3.pel 2 - peranti s.konduktor - diod.ppt

1

Chapter 2

DIODE

EE201 SEMICONDUCTOR DEVICES

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Physical

Symbol

AnodKatod Katod Anod

N-type material is called Katod (K) P-type material is called Anod (A)

N P

a. Physical structure and symbol

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i. Anod is connected to positive supplyii. Katod is connected to negative supply

i.Forward bias

+ve

-ve

K A

+ve

-ve

K A

ON

When forward biased, current can flow through the diode

Like a switch is switched on.

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+ve

-ve

K A

Reverse bias:– i. Anod is connected to negative supplyii. Katod is connected to positif supply

+ve

- ve

K A

Suis OFF

Current cannot flow through the diode

Like a switch is switched off

ii.Reverse bias

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iii. I-V characteristic curve for silicon diode

VD

Forward voltage

IS (μA)

Reverse current

VSReverse voltage

ID (mA)

Forward current

Silikon – 0.7v

Brekdown voltage

Knee voltage

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a. Knee voltage(threshold voltage)

Voltage level where the increment of current happens. When the applied forward biased voltage reach the barrier voltage.

Knee voltage for diode Si – 0.7v

Ge – 0.3v

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b. Forward current(Id)(milliampere)

Amount of current that can be handled savely when the forward voltage is supplied

Measured in milliampere (mA)

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Very small current or leakage current when the diod is reversed biased.

Measured in micro ampere (µA)

c. Reverse current (microampere)

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d. Breakdown voltage

Definition:-

Voltage level where the increment of reverse current happens (in microampere)

Big current value exceeds the breakdown level can burn the p-n junction and damage it.

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e. Burning Level (when Id, Vd exceeds P mak)

Power(P) that exceeds the max power of the diode during forward biased.

P maks is produced from Id and Vd, Vd is a constant

Normally P is represented by maksimum current (Id)

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Difference between silicon and germanium diod

1. Knee voltage for diode Ge is 0.3V

2. Diode Ge need only a smaller forward biased voltage and let the current pass compared to diode Si (0.7V)

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Difference between silicon and germanium diode

3. Forward current increment rate after exceeds the knee voltage is slower compared to forward current diode Si

4. Reverse current or leakage current for diode Ge is bigger compared to diode Si

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Ideal diode concept

When diode operates, the characteristic inside the diode makes the analysis work of electronics circuits difficult. The characteristics are:-a. Barrier voltage

b. Forward current

c. Reverse current (leakage current)

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2. ZENERDIODE

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Schematic symbol for zener diode

Operates in the reverse biased zone Diode zener voltage rate is from

2.4v to 200v with power rate 1/4w to 50w

Used to set a reference point for some of dc output voltage

Katod Anod

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Vs Vd

Is

Id

Characteristic curve for zener diode

Breakdown voltage or zener voltage

(Vz)

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Unregulated input voltage

R1

DzRegulated output

Usage :-

Stabilizing the output voltage even there is a change in input voltage

Voltage regulator / voltage stabilizer

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Advantages of zener diode

1. AT the zener voltage level and beyond it, it able to conduct the high reverse current without damage.

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Advantages of zener diode

2. AT the zener voltage level and beyond it, the voltage across the diode – constant and same as the zener voltage. Only change in the current value.

3. Operates in breakdown zone.

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3. Light Emitting Diode(LED)

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iv. Schematic symbol for LED

Katod

Anod

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Characteristics of LED

1. Used as indicator

2. Like diode, LED operates when receives forward bias voltage. Electrons from N-type will combine with hole in P-type.

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3. Semiconductor – Silicon @ Germanium, this

combination will produce heat.

semiconductor – Galium Arsenide (GaAs), Galium Phosphate (GaP) or Galium Arsenide Phosphate (GaAsP), this combination will produce light


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The light colours depends on the type of the material.

a.GaAs = infra red

b.GaP = red @ green

c.GaAsP = red @ yellow


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5. a. Operates at low voltage between 1 to 4V and let current 10 to 40mA flow.

b. Breakdow voltage – low, 3 to 5V

c. higher voltage or current can destroy the LED

d. the LED brightness depends on the current value

•RECTIFIERRECTIFIER

Use one or more diode

Cut ½ –ve cycle or ½ +ve cycle

Convert a.c. to pulses rippling d.c.

