electronics and microelectronics b ae4b34em...4. lecture • diodes - applications • rectifiers in...

4. lecture

• Diodes - applications

• Rectifiers

Jiří Jakovenko – Electronics and Microelectronics - Department of Microelectronics – CTU

Electronics and Microelectronics AE4B34EM

Junction breakdown or reverse breakdown

An applied reverse bias (voltage) will result in a small current to flow through the device.

At a particular high voltage value, which is called as breakdown voltage VB, large currents start to flow.

If there is no current limiting resistor which is connected in series to the diode, the diode will be destroyed.

There are two physical effects which cause this breakdown:

1) Zener breakdown is observed in highly doped p-n junctions and occurs for voltages of about 5 V or less.

2) Avalanche breakdown is observed

in less doped p-n junctions.


Junction breakdown or reverse breakdown

Avalanche breakdown mechanism occurs when electrons and holes moving through the DR and acquire sufficient energy from the electric field to break a bond i.e. create electron-hole pairs by colliding with atomic electrons within the depletion region.

The newly created electrons and holes move in opposite directions due to the electric field and thereby add to the existing reverse bias current. This is the most important breakdown mechanism in p-n junction.

Zener breakdown occurs at highly doped p-n junctions with a tunneling mechanism. In a highly doped p-n junction the conduction and valance bands on

opposite side of the junction become so close during the reverse-bias that the electrons on the p-side can tunnel from directly VB into the CB on the n-side.


Diode Circuit Models

The Ideal Diode Model The diode is designed to allow current to flow in only one direction. The perfect diode would be a perfect conductor in one direction (forward bias) and a perfect insulator in the other direction (reverse bias). In many situations, using the ideal diode approximation is acceptable.

Example: Assume the diode in the circuit below is ideal. Determine the value of ID if a) VA = 5 volts (forward bias) and b) VA = -5 volts (reverse bias)

+

_ VA

ID

RS = 50

a) With VA > 0 the diode is in forward bias and is acting like a perfect conductor so:

ID = VA/RS = 5 V / 50 = 100 mA

b) With VA < 0 the diode is in reverse bias and is acting like a perfect insulator, therefore no current can flow and ID = 0.


The Ideal Diode with Barrier Potential This model is more accurate than the simple ideal diode

model because it includes the approximate barrier potential voltage. Remember the barrier potential voltage is the voltage at which appreciable current starts to flow.

Example: To be more accurate than just using the ideal diode model include the barrier potential. Assume V = 0.3 volts (typical for a germanium diode) Determine the value of ID if VA = 5 volts (forward bias).

+

_ VA

ID

RS = 50

With VA > 0 the diode is in forward bias and

is acting like a perfect conductor so write a

KVL equation to find ID:

0 = VA – IDRS - V

ID = VA - V = 4.7 V = 94 mA

RS 50

V

+

V

+



The Ideal Diode with Barrier Potential and

Linear Forward Resistance

This model is the most accurate of the three. It includes a linear forward resistance that is calculated from the slope of the linear portion of the transconductance curve. However, this is usually not necessary since the RF (forward resistance) value is pretty constant. For low-power germanium and silicon diodes the RF value is usually in the 2 to 5 ohms range, while higher power diodes have a RF value closer to 1 ohm.

Linear Portion of transconductance curve

VD

ID

VD

ID

RF = VD

ID

+ V

RF



Rs D

D+Rs

The Ideal Diode with Barrier Potential and Linear Forward Resistance

Example: Assume the diode is a low-power diode with a forward resistance value of 5 ohms. The barrier potential voltage is still: V = 0.3 volts (typical for a germanium diode) Determine the value of ID if VA = 5 volts.

+

_ VA

ID

RS = 50

V

+

RF

Once again, write a KVL equation for the circuit: 0 = VA – IDRS - V - IDRF

ID = VA - V = 5 – 0.3 = 85.5 mA

RS + RF 50 + 5



Values of ID for the Three Different Diode Circuit Models

Ideal Diode

Model

Ideal Diode

Model with

Barrier

Potential

Voltage

Ideal Diode

Model with

Barrier

Potential and

Linear Forward

Resistance

ID 100 mA 94 mA 85.5 mA

These are the values found in the examples on previous slides where the applied voltage was 5 volts, the barrier potential was 0.3 volts and the linear forward resistance value was assumed to be 5 ohms.



