Chapter 2 - Ele232_edited

Chapter 2 : DIODE APPLICATIONSp

Norsabrina SihabNorsabrina SihabFaculty of Electrical Engineering,Universiti Teknologi MARAPulau PinangTel : 04-3823355Email : [email protected]

1

Chapter 2 – Diode Applications

Learning Outcome

2

Learning Outcome

At the end of this chapter, students able to:Explain and analyze the operation of half-wave rectifierExplain and analyze the operation of full-wave rectifierp a a d a a y e t e ope at o o u a e ect eExplain and analyze the operation of clipperExplain and analyze the operation of clamperExplain and analyze the operation of diode multiplierExplain and analyze the operation of diode multiplierExplain and analyze the operation of voltage regulatorExplain the basic DC power supply components

Electronics 1Norsabrina Sihab Updated May 2011

Chapter 2 – Diode Applications3

IntroductionIntroductionIt can conduct current in only ONE way direction and can act as switch (ON/OFF).2 di d diti ON & OFF t t2 diode conditions – ON & OFF state.2 basic conditions for diode in ON state.

Diode must in forward bias conditionl l b h h d d lVoltage supply, Vi must be greater than the diode voltage, VD

(Vi>VD)VSi=0.7V, VGe=0.3V and Videal diode=0V



IntroductionDiode in OFF state – act as open circuit. So I=0A.

Introduction

Diode in Series with DC SupplyCheck diodes whether ON or OFFRedraw diode equivalent circuit including others component.

Diode in Series with DC Supply


q g pApply KVL to determine current or voltage


Example 1Determine ID, VR and Vo.

Example 2Determine V V2 V and IDetermine ID, VR and Vo.

10VGe Si

+ Vo

5.6kID+ VR

Determine V1, V2, Vo and ID

Solution-5V

-

Solution



Diode in Parallel with DC supplyDiode in Parallel with DC supplyExample 3

Determine ID, Ix and Vo

+20V

IX

+ Vo

ID Si Ge

+ Vo

2.2k

- 5V

Solution



Diode As RectifierDiode As RectifierRectifier – convert AC to DC Voltage (not pure DC). Allow current flow in one direction only.2 types – 1) half wave rectifier, 2) full wave rectifier

1) Half Wave Rectifier)Normally used in non-critical low current applicationsMade up of a diode, D and a resistor, RHas abilit to cond ct c ent in one di ection and block c ent inHas ability to conduct current in one direction and block current in the other direction.

+

Vi

+

Vo

-

R


-

Basic rectifier circuit


Half Wave Rectifier (contd)Half Wave Rectifier (contd)Half-wave average voltage, Vdc

Is determine by calculate the area under the curve and divide it by the period of rectified waveformby the period of rectified waveform.

T

idc dttVT

V

1

).(1

20

= ∫π π

Vi

Vm

[ ]m

m

V

dtdttV

cos2

.0.sin21

0

0

−=

+= ∫ ∫

θπ

ωπ

π

π

[ ]

m

m

V

V )0cos(cos2

2

=

−−−= ππ

πVo

Vm

Vdc=0 318V

mdc VV 318.0=

=π

Where V i i ( k) l f AC lt

Vdc 0.318V

T


Vm is maximum (peak) value of AC voltageVdc is average value of rectified voltage


Half Wave Rectifier - OperationsHalf Wave Rectifier Operations

Positive half cycle of Vi

Negative half cycle of Vi



Half Wave Rectifier - Peak Inverse Voltage (PIV)

PIV l k k l ( )Also known as Peak Reverse Voltage (PRV)

Is maximum voltage across the diode in the direction to block current flow.Occurs at the peak of the negative alternation of the input cycle when diode is reverse biased.

KVL: Vm- Vo- PIV = 0Vm = PIV



2) Full Wave RectifierThe rectification process can be improved by using more diodes in a full-wave rectifier circuit.It can improve 100% of the DC level obtained from a sinusoidal

2) Full Wave Rectifier

It can improve 100% of the DC level obtained from a sinusoidal input. Full-wave rectification produces a greater DC output:

V )(⎟⎞

⎜⎛

mdc

mm

dc

VV

VVV

636.0

)318.0(22

=

=⎟⎠⎞

⎜⎝⎛=π

2 types – a) center tapped transformer, b) bridge network.



