2.diode applications

7/31/2019 2.Diode Applications

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Diode ApplicationsBEE2213 Analog Electronics I

Rosdiyana Samad 2012, FKEE, UMP


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Series Parallel Configuration

Diode can be applied to any circuits

Usually diode is represented as an approximated (simplified)

model diode

To keep the calculation simple, just use the Kirchoffs voltage& current law

Hint: it is easier to use nodal analysis technique for circuitrepresentation

Important: strong knowledge in CIRCUIT THEORY!!!!!


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Review Diode in series circuit: forward bias

The diode is forward bias.

VD = 0.7V (or VD = E if E


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Review

Diode in series circuit: reverse bias

The diode is reverse biased.

VD = E

VR= 0V

ID = IR= IT = 0A


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Review An open circuit can have any voltage across its terminals,

but the current is always 0A.

A short circuit has a 0V drop across its terminals, but thecurrent is limited only by the surrounding network.

Source notation :

E=10V

+10V

E= -5V

-5V

E=+10V E= -5V


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Review

Determine ID, VD2 and Vo for the circuit.

+20V

Remember, the combination ofshort circuit in series with an

open circuit always results inan open circuit and ID=0A.

+20V


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Problem 2.5a

FindI

The circuit:


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Problem 2.5a

For Si, VD = 0.7 V

Notice that the diode is in reverse-biasconfiguration

So, no current will flow,I= 0 A


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Problem 2.5b

FindI

The circuit:


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Problem 2.5b

Solution:

Using nodal analysis, node Vis equal to thevoltage supplied, so V= 20


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Problem 2.5b

Solution: Using the simple Ohms law:

A965.020

7.020

R

VI


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Problem 2.7a

FindI

The circuit:


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Problem 2.7a Solution:

One of the diode is in reverse-bias resulting inopen circuit for that part


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Problem 2.7a

So, by using the simple Ohms law:

A1

10

10

R

VI


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Problem 2.7b Find Vo

The circuit:

Exercise!


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Problem 2.11a

Find Vo &I:

The circuit:

Exercise!


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Rectification Rectify means improvement, cure healing

(pembaikan, penambahbaikan)

For a sinusoidal waveform or any supply thathas a variation of input value, diode can be usedfor rectification

Rectification are used to modified the inputvalue to become only the signal that we want


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Review: Diodes in AC Circuits

Inputs: -Sinusoidal waveform -Square wave

This circuit is called half-wave rectifier, which generate

waveform vo that will have an average value of particular usein the ac-to-dc conversion process.

The diode only conducts when it is in forward bias, therefore only half of the AC cycle passes through the diode.


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Half-Wave Rectification

For a full cycle of a sinusoidal or continuous waveform,only half of the waveform is taken to be rectified


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For the period 0 T/2, the sinusoidal input will give aforward bias supply to the circuit

The diode will on and current will pass through

Assume that the diode is ideal


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For the period T/2 T, the sinusoidal input will give areverse bias supply to the circuit

The diode will off and no current can pass through

Assume that the diode is ideal


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For a continuousperiodic waveform, therectified waveform will

become:

Where as:

mdcVV

318.0


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Problem 2.25

Sketch Vo and determine Vdc:


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Problem 2.25

Solution:

To obtain Vm from Vrms:

The output Vo will be:

Vdc will be:

V56.155)110(2

2

rmsm

VV

V47.49)56.155(318.0

318.0

mdc

VV


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Problem 2.26

Sketch VoExercise!


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PIV or PRV

Peak Inverse Voltage (PIV) or Peak ReverseVoltage (PRV)

It is a rating to make sure for the reverse-biasoperation, the diode didnt enter the Zenerregion

PIV is set according to the circuit and the input

voltage

mVratingPIV

for half-wave rectifier


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Full-Wave Rectification

The whole cycle of input signal is used andrectified

Two commonly types of full-wave rectifier:

Bridge Network Center-Tapped (CT) Transformer

The dc level from a sinusoidal input can beimproved 100%. So the Vdc becomes:

mdcVV 636.0


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Full-Wave Rectifier: Bridge Network

The most commonly bridge networkconfiguration are build with 4 diodes


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For the positive input supply, the current willtake the route as shown below, and the outputwill becomes:


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For the negative input supply, the current willtake the route as shown below, and the output

will becomes:


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Combine both of the output becomes:


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Animation on the full-wave rectifier for bridgenetwork (ideal condition):


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Animation on the full-wave rectifier for bridgenetwork (non ideal condition):


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Due to the maximum voltage from the inputsupply is Vm, to keep the diode away from the

Zener region, the PIV rating is:

mVratingPIV

for full-wave rectifier: bridge network


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Animation of full bridge rectifier


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Determine the output waveform for the network below andcalculate the output dc level.

Example: Full-Wave Rectifier


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Conduction path for the +ve region



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Conduction path for the -ve region



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Full-Wave Rectifier: Center-Tapped

(CT) Transformer It is constructed with 2 diodes and a center-tapped

transformer

The transformer ratio is 1:2


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(CT) Transformer

For the positive input supply:


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(CT) Transformer

For the negative input supply:


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Animation of center-tapped transformer rectifier


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Show the voltage waveform across the secondary windingand across R when an input sinusoidal is applied to theprimary winding.



