EE 21-Lecture 3- Diode Applications

8/3/2019 EE 21-Lecture 3- Diode Applications

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Diode Circuits andApplicationsEE 21-Fundamentals of Electronics

EE21Slides(AAMS)

Topics

Series and parallel diode circuits (DC voltages)

Clippers / Limiters

Clampers

Rectifiers

Zener Diodes

Light-emitting Diodes

Power supply construction EE21Slides(AAMS)

Solving diode circuits

1. Determine the state of the diode/s in

the circuit (FB/RB).

2. Substitute the equivalent circuit per

diode (depending on the specified

approximation)*

3. Solve the desired quantities usingknown circuit analysis methods

EE21Slides(AAMS)

Series diode configurations

For the circuit shown

(a) determine VD, IR,and VR by 2

ndapproximation.

(b) Repeat (a) using 3rdapproximation (letrd=0.3).

(c) Repeat (a) with thediode reversed (2ndapproximation)

Example 1 EE21Slides(AAMS)


Answers:

(a) VD=0.7 VoltsVR =7.3 VoltsID = 3.32 mA

(b) VD=0.7 VoltsVR =7.3 VoltsID = 3.32 mA

(c) VD= 8 VoltsVR = 0 VoltsID = 0 A

Example 1


This is the same

circuit as the previousexample, except that

the source voltage is

now 0.5 volts.

Determine VD, VR, and

ID using 2nd

approximation. Example 1.5.EE21Slides(AAMS)


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Industry notation (grounding

notation?...)source notation?? Will be significant

later on (so get used

to it) Uses a reference

ground.

Analysis is much

easier

Tipid ng linya!

Masasanay din kayo.

EE21Slides(AAMS)


Determine ID, IR and

VO for the given

circuit. Use 2ndapproximation.

Answers:

Vo = 11 Volts

ID = IR = 1.96 mAExample 2.

EE21Slides(AAMS)


Determine ID, VD2

and VO for the given

circuit.

Answers:

ID = 0 A

VO = VR = 0 V

VD2 = 12 Volts

Example 3.EE21Slides(AAMS)


Determine I, V1, V2,

and VO for the series

dc configuration

below.

Answers:

I = 2.072 mA

V1 = 9.74 V

V2 = 4.56 V

Vo = -0.44 V

EE21Slides(AAMS)

Parallel diode configurations

For the circuit

below, determine VO,I1, ID1, and ID2.

Answers:

V0 = 0.7 V

I1 = 28.18 mA

ID1=ID2=14.09 mA


Determine the

current I for thegiven circuit.

Answer:

I = 6.95 mA

EE21Slides(AAMS)


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Determine the voltage

Vo for the given

circuit. Use 2ndapproximation.

Answer:

Vo = 11.3 Volts

Boylestad, R, Nashelsky, L. 2007.

EE21Slides(AAMS)

Series-parallel combination

For the circuit

shown, determine

the currents I1, I2,and ID2.

Answers:

I1 = 0.212 mA

I2 = 3.32 mA

ID2 = 3.108 mA

Series-parallel combination.EE21Slides(AAMS)

Multiple-diode cases with

uncertainty Sometimes we cant easily tell a diodes state

For some cases, it may be necessary to make a

guess on the state of the diode

After making initial guesses, prove/disprove

by performing equivalent circuit analysis

Multiple-diode cases with

uncertainty Recall that a forward-biased diode has to have

a current in the direction of the diode arrow

A reverse-biased diode has zero current but it

may have an open-circuit voltage VD.

Once proven/disproven, change initial guesses

and solve the equivalent circuit.

Example

DETERMINE:

a. Correct diode

states

b. V1 and V2.

Parallel diode application:

AND/OR gatesOR Gate AND Gate

EE21Slides(AAMS)


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Rectifiers

DC value = average value; for a sine wave, the

average value = 0.

Diodes are used as rectifiers in AC-to-DCconversion processes

EE21Slides(AAMS)

Half-wave rectifier analysis:

Positive cycle of input Diode: FB

Vo = Vi, thus

copying the shape ofthe input waveform

in consideration

EE21Slides(AAMS)

Half-wave rectifier analysis:

Negative cycle of input At negative cycle,

diode: RB

Vo = 0 (open circuit;

zero current

through R)

EE21Slides(AAMS)

Input vs. Output waveforms

From a zero DC

level, half-wave

rectification yields

us a DC or average

value

VDC = 0.318 Vm

Question: Where

did 0.318 come

from??

