ece101 modules 9 zener diodes

ZENER DIODESModule 9

By

Dr. Bernie Redoa

Reference: Electronic Devices and Circuit Theory by Boylestad & Nashelsky, 10th Ed

1. Interpret the characteristic curves of a zener diode.

2. Draw the equivalent circuit of a zener diode.

LEARNING OUTCOMES:

2. Draw the equivalent circuit of a zener diode.3. Explain how a zener diode produces a

constant level of dc voltage during reverse bias condition.

4. Solve and analyze circuits with zener diodes.

I. REVIEW: Diode Characteristics

3

Zener region is where the characteristic drops in an almostvertical manner at a reverse-biaspotential denoted Vz. The curve drops down and away from the horizontal axis.

I. What is the Zener Region?

4

away from the horizontal axis.

The current in the zener regionhas a direction opposite to that ofa forward biased diode.

This region of uniquecharacteristics is employed in the design of Zener diodes.

I. Comparison: Diode Conduction DirectionID

+ ++

VD

IR

VR

IZ

VZ

5

(a) Zener Diode

(b) Semiconductor Diode

(c) Resistive Element

- --

For semiconductor diode, the ON state will support a current in the direction of the arrow in the symbol, for Zener diode the direction of conduction is opposite to that of the arrow in the symbol.

The polarity of VZ and VD are the same as would be obtained if eachwere a resistive element.

I. Comparison: Diode Conduction Direction The location of the Zener region can be

controlled by varying the doping levels.

An increase in doping that produces anincrease in the number of addedimpurities, will decrease the Zener

6

+

-

IZ

VZ

impurities, will decrease the Zenerpotential.

Zener diodes are available having Zenerpotentials of 1.8 V to 200 V withpower ratings from W to 50 W.

Because of its excellent temperature and current capabilities, silicon is the preferred material in the manufacture of Zener diodes.

I. Zener Diode Characteristics+

-

rZ

0.7 V+

- 0.7 V+

-

+

-

rZ

IZ

0.7 VVZ VR

7

+

-

rZ

0.7 V+

-

+

-

VZ

VZ

0.7 V

10 A = IR0.25 mA = IZK

IZT = 12.5 mA

IZM = 32 mA

VZ VR

rZ = 8.5 = ZZT

rZ = VZ / IZ

The Zener potential of a Zener diode is very sensitive to the temperature of operation.

The temperature coefficient can be used to find the change in Zenerpotential due to a change in temperature using the following equation.

I. Zener Diode Characteristics

VZ / VZ X 100% / C

8

where:T1 is the new temperature levelT0 is room temperature in an enclosed cabinet (25C)TC is the temperature coefficientVZ is the nominal Zener potential at 25C

TC = VZ / VZT1 - T0

X 100% / C

EXAMPLE: Analyze the 10-V Zener diode with temperature coefficient of +0.072 (%/ C) if the temperature increased to 100 C (the boiling point of water).

SOLUTION: Substituting to the equation, we obtain

I. Zener Diode Characteristics

V = TCVZ (T - T )

9

VZ = 0.54 V

(0.072% / C) (10 V) (100 C - 25 C )100%

VZ = TCVZ (T1 - T0 )100%

VZ =

VZ = VZ + 0.54 V = 10.54 V

The analysis of networks employing Zener diodesis quite similar to the analysis of semiconductor diodes in previous discussions:

II. Zener Diode Circuit Analysis

11

1. Determine the state of the diode.2. Substitute the appropriate model.3. Determine other unknown quantities of the

network.

EXAMPLE: Zener Diodes1. Determine the references

voltages provided by the network which uses a white LED to indicate that the power is on. What is the level of current through the

White

R 1.3 k

Vo2

40 V

12

level of current through the LED and the power delivered by the supply. How does the power absorbed by the LED compare to that of the 6-V Zener diode?