2a) Full Wave Center Tapped Rectifier) ppRequires

Two diodesCenter-tapped transformer



Full Wave Center Tapped Rectifier - Operationspp pPositive half cycle of Vi – D1 ‘ON’ & D2 ‘OFF”

Vm

Vdc=0.636Vm

Vo

Negative half cycle of Vi – D1 ‘OFF’ & D2 ‘ON”T/2 T

Negative half cycle of Vi D1 OFF & D2 ON



Center Tapped Peak Inverse Voltage (PIV)KVL : - Vm + PIV – Vm=0

PIV = 2Vm

Center Tapped Peak Inverse Voltage (PIV)

-- PIV +

Vi=Vm

+R

Vi Vm

- Vm +

Example 4Calculates:Calculates:

i) the DC voltage obtained from a center tapped full wave rectifier for which the peak of rectified voltage is 100Vii) h PIV d l d h di dii) the PIV developed across the diode



2b) Full Wave Bridge Rectifier

Require four diodes, transformer and resistorVDC = 0.636 Vm

Advantages of FWBRnot used of the center tapped transformer and it requires anot used of the center-tapped transformer and it requires a maximum voltage of Vi across the transformer.PIV required of each diode is half of the center tapped full wave circuit


circuit.


Full Wave Bridge Rectifier (1) - OperationsFull Wave Bridge Rectifier (1) OperationsPositive half cycle of Vi : D2 & D3 – ‘ON’

D1 D2

D3 D4

Negative half cycle of Vi : D1 & D4 – ‘ON’

D1 D2D1 D2


D3D4


Full Wave Bridge Rectifier (2) - Operations

Positive half cycle of Vi : D1 & D2 – ‘ON’

Negative half cycle of Vi : D3 & D4 – ‘ON’



Full Wave Bridge Rectifier - Peak Inverse Voltage (PIV)Full Wave Bridge Rectifier Peak Inverse Voltage (PIV)

KVL : PIV- Vm=0m

PIV = Vm

Summary of Rectifier CircuitsSummary of Rectifier CircuitsSummary of Rectifier CircuitsSummary of Rectifier CircuitsRectifierRectifier VVDCDC PIVPIV

H lf W R tifi V 0 318(V ) VHalf Wave Rectifier VDCDC = 0.318(Vm) Vm

Bridge Rectifier VDCDC = 0.636(Vm) Vm


Center-Tapped Transformer Rectifier VDCDC = 0.636(Vm) 2Vm


Power Supplies (Voltage Regulators)Power Supplies (Voltage Regulators)Transformer – To step up or step down or isolate the input voltageRectifier – Diodes that convert the AC to DC voltage (half wave or full wave)full wave)Filter – Example capacitor. Use to filter the rectified signal from rectifier to produce a DC voltage with some ripples or AC voltage variationvariation.Regulator – To maintain a constant output voltage with less ripple but remain the same DC value.



FiltersFilters Is smoothing circuit which to obtain a smoother DC signalTypes – 1) RC filter, 2) L filter, 3) LC filter, 4) π-type filter

Basic Capacitor FilterBasic Capacitor FilterCapacitor is most commonly used of basic filter.Simply capacitor, C connected in parallel with the load resistance.



Ripples VoltageRipples VoltageRipples voltage occur when capacitor quickly charges at beginning of a cycle and slowly discharges after the positive peak (when reverse bias))DC voltage derive from an AC source signal after rectifying and filtering will have some ripples/variations.Output voltage known as ripples voltage, VrOutput voltage known as ripples voltage, Vr

Vr(p-p) – peak-to-peak ripples voltage


Approximate output voltage of capacitor filter circuit


Ripples Voltage (contd)DC voltage derive from an AC source signal after rectifying and filtering will have some ripples/variations.

Ripples Voltage (contd)

VVVd −=1

)(

FWfC

IV

VVV

dcm

pprmdc

→−=

= −

4

2 )(

AC DC

HWfC

IV dcm →−=

2 22: mp

rmsVV

Vwhere ==

dcdc VIV ==

Vr – variation in Vo due to charge and discharge

3434

)()(

)(

ppr

Lrmsr

VV

fCRfCV

−=

==


32)(rmsrV


Ripples Factor, rSmaller ripple respect to dc level meaning better filter circuit.

pp ,

Ripple factor, r

%100% XvoltageDC

rmsinvoltagerippler =32

)()(

pprrmsr

VVwhere −=

%100)( XV

VvoltageDC

dc

rmsr=

32

Ripples factor, r is also depends to the load, RL

Light load Vdc ≅ Vm. Therefore very small ripple factor and capacitor filter provides large DC voltage V


capacitor filter provides large DC voltage, Vdc.Heavy load Vdc < Vm. Therefore bigger ripple factor and capacitor filter provides smaller DC voltage, Vdc.


