Semiconductor Diodes,diode approxim.pptx

Semiconductor applications

1. Module 1 Semiconductor Diodes :Introduction, PN Junction diode, Characteristics and Parameters, Diode Approximations, DC Load Line. Diode Applications : Introduction, Half Wave Rectification, Full Wave Rectification, Full Wave Rectifier Power Supply: Capacitor Filter Circuit, RC π Filter. Zener Diodes : Junction Breakdown, Circuit Symbol and Package, Characteristics and Parameters, Equivalent Circuit, Zener Diode Voltage Regulator.

2. Semiconductor Diodes

3. Introduction The term diode identifies a two-electrode or two-terminal device. A Semiconductor diode is simply a p-n junction with a connecting Lead on each side. A diode is a one-way device offering a low Resistance when forward biased and behaving almost as an open Switch when reverse biased.

4. An approximately constant voltage drop occurs across a forward-biased diode. Diode (forward and reverse) characteristics are graphs of corresponding current and voltage levels. For precise circuit analysis, dc load lines are drawn on the diode forward characteristic. Some diodes are low-current devices for use in switching circuit . High-current diodes are most often used as rectifiers for ac to dc Conversion. Zener diodes are operated in reverse breakdown because they have a very stable breakdown voltage.

5. p-n Junction Diode A p-n junction has the ability to permit essential current flow when forward-biased and to block current when reverse-biased. Thus it can be used as a Switch: on when forward-biased , and off when biased in reverse. A p-n junction provided with copper wire connecting leads becomes an electronic device known as a diode. Circuit symbol (or graphic symbol) for a diode is an arrowhead and bar. The arrowhead indicates the Conventional direction of current flow when the diode is forward biased. The p-side of the diode is always the positive terminal for forward bias and is termed the anode. The n-side called the cathode is the negative terminal.

6. Continued… A p-n junction diode can be destroyed by a high level of forward current overheating the device. It can also be destroyed by a large reverse voltage causing the junction to break down. In general physically large diodes pass the largest currents and survive the largest reverse voltages. Small diodes are limited to low current levels and low reverse voltages.

7. Maximum current is 100mA and reverse voltage is 75V and reverse current is less than 1µA Maximum current is 400mA and reverse voltage is 200V

8. Characteristics and Parameters Typical forward and reverse characteristics for a silicon diode. There is a substantial forward current (IF) when the forward voltage (VF) exceeds approximately 0.7 V.

9. ‘Typical forward and reverse characteristics for a germanium diode. Substantial forward current (IF) “flows when ‘the ‘forward voltage (VF) exceeds approximately 0.3v

10. Diode Parameters VF- forward Voltage drop IR- reverse saturation current VBR -reverse breakdown voltage rd -dynamic resistance IF(max) -maximum forward current Some of the parameters can be determined directly from the diode characteristics. For the silicon diode characteristics VF= 0.7 V, IR = 100 nA , and VBR= 75 V. Forward resistance: It is a constant resistance of the diode at a particular constant forward current.

11. Continued… The dynamic resistance of the diode is the resistance offered to changing levels of forward voltage. The dynamic resistance, also known as the incremental resistance or ac resistance, is the reciprocal of the slope of the forward characteristics beyond the knee. rd=∆ Vf /∆If The dynamic resistance can also be calculated from the equation rd =26mV/If

12. Example 1.Determine the dynamic resistance at a forward current of 70 mA for the diode characteristics given . Using Equation, estimate the diode dynamic resistance.

13. Solution

14. Diode Approximations IDEAL DIODE CHARACTERSTICS As already discussed, a diode is essentially a one-way device. It has low resistance when forward biased, and a high resistance when biased in reverse. An ideal diode (or perfect diode) have zero forward resistance and zero forward voltage drop. It has an infinitely high reverse resistance, which would result in zero reverse current.

15. Continued… An ideal diode does not exist, there are many applications where diodes can be assumed to be near-ideal devices. In circuits with supply voltages much larger than the diode forward voltage drop, VF can be assumed to be constant without introducing any serious error. Also, the diode reverse current is normally so much smaller than the forward current that the reverse current can be ignored. These assumptions lead to the near-ideal.

