find the Q-Point (ID, VD) for the following diodes assuming ideal diode model.



Analysis using Constant Voltage Drop Model for Diode

Analysis:Since 10V source is forcing positive currentthrough diode assume diode is on.

v D = V on for i D >0 andi D

= 0 for v D

< V on

.

mA94.0k 10

V)6.010(

k 10

V)10(

=Ω

−=

Ω

−= on

D

V I

Find the Q-Point (ID, VD) for the following diode assuming CVD, with VDO=0.6V.



Two-Diode Circuit Analysis

Analysis: Since 15V source is forcing positivecurrent through D 1 and D 2 and -10V source isforcing positive current through D 2, assume bothdiodes are on.

Since voltage at node D is zero due to shortcircuit of ideal diode D 1,

Q-points are (-0.5 mA, 0 V) and (2.0 mA, 0 V)

But, I D1 <0 is not allowed by diode, so try again.

mA50.1k 10

V)015(1 =Ω−= I mA00.2k 5

V)10(02 =Ω−−= D

I

211 D I

D I I += mA50.025.1

1−=−=∴

D I

Find the Q-Point (ID, VD) for the following diodes assuming ideal diode model.



Two-Diode Circuit Analysis (cont.)

Since current in D 1 is zero, I D2 = I 1,

Q-points are D 1 : (0 mA, -1.67 V):off

D 2 : (1.67 mA, 0 V) :on

Since current in D 2 but thatin D 1 is invalid, the second

guess is D 1 off and D 2 on.

V67.17.16151

000,10151

mA67.115,000

V251

0)10(2

000,51

000,1015

−=−=−=

=Ω=∴

=−−−−

I DV

I

D I I

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