VZ1 6 V

VZ2 3.3 V

Si

Vo1

EXAMPLE: Zener Diodes (ANSWER)1. Determine the references

voltages provided by the network which uses a white LED to indicate that the power is on. What is the level of current through

White

R 1.3 k

Vo2

40 V

ANSWERS:

Vo1 = 4 V

Vo = 10 V

13

the level of current through the LED and the power delivered by the supply. How does the power absorbed by the LED compare to that of the 6-V Zener diode?

VZ1 6 V

VZ2 3.3 V

Si

Vo2

Vo1

Vo2 = 10 VIR = ILED = 20 mAPs = 800 mWPLED = 80 mWPZ = 120 mW

2. The network is designed to limit the voltage to 20 V during the positive portion of the applied voltage and to 0 V for the negative excursion of the applied voltage. Check its operation and plot the waveform of the voltage across the system for the applied signal. Assume the system has a very high input resistance so it will not affect the behavior of the network.

R

EXAMPLE: Zener Diodes

Vi

14

+

-

vi 20 Vt

Vi

0

60 V

-60 V

SystemVZ

Si

2. The network is designed to limit the voltage to 20 V during the positive portion of the applied voltage and to 0 V for the negative excursion of the applied voltage. Check its operation and plot the waveform of the voltage across the system for the applied signal. Assume the system has a very high input resistance so it will not affect the behavior of the network.

EXAMPLE: Zener Diodes (ANSWER)

VoRVi

15

t0

60 V

-60 V

R

+

-

vi 20 V SystemVZ

Si

t

Vi

0

60 V

-60 V

20 V

III. Zener Diode Circuit AnalysisA. Simple Zener Diode Regulator Circuit Vi and R fixed

R

VR+ -REQUIRED:Determine VL, VR, IZ and PZ

16

+

-

viVZ = 10 V

+

-

IZ

RL 3 k

PZM = 30 mW

16 V

1 k

VL

+

-


STEP 1: Determine the state of the Zener diode by removing it from the network and calculating the voltage across the resulting open circuit:

RIR IL SOLUTION:

By Voltage Divider Rule:

17

V = 12 V

+

-

vi

V+

-

IZ

RL 3 k

PZM = 30 mW

16 V

1 k

VL

+

-

By Voltage Divider Rule:

V = VL = RLVi

R + RL(3 k) (16 V)

1 k + 3 k V =

If V VZ, the zener diode is ONIf V < VZ, the zener diode is OFF

Since V=12 V is greater than VZ=10 V, diode is in the ON state


STEP 2: Substitute the appropriate equivalent circuit and solve for the desired unknowns:

RIR IL VR = Vi - VL = 16 V - 10 V = 6 V

VR+-

SOLUTION:VL = VZ = 10 V

18

+

-

vi+

-

IZ

RL 3 k

PZM = 30 mW

16 V

1 k

VL

+

-

VR = Vi - VL = 16 V - 10 V = 6 V

IZ = 2.67 mA

VZ 10 V

IL = = = 3.33 mA10 V

3 k

VLRL

6 V

1 k

VRR

IR = = = 6 mA

IZ = IR - IL = 6 mA - 3.33 mA

PZ = VZ IZ = (10 V) (2.67 mA ) = 26.7 mW

III. Zener Diode Circuit AnalysisB. Simple Zener Diode Regulator Circuit

Fixed Vi , Variable R

R

VR+ -REQUIRED:Determine the range of RLand I that will result in V

IR IL

19

+

-

viVZ = 10 VIZM = 32 mA

+

-

RL 50 V

1 kand IL that will result in VRLbeing maintained at 10 VIZ

STEP 1: Determine the minimum load resistance that will turn the zenerdiode ON (i.e., the value of RL that will result in load voltage VL = VZ)

SOLUTION:R

VR+ - IR IL

V = V = RLVi



20

RLmin = 250 If RL > RLmin, the zener diode is ON

+

-


+

-

RL 50 V

1 k IZ

Solving for RL :