ID (mA)

VD (Volts)

2

4

6

8

10

12

0.2 0.4 0.6 0.8 1.0 1.2 1.4

The transconductance curve below is for a Silicon diode. The Q point in this example is located at 0.7 V and 5.3 mA.

4.6

0.7

5.3

Q Point: The intersection of the load

line and the

transconductance curve.

The Q Point


+

_ VA

= 6V

ID

RS = 1000

V

+ Rectifiers


TYPES OF RECTIFIERS

Rectifier

Half-wave Rectifier Full-wave Rectifier

Centre-tape

full-wave rec.

Full-wave

Bridge rec.

Jiří Jakovenko – Electronics and Microelectronics - Department of Microelectronics – CTU Jiří Jakovenko – Electronics and Microelectronics - Department of Microelectronics – CTU

Half-wave rectifier

DC

230 V RMS 50 Hz

diode

RL Rectifier vL

1:a

Transformer Load

+

-

t

v L

t 0

DC

Average

VMax VRMS 2Vm for sineorcosinewave

VM VRMS 2 a

VL is the DC part of vL

aVL

.2230

Average value of half-wave rectified

π

V

θcos2π

V

dθ0dθθsinV2π

1

dtVT

1V

P

π

0

P

2π

0p

T

0DC


RMS Value of half-wave rectified

4

V

θsinθπ4

V

dθ2θcos12

1

π2

V dθθ

π2

V

dθ0dθθsinVπ2

1

dtVT

1V

P

π

0

P

P π

0

P

2π

0

2

p

T

0

2

rms

2

2

0

2

2

2

2

2

22

1

sin


VDC

Vrms

t0

t1

t2

t3

Vp

VO

t

Half-wave rectified


VDC – Average value Vrms – RMS value

A center-tapped full-wave rectifier



diode 1

RL

vL

1:a

+

diode 2

100 RMS 50 Hz

-

D1 D1 D1 D2 D2

vL

t(ms)

1/50 1/100

VL

DC

0

aVL

21002

L

LR

aI

21002

Full-wave rectifier Average Value of center-tapped full-wave rectifier

π

V2

θcosπ

V

dθθsinVπ

1

dtVT

1V

P

π

0

P

π

0p

T

0DC


RMS Value of center-tapped full-wave rectifier

2

V

θsinθπ2

V

dθ2θcos12

1

π

V

dθθsinVπ

1

dtVT

1V

P

π

0

P

P

π

0

2

p

T

0

2

rms

2

2

0

2

2

2

22

1


Full-wave Bridge rectifier




diode-1

diode-2

diode-4

RL

vL

100 RMS

50 Hz

+ -

diode-3

t(s) 1/50 1/100

VL

DC or

Average

D1 D3

D4 D2

0


aVL

21002

L

LR

aI

21002


Low Pass Filter To smooth the bumps (ripple)


RL

diode

Rectifier vL

1:a

Transformer Load 230 V RMS

50 Hz +

-

C

Filter

iC iRL

Diode; Reverse Biased RL

vL

-

C

+

-

Low Pass Filter To smooth the bumps (ripple)

Capacitor

Charge

Capacitor

discharge

t(s) 0

v L

VL

DC or

Average

∆V

same filter capacitor and load and derived from the same sinusoidal input voltage

Comparison of ripple voltages for half-wave and full-wave rectified


t(s) 0

v L

VL C1

C2 >C1

∆V1

t1 t2 t3

T

iD

VL C2

C1

Vm

ID DC

∆V2

T

t3 t1 t2

VL Vm(1T

2RLC)

v Vm 1 eT /(RLC )

v VmT

RLC

If T <<< RLC

T = T/2 for Full

wave


Full wave:

vVmT

RLC

If T <<< RLC

Full Wave: vL

t(ms) 1/50 1/100

DC

Effects of RL

and C

C=1000µF C=470µF

C=100µF

R=1500Ω

R=1000Ω

R=500Ω

(a) RL fixed (b) C fixed


COMPARISON OF RECTIFIERS

Half-wave Centre-tap Bridge type

No. of diode 1 2 4

Transformer

necessary

No Yes No

Maximum

efficiency

40.6% 81.2% 81.2%


Rectifier

Half-wave

or

Full-wave

Low pass

Filter

Voltage

regulator

RL

-

+

IL

Power Supply

VBR

electronics and microelectronics b ae4b34em...4. lecture • diodes - applications • rectifiers in...

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