ELE232 - Chapter 2 - Diode [Compatibility Mode]-4.pdf

7/26/2019 ELE232 - Chapter 2 - Diode [Compatibility Mode]-4.pdf

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Chapter : DIODE

Norsabrina Sihab

Faculty of Electrical Engineering,

Universiti Teknologi MARA

Pulau Pinang

Tel : 04-3823355

Email : [email protected]

Chapter 2 Diode

ELE232 Electronics 1NorsabrinaSihab Updated Nov 2013

Learning Outcome

At the end of this chapter, students able to: Describe a structure of diode, pn junction biasing, I-V

characteristics and diode equivalent circuit

Explain and analyze the diode series/parallel configurationwith DC supply

Explain and analyze the operation of clipper

Explain and analyze the operation of clamper

Explain and analyze the operation of voltage multiplier

Explain and analyze the characteristic and application ofzener diode as voltage regulator

2

Chapter 2 Diode


DiodeDiode is created when a pn-junction forms at the boundarybetween the two regions that is n-type and p-type material.

P-region has many holes (majority carriers) and only fewthermally generated free electrons (minority carriers).

N-region has many free electrons (majority carriers) and only afew thermally generated holes (minority carriers).

3

Figure 2.1 - The basic diode structure at the instant of junction formationshowing only the majority and minority carriers.

Chapter 2 Diode


4

Diode

Depletion Layer -> Area around a p-n junction is called depletionlayer or region which is depleted of free carriers. Free electrons inthe n-region are aimlessly drifting in all directions. At the instant of

p-n junction formation, the free electrons near the junction in the nregion to diffuse across the junction into the p region where theycombine with holes near the junction. When the p-n junction isformed, the n region loses free electrons as they diffuse across the

junction. This creates a layer of positive charges (pentavalent ions)near the junction. As the electronics move across the junction, thep region loses holes as the electrons and holes combine. Thiscreates a layer of negative charges form the depletion region.When at equilibrium the depletion region widened, no electrons can

across the p-n junction.


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Chapter 2 Diode


5

Diode

Figure 2.2 Formation of depletion region

Chapter 2 Diode


6

Diode

Barrier potential >Any time there is a positive charge and anegative charge near each other, there is a forcing acting on the

charges. In the depletion region there are many positive charges

and many negative charges. The forces between the opposite

charges form an electric field. This electric field is a barrier to the

free electrons in n region and external energy must be applied to

get electrons to move across barrier of electric field. The potential

difference of the electric field across the depletion region is the

amount of voltage required to move electrons through the electric

field. The potential difference is called barrier potential.

Chapter 2 Diode


7

Biasing A Diode

Diode is a 2-terminal device that make from p-type and n-typematerials. Ideally conducts current in only one direction.

3 operating conditions:

No bias - No external voltage is applied: VD = 0V, No current isflowing: ID = 0A, only a modest depletion layer exists

Forward bias - External voltage is applied across the p-njunction in the same polarity as the p- and n-type materials.

Reverse bias - External voltage is applied across the p-njunction in the opposite polarity of the p-type and n-typematerials.

Figure 2.3 Diode during forward and reverse bias

Chapter 2 Diode


8

Biasing A Diode - Forward Bias

Condition that allows current through the p-n junction

External voltage or VBIAS connected to the p region (+VBIAS) and n

region (-VBIAS) where VBIAS > VB (barrier potential)Positive terminal of VBIAS will push the holes in the p-region towardsthe p-n junction. Recombination occurs and number of negativeions (acceptors) in the p-region near the junction decreases.

Negative terminal of VBIAS will push the free electrons in n-regiontowards the junction. Recombination with positive ion and numberof positive ion decreases.

As a result, the number of positive and negative ions decrease sothe width of depletion layer become narrow. e - in n-region easily

move to the p-type. So large number of majority carrier flow acrossthe junction.