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The transformer turns ratio = 0.5.

The total peak secondary voltage

is;

Vp(sec) = nV

p(pri) = 0.5(100)=50V.

There is a 25 V peak across each

of the secondary with respect to

ground.



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Note: Vm = peak of the AC voltage. Be careful, in the center tappedtransformer rectifier circuit the peak AC voltage is the transformersecondary voltage to the tap.

Rectifier Circuit Summary


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Problem 2.28

The circuit:

1 k

120Vrms

All diodes are silicon


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Problem 2.28a

Determine DC voltage for output

Solution:

Vm:

Vo:

So, Vdc:

V71.16912022 rmsm

VV

V31.1687.07.071.1690 V

)31.168(636.0.6360 0 VVdc


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Problem 2.28b

Determine the required PIV rating for eachdiode from problem 2.8

Solution: PIV:

V01.1697.031.168(load)PIV Dm

VV


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Problem 2.28c

Find the maximum current through each diode

Solution:

ID(max):

mA31.168

1

31.168(max)0(max)

kR

VI

L

D


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Problem 2.28d

What is the required power for each diode?

Solution:

mW82.1787.031.168(max) mVIP DD


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Problem 2.31

Sketch Vo and determine Vdc

The input and circuit:

Exercise!


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Clippers

Configuration that employ diodes to clip awaya portion of an input signal without distortingthe remaining part of the applied waveform

Mainly, there are two types of configuration

Series

Parallel


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Clippers

Example of series configuration and the output waveform:


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Clippers

Example of parallel configuration and the output waveform:


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Clippers

Notice something?

Is the configuration similar to something?

Half-wave rectifieris a part of

CLIPPERSconfiguration


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Example 2.18

Sketch vo


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For positive input cycle:

The output will be thesum ofviand +5V

Example 2.18

The output waveform willbecome:

50 ivv


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Example 2.18

The output waveform will becomes:


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Example 2.18

By comparing the input with the whole output:

E l 2 18


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Repeat previous example for the square-wave input.

Example 2.18


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Example 2.20

Sketch vo


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Example 2.20

For positive input cycle:

For vi 4 - For vi 4

i

vv 0

+

-

40v


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Example 2.20

The output waveform will become:


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Example 2.20

The output for negative input cycle will always +4V due to theexternal supply of 4V series with the diode

The diode will always be in the on mode

The circuit becomes: - The output waveform:

40 v


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Example 2.20 By comparing the input

with the whole output:


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Problem 2.35a

Sketch vo for the network and input given:


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Problem 2.35a

Solution:

We will examine the positive input cycle and thenegative input cycle separately and combine them

at the end Due to diode and 4-V supply is connected in

series, the transition level will be 4 + 0.7 = 4.7 V

Two conditions for each half cycle:

Input condition before 4.7 V (transition level)

Input condition after 4.7 V (transition level)


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Problem 2.35a

At positive input cycle before 4.7 V:

open-circuit(no current flow)

vi = 0 4.7

&vi = 4.7 0


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Problem 2.35a

At positive input cycle after 4.7 V:

curr

entflow

vi = 4.7 8

&vi = 8 4.7

0.7 V


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Problem 2.35a

At negative input cycle before 4.7 V:

open-circuit,

diode inreversed-bias

(no current flow)

vi = 0 4.7

&vi = 4.7 0

+4.7 V


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Problem 2.35a At negative input cycle after 4.7 V:

open-circuit,

diode inreversed-bias

(no current flow)

vi = 4.7 8

&vi = 8 4.7

+4.7 V


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Problem 2.35a

By adding all the output sketches:


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Problem 2.35b

Sketch vo for the network and input given:

Exercise!


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Clampers

Construct of a diode, a resistor and a capacitor

It will shift the waveform to a different level

without changing the appearance of theoriginal input signal

The capacitor and resistor ( ) must belarge to ensure it doesnt discharge during theinterval that the diode is non-conducting

RC


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Clampers

The circuit:


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Clampers

Steps for clampers analysis:1. Start the analysis with the condition where the

diode is in forward bias

2. The capacitor will charge up instantaneouslyduring the interval of +ve or ve input supplywhere the diode is in forward-bias condition

3. The capacitor will discharge during the nextinterval of +ve or ve input supply where thediode is in reverse-bias condition

4. Check that the total swing of the output is thesame with the input


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Example 2.22

Sketch vo


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Example 2.22

Just to check whether the capacitor is appropriate forclampers configuration:

For the input given:

For every half interval (+ve or ve input cycle):

This shows that the capacitor is capable of charging anddischarging according to the clampers configurationrequirement

ms10)1.0)(100( kRC

ms110001

f1T

ms5.0

2

ms1

2

T

2

T


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Example 2.22

To start the analysis with the diode in forward-bias mode, thenegative input cycle has to be inserted first into the circuit

The circuit: - The output waveform:

The capacitor will charge up to 25V


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Example 2.22

For the next half input cycle that is the +ve cycle:

The circuit: - The output waveform:

The capacitor will discharge the

voltage of 25V


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Example 2.22

The whole output waveformwill become:

Checking the total swing of theoutput must match the input:

The total swing of the output is the same with the input that is 30 V


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Problem 2.37a

Sketch vo for the network and input shown:


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Problem 2.37a

By examining the circuit, the diode is in forward-bias at negativeinput cycle

As for that the current will ONLY flow through the diode and thecapacitor is charge to 20 V

c

urrentflow

20 V

+


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Problem 2.37a

For the next positive input cycle, the diode is in reverse-bias, so thecurrent will flow through the resistor and the capacitor discharge(providing extra supplies to the circuit)

currentflow

20 V


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Problem 2.37a

Combine all the output sketches:


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Problem 2.37b

Sketch vo for the network and input shown:

Exercise!


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Problem 2.39

For the network given:

a) Calculate 5

b) Compare 5 to half the period of the applied signal

c) Sketch v o

Exercise!


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Problem 2.39

c) Due to the diode is in forward-bias for positive input cycle,therefore the positive input cycle will be examine first. Thecapacitor will charge with 10 0.7 + 2 = 11.3 V

+

0.7 V

+ 11.3 V

currentflo

w


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Zener Diodes

The application of Zener diodes have been explained inSubtopic 1.3

The analysis of Zener diodes can be divided into 3

categories: Fixed ViandRL Fixed Vi, variableRL Variable Vi, fixedRL

To make the analysis simple, the analysis will be explaindirectly from the examples


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Example 2.26a (Fixed Vi and RL)

Determine VL, VR andIZ


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To check whether VZis in the on or off mode, the value ofVLmust be determine first

To do that, take out the Zener diode from the diode

The circuit become:


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By doing a nodal analysis for the node VL

As we can see, the value ofVL is smaller than VZ, so the Zener diodeis in the off mode

Which will result in:

And:

V73.8

2.11

16

L

LL

V

k

V

k

V

A0ZI

V27.773.816

R

LRi

V

VVV


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Example 2.26b (Fixed Vi and RL)

Repeat Example 2.26a withRL= 3k

3 k


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The same analysis is repeated from Example 2.26a where the Zenerdiode is taken out to examine the value ofVL

The circuit becomes:

3 k


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By doing a nodal analysis for the node VL

As we can see, the value ofVL is larger than VZ, so the Zener diode isin the on mode

When the Zener diode is in the on mode, it will maintain thevoltage of 10V. Because of that VL becomes:

And VR becomes:

V12

31

16

L

LL

V

k

V

k

V

V01ZLVV

V61016 RV


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Using current divider theory:

mA67.2

3

10

1

6

kk

R

V

R

V

III

L

LR

LiZ


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Example 2.27 (Fixed Vi, Variable RL)

a) Determine the range ofRL andIL that will result in VL beingmaintained at 10 V

b) Determine the maximum wattage rating of the diode


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a) To maintain VLat 10 V, the Zener diode must be in the on mode

ForIZM= 32 mA, the current at load:

The load would be:mA8

321

1050

mk

IIIZMRL

k25.18

10

mI

VR

L

L

L


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Example 2.27 (Fixed Vi, Variable R

L)

ForIZ(min), the Zener diode are assume off but the voltage VZaremaintained at 10 V

The load current would be:

The load would be:mA40

1

1050

k

IIRL

25040

10

mI

VR

L

L

L


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Retrieve back all the IL and RL value:

250mA40

k25.1mA8

LL

LL

RI

RI

IL (min)

IL (max) RL (min)

RL (max)


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Example 2.27 (Fixed Vi, Variable R

L)

Plotting VL vs RL: Plotting VL vs IL:


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b) To calculate the maximum wattage (power) ofthe Zener diode:

mW320)32)(10(max mIVP ZMZ


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Example 2.28 (Variable Vi, Fixed RL)

Determine the range ofVithat will maintain the Zener diode in theon mode


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Example 2.28 (Variable Vi, Fixed R

L)

To maintain Zener diode in on mode, VZmust equal to VL:

Taking the maximum current of the Zener diode, input currentbecomes:

The input voltage will become:

V20LZ

VV

mA67.762.1

2060

k

m

IIILZMR

V87.36

)220)(67.76(20

i

Ri

V

mRIV


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Example 2.28 (Variable Vi, Fixed RL)

ForIZ(min), the Zener diode are assume off but the voltage VZaremaintained at 20 V

Using nodal analysis at node VL:

Retrieve back all the value ofVi:

V67.23

2.120

22020

i

i

V

k

V

V67.23V87.36 ii

VV

Vi(max) Vi(min)


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Example 2.28 (Variable Vi, Fixed R

L)

Plotting VL versus Vi:


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Problem 2.43a

Question:

Design the network given to maintain VL at 12 V for a loadvariation (IL) from 0 mA to 200 mA. That is determine Rs andVZ.

Exercise!


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Problem 2.43b

Question:

Determine PZ(max) for the Zener diode in Problem2.43a

Exercise!

2.diode applications

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