Department of Electronics and Communication Engineering, School Of Engineering, Manipal University,...

Department of Electronics and Communication Engineering, School Of Engineering, Manipal University, jaipur

PN JUNCTION DIODEPN JUNCTION DIODE

(Characteristics and Applications)


• A diode is simply a pn junction, but its applications are extensive in electronic circuits.

• Three important characteristics of a diode are:Forward voltage drop.Reverse voltage drop.Reverse breakdown voltage.

PN JUNCTION DIODE


A diode is a two-terminal electronic

device

consisting of a single p-n junction. This

p-n

junction is usually created on a single

block

of silicon by doping the block with

donor and

acceptor dopants at opposite ends.

PN JUNCTION DIODE


A diode is a rectifier, allowing current

to pass in one direction but not in the

opposite direction.

PN JUNCTION DIODE


PN JUNCTIONPN JUNCTION

v

v

- -- -- -- -- -

+ ++ ++ ++ +

+ +

P Type

N Type

Free ElectronsHoles

Depletion regionPotential at this junction is called as Barrier Potential


The combination of Electrons and

holes

depletes the holes in the P region

and

electrons in the N region near the

junction

PN JUNCTION


The device is called as PN junction Diode

P N

Anode Cathode

Diode Symbol

AnodeCathode


Forward Biasing of PN – Junction Diode

Variable DC Voltage

Connect P type to positive

of the dc source

Connect N type to negative

of the dc source


Forward biasing the p-n junction

drives holes to the junction from the p-

type material and electrons to the

junction from the n-type material.

At the junction the electrons and

holes combine so that a continuous

current can be maintained.

Forward Biasing of PN – Junction Diode


Reverse Biasing of PN – Junction Diode

Variable DC Voltage

- +

Connect N type to positive

of the dc source

Connect P type to negative

of the dc source


The application of a reverse voltage to the p-n junction will cause a transient current to flow as both electrons and holes are pulled away from the junction.

When the potential formed by the widened depletion layer equals

the applied voltage, the current will cease except for the small thermal current.

Reverse Biasing of PN – Junction Diode


Diode is unidirectional, i.e. current

flows in only one direction (anode to

cathode internally). When a forward

voltage is applied, the diode conducts;

and when a reverse voltage is applied,

there is no conduction.


Vknee

mA

(V)

μA

(V)

VI Characteristics of PN Junction Diode

- +


1 )( eII T

D

η V

V

0D

The Diode Current EquationThe Diode Current Equation

Reverse saturation Reverse saturation currentcurrent

Applied Applied Forward Forward VoltageVoltage

TemperatureTemperatureEquivalent of VoltEquivalent of Volt


Tutorials

1. A germanium diode has reverse saturation current of 0.19μA. Assuming η =1, find the current in the diode when it is forward biased with 0.3 V at 27oC. (Ans: 19.5mA)

2. The forward current in a Si diode is 15 mA at 27oC. If reverse saturation current is 0.24nA, what is the forward bias voltage?

(Ans: 0.93V)

3. A silicon diode is reverse biased with 5V at room temperature. If reverse sat current is 60 pA, what is the diode current?

4. A germanium diode carries a current of 10mA when it is forward biased with 0.2V at 27oC. (a) Find reverse sat current. (b) Find the bias voltage required to get a current of 100mA.

(Ans: 4.42μA, 0.259V)


The reverse break down in diodes can occur due to two mechanisms, each of them require critical electric field at the

depletion region of the diode.

They are

Zener breakdown

Avalanche Multiplication

The Diode BreakdownThe Diode Breakdown


Zener breakdownZener breakdown

When the doping is very high (≥ 1025 atoms/m3).

The depletion region is very narrow.

which results in tunneling of electrons from p-type valance

band to the n-side conduction band constitutes a reverse current

from n to p, this is called Zener effect.

The basic requirement for the tunneling current is a large

number of electrons separated from a large number of empty

states by a narrow potential barrier.


The electric field resulting due to the depletion region causes field

emission where by the force on outer orbit electrons due to field

is very high that they are pulled out from the parent nucleus to

become free carriers.

This ionization by electro-static attraction is known as “Zener

breakdown” and causes an increase in the free carriers density

and hence an increase in the reverse current of the junction. Only

for the lower level of reverse voltage the zener effect is exhibited.


When the diode is reverse biased, carriers acquire sufficient

energy from the thermal energy and along with the applied

reverse bias results in the high electric field in the depletion

region.