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Chapter 2 Diode


9

Biasing A Diode - Forward Bias

Figure 2.4 A forward biased diode

Chapter 2 Diode


10

Biasing A Diode - Reverse Bias

Condition that prevents current through the p-n junction

External voltage or VBIAS connected to the p region (-VBIAS) and nregion (+ VBIAS) where VBIAS < VB (barrier potential)

Positive terminal of VBIAS will pulls the free electrons away from p-n

junction and positive ions (donors) in n-region increase.

Negative terminal of VBIAS will pulls the free holes from p-region andnumber of negative ions (acceptor) in p-region increase.

As a result, the number of positive and negative ions increases sothe width of depletion layer become widen.

Due to widening depletion region, the p-n junction act like a verypoor conductor and allow minority carrier flows (A). It calledreverse current or leakage current.

Chapter 2 Diode


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Biasing A Diode - Reverse Bias

Figure 2.5 A reverse biased diode

Chapter 2 Diode


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Biasing A Diode

Reverse Saturation Current -> Also called as leakage currentwhere current in reverse biased condition. The extremely smallcurrent (can be neglected), Is that exist in reverse bias after the

transition current dies out. It caused by the minority carriers in then-region and p-regions that produced by thermally generated EHP.

Figure 2.6 I-V characteristics for diode


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Chapter 2 Diode


Biasing A Diode - Breakdown Voltage

Breakdown Voltage > If the external reverse bias voltage isincreased to a value called the breakdown, the reverse currentwill drastically increase. The reverse-bias potential that resultsin this dramatic change in characteristics is called zener voltage,

Vz.

Avalance region (Vz) can be brougt closer to the vertical axis byincreasing the doping levels in the p-type and n-type materials.

As Vz decreases to very low levels, such as -5V, othermechanism, called Zener Breakdown, will contribute to thesharp change in the characteristics. It occurs because there is a

strong electric field in the region of the junction that can disruptthe bonding forces within the atom and generate carriers.Although the zener breakdown mechanism is a significantcontributor only at lower levels of Vz, this sharp change incharacteristic of a p-n junction called zener diodes.

13Chapter 2 Diode


14

Biasing A Diode - Breakdown Voltage

The maximum reverse-bias potential can be applied at thisregion called Peak Inverse Voltage (PIV) or Peak Reverse

Voltage (PRV) or Breakdown Voltage (VBR).

Diode that operating in this region is called Zener Diode whichnormally used as a voltage regulator.

Chapter 2 Diode


Diode Model

Has 2 terminal Anode and Cathode

The ideal diode in the conduction region (ON State)

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Figure 2.7 Diode structure and symbol

Figure 2.8 Diode ON state

Chapter 2 Diode


The ideal diode in the non-conduction region (OFF State)

Diode resistance levels - Semiconductors act differently to DCand AC currents. There are three types of resistances:

1. DC, or static resistance

2. AC, or dynamic resistance3. Average AC resistance

Diode Model

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Figure 2.9 Diode OFF state


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Chapter 2 Diode


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Diode Model

1. DC or Static Resistance, RDFor a specific applied DC voltage V

D

, the diode has a specific currentID, and a specific resistance RD.

D

DD

I

VR

Figure 2.10 Static Resistance

Chapter 2 Diode


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Diode Model

2. AC or Dynamic, Resistance In the reverse bias region:

In the forward bias region:

The resistance depends on the amount of current (ID) in thediode.

The voltage across the diode is fairly constant (26mV for25C).

rB ranges from a typical 0.1 for high power devices to 2 for

low power, general purpose diodes. In some cases rB can beignored.

The resistance is essentially infinite. The diode acts like an open.

rd

BD

d rI

mV26r

Chapter 2 Diode


Diode Model

3.Average AC Resistance

AC resistance can be determined by selecting two points on thecharacteristic curve developed for a particular circuit.

19

point)to(pointI

Vr

d

dav

Chapter 2 Diode


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Diode Equivalent Circuit

1) Piecewise Linear Equivalent Circuit

Total forward voltage, VD across the diode must be greater than VTbefore the ideal diode in the equivalent circuit will forward bias.


