Chapter 1 Diodes.pdf

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DIODES

EDB 2034

Analogue Electronics

DIODES

Lecturer : Dr. Azrina Abd AzizPhone : 05-368 7850Office : 23-03-36

Email : [email protected]



DIODES

Coverage :

• Introduction to Diode

• The ideal diode

• i – v Characteristic of a Real Diodes

• Analysis of Diode Circuits

• Diode Small Signal Model

• Operation in the reverse breakdown region-Zener Diode

• Rectifier Circuits

• Limiting and Clamping Circuits



DIODES

•

A diode is a solid-state device that allows current to flow in only onedirection, a process known as rectification.• Diodes are a fundamental component of many electrical circuits used

as rectifier, voltage regulator, switching.

P N



DIODES

Diode is formed by joining 2 elements which are P-type and N-typesilicon materials

The symbol of diode is

Physical Structure



Types of Diode



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Triple Power Supply System



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Introduction to Diode : Movie



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The Ideal I-V Characteristic

Open Circuit Short Circuit

When the diode is connected to a voltage supply, the voltage is called as biasvoltage



DIODES

The Ideal Diode : Example

The Rectifier Circuit : (Applicable in power supply circuit for AC to DCconversion)

Positive cycle(Forward Bias)

Negative cycle(Reverse Bias)

Rectifier circuit rectifiesthe input signal



DIODES

Example 2:

Find the value of I and V for the following circuits. Use ideal diode.

+5V

2.5 K

V

(a) 2 mA, 0 VI

(a)

(b) 0 mA, 5 V

+5V

2.5 K

V

I

(b)



DIODES

i – v Characteristic of Real Diodes (Actual Case For Silicon Diode)

The characteristic curve consistsof three regions :

Forward-bias regionv > 0 V

Reverse-bias regionv < 0 V

Breakdown regionv

< -V ZK

Cut-in voltage @ VF Full conducting voltage

Nonlinear IV response



DIODES

i – v Characteristic of a Real Diodes : The Forward-Bias

The i – v relationship is given by Shockley’sEquation

/1T

v nV si I e

where,

I s Saturation Current (constant at a giventemp. normally 10-15 A)(Double every 5oCrise).

n constant = 1 depending on material

V T

Thermal Voltage given by

T kT

V q

where,k = Boltzmann’s constant = 1.38 x 10-23 J/K T = Temperature (Kelvin = 273 + oC)

q = Electron charge = 1.60 x 10-19 C

V T = 25 mV at 20oC

(1.0)

(1.1)



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Effect of temperature on Forward Bias

I-V curves

shifts to the

left at the

rate of -2.5mV per OC

change in

temperature



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The Forward-Bias Continue …

Specifically for i >> I s Eqn (1.0) can be reduced to

/T

v nV si I e or lnT

s

iv nV

I (1.3)

The Reverse-Bias Region:

During reverse-bias,

si I

I S for reverse-bias doubled for every 10oC.

(1.4)

(1.2)



DIODES

Effect of temperature on Reverse Bias

Reverse

saturation

current

doubles forevery 10oC

rise in

temperature



DIODES

Example:

The diode in the circuit below is a large, high-current device whosereverse current is reasonably independent of voltage. If V = 1 V at 20oC,find the value of V at 40oC.

+9V

1 M V

I s = V20oC /R = 1 uA

I s(40 o C) = 4 uA

V (40

o

C) = (4 uA)(1 M)= 4 V.



DIODES

The Breakdown Region

Breakdown occurs when the magnitude of the reverse voltage exceeded athreshold value known as the breakdown voltage, denoted by V ZK .