An electron entering from the p-side may be accelerated to high kinetic

energy to cause ionizing collision. This ionizing collision results in the

breakage of covalent bonds, and generate new electron-hole pair. The

original electron and generated electron are both swept to the n-side of

the junction and generated hole is swept to the p-side.

Avalanche MultiplicationAvalanche Multiplication


The generation of electron-hole pair results in the generation of

enormous energy by the process called fission.

The liberated fission energy, along with the applied potential and

the thermal energy colloid with other non-ionized bonds.

This collision and generation of new electron-hole pairs are

continuous and multiplicative, which results in a large amount of

charge carriers and thus an increase in the reverse current.


In other words we can say, This phenomenon generally occurs in

wider depletion region, where the electric field strength is not

strong enough to produce zener breakdown.

Instead, free electrons (minority carriers) accelerated by the

electric field collide with electrons in the covalent bonds and

breaks it. Thus, an electron-hole pairs are accelerated by the field,

resulting in more collision, creating more free electrons. The

process quickly avalanches to produce the abrupt rise in current.


Since the reverse saturation current is temperature dependent parameter,

the reverse saturation current approximately doubles for every 10o C rise in

temperature. Let I01 is the reverse saturation current at temperature T1 and I02 is

the reverse saturation current at temperature T2, where T2 > T1. The rise in

reverse saturation current is given by the relation.

Effect of Temperature on the Reverse currentEffect of Temperature on the Reverse current


A Silicon diode has a saturation current of 1pA at 20 Deg C. Find Diode bias voltage when diode current is 3mA. Diode bias current when the temperature is 100 Deg C assuming the diode voltage to be constant.

Solution:Given The diode current ID=3mA,Reverse saturation current IO= 1 pA, Temperature T=20 Deg C = 273+20 = 293 KThe diode is silicon η=2The equation for the diode current ID is given by


The diode bias voltage

The diode current when the temperature is 100OC

The temperature is raised to 100OC (So the reverse saturation current I0 changes) use the relation.

p Axx 2 5 6)2(1 01)2(1 012II 81 21 0/)2 01 0 0(1 2) / 1 0T( T0 10 2

12



Diode resistances

• Two types of resistances are defined for a diode :

• Static or DC resistance:

– It is simply the ratio of diode voltage and diode current

– The dc resistance at the knee and below will be greater than the resistance at the linear section of characteristics

– The dc resistance in the reverse bias region will naturally be quite high

D

DD I

VR


Diode resistances

• Determine the dc resistances at the three different operating points shown

Ans:– A: 40 Ω

B: 250 Ω

C: 10 MΩ(μA)

(mA)

V (volts)

I

1μA

0.8V0.5V

20mA

2mA–10V

C

A

B


Diode resistances

• Dynamic or AC resistance– Often sinusoidal voltages are applied to diode– So the instantaneous operating point moves up and down in

the characteristic curve– So DC resistance is not a suitable parameter– Instead, AC resistance is used– It is the change in the diode voltage divided by the

corresponding change in the diode current, where the change is as small as possible

D

Dd I

Vr


Diode resistances

• Determine the AC resistances at operating points A and B

• Ans:A: 2 Ω

B: 25 Ω

V (volts)

0.80.78

I (mA)30

20

0.6 0.7

4

A

B


Diode resistances

• AC resistance is nothing but reciprocal of the slope of the tangent line drawn at that point

• Derivative of a function at a point is equal to the slope of the tangent line at that point

oVVo

DD

D

IeIdV

dI

dV

dTD /)(

T

oD

D

D

V

II

dV

dI

D

T

oD

T

D

D

D

Dd I

V

II

V

dI

dV

I

Vr


Diode resistances

• Dynamic resistance can be found using previous equation, no need of characteristic curve

• Dynamic resistance in reverse region is very high, since slope of characteristic curve is almost zero

• The resistance calculated using equation does not include the resistance due to the metal contact (usually less than 0.1 Ω)


Find the static and dynamic resistance of a p-n junction germanium diode if the temperature is 27 Deg C and IO=1μA for an applied forward bias of 0.2V.