ExerciseExercise

1. What is the ripple factor of a sinusoidal signal having peak ripple (Vr(p)) of 2V on an average (Vdc) of 50V? Answer : 2.31%

2. The 35 Vrms ac voltage Vs is derived from an ideal 60Hz main transformer. It is connected to a half-wave rectifier and a 220μF capacitor to form a dc power supply. If the load RL draws an average current 0.15A, determine:

i. the peak-to-peak capacitor ripple voltage Answer : Vr(pp) = 5.6V

ii. the average or dc voltage across the load Answer : Vdc = 43.86V

iii. the percentage ripple factor, %r. Answer : 12.8%



CLIPPERSCLIPPERSBasically to clipped-off/eliminate a portion of an AC signal voltage above or below specific range.HW rectifier is a basic clipper.Functions:1. Altering the shape of the output waveform2. Circuit transient protection3. Detection2 types : 1) series clipper 2) parallel (shunt) clipper2 types : 1) series clipper, 2) parallel (shunt) clipper

1) Series Clipper2 types : a) negative series clipper b) positive series clipper2 types : a) negative series clipper, b) positive series clipperThe diode in a series clipperseries clipper circuit “clips” any voltage that does not forward bias it:

A bi i l it


A reverse-biasing polarityA forward-biasing polarity less than 0.7V for a silicon diode


1a) Negative Series Clipper1a) Negative Series ClipperClipped off half negative cycle. Diode forward bias during positive cycle of Vi.V is transition voltage (V V +V )VT is transition voltage. (VT=VDC+Vdiode)

During positive half cycleVT=Vdc+VD=4Vif Vi ≤ VT diode will OFFif Vi ≤ VT diode will OFF. Vo=0V.If Vi > VT diode will ON. KVL : Vi – 4 – Vo =0.


oVo=Vi-VT=16V


1a) Negative Series Clipper (contd)

During negative half cycle.d f ll l f

1a) Negative Series Clipper (contd)

Final outputDiode is OFF for all value of Vi.VO=0V.

Vi

20

VT=4V

- 20

VT 4V

Vo

166



1b) Positive Series Clipper1b) Positive Series ClipperClipped off half positive cycle. Diode forward bias during negative cycle of Vi.

During positive half cycleDiode is OFF for all value of Vi.VO=0V.



1b) Positive Series Clipper (contd)

During negative half cycle.V 4 V 4V

1b) Positive Series Clipper (contd)

Final outputVT=- 4-VD=- 4Vif lVil ≤ lVTl diode OFF. Vo=0V.If lVil > lVTl diode ON.

Vi

20

KVL : Vi +Vo- 4 =0. Vo= - Vi+4=-20+4=-16V

- 20

VT= - 4V

Vo

-16



2a) Negative Parallel ClipperThe operation is opposite series clipper.

2a) Negative Parallel Clipper

Vi

20

R

+ +20

Si

5V

+

Vi

-

+

Vo

-

During positive half cycle

- 20

Diode is OFF for all value of Vi.VO=Vi=20V

Vi

R

+ +i

20 Si

5V

+

Vi

-

+

Vo

-



2a) Negative Parallel Clipper (contd)2a) Negative Parallel Clipper (contd)During negative half cycle

VT = -0.7-5 = -5.7Vf l l l l d d

Final output

if lVil ≤ lVTl diode OFF. Vo=Vi

If lVil> lVTl diode ON. i T

KVL : Vo +0.7+5 =0Vo = VT = -5.7V

Vi

0.7V

R

- +

- 20 5V

Vi

+

Vo

-



2b) Positive Parallel Clipper2b) Positive Parallel ClipperVi

20

Si

R

+ +

- 20

Si

5V

Vi

-

Vo

-

During positive half cycleVT- Vdc- VD = 0.

VT=5.7VVi

R

+ +T

If Vi ≤ VT diode OFF. Vo=Vi.If Vi > VT diode ON. KVL : V -0 7-5 =0

20

0.7

5V

+

Vi

+

Vo

KVL : Vo-0.7-5 =0. Vo=5.7V - 20

- -



2b) Positive Parallel Clipper (contd)During negative half cycle.

Diode is OFF for all value of Vi.