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Chapter 2 Diode


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2.Simplified Equivalent Circuit (Approximate)

Total forward voltage, VD across the diode must be greater than VTbefore the ideal diode in the equivalent circuit will forward bias.

Chapter 2 Diode


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3. Ideal Device

The barrier potential is negligible, hence once the circuit ON or

short at zero potential current will flow significantly and VD=0V.


Chapter 2 Diode


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Diode As A Switch

It can conduct current in only ONE way direction and can act asswitch (ON/OFF).

2 diode conditions ON & OFF state.

2 basic conditions for diode in ON state.

Diode must in forward bias condition Voltage supply, Vi must be greater than the diode voltage, VD

(Vi>VD)

VSi=0.7V, VGe=0.3V and Videal diode=0V

Chapter 2 Diode


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

Diode As A Switch


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Chapter 2 Diode


Diode in Series with DC Supply

Check diodes whether ON or OFF

Redraw diode equivalent circuit including others component.Apply KVL to determine current or voltage

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Example 1

Determine ID, VRand Vo.

-5V

10V

Ge Si

+ Vo

5.6kID

+

VR-

Chapter 2 Diode


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Diode in Parallel with DC supply

Example 2

Determine ID, Ix and Vo

+20V

+ Vo

2.2k

IDSiGe

- 5V

IX

Chapter 2 Diode


Exercise

27Chapter 2 Diode


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Diode Application : Clipper

Basically to clipped-off/eliminate a portion of an AC signal voltageabove or below specific range.

HW rectifier is a basic clipper.

Functions:

1. Altering the shape of the output waveform

2. Circuit transient protection

3. Detection

2 types : 1) series clipper, 2) parallel (shunt) clipper

1) Series Clipper

2 types : a) negative series clipper, b) positive series clipper

The diode in a series clipper circuit clips any voltage that does

not forward bias it:

A reverse-biasing polarity

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


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Chapter 2 Diode


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1a) Negative Series ClipperClipped off half negative cycle. Diode forward bias during positivecycle of Vi.

VT is transition voltage. (VT=VDC+Vdiode)

During positive half cycleVT=Vdc+VD=4Vif Vi VT diode will OFF.

Vo=0V.If Vi > VT diode will ON.

KVL : Vi 4 Vo =0.Vo=Vi-VT=16V

Chapter 2 Diode


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During negative half cycle.

Diode is OFF for all value of V i.

VO=0V.

1a) Negative Series Clipper (contd)

Vi

20

- 20

Vo

16

VT=4V

Final output

Chapter 2 Diode


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1b) Positive Series Clipper

Clipped off half positive cycle. Diode forward bias during negativecycle of Vi.

During positive half cycle

Diode is OFF for all value of Vi.

VO=0V.

Chapter 2 Diode


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VT=- 4-VD=- 4V

if lVil lVTl diode OFF. Vo=0V.

If lVil > lVTl diode ON.KVL : Vi +Vo- 4 =0.

Vo= - Vi+4=-20+4=-16V

1b) Positive Series Clipper (contd)

Final output

Vi

20

- 20

Vo

-16

VT= - 4V


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Chapter 2 Diode


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The operation is opposite series cl ipper.



VO=Vi=20V

2a) Negative Parallel Clipper

Vi

20

- 20

Si

5V

R

+

Vi

-

+

Vo

-

Vi

20 Si

5V

R

+

Vi

-

+

Vo

-

Chapter 2 Diode


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

VT = -0.7-5 = -5.7V

if lVil lVTl diode OFF.

Vo=ViIf lVil> lVTl diode ON.

KVL : Vo +0.7+5=0

Vo = VT = -5.7V

Vi

- 20

0.7V

5V

R

-

Vi

+

+

Vo

-

Final output

Chapter 2 Diode


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2b) Positive Parallel Clipper


VT- Vdc- VD = 0.

VT=5.7V

If Vi VT diode OFF. Vo=Vi.

If Vi > VT diode ON.