SolutionGiven Applied forward voltage= 0.2 V Reverse saturation current IO=1uA, Temperature T=27 Deg C = 273+27 =

300 K The diode is Ge η=1



Diode Equivalent Circuit

• Diode is often replaced by its equivalent circuit during circuit analysis and design

• Equivalent circuit is obtained by replacing the characteristic curve by straight-line segments

Vγ

1/RF

RR = VγRF

A K

A K

A K

Forward bias

Reverse bias



• As further approximation, we can neglect the slope of the characteristic i.e., RF = 0

Vγ

RR =

RF = 0

Vγ

A K

A K

A K

Forward bias

Reverse bias



• As third approximation, even the cut-in voltage can be neglected (Ideal diode)

Vγ = 0

RR =

RF = 0

A K

A K

A K

Forward bias

Reverse bias


This is defined as the voltage that the diode has to withstand under reverse biased condition.

Peak Inverse Voltage (PIV)

Department of Electronics and Communication Engineering, Manipal University, jaipur

Zener Diode

• Zener diode is heavily doped P-N junction diode

• Optimized to operate in reverse breakdown region

• Each zener diode has specific breakdown voltage (VZ). Value of VZ depends on doping level (inversely proportional)

• Zener diodes are available with VZ ranging from 1.8V to 200V, power ratings from 250mW to 50W

• Symbol of zener diode:

Anode Cathode

P N


A Zener diode is a type of diode that permits current

to flow in the forward direction like a normal diode, but

also in the reverse direction if the voltage is larger than

the rated breakdown voltage or "Zener voltage".

Zener DiodesZener Diodes


A conventional solid-state diode will not let current flow

if reverse-biased below its reverse breakdown voltage.

By exceeding the breakdown voltage, a conventional

diode is destroyed in the breakdown due to excess

current which brings about overheating.

In case of forward-bias (in the direction of the arrow),

the diode exhibits a voltage drop of roughly 0.6 volt for a

typical silicon diode. The voltage drop depends on the

type of the diode.


A Zener diode exhibits almost the same properties,

except the device is especially designed so as to have a

greatly reduced breakdown voltage, the so-called Zener

voltage.

A Zener diode contains a heavily doped p-n junction

allowing electrons to tunnel from the valence band of the

p-type material to the conduction band of the n-type

material.

A reverse-biased Zener diode will exhibit a controlled

breakdown and let the current flow to keep the voltage

across the Zener diode at the Zener voltage


For example, a 3.2-volt Zener diode will exhibit a voltage drop of 3.2 volts if reverse biased. However, the current is not unlimited, so the Zener diode is typically used to generate a reference voltage for an amplifier stage, or as a voltage stabilizer for low-current applications.


Zener Diode characteristics



• V-I characteristics:

VZ

IZK

IZM

I

V



• IZK or IZmin – Minimum current necessary to maintain breakdown• IZM or IZMax – Maximum current that can be safely passed through

the zener diode• PZM or PZMax – Maximum power dissipation across zener diode• PZM = VZ.IZM

• Zener diode is always connected such that it is reverse biased, and it is in zener breakdown region

VZ

+

–

IZ



• V-I characteristics:

– When zener diode is forward biased, it acts like ordinary diode – i.e., until certain voltage Vγ is reached, current is zero, then afterwards, current rises exponentially

– When zener diode is reverse biased, until the breakdown voltage is reached, current is zero or negligible

– When reverse voltage equals zener voltage, current rises exponentially in reverse direction

– After the breakdown has occurred, voltage across zener diode remains almost constant at VZ, only the current increases with the increase in applied reverse bias.



• Equivalent circuits of zener diode

Forward Reverse Breakdown

• Note: RZ is usually very small, can be neglected

Vγ

RFRR ≈ RZ

VZ

–

+ –

+N

P

N N N

P P P

Zener Voltage regulator

• Zener diodes are widely used to regulate the voltage across a circuit.

• When connected in parallel with a variable voltage source so that it is reverse biased, a zener diode conducts when the voltage reaches the diode's reverse breakdown voltage. From that point it keeps the voltage at that value.


Zener Voltage Regulator


The circuit holds the voltage across the load RL almost equal to the voltage across zener VZ even after the input Vin and load resistor RL undergo changes.

If the unregulated dc voltage Vin rises, the current through R increases.


• If the load requires more current when RL is decreased, the zener diode can supply the extra current without affecting the load Voltage.

• Let I be the current through the resister R , we can write ,


The power dissipated in the diode is Pz=IZ Vz

The selection of Rs is very important here. We have,

After substituting the value of I we get,


• For Line regulation RL is constant.

• is also constant.

