diodes. conduct electricity in one direction only silicon and germanium are the main semiconductor...

TEC 284Diodes

Diodes

Conduct electricity in one direction only

Silicon and Germanium are the main semiconductor materials used in manufacturing diodes, transistors and integrated circuits

Semiconductor material is refined to make it pure

Minute controlled amounts of impurity are then added in a process called doping

Silicon Crystal

Silicon crystal in its pure form is an insulator

No free electrons It has a lattice structure Each atom forms four covalent bonds with

neighboring atoms

Diodes

Two Dopant Types N-type (Negative) – Free flowing

electrons are added to the silicon structure

P-type (Positive) – Lack of electrons creates holes or slots which allow spaces for electrons to migrate to

Doping Silicon – N type

This is an N type semiconductor region due to the excess of electrons

Source: http://hyperphysics.phy-astr.gsu.edu/hbase/solids/dope.html#c3

Addition of impurities such as antimony, arsenic or phosphorous adds free electrons and greatly increases the conductivity of silicon

Doping Silicon – P type

This is an P type semiconductor region due to the deficiency of electrons

Source: http://hyperphysics.phy-astr.gsu.edu/hbase/solids/dope.html#c4

Addition of impurities such as boron, aluminum or gallium creates deficiencies of electrons or “holes”

PN junction

When a semiconductor chip contains a N doped region adjacent to a P doped region a diode junction or PN junction is formed

Junctions can be made from silicon or germanium, but the two are not mixed when making PN junctions

In a PN junction P material is called the anode and N material is called the cathode

PN Junction

Electric current will flow through a PN junction in one direction only

Forward Biased Diode

When diode is connected so that the current is flowing it is said to be forward biased

For forward biased diodes, anode is connected to a higher voltage than the cathode

Reverse Biased Diode

When diode is connected so that the current is not flowing it is said to be reverse biased

For reverse biased diodes, cathode is connected to a higher voltage than the anode

The diode cannot conduct

Diode Assumptions

In many circuits, the diode is considered to be a perfect diode to simplify calculations

A perfect diode means a zero voltage drop in the forward direction and no current conduction in the reverse direction

A forward based diode can be compared to a closed switch

A reversed based diode can be compared to an open switch

I-V curve for a diode

I-V characteristics for a P-N junction diode

Image Source : http://en.wikipedia.org/wiki/Diode

Knee voltage

For diodes there is a range or region where the diode resistance changes from high to low

This is called the knee regionThe voltage at which the diode

“turns on” is called the knee voltageFor most silicon diodes, this is about

0.7 VFor germanium diodes, it is approx

0.3 V

Knee voltage

Assumptions when using imperfect diodes

The voltage drop across the diode is either 0.7 V or 0.3 V. In some instances when voltage are large, it is assumed that diodes are perfect and when conducting, the voltage drop across the diode is oV. Diodes are assumed to be perfect when voltage supply is 10V or more

Excessive current is prevented from flowing through the diode by using a resistor in series with the diode

Calculation

Calculate the current flowing through the diode below. Assume VD = 0.7 V

Answer

I = (VS – VD) / VR

I = (5 – o.7) / 1000 = 4.3 mAThis is a series curcuit I is the same everywhere in the

circuit

Power dissipation in diodes

When current flows through a diode there is heating and power dissipation just like in a resistor

Voltage drop for a silicon resistor is assumed to be 0.7 V

For a silicon diode with 100 mA flowing through it

P = IV = .1 x .7 = 70 mW

Question

If a silicon diode has a max power rating of 2 Watts, how much current can it safely pass?

P = IV I = P / V I = 2 /0.7 = 2.86A

Finding the current through a diode

Find ID in the circuit above (the current through the diode)

ID

5 V

Steps• Find I2

• Find VR

• Find IT

• Find ID

Finding the current through a diode

I2 = (VD / R2) = 0.7/70 = 10 mAVR = (5 – VD) = (5 – 0.7) = 4.3 V IT = VR / R1 = 4.3 / 43 = 0.1 A = 100

mA ID = IT – I2 = 100 – 10 = 90 mA

ID

5 V

Steps• Find I2

• Find VR

• Find IT

• Find ID

Diode Breakdown

If you place the diode in the circuit backwards (reverse biased) then almost no current flows.

Diode Breakdown

The I-V curve for a perfect diode would be zero current for all values

For a real diode, when a certain voltage is reached, the diode “breaks down” and allows a large amount of current to flow

If this is allowed to continue, the diode will burn out

You can avoid burnout by limiting the current with a resistor

Diode Breakdown

Breakdown is not catastrophic and does not destroy the diode

The diode will recover and operate normally provided the current is limited to prevent burn out

Breakdown Voltage is called the Peak Inverse Voltage (PIV) or Peak Reverse Voltage (PRV)

Breakdown voltage varies from one type of diode to another

Zener Diode

Diodes can be manufactured so that breakdown occurs at lower and more precise voltages

These are called Zener diodesAt the Zener voltage a small current

will flow through the diodeThe current must be maintained to

the keep the diode at the zener point

Zener Diode

Zeners are used to maintain constant voltage at some point in a circuit

Zener Applications

Lamp requires 20 V and 1.5 A Power source is a generator at 50 V Generator voltage may fluctuate causing lamp to

get brighter and dimmer. Resistor calculation R is no longer valid for fluctuations.

This may be unacceptable behavior

Zener Applications

Alternative arrangementZener diode used to maintain

constant voltage across the lamp

Zener Considerations

Choose values of resistors that would prevent the Zener diode from burning out

This max current rating of the zener diode is given in the specs or can be calculated

P = I V

e.g A zener diode rated at 20 W which has a breakdown voltage of 10 V will allow at max 2 A to flow through it

Zener Calculation

Find the current going through the Zener diode above

IZ

Zener Calculation

IR = (50 – 20)/ 7.5 = 4 A

IZ = IR – IL IR = 4 – 1.5 = 2.5 A

IR

IZ

Zener Calculation

What happens if the input voltage changes to 35 V?

IR = (35 – 20)/ 7.5 = 2 A

IZ = IR – IL IZ = 2 – 1.5 = 0.5 A

IR

IZ

Zener behavior

When the input voltage changes from 50 V to 35 V (as in the case when the voltage on a generator fluctuates), all other values in the circuit remains the same

Only the current flowing through the zener diode changes

When determining the value of R for your circuit, start off with the worst case possible voltage (in this case 35 V) which allows 1.5 A to flow through the lamp