© Electronics ECE 1231 Recall-Lecture 4 Current generated due to two main factors Drift –...

© Electronics© Electronics ECE 1231ECE 1231

Recall-Lecture 4

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 4 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


PN Junction DiodePN Junction Diode The basic PN junction diode circuit symbol, and

conventional current direction and voltage polarity.

The graphs shows the ideal I-V characteristics of a PN junction diode.

The diode current is an exponential function of diode voltage in the forward-bias region.

The current is very nearly zero in the reverse-bias region.


PN Junction DiodePN Junction Diode Temperature Effects

Both IS and VT are functions of temperature.

The diode characteristics vary with temperature.

For silicon diodes, the change is approximately 2 mV/oC.

•Forward-biased PN junction characteristics versus temperature. •The required diode voltage, V to produce a given current decreases with an increase in temperature.

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 to use these models?


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, current and the

power dissipated by the diode using

piecewise linear model.

Assume piecewise linear diode parameters of

V = 0.6 V and rf = 10 Ω.

Solution:

The diode current is determined by:

Power dissipated: VD x ID = 0.622 x 2.19 = 1.36 mW


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)


DC Load Line

A linear line equation ID versus VD

Obtain the equation using KVL


2 2

Use KVL: 2ID + VD – 5 = 0ID = -VD + 5 = - VD + 2.5

V

The value of ID at VD = V

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

© Electronics ECE 1231 Recall-Lecture 4 Current generated due to two main factors Drift –...

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