• And Vin varies between Vin(min) to Vin(max)

• For Load Regulation, Vin is constant and RL varies between RLmin and RLmax and load current is given by,

•


• When Vin=Vin(min),and IL is constant then,

• Similarly when Vin=Vin(max) we have,


Lz(max)max III

• The selected R must be small enough to permit minimum zener current to ensure that the diode is in its breakdown region.

• That is R must be small enough to ensure that minimum current IZ(min ) flows under worst condition.


This is when Vin falls to its smallest possible value Vinmin and IL is its largest possible value ILmax (Load Regulation). •At the same time R must be selected large enough to ensure that the current through the zener diode should not exceed the maximum zener current Izmax.

•so that power dissipation in the diode will not will not exceed Pz.


• That is the condition when Vin rises to the value of Vinmax and load current IL to its minimum Ilmin.

• So we can write,


Lminzmax

Zin(max)

II

VVR

• Applications:

As Voltage regulators

As Voltage Limiters

Wave shaping

Protection diode

Fixed reference voltage


LED

• The increasing use of digital displays in calculators, watches and all form of instrumentation has contributed to an extensive inherent in structures that emit light when properly biased.

• The two types of displays commonly used are light emitting diode (LED) and liquid crystal display(LCD).


LED• Light emitting diode is a diode that gives of visible or

invisible (infrared) light when energized.

• In any p-n junction there is a recombination of holes and electrons.

• During this process energy possessed by the free electron is transferred to another state, some of this energy is transferred into heat and some in the form of photons.

• In silicon and germanium greater percentage is converted into heat and the emitted light is insignificant.


• Diodes constructed of GaAs emit light in the infrared zone. Even though the light is not visible, they have numerous applications like, security systems, industrial processing, optical coupling etc.

• By using elements like gallium, arsenic and phosphorous LEDs producing red, green, yellow, blue, orange or infrared (visible). LED’s have replaced incandescent lamps in many applications because of their low voltage, long life, and fast on-off switching.


Problem

• In a zener voltage regulator if Vz=10V, Rs=1KΩ, RL=2KΩ. If the input voltage Vin varies from 22 to 40 V, find the maximum and minimum values of zener current.

Solution:


Lz(min)min III 7mA5x1012x10III 33Lminz(min)

Similarly,

Using above relation we can find


Lz(max)max III

25mAIII Lmaxz(max)

Problem

2) Design a Zener voltage regulator for the following specifications:

o/p voltage=5v,

i/p voltage, Vin =

Load current, IL=20 mA

Zener power, PZ(max)= 500 mw

IZ(min)=2mA.


Solution:


(max)(max) 500 ZZZ IVmwP

mAI Z 100(max)


8.181202

59

mAmARS

Lminzmax

Zin(max)

II

VVR

3.8320100

515

mAmARS

8.1813.83 SR 120SR

Problem1. A germanium diode carries a current of 10mA when a forward bias of 0.2V is applied across it at 27 Deg C.

a)Find the reverse saturation current.

b)Calculate the bias voltage needed for diode currents of 1mA and 100mA.

Answer:

a)I0= 4.42 µA

b) V=0.141,0.259


Problem

2. A silicon diode has a saturation current of 12nA at 20 deg C.

a)Find its current when it is forward biased by 0.65V.

b)Find the current in the same diode when the temperature rises to 100 Deg C.

Answer:

a) 4.57 mA

b) I0=3.072 µA, I=74.39 mA


Problem

3) The zener diode in a regulator circuit has a breakdown voltage of 15V and power rating of 0.5 W. if input voltage =40V, what is the minimum value of R that prevents zener diode from being destroyed.


Assignment

1) In a Zener diode regulator circuit, VL =15V,maximum load current is 100 mA, input voltage range is 18V to 20V, R =10 ohm.

Determine,

a) Maximum power dissipated by R

b) Minimum diode current.

c) The power that must be dissipated by R if the output is accidentally short circuited.



Assignment

1. A Silicon diode has a saturation current of 0.1pA at 20OC. Find its forward voltage when the current is 0.3mA.

2. A Germanium diode has IO=10μA. Determine its forward voltage when it is carrying 50mA of current. Compute the dynamic resistance at this operating point.

3. A Silicon diode at room temperature conducts 5mA at 0.7V. If the voltage increases to 0.8V. Find reverse saturation current.

4. Calculate the factor by which reverse saturation current IO of Germanium diode is multiplied when the temperature increases from 25 to 100OC.

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