Ripple d.c. output

a.c. component exist in d.c.

•A.c. input with lowered voltage

3 types of rectifier:-

i. half wave

ii. Full wave

iii. Bridge

i. Half wave rectifier

•Circuit

•V A.C

240v 50hZ

•D1

•RL

•Transformer •+

ve

•- ve

•+

•i/p •o/p

•+ve

•0

•+•D1

•A

•B

•+ve

•-ve

•+ve

•0

•D1 ON

•RL

•input ½ +ve

•Output ½ +ve

•Circuit operation


•During ½ +ve cycle

•i. Terminal A is +ve, terminal B is – ve

•ii. Anod D1 gets +ve voltage

•iii.D1 in forward biased




•iv. D1 let current pass through it.

•v. Voltage drop exist across RL

•iv. Voltage drop RL is the output voltage




•+

•D1

•A

•B

•+ve

•-ve

•D1 OFF

•RL

•Input ½ -ve

•0

•+ve

•0

•- ve

•No output


•During ½ -ve cycle


•i. Terminal A is -ve, terminal B +ve•ii.Anod D1 gets –ve voltage•iii. D1 in reverse biased




•i. D1 blocks the current flow through it•ii.No voltage drop across RL because no current flow •iii. Output voltage is zero




Output voltage

• Occurs during positive half cycle only

• Voltage drop across diode is 0.7V (assume silicon diode) , output voltage is :-

Vo = Vi - 0.7

•+ve

•-ve

•D1 •A

•B

•RL

•Input

•output

•D1 ON

•D1 OFF


Output voltage

Output frequencySame as input:- 50Hz

•ExampleA half wave rectifier gets input voltage A half wave rectifier gets input voltage 20Vp-p, 50Hz. With the assumption that 20Vp-p, 50Hz. With the assumption that is no voltage drop across diode, calculate is no voltage drop across diode, calculate :- :-

i.i. Rectifier output voltageRectifier output voltage

ii.ii. Output frequencyOutput frequency

•D1

•RL

•+ve

•- ve

•+

•Example

•+10V

•-10V

•i/p •o/p

•Vm = 20Vp-p

• = 10Vp-p

•solution

•Output voltage = 10V

•Output frequency = input frequency

• = 50Hz

•12

v

•12

v •D1

•+ve

•- ve

•RL

•D2

•0v

•+

ii. Full wave rectifier

•V A.C 240v 50Hz

•Circuit

•D1

•+ve

•- ve

•RL

•+•0v

•+•0

•0

•D2

•D2

•+

•RL

•0v

•A

•D1

•D1 ON

•D2 OFF

•A

•B

•+•Switched On

•RL

•0v

•B•Switched Off

•Circuit operation •During ½ +ve cycle

• Terminal A is +ve, terminal B is –ve

• D1 forward biased, D2 reverse biased

• Current can flow through D1 like a switch is switched on




• D2 blocks current flow, switch off

• Current flow through D1, RL and return to terminal 0

• Voltage drop across RL

• Voltage drop across RL is the output voltage




•D1

•+ve

•-ve

•RL

•0v

•0

•+•0

•D2

•D2

•RL

•0v

•A

•D1

•D2 ON

•D1 OFF

•+•RL

•0v

•B

•Switch Off

•+

•A

•C

•B

•+


•Switch On


• Terminal A is -ve, terminal C is +ve

• D2 forward biased , D1 reverse biased

• Current can flow through D2 like a switch is switched on

i. Full wave rectifier



• D1 blocks current flow, switch off

• Current flow through D2, RL and return to terminal 0

• Voltage drop across RL

• Voltage drop across RL is the output voltage

i. Full wave rectifier

•During –ve ½ cycle


• Occurs in both cycle

• Voltage drop across diod is 0.7V (assume silicon diode), output voltage is :-

Vo = VAB - 0.7

Output voltage

Output frequency Double the input frequency :- 100Hz

•Example

A full wave rectifier with input voltage 20Vp-p, A full wave rectifier with input voltage 20Vp-p, 50Hz. Transformer with turns ratio 2 : 1. with 50Hz. Transformer with turns ratio 2 : 1. with assumption no voltage drop across diode, assumption no voltage drop across diode, calculate :-calculate :-

