Diodes. Conduct electricity in one direction only Silicon and Germanium are the main semiconductor...
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Transcript of 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