1. Diodes
TEC 284

2. 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

3. 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

4. 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

5. Doping Silicon – N type
Addition of impurities such as antimony, arsenic or phosphorous adds free electrons and greatly increases the conductivity of silicon
This is an N type semiconductor region due to the excess of electrons
Source:

6. Doping Silicon – P type
Addition of impurities such as boron, aluminum or gallium creates deficiencies of electrons or “holes”
This is an P type semiconductor region due to the deficiency of electrons
Source:

7. 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

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

9. 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

10. 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

11. 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

12. I-V curve for a diode I-V characteristics for a P-N junction diode
Image Source :

13. 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

14. Knee voltage

15. 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

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

17. 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

18. 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

19. 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

20. Finding the current through a diode
Steps
Find I2
Find VR
Find IT
Find ID
5 V ID
Find ID in the circuit above (the current through the diode)

21. Finding the current through a diode
Steps
Find I2
Find VR
Find IT
Find ID
5 V ID
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

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

23. 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

24. 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

25. 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

26. Zener Diode
Zeners are used to maintain constant voltage at some point in a circuit

27. 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

28. Zener Applications Alternative arrangement
Zener diode used to maintain constant voltage across the lamp

29. 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

30. Zener Calculation Find the current going through the Zener diode aboveIZ
Find the current going through the Zener diode above

31. Zener Calculation IR = (50 – 20)/ 7.5 = 4 A IZ = IR – IL

32. 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

33. 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