i.i. Rectifier output voltageRectifier output voltage

ii.ii. Output frequency Output frequency

•Vi = 20Vp-p• = 10Vp-p

•Solution

•Output frequency = 2 X input frequency = 100Hz•=

•VAC

•Vi

•Ns

•Np

•VAC •=•Ns

Np

•x Vm

•1•2

•= •x 10Vp

•= •5 Vp

•VAB •= •½ VAC•2.5 Vp•=

•So Vk = VBC

• = 2.5 Vp

•D1

•D3 •D2

•D4

•240VAC 50Hz

•Input voltage •Output

voltage

•Circuit

iii. Bridge rectifier

•D1

•D3 •D2

•D4


•A

•B

•Input voltage

•+

•0

•+

•Output voltage

•+

•0

•D1&D3 ON

•D2&D4 OFF

•+

•Circuit operation •During ½ +ve cycle

•D1

•D3

•A

•B

•Input voltage

•+

•0

•+

•Output voltage

•+

•0

•D1&D3 ON

•D2&D4 OFF

•+

iii. Bridge rectifier •Circuit operation •During ½ +ve

cycle

• Bridge rectifier


Circuit operation

Terminal A is +ve, terminal B is –ve

Anod D1 is +ve, Katod D3 is –ve

D1 and D3 forward biased.

Let current pass through

• rectifier Tetimbang

•Ketika ½ cycle +ve

Kendalian Litar

Terminal A is +ve, terminal B is –ve

Anod D2 is -ve, Katod D4 is +ve

D2 and D4 reverse biased

Block current flow

•D1

•D3 •D2

•D4

•A

•B

•Input voltage •+

•Output voltage

•+

•0

•D2&D4 ON

•D1&D3 OFF•+

•0

iii. Bridge rectifier•Circuit operation •During ½ +ve

cycle

•D1

•D3 •D2

•D4

•A

•B

•Input voltage •+

•Output voltage

•+

•0

•D2&D4 ON

•D1&D3 OFF•+

•0

iii. Bridge rectifier•Circuit operation •During ½ -ve

cycle

Terminal A is -ve, terminal B is +ve

Anod D2 is +ve, Katod D4 is –ve

D2 and D4 forward biased

Let current pass through


cycle

Terminal A is -ve, terminal B is +ve

Anod D1 is -ve, Katod D3 is +ve

D1 and D3 reverse biased

Block current flow


cycle

•Vo = VA-B - 1.4V

Occurs in both cycles

At one cycle, voltage drop across 2 diodes

So, the total voltage drop is 1.4V (assume silicon diode),

Output voltage is :-

Output voltage

Output frequency

•Output frequency = 2 X input frequency

•A bridge rectifier with input voltage 20 Vp-p 50 Hz. A transformer with turns ratio 2:1. With assumption that no voltage drop across diodes, calculate:-

i. Output voltage rectifier

ii. Output frequency

•Question Question

i. rectifier output voltage

•VinVin == 20Vp-p20Vp-p

•VA-B

•Vin

•== 10 Vp10 Vp

•=•Ns

•Np

•VA-B •= •Ns

•Np• x Vm

•xx 1010 Vp Vp•1

•2•=

•= •5Vp•Vout = VA-B

• = 5Vp

ii. Frekuensi isyarat keluaran

•2 x input frequency

•= 100 Hz

SEKIAN

SEMOGA

BERJAYA

DAN

DIBERKATI

3.pel 2 - peranti s.konduktor - diod.ppt

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