When breakdown occurs: reverse current increases rapidly

destructive without external limiting circuit



DIODES

Analysis of Diode Circuits

Usually consider during forward-bias operation.Two modes of diode analysis: DC Analysis (Graphical, piecewise linear & constant-voltage drop

model) AC Analysis (small-signal model)

A simple diode circuit as shown has twounknown ID and VD which can be obtainedfrom

DC ANALYSIS:

/ D T

V nV D s I I e (1.5)

DD D D

V V I

R

(1.6) Using KVL

Diode AC



DIODES

Solution to Eqn. (1.5) & (1.6) can be obtained using:

a)Graphical AnalysisUsing Eqn. (1.5) and (1.6), we can draw the following graph

From Eqn. (1.6) by lettingID = 0 and then VD = 0

Using Eqn. (1.5)

Using Eqn. (1.6)

0; DD D D D DD

V V I V V

R

0; DD DD

D DV V

I I R R

DD D D

V V I

R



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b) Piecewise Linear Model Also known as “battery-plus-resistance” model.

Consists of two lines A and B. Line A with 0 slope while line B with 1/rD

slope. This model can berepresented by

00; D D Di v V

0 0/ ; D D D D D Di v V r V V

Where V D0 and r D will be provided



DIODES

Piecewise-linear model of the diode forward characteristic and its equivalentcircuit representation



DIODES

Example:

Utilizing the piecewise-linear model for the diode circuit below with itsparameters given as R = 1 k , VDD = 5 V, V D0 = 0.65 V and r D =20, find I D and V D .

0 4.26 DD D D

D

V V I mA

R r

0 0.735 D D D DV V I r V



DIODES

c) Constant-Voltage Drop ModelSimpler diode equivalent model. A vertical straight line is used toapproximate the fast-rising part of the exponential curve. V D is taken tobe 0.7 V.



DIODES

Example:

Utilizing a constant-voltage drop model for the diode circuit belowwith its parameters given as R = 1 k and VDD = 5 V, find I D and V D .

0.74.3 DD

DV

I mA R

0.7 DV V



DIODES

AC ANALYSIS: Diode Small-Signal Model

Small-signal equivalent model is required when a diode is bias in forwarddirection and a small ac signal superimposed on the dc quantities.

0 D D D d v V i r

0 D D d d V I i r

0 D D d d d V I r i r

D d d V i r

Where,r d diode small-signal resistance

T d

D

nV r

I

Diode DC

O S



DIODES

Diode circuit

Example

Equivalent model

dc signal quantities ac signal quantities

d d s

d

r v v

R r

DIODES



DIODES

From previous slide, assuming that VDD = 10 V with superimposed of 60 Hz1-V peak amplitude of an ac signal, find the peak-to-peak signal , v d acrossthe diode. Let V D = 0.7 V, R = 10 k , n = 2, and V T = 25 mV.

10 0.70.93

10 D

I mAk

2(25 )53.8

0.93

T d

D

nV mr

I m

53.82(1) 10.710 53.8d d s

d

r v v mV R r k

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DIODES

Operating In The Reverse Breakdown-Zener Diodes

Breakdown diodes or Zener diodes are manufactured to operate specificallyin the breakdown region to limit the voltage. Most of this diodes are used inthe design of voltage regulators.

V z is specified atthe test current

I ZT

r z dynamic resistance (m to )V Z voltage across zener diode

z V r I

V zo where 1/rz

intersectsvoltage axis

I-V almoststraight lineabove knee-

voltage, V ZK

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DIODES

Model of zener diode:

Z ZO z Z V V r I

The equivalent circuit model can beanalytically described by

Which applies for IZ>IZK and VZ >VZO

DIODES



DIODES

Example :

The 6.8 V zener diode for the circuit below is specified to have V Z = 6.8 V atI Z = 5 mA, r z = 20 , and I ZK = 0.2 mA. The supply voltage V+ is nominally10 V but can vary by + 1V.

V+ = 10 + 1 V

R = 0.5 k

V0 R L

a) Find V0 with no load and with V+ at itsnominal value.

b) Find the change in V0 resulting from the+1 V change in V+.

c) Find the change in V0 resulting fromconnecting a load resistance R L = 2 k /0.5k .

d) What is the minimum value of R L forwhich the diode is still operates in thebreakdown region?.