EE21Slides(AAMS)

Non-ideal diode analysis

The diode thresholdvoltage results in aminimum valuerequirement for FBoperation

Similarly, the peakrectified voltage isnow reduced by thevalue of VT. Also,

VDC = 0.318(VM-VT)

EE21Slides(AAMS)

Peak Inverse Voltage (PIV)

Tested in the

reverse bias regionof operation

If PIV is exceeded,

diode enters the

Zener avalanche

region

Ex. for HW rectifier

(ideal diode),

VmPIV EE21Slides(AAMS)


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Full-wave rectifier

Bridge Type Utilizes four diodes (some circuits may have

a bridge type but using less diodes)

Most familiar network

EE21Slides(AAMS)

FWBT Rectifier Analysis:

Positive Cycle of input Diodes D2 and D3

are forward-biased

(on) while diodesD1 and D4 are

reverse biased

(off)

Diode pairs switch

states on negative

cycleEE21Slides(AAMS)


Positive Cycle of input Equivalent circuit

(and conduction

path)

Vo = Vin (takes the

shape of the input)

EE21Slides(AAMS)


Negative Cycle of input Equivalent circuit

(and conduction

path)

Vo = Vi (again, takes

the shape of input)

EE21Slides(AAMS)

Input vs. output waveforms

From a zero DC

level, full-waverectification yields

us a DC or average

value

VDC = 0.636 Vm

(twice that of a half-

wave rectifier. How

did it differ?)EE21Slides(AAMS)

Non-ideal diode analysis

Conduction path

simply encounters

voltage drops (2nd

approximation)

Threshold voltage VTcauses the offset in

conduction and peak

rectified voltage

VDC = 0.636(VM-VT)

EE21Slides(AAMS)


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Full-wave rectifier Center

Tapped Transformer Type Uses only two

diodes but requires

a transformer toestablish the input

signal

EE21Slides(AAMS)

FWCT Rectifier Analysis:

Positive cycle of input Corresponding

polarities result in

one diode forwardbiased and the other

diode reverse biased

Vo = Vi (but take

note of the

transformer ratio)

EE21Slides(AAMS)

FWCT Rectifier Analysis:

Negative cycle of input Diode states become

opposite that of the

positive cycle states

Vo = Vi (but take

note of the

transformer ratio)

EE21Slides(AAMS)

FWCT Rectifier PIV

Assuming 1:2 turns

ratio, VSEC = VM,

therefore,

VmPIV 2

EE21Slides(AAMS)

Example:

Modified Bridge Type Rectifier Sketch the input-output waveform for the network

given. Also, determine the DC voltage availableand the required minimum PIV for each diode.Assume a sinusoidal, 10 VPEAK input.

EE21Slides(AAMS)

Clippers

Clips a portion of the input signal without

distorting the rest of the waveform

May clip either the positive or the negative

cycle (or both)

Two types: Series and Parallel

Also known as a limiter

EE21Slides(AAMS)


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Series Clipper

The diode is in

series with the load

Simplest example:

HW rectifier Series (negative) clipper

EE21Slides(AAMS)

Series clipper waveforms

EE21Slides(AAMS)

Example: Biased Limiter

Using the ideal approximation, sketch the

waveform of the output voltage Vo. Match it

with the input voltage waveform.

EE21Slides(AAMS)

Example: with square-wave

input Repeat the example, this time for the given

square-wave input.

EE21Slides(AAMS)

Assignment:

Repeat the previous examples using second

approximation and a silicon diodewith VT = 0.7 Volts.

EE21Slides(AAMS)

Parallel Clipper

EE21Slides(AAMS)


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Example: Biased limiter

Sketch the output waveform for the given clipper

circuit. Use (a) ideal approximation (b) second

approximation and a silicon diode (VT = 0.7 V)

EE21Slides(AAMS)

Example:

Parallel combination limiter Sketch the waveforms for VO and irfor the

combination limiter circuit shown.

EE21Slides(AAMS)

Clampers

Clamps the input signal to another axis

Involves a charging/discharging capacitor (time

constant=RC is important!)

The shape and peak-to-peak values of the input

waveform are retained after clamping

Analysis is always started on first forward bias

EE21Slides(AAMS)

Clamper Analysis: things to

remember The time constant elements must be chosen

in such a way that = RC is significantlylarger than half the input signals period (i.e.positive or negative cycle)

If RC components are not chosen accordingly,waveform shape becomes tapered

For clamper design, start on the first forwardbias

Basic clamper circuit

Take note of the capacitor-diode

connections: (-) of capacitor to anode OR(+) of capacitor to cathode.

Basic clamper circuit: First

Forward Bias Analysis Diode (ideal) shorts,

creating a loop andin turn charging C to

a voltage VC = V.

Also,

VO = 0.