2b) Positive Parallel Clipper (contd)Final output

VO=Vi. Vi

20

VT

- 20

Vo

20

- 20



Test 1 Apr 2010 Q2cTest 1 Apr 2010 Q2c



Combination of Negative and Positive Parallel ClipperCombination of Negative and Positive Parallel ClipperVi

10

Si

R

+ +

Ge

10

Si

5V

Vi

-

Vo

-

Ge

7.7V

During positive half cycleDG OFF for all value of Vi

- 10

DGe OFF for all value of Vi

DSi ON conditionallyVT=VDSi+5=5.7VIf V ≤ V D OFFIf Vi ≤ VT DSi OFF. Vo=Vi.If Vi > VT DSi ON. Vo=5.7V


o


Combination of Negative and Positive Parallel Cli ( d)Clipper (contd)

Final Output

Vi

10

V 5 7V

p

During negative half cycle 10

VT=5.7V

VT= - 8V

During negative half cycleDSi OFF for all value of Vi

DGe ON conditionallyVT = -VDGe-7.7 = -8V

- 10

Vo

10T DGe 8

If |Vi| ≤ |VT| DGe OFF. Vo=Vi.If |Vi| > |VT| DGe ON. -8V

5.7V


| i| | T| Ge Vo+0.3+7.7=0Vo=-8V

- 10


Test 1 August 2009 Q3b

37




Summary of Clipper CircuitSummary of Clipper Circuit



CLAMPERSCLAMPERSTo clamp or shift a signal to a different DC levelCircuit consist of C,D and R

1) Negative Clamper

During positive half cycleDuring positive half cycleStep 1: Find polarity of VC

Step 2: Determine VO using KVL at o/pV V V 0Vo – VD+VDC= 0Vo=0.7-5= - 4.3V

Step 3: Determine value of VC


Vi-Vc-Vo=0Vc=24.3V


1) Negative Clamper (contd)1) Negative Clamper (contd)During negative half cycle

Vi

Final Output

20

Step 1: Determine Vo using KVL at i/p Vo

- 20

Vi+Vo+Vc=0Vo=–Vi–Vc= – 20 – 24.3= – 44.3V

- 4.3V

- 44.3



2) Positive Clamper2) Positive ClamperVi

20

Si

++

C

- 20

5V

RVi

-

Vo

-

During negative half cycle (because Diode ON at this cycle)Step 1: Find polarity of VCStep 2: Determine VO using KVL at o/p

Vo+ VD- VDC= 0Vo= 5 - 0.7= 4.3V

Step 3: Determine VC using KVL at i/pV+V V V 0Vi+VDC–VD–VC=0VC=Vi+VDC–VD=24.3V



2) Positive Clamper (contd)2) Positive Clamper (contd)During positive half cycle

Step 1: Find polarity of Vo Vi

Final Output

Vi+Vc-Vo=0Vo=20+14.3= 44.3V

i

20

V

- 20

Vo

44.3

4.3



Example – Design a ClamperDesigned a clamper circuit to produce output voltage, Vo. Use silicon diode in your design.

KVL: +Vo-0.7-5=0Vo=+5.7V

KVL:Vi – Vc – Vo=0Vc=Vi-Vo=15 - 5.7=9.3V

SolutionD i iti l

During negative cycle

During positive cyclePropose design clamper circuit which D ‘ON’ during positive cycle V =5 7Vcycle. Vo=5.7V

KVL:+Vi+Vo+Vc=0


i o c

Vo= - 15 - 9.3= - 24.3V


Summary of Clamper CircuitSummary of Clamper Circuit




45




Voltage MultiplierVoltage Multiplier

Function – use clamping action to increase peak rectified voltage without the necessity of increasing the transformer’s voltage rating. A voltage doubler is similar to the peak-to-peak detector but uses rectifier diodes instead of small-signal diodes.Types – Voltage Doubler (multiply the input peak by factors of 2), Voltage Tripler (multiply the input peak by factors of 3) and Voltage Quardrupler (multiply the input peak by factors of 4) Application – in high voltage, low current, high frequencies. Eg Chathode-ray tubes (CRTs), particle accelerators etc.



Voltage DoublerHalf-wave voltage doubler

Voltage Doubler

D1VP

C2D2 2VP

C1

• During positive cycle : D1 on, D2 Off. C1 charged to the Vp.

• During negative cycle : D2D ff C ’ di hon, D1 off. C1 can’t discharge

so Vp on C1 adds the supply voltage to charge C2 to approximately 2Vp.


approximately 2Vp.


T1 Feb09 Q3aT1 Feb09 Q3a



Full-wave voltage doublerD

VDC1

D1

VP

VP

2VP

D21

C2P

• During positive cycle : D1 on, D2 Off. C1 charged to the Vp.

• During negative cycle : D2on, D1 off. C2 charges to approx V Output across theapprox Vp. Output across the two series capacitors approximately 2Vp.