DIODES



DIODES

Solution :

Given VZ = 6.8 V at I Z = 5 mA, r z = 20 , I ZK = 0.2 mA, V+ = 10 V + 1V.

Iz

V+

R

V0

a) With no load connected across zener diode,

0 10 6.7 6.350.5 0.02

Z

z

V V Z R r

I mAk

6.8 (20)(5 ) 6.7 ZO Z z Z V V r I m V

0 6.7 (6.35)(0.02) 6.83O Z z Z V V r I V

b) For a + 1V change in V+,

020

1 38.5500 20

z

z

r

R r V V mV

DIODES



DIODES

Iz

V+

R

V0 R L

IL

c) When a load resistance of 2 k connected across zener diode,

The load current, I L is

/ 6.8 / 2 3.4 L Z L I V R V k mA

The change in zener current istherefore

3.4 z I mA

Finally

0 20( 3.4) 68 z z V r I mV

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DIODES

V0 when R L = 0.5 k can be calculated as,

The new load current, I L is

/ 6.8 / 0.5 13.6 L Z L I V R V k mA

Since the current passed through R to zener is only limited to 6.4 mA, zener willbe cut off and no longer operate in the breakdown region. V0 can be calculatedusing voltage divider formed by R L and R.

00.5

10 50.5 0.5

L

L

RV V V

R R

DIODES



DIODES

d) Minimum value of R L for which the diode still operates in the breakdown region.

For the zener to be at the edge of the breakdown region,

0.2 Z ZK I I mA

0 6.7 (0.2 )(20) 6.7O Z Z z Z V V V r I m V

The lowest possible current supplied through R is given by

(10 1) 6.7 / 0.5 4.6 I k mA

Thus the load current is given by

4.6 0.2 4.4 L I m m mA

Which corresponds to R L of

6.71.5

4.4

L R k

DIODES



DIODES

RECTIFIER CIRCUITS

Important applications of diodes Apply in power supply building block Transformer is used to step down the input voltage by v s

(rms)= 120(N2 /N1) Diode rectifier converts sinusoid signal to unipolar output. Filter produce ripple output signal. Voltage regulator remove the ripple signal to produce a clean

DC signal.

Fig. 3.36 Block diagram of a dc power supply.

DIODES



DIODES

RECTIFIER CIRCUITS: Half-Wave Rectifier

Half-wave rectifier utilizes alternate half cycles of the input signal v s . From the equivalent model, we can write the following equations:

piecewise-linearmodel

0 00, s Dv v V

0 0 0, s D s D D

Rv v V v V R r

Since rD<<R therefore,

0 0 s Dv v V

DIODES



DIODES

The input-output waveforms from the rectifier circuit by assuming that rD<<R is

as follows:

Diode selection for the rectifier depends on the current and the peak-inversevoltage (PIV) which can withstand without breakdown. (The largest inversevoltage that appears across the diode)

0 0 s Dv v V

s PIV V

DIODES



DIODES

RECTIFIER CIRCUITS: Full-Wave Rectifier

Full-wave rectifier utilizes both half cycles of the input signal v s . Can be implemented using a transformer with center-tapped output (two output

voltages).

During positive cycle, diode D1 will conduct and diode D2 will be reverse biasedand vice versa.

The PIV of the diode is given by:

0 0( _ _ 2) ( ) 2

o s s s D s D PIV across diode D v v V V V V V

Positive cycle

Negative cycle

0o s Dv V V

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The output waveform

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DIODES

RECTIFIER CIRCUITS: The Bridge Rectifier

Similar configuration as the Wheatstone Bridge used to measure resistance change. Does not require center-tapped transformer. During positive cycle, vs is positive, current passed through diode D1, R, and D2.

Diode D1 and D2 conduct while diode D3 and D4 cut-off. During negative cycle, vs is negative, current passed through diode D3, R, and D4.