VL = VZ = R + RL

RVZ

Vi - VZRLmin =

(1 k) (10 V) 50 V - 10 V

RLmin =

SOLUTION:

RVR+ - IR IL VR = 40 V

VR = Vi - VZ = 50 V 10 V Voltage across the resistor R:



21

If RL > RLmin, the zener diode is ON

+

-


+

-

RL 50 V

1 k IZ

To get the minimum level of IL

VR = 40 V

ILmin = 8 mA

To get the magnitude of IR40 V

1 k=

VRR

= 40 mAIR =

ILmin = IR - IZM = 40 mA 32 mA

SOLUTION:

RVR+ - IR IL

Determining the maximum value of RL10 V

8 mA=

VZI

RLmax =



22

If RL > RLmin, the zener diode is ON

+

-


+

-

RL 50 V

1 k IZ

To get the Pmax

Pmax = 320 mW

Pmax = VZ IZM = (10 V) (32 mA)

8 mAILmin

RLmax = 1.25 k

REQUIRED:Determine the range of values of V that will

III. Zener Diode Circuit AnalysisC. Simple Zener Diode Regulator Circuit

Fixed RL , Variable Vi

R

+ I220

VR+ -IR IL

23

values of Vi that will maintain the Zener diode on the ON state.

+

-

vi

+

-

IZ

RL 1.2 k

220

VL

+

-

VZ = 20 VIZM = 60 mA

For fixed values of RL, the voltage Vi must be sufficiently large to turn the zener diode ON.



RVR+ -

IR ILVL = VZ =

RLVi

R + RL

24

+

-

vi

+

-

IZ

RL 1.2 k

220

VL

+

-


Vimin = (RL + R) VZ

RL

Vimin = 23.67 V

R + RL

(1200 + 220 ) (20 V) 1200

Vimin =



RVR+ -

IR IL=

20 V

1.2 k

VL

RL=

VZ

RLIL =

25

+

-

vi

+

-

IZ

RL 1.2 k

220

VL

+

-


IRmax = 76.67 mA

1.2 kRL RL

IL = 16.67 mA

IRmax = 60 mA + 16.67 mA

IRmax = IZM + IL



RVR+ -

IR IL

Vimax = IRmax R + VZ

= 16.87 V + 20 V

= (76.67 mA)(0.22 k) + 20 V

26

+

-

vi

+

-

IZ

RL 1.2 k

220

VL

+

-


Vimax = 36.87 V

VL

Vi

20 V

36.87 V23.67 V

A diode that gives off visible or invisible (infrared) light when energized

I. Light Emitting Diodes

Light Emitting Diodes

COLOR CONSTRUCTION FORWARD VOLTAGESample Applications:

28

COLOR CONSTRUCTION VOLTAGEAmber AlInGaP 2.1Blue GaN 5.0

Green GaP 2.2Orange GaAsP 2.0

Red GaAsP 1.8White GaN 4.1Yellow AlInGaP 2.1

Displays in calculators, watches, parking light forcar, miniature flashlight, traffic light.

A semiconductor p-n junction device whoseregion of operation is limited to the reverse-biasregion. The application of light to the junction willresult to the transfer of energy, resulting to the

II. Photodiode

29

result to the transfer of energy, resulting to the increased level of reverse current.

Sample Applications:Alarm system ; Count items in a conveyor

A special-application diode used in systems requiring

III. Schottky Barrier DiodeIdeal diode

Cj

30

very high frequency range and quick response time.

Sample Applications:Low voltage/high-current power supplies, ac-dc

converters, radar systems, Schottky TTL logic forcomputers, mixers and detectors in communicationequipment, instrumentation and analog-to-digitalconverters.

Knowledge will give you POWER,

32

POWER,

but character RESPECT

- Bruce Lee

ece101 modules 9 zener diodes

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