Voltage Doubler (contd)Voltage Doubler (contd)



Voltage TriplerVoltage Tripler

By connecting another diode-capacitor section to the voltage doubler creates a voltage tripler.First two sections act a doubler.Positive cycle : C1 charge to Vp thru D1.Negative cycle : C2 charge 2Vp thru D2.Next positive cycle : C3 charges to 2Vp thru D3.Tripler output is taken across C1 and C3.



Voltage QuadruplerVoltage Quadrupler

By connecting another diode-capacitor section.First two sections act a doubler.Positive cycle : C1 charge to Vp thru D1.Negative cycle : C2 charge 2Vp thru D2.p

Next positive cycle : C3 charges to 2Vp thru D3.Next negative cycle : C4 charges to 2Vp thru D4 Quardrupler output is taken across C2 and C4.



ExerciseExercise

Figure below shows another kind of power supply known as voltage lti limultiplier.i. Determine the peak values of output voltages available at the

two terminals indicated as Vo1 and Vo2 if the secondary volatage of the transformer is 120 Vrms.

ii C l l t th i d PIV ti f h di d i thi ltii. Calculate the required PIV rating of each diode in this voltage multiplier circuit.

Show suitable equivalent circuits and equations to justify the calculations. Assume ideal diodes.

(8 marks)



ZENER REGULATORZENER REGULATORIDZ is opposite from ID which is designed to work in reverse bias.

+

V

Application : Regulator -

VZ > V > 0DZ will OFF



ZENER REGULATOR (contd)ZENER REGULATOR (contd)Simplest regulator as shown in figure below.3 conditions of Vi and load resistance, RL to maintain designed zener voltage:1. Vi and RL fixed2. Vi fixed and RL variablei L

3. Vi variable and RL fixed



1. Fixed Vi and Fixed RL1. Fixed Vi and Fixed RLStep 1: Determine the state of the zener diode by removing it from the network Calculate voltage across the

Step 2: Substitute appropriate equivalent circuit

network. Calculate voltage across the resulting open circuit.

zL VV =

LiRL

LRZ

LZR

VVVIdVIh

IIIIIIKCL

−−−−=

+=)1(

:

)(

:

ONDVVifRRVRVV

VDR

L

iLL +==

zzz

LiRR

L

LL

VIPdiodezenerbydissipatedpower

RRIand

RIwhere

=

===

:


)()(

OFFDVVifONDVVif

ZZ

ZZ

<≥ zzz


2. Fixed Vi and Variable RL2. Fixed Vi and Variable RLSpecific range of RL to turn ON DZ

LL maximumisIthereforeminimum,isRsince

== ZLL

VVI

maxL

ZLmin

minmax

RVIand =

LLL RR

I

VDR :

iLLZ

L

iLL

VRRRVRRVRVV

VDR

=++

==

)(

:Once DZ ON, VR remains fixed

RR fixedalsoIfixed,isVsince=− RZi VVV

iLL

Z

iLLZ

R

VRRRV

VRRRV

=+

+

)(

)(R

:RV

+=

=

LZR

R

IIIKCL

I

Z

Zi

L

iL

Z

VVV

RR

VRRV

−=

=+ )1(

RLminZmax

LmaxZmin

constantisIbecauseIwhenIandIwhenIresulting

−= LRZ IIIso


Zi

ZL

ZL

VVRVRso−

=min Lmin

ZLmaxZMRLmin I

VR&I-II ==


3. Fixed RL and Variable Vi3. Fixed RL and Variable ViVi must be sufficiently large to turn DZ

Mi lt t t ON i

The max of Vimax is limited by the max zener current, IZM

Min voltage to turn ON is Vi=Vimin

VDR : ZRi

LZMR

VVV

III

+=

+=

maxmax

max

L

iLZL RR

VRVV

VDR

+==

:

ZL

i

iLLZ

VR

RRV

VRRRV+

=

=+

min

)(

LR



ExerciseExercise1) Determine VL,VR, IZ and PZ.Answer: 10V, 10V,6.3mA, 63mW

3) Determine range of Vi that will maintain zener diode in ON state A 23 67V 36 8VON state. Answer: 23.67V ~ 36.8V

2)

a) Determine the range of RL and IL that will result in VRL being maintained at 10V Answer: 250 Ω ~ 1 2kΩ 8mA ~ 40mA


1.2kΩ, 8mA ~ 40mA

b) Determine the max wattage rating of DZ. Answer: 32mW

Chapter 2 - Ele232_edited

Documents

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operationspositive

half wave

dc voltage

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negative half

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