16. Problem A silicon diode is used in the circuit shown below. Calculate the diode current.

17. Piecewise Linear Characteristic When the forward characteristic of a diode is not available, a straight-line approximation called the piecewise linear characteristic may be used. To construct the piecewise linear characteristic, VF is first marked on the horizontal axis. Then, from VF straight line is drawn with a slope equal to the diode dynamic resistance.

18. Example Construct the piecewise linear characteristic for a silicon diode which has a 0.25 dynamic resistance and a 200mA maximum forward current.

19. DC Equivalent Circuits An equivalent circuit for a device is a circuit that represents the device behavior. Usually, the equivalent circuit is made up of a number of components, such as resistors and voltage cells. A diode equivalent circuit may be substituted for the device when investigating a circuit containing the diode. Equivalent circuits may also be used as device models for computer analysis.

20. Continued… A forward-biased diode is assumed to have a constant forward voltage drop (VF) and negligible series resistance. In this case the diode equivalent circuit is assumed to be a voltage cell with a voltage VF. This simple dc - equivalent circuit is quite suitable for a many diode applications.

21. Continued… A more accurate equivalent circuit includes the diode dynamic resistance (rd) in series with the voltage cell.

22. Example Calculate IF for the diode circuit shown in the figure, assuming that the diode has VF = 0.7 V and rd = 0. Then recalculate the current taking rd = 0.25 Ω .

23. Solution

24. DC LOAD LINE ANALYSIS DC Load line: Figure shows a diode in series with a 100 Ω resistance and a supply voltage. The polarity of E is such that the diode is forward biased, so a diode forward current (IF) flows. As already discussed, the circuit current can be determined approximately by assuming a constant diode forward voltage drop (VF). When the precise levels of the diode current and voltage must be calculated, graphical analysis (also termed dc load line analysis) is employed.

25. Continued… For graphical analysis, a dc load line is drawn on the diode forward characteristics. This is a straight line that illustrates all dc conditions that could exist within the circuit. Because the load line is always straight, it can be constructed by plotting any two corresponding current and voltage points and then drawing a straight line through them. To determine two points on the load line, an equation relating voltage, current, and resistance is first derived for the circuit. E=(IF R1)+VF. Any convenient two levels of IF can be substituted into the equation to calculate the corresponding VF levels, or vice versa. It is convenient to calculate VF with IF=0, and to determine IF when VF=0.

26. Example Draw the dc load line for the circuit shown in fig on the diode forward characteristics.

27. Q-POINT The relationship between the diode forward voltage and current is defined by the device characteristic consequently; there is only one point on the dc load line where diode voltage and current are compatible with the circuit conditions. That is Q point termed as the quiescent point or dc bias point, where the load line intersects the characteristic.

28. Continued… This may be checked by substituting the levels of IF and VF at point Q. From the Q point IF= 40 mA and VF= 1V. E = (IF R1) + VF. Therefore, E = (40 mA x l00Ω)+1V=5 V So, with E =5 V and R, = 100 Ω , the only levels of IF and VF are IF= 40 mA and VF= 1 V.

29. Calculating Load Resistance and Supply Voltage When designing a diode circuit, It may be required to use a given supply voltage and set up a specified forward current. In this case, points A and Q are first plotted and the load line is drawn. R1, is then calculated from the slope of the load line. The problem could also occur in another way. For example, R1 , and the required IF are known, and the required supply voltage is to be determined. This problem is solved by plotting point Q and drawing the load line with slope 1/R1. The supply voltage is then read at point A.

30. Numerical Using the device characteristics determine the required load resistance for the circuit shown below to give IF = 30 mA .

31. 2.Determine a new supply voltage for the circuit shown below to give a 50 mA diode forward current when R1 = 100 Ω

32. Solution From Equation VF = E — (IF R1) Substituting IF = 0, VF =E-0=5V. 1.Plot point A on the diode characteristic at IF=0 and VF=5V 2.Now plot point Q at IF = 30mA Draw the dc load line through points A and Q. From the load line