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Diodes. TEC 284. 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
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Diodes TEC 284
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 region • The voltage at which the diode “turns on” is called the knee voltage • For most silicon diodes, this is about 0.7 V • For germanium diodes, it is approx 0.3 V
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 mA • This 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) • Steps • Find I2 • Find VR • Find IT • Find ID 5 V ID
Finding the current through a diode • I2 = (VD / R2) = 0.7/70 = 10 mA • VR = (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 • Steps • Find I2 • Find VR • Find IT • Find ID 5 V 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 diodes • At the Zener voltage a small current will flow through the diode • The 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 arrangement • Zener 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