Recall-Lecture 3

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Recall-Lecture 3

Current generated due to two main factors Drift – movement of carriers due to the existence of

electric field Diffusion – movement of carriers due to gradient in

concentrations


Recall-Lecture 3 Introduction of PN junction

Space charge region/depletion region Built-in potential voltage Vbi

Reversed biased pn junction no current flow

Forward biased pn junction current flow due to diffusion of carriers.

Vbi

Analysis of PN Junction Diode in a Circuit


CIRCUIT REPRESENTATION OF DIODE

i

vD

The I -V characteristics of the ideal diode.

V = 0V

Conducting state

Reverse bias

Reverse biasedopen circuit

Conducting state short circuit


CIRCUIT REPRESENTATION OF DIODE – Piecewise Linear Model

i

vDV

Conducting state

Reverse bias

VD = V for diode to turn on.

Hence during conducting state:

=

V

Represented as a battery of voltage = V



i

vDV

Conducting state

Reverse bias

CIRCUIT REPRESENTATION OF DIODE – Piecewise Linear Model

VD ≥ V for diode to turn on.

Hence during conducting state:


=

V

Represented as a battery of voltage = V and forward resistance, rf in series rf

+ VD -


Diode Circuits: DC Analysis and ModelsDiode Circuits: DC Analysis and Models

Example

Consider a circuit with a dc voltage VPS applied across a resistor and a diode.

Applying KVL, we can write,

or,

The diode voltage VD and current ID are related by the ideal diode equation: (IS is assumed to be known for a particular diode)

Equation contains only one unknown, VD:

Why do you need the Piecewise Linear Model?


Diode Circuits: Direct ApproachDiode Circuits: Direct Approach

Question

Determine the diode voltage and current for the circuit.

Consider IS = 10-13 A.

and

ITERATION METHOD


Diode Circuits: Using ModelsDiode Circuits: Using Models

Example

Determine the diode voltage and current using a piecewise linear model.

Assume piecewise linear diode parameters of Vf = 0.6 V and rf = 10 Ω.

Solution:

The diode current is determined by:


DIODE DC ANALYSIS Find I and VO for the circuit shown below if the diode cut

in voltage is V = 0.7V

I = 0.2325mA

Vo = -0.35V

5 V 5 V

D 1

2 0 k 2 0 k+

VO

-

I


Find I and VO for the circuit shown below if the diode cut in voltage is V = 0.7V

2 V 8 V

D 1

5 k 2 0 k+

VO

-

I

I = 0.372mA

Vo = 0.14V


Example 2

a) Determine ID if V = 0.7VR = 4k

b) If VPS = 8V, what must be the value of R to get ID equal to part (a)

DIODE AC EQUIVALENT


Sinusoidal Analysis

The total input voltage vI = dc VPS + ac vi

iD = IDQ + id

vD = VDQ + vd

IDQ and VDQ are the DC diode current

and voltage respectively.


Current-voltage Relation

The relation between the diode current and voltage can be written as:

VDQ = dc quiescent voltage

vd = ac component

The -1 term in the equation is neglected.

The equation can be written as:

Diode Circuits: AC Equivalent CircuitDiode Circuits: AC Equivalent Circuit

If vd << VT, the equation can be expanded into linear series as:

The DC diode current Is:


iD = ID [ 1 + vd/VT]

iD = ID + ID vd / VT = ID + id

where id = ID vd / VT

using Ohm’s law:

I = V/R hence, id = vd / rd compare with id = ID vd / VT

which reveals that rd = VT / ID

CONCLUSION: During AC analysis the diode is equivalent to a resistor, rd


DC equivalent AC equivalent

rd

idIDQ

VDQ = V


Example 1Analyze the circuit (by determining VO & vo ).

Assume circuit and diode parameters of

VPS = 5 V, R = 5 kΩ, Vγ = 0.6 V & vi = 0.1 sin ωt

VDQ = V

IDQ


rd

id

DC ANALYSIS

DIODE = MODEL 1 ,2 OR 3

CALCULATE DC CURRENT, ID

CALCULATE rd

AC ANALYSIS

DIODE = RESISTOR, rd

CALCULATE AC CURRENT, id


EXAMPLE 2

Assume the circuit and diode parameters for the circuit below are

VPS = 10V, R = 20k, V = 0.7V, and vi = 0.2 sin t. Determine the current, IDQ and the time varying current, id

Recall-Lecture 3

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