KVL : Vo-0.7-5 =0.

Vo=5.7V

Vi

20

- 20

Si

5V

R

+

Vi

-

+

Vo

-

Vi

20

- 20

0.7

5V

R

+

Vi

-

+

Vo

-

Chapter 2 Diode


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VO=Vi.

2b) Positive Parallel Clipper (contd)

Vi

20

- 20

Vo

20

- 20

VT

Final output


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Chapter 2 Diode


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Combination of Negative and Positive Parallel Clipper


DGe OFF for all value of ViDSi ON conditionally

VT=VDSi+5=5.7V

If Vi

VT

DSi

OFF.Vo=Vi.

If Vi > VT DSi ON.Vo=5.7V

Vi

10

- 10

Si

5V

R

+

Vi

-

+

Vo

-

Ge

7.7V

Chapter 2 Diode


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Combination of Negative and Positive ParallelClipper (contd)

During negative half cycleDSi OFF for all value of ViDGe ON conditionally

VT = -VDGe-7.7 = -8V

If |Vi| |VT| DGe OFF.Vo=Vi.

If |Vi| > |VT| DGe ON.Vo+0.3+7.7=0

Vo=-8V

Vi

10

- 10

Vo

10

- 10

VT=5.7V

VT= -8V

-8V

5.7V

Final Output

Chapter 2 Diode


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Summary of Clipper Circuit

Chapter 2 Diode


Exercise

40


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Chapter 2 Diode


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Diode Application : ClamperFunction -> To clamp or shift a signal to a different DC level

Circuit consist of C,D and R

1) Negative Clamper


Step 1: Find polarity of VCStep 2: Determine VO using KVL at o/p

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

Step 3: Determine value of VCVi-Vc-Vo=0

Vc=24.3V

Chapter 2 Diode


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1) Negative Clamper (contd)

During negative half cycle

Step 1: Determine Vo using KVL at i/p

Vi+Vo+Vc=0

Vo=ViVc= 20 24.3= 44.3V

Vo

- 44.3

- 4.3V

Vi

20

- 20

Final Output

Chapter 2 Diode


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2) Positive Clamper

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/pVi+VDCVDVC=0VC=Vi+VDCVD=24.3V

Vi

20

- 20

Si

5V

R

+

Vi

-

+

Vo

-

C

Chapter 2 Diode


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2) Positive Clamper (contd)


Step 1: Find polarity of VoVi+Vc-Vo=0

Vo=20+14.3= 44.3V

Vo

4.3

44.3

Vi

20

- 20

Final Output


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Chapter 2 Diode


45

Example Design a Clamper

Designed a clamper circuit toproduce output voltage, Vo. Usesilicon diode in your design.

Solution

During positive cycle

Propose design clamper circuitwhich D ON during positivecycle. Vo=5.7V

KVL: +Vo-0.7-5=0

Vo=+5.7V

KVL:

Vi Vc Vo=0

Vc=Vi-Vo=15 - 5.7=9.3V

During negative cycle

KVL:

+Vi+Vo+Vc=0

Vo= - 15 - 9.3= - 24.3V

Chapter 2 Diode


46

Summary of Clamper Circuit

Chapter 2 Diode


Exercise

47Chapter 2 Diode


Exercise

48


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Chapter 2 Diode


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Diode Application : Voltage Multiplier

Function use clamping action to increase peak rectifiedvoltage without the necessity of increasing the transformers

voltage rating. A voltage doubler is similar to the peak-to-peakdetector but uses rectifier diodes instead of small-signal diodes.

Types Voltage Doubler (multiply the input peak by factors of2), 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. EgChathode-ray tubes (CRTs), particle accelerators etc.