Diode D3 and D4 conduct while diode D1 and D2 cut-off. PIV of each diode (example diode D3) is given by

0 02 s Dv v V

3( ) 2( ) D reverse o D fwd v v v 0 0 02 s D D s D PIV v V V v V

DIODES



DIODES

Rectifier with a Filter Capacitor

Pulsating signal from the rectifier circuit is not suitable for applications in electroniccircuit. Filter capacitor in parallel with the load reduces substantially the variationeffect.

Assuming the diode to be IDEAL, during positive cycle, the capacitor, C will becharged to the peak of the input voltage, VP. During negative cycle, the diode is off.The capacitor discharges through the resistor R until the next positive cycleappears.

To keep the output voltage, vo from decreasing too much, the time constantRC >> T

DIODES



DIODES

Pulsating signal from the rectifier circuit is not suitable for applications in electroniccircuit. Filter capacitor in parallel with the load reduces substantially the variationeffect.

Assuming the diode to be IDEAL, during positive cycle, the capacitor, C will be

charged to the peak of the input voltage, VP. During negative cycle, the diode is off.The capacitor discharges through the resistor R until the next positive cycleappears.

To keep the output voltage, vo from decreasing too much, the time constant RC >>T.

DIODES



DIODES

The load current is given by

And the diode current during positive cycle is

/ L oi v R

L D c L L

dvi i i C i

dt

During the diode-off interval and at the end of the discharge interval, the outputvoltage are given by

/t RC o P v V e

/t RC

o P r P v V V V e

Solving 2nd equation will give us the solution of the peak-to-peak ripple voltage

/1 1 1 /t RC P r P P P

V T V V e V T RC V

RC fRC

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DIODES

Example :

Consider a peak rectifier fed by a 60-Hz sinusoid having a peak value VP = 100 V.Let the load resistance R = 10 k . Find the value of the capacitance C that willresult in peak-to-peak ripple of 2 V.

P r

V V

fRC

3

10083.3

2 60 10 10

P

r

V C F

fRV

DIODES



DIODES

LIMITING AND CLAMPING CIRCUITS : Limiter Circuits

vo

vi

L+

L-

L+ /K

L- /K

t L

+

L-

Limiter circuit is used to clipped-off the unwanted part of the signal described by

/ /i L K V L K

Where the linear relation between vi and vo is given by

o iv Kv

If double limiter circuit is used, both positive and negative peaks will be clipped-off.

DIODES



DIODES

Assuming we used constant-voltage model, for circuit (a), for vi< 0.5 V, the diode iscut off and has no current flow. When the voltage rises above 0.7 V, the diode isfully on with the voltage across the diode is 0.7 V, thus limited the positive peak

output to only 0.7 V. For circuit (b), during positive cycle, the diode is reverse bias with small leakage

current. During negative bias with vi> -0.5 V, the diode is cut off while the output islimited to 0.7 V for vi 0.7 V.

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DIODES

Example:

Sketch the output voltage of the following limiter circuit.

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DIODES

LIMITING AND CLAMPING CIRCUITS : Clamper Circuits

Clamper circuits are circuit that clamp the input signal to a different DC level, but

the peak-to-peak swing of the applied signal will remain the same.

C

Vi VoR

When vi > 0.7 V, (Diode ON), using KVL

0.7 0 0.7i c c iV V V V V V

0.7o DV V V

When vi = 0 V or less than 0.7 V,

0 0c o o cV V V V

When vi < 0 V, (Diode OFF)

0i c o o i cV V V V V V

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DIODES

Clamper Circuit : Example

For the previous clamper circuit, let Vi= 5 sin (wt) and V

D= 0.7 V. Sketch the

output signal Vo.

t

0.7V

vi

-9.3V

When Vi > 0 V

0.7 5 0.7 4.3c iV V V V V

0.7o DV V V

When Vi = 0 V

4.3o cV V V

When Vi < 0 V

5 4.3 9.3o i cV V V V

t

-4.3V

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