© electronics ece 1231 recall-lecture 4 current generated due to two main factors drift –...

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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

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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

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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.

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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

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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

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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

Reverse biasedopen circuit

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i

vDV

Conducting state

Reverse bias

CIRCUIT REPRESENTATION OF DIODE – Piecewise Linear Model

VD ≥ V for diode to turn on.

Hence during conducting state:

Reverse biasedopen circuit

=

V

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

+ VD -

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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?

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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

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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

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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

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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

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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)

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DC Load Line

A linear line equation ID versus VD

Obtain the equation using KVL

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2 2

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

V

The value of ID at VD = V

DIODE AC EQUIVALENT

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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.

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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:

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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

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DC equivalent AC equivalent

rd

idIDQ

VDQ = V

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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

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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

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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