# Recall-Lecture 3 - PowerPoint PPT Presentation

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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

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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.

Reverse bias

Conducting state

V = 0V

Reverse biased

open circuit

Conducting state

short circuit

### CIRCUIT REPRESENTATION OF DIODE – Piecewise Linear Model

i

vD

Represented as a battery of voltage = V

=

V

Reverse bias

Conducting state

Reverse biased

open circuit

V

VD =V for diode to turn on.

Hence during conducting state:

i

vD

Represented as a battery of voltage = V and forward resistance, rfin series

+ VD -

=

rf

V

CIRCUIT REPRESENTATION OF DIODE – Piecewise Linear Model

Reverse bias

Conducting state

Reverse biased

open circuit

V

VD ≥V for diode to turn on.

Hence during conducting state:

Diode Circuits: DC Analysis and Models

Example

Consider a circuit with a dc voltage VPSapplied 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 Approach

Question

Determine the diode voltage and current for the circuit.

Consider IS = 10-13 A.

and

ITERATION METHOD

Diode 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

I

• Find I and VOfor the circuit shown below if the diode cut in voltage is V = 0.7V

+

VO

-

I = 0.2325mA

Vo = -0.35V

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

I

+

VO

-

I = 0.372mA

Vo = 0.14V

Example 2

Determine ID if V = 0.7V

R = 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 = dcVPS + acvi

iD= IDQ + id

vD = VDQ+ vd

IDQ and VDQ are the DC diode current

and voltage respectively.

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

The DC diode current Is:

Diode Circuits: AC Equivalent Circuit

• 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:

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

VDQ = V

rd

IDQ

id

DC equivalent

AC equivalent

VDQ = V

IDQ

Example 1

Analyze 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

rd

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 quiescent current, IDQ and the time varying current.