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Basic clamper circuit: next (-)

cycle analysis Diode = RB

Total voltage across

load is the inputvoltage plus the

stored VC. Therefore,

VO = (V + VC)

Input vs. Output waveform

Note the conservation ofpeak-to-peak voltages:

Input: V ( V) = 2V Output: 0 ( 2V) = 2V

Waveform is clampedto a new axis Vo = V

How about for 2ndapproximation?

Example: Biased Clamper

Sketch the output waveform of the following

circuit. First use ideal approximation, then

second approximation with a silicon diode.

Clamping a sinusoidal input

Treat input as though it was square

Use the square wave as a mold or envelope

for the sinusoidal input.

Example: Clamper Design

Design a clamper to perform the function

indicated. Choose component values thatwould satisfy the time constant requirement

of the circuit.

Zener Diodes

Special-purpose

diodes thatspecifically work on

the Zener region

(reverse bias)

Used in regulation

circuits or as a

reference voltage

http://ielectronicnews.com/2011/09

/zener-diodes/

A K


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CharacteristicCurve

If VAK > VT (0.7 for Si), diode

operates in the usual forward

bias condition.

If VAK < 0.7 but |VAK| < |Vz|,

diode is in reverse bias (open)

If |VAK|> |VZ|, diode is

operation on the zener region

and is equivalent to a source

with voltage = zener voltage

Vz.

If VAK> VT:

If VAK< VT but |VAK| < |VZ|

If VAK< VT and |VAK| > |VZ|

Conditions and Equivalents

Reverse Bias,

OPEN CKT

Reverse Bias,

ZENER REGION

Fixed Vin & RL

Fixed Vin, variable RL

Fixed RL, variable Vin

Three Cases of Zener Diode

CircuitsCommon required quantities

Pz, Power dissipated by zener diode

Iz, Zener diode current

IL, load current through resistor

Range of values of IL, RL and VIN toensure zener region operation

Basic Zener Circuit

Determine expressions for Pz, Iz, and IL

Vin is a DC source input, and note that the

zener diode is connected in a reverse-bias

manner so as to simplify computations.

General method of solution:

Determine the state of the diode by using VDR:

If Vd is at least equal to Vz, then zener diode is on

(on: operating on [desired] zener region). Otherwise,

substitute corresponding model.

Equivalent ckt if

Zener diode is on:

L

L

LRRs

RViVVd


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Equivalent circuit: Zener

diode on

Applying KCL : IRs = IZ + IL, therefore IZ = IRs IL

Also, , and VL = Vz. Therefore,

Finally, Pz = VzIz. (computed value must be less than Pzmax)

L

L

L

R

VI

LL

L

RsR

VzVi

R

VViI

Example 1

Determine VL, VRS, IZ and PZ for RS = 1k and RS = 3k.

Zener diode parameters: Vz = 10 V, Pzm = 30 mW.

Some circuits require changing the input voltage Vi

and/or the load resistor RL to ensure zener diodes

correct operation.

If RL is too small, the Zener diode will remain in

off state. If RL is too large, the maximum zener

power or current may be exceeded.

Same principles can be considered for the supply

voltage Vi.

Fixed Vin, Variable RL /

Fixed RL, Variable VinFormula Considerations

Given the basic zener circuit, let us derive

expressions for the minimum and maximum

possible values of RL and Vin, given that wevary either the load resistance or the input

voltage while keeping the other constant.

RL for turn-on voltage of zener: VL = VZ. By VDR,

minimum resistance can be found by manipulating

to obtain:

with RLmin comes ILmax (low resistance -> high current):

Formula Considerations: Fixed Vi,

Variable RL

min

min

L

L

ZRRs

RViV

VzVi

VzRsR

Lmin

minmin

max

L

Z

L

L

LR

V

R

VI

By KVL , VRS = Vi VZ. Also, IRS = VRS/RS

By KCL: IZ = IRS IL., and IL = IRS IZ

Note: Iz is minimum when IL is maximum, and vice

versa.

Therefore: ILmin = IRS IZmax

And finally,

Formula Considerations: Fixed Vi,

Variable RL

min

max

L

Z

LI

VR


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From the circuit, ,

and by manipulation,

the minimum required Vin is given by:

The zener diodes maximum current, IZM, limits Vimax:

IRSmax = IZmax + IL

Vimax = VRSmax + Vz

Vimax = IRmaxRS + Vz

*note: in operation, IL is constant = VL/RL

Formula Considerations: Fixed RL,

Variable Vi

L

L

ZRRs

RViV

L

L

iR

RRsVzV

min

Example 2

Determine the range of RL and IL that will keep

the voltage at RL maintained at 10 volts. Also,

find the maximum wattage rating, PZmax, of the

zener diode.

Example 3

Determine the range of Vi that will keep the zener diode

in the ON state.

Documents

EE 21-Lecture 3- Diode Applications