Chapter 2 Diode


50

Half-wave voltage doubler

Half Wave Voltage Doubler

C2D2 2VP

C1

D1VP

1st negative half cycle

D1 ON while D2 OFF

C1 charging up quickly tothe peak value of input

KVL : -Vi+VD+Vc1=0

So VC1 = VpVD orapproximately Vp

Chapter 2 Diode


Half Wave Voltage Doubler (contd)

1st positive half cycle:

D2 ON while D1 OFF

C2 charging up quickly to the peak value of input

So KVL : -Vp -VC1 +VD+ VC2 =0

Vo = VC2 = VP+VC1VD or approximately 2Vp

51Chapter 2 Diode


52

Full-wave Voltage Doubler

1st positive half cycle:

D1 ON while D2 OFF

C1 charging up quickly to the peak

value of inputKVL : -Vp +VD+ VC1 =0

VC1 = VPVD or approximately Vp


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Chapter 2 Diode


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Full-wave Voltage Doubler (contd)

1st negative half cycle

D2 ON while D1 OFF

C2 charging up quickly to the peakvalue of input

KVL : +Vi+VC2+VD=0

VC2 = VpVD or approximately VpThus the output Vo=VC1+VC22Vp

Chapter 2 Diode


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Voltage Tripler

By connecting another diode-capacitor section to the voltage doublercreates 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.

Chapter 2 Diode


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Voltage 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.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.

Chapter 2 Diode


Exercise

56


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Chapter 2 Diode


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Zener DiodeIDZ is opposite from ID which isdesigned to work in reversebias.

Application : Voltage Regulator

Chapter 2 Diode


58

Zener Diode Application : Voltage Regulator

Simplest regulator as shown in figure below.

3 conditions of Vi and load resistance, RL to maintain designedzener voltage:

1. Vi and RL fixed

2. Vi fixed and RL variable

3. Vi variable and RL fixed

Chapter 2 Diode


59

1. Fixed Vi and Fixed RLStep 1: Determine the state of thezener diode by removing it from thenetwork. Calculate voltage across theresulting open circuit.

)(

)(

:

OFFDVVif

ONDVVif

RR

VR

VV

VDR

ZZ

ZZ

L

iL

L

Step 2: Substitute appropriateequivalent circuit

zzz

LiRR

L

LL

LRZ

LZR

zL

VIPdiodezenerbydissipatedpower

R

VV

R

VIand

R

VIwhere

III

IIIKCL

VV

:

)1(

:

Chapter 2 Diode


60

2. Fixed Vi and Variable RLSpecific range of RL to turnON DZ

maxL

ZLmin

LL

R

VIand

maximumisIthereforeminimum,isRsince

min

max

L

Z

L

LL

R

V

R

VI

Zi

ZL

Z

Zi

L

i

L

Z

iL

L

Z

iLLZ

L

iLL

VV

RVRso

V

VV

R

R

VR

RV

VRRR

V

VRRRV

RR

VRVV

VDR

min

)1(

)(

)(

:Once DZ ON, VRremains fixed

Lmin

ZLmaxZMRLmin

RLminZmax

LmaxZmin

R

RR

I

VR&I-II

constantisIbecauseIwhenIand

IwhenIresulting

R

V

fixedalsoIfixed,isVsince

LRZ

LZR

R

RZi

IIIso

IIIKCL

I

VVV

:


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Chapter 2 Diode


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3. Fixed RL and Variable ViVi must be sufficiently largeto turn DZMin voltage to turn ON is

Vi=Vimin

Z

L

Li

iLLZ

L

iLZL

VR

RRV

VRRRV

RR

VRVV

VDR

min

)(

:

The max of Vimax is limited bythe max zener current, IZM

ZRi

LZMR

VVVIII

maxmax

max

Chapter 2 Diode


62

Exercise1) Determine VL,VR, IZ and PZ.

Answer: 10V, 10V,6.3mA, 63mW

2)

a) Determine the range of RL andIL that will result in VRL beingmaintained at 10VAnswer: 250 ~1.2k, 8mA ~ 40mA

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

3) Determine range of Vi thatwill maintain zener diode inON state.Answer: 23.67V ~ 36.8V

Chapter 2 Diode


Exercise

63

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ELE232 - Chapter 2 - Diode [Compatibility Mode]-4.pdf