Diode Applications rectifires, and .pptx

Diode applications

1. Diode Applications

2. Introduction One of the most important applications of diodes is rectification: conversion of a sinusoidal ac waveform into single-polarity half cycles. Rectification may be performed by half-wave or full-wave rectifier circuits. A dc power supply converts a sinusoidal ac supply to dc by rectification and filtering. The filtering process normally involves the use of a large reservoir capacitor, which charges to the peak input voltage level to produce the dc output.

3. Rectifier A rectifier is a device that converts alternating current (AC) to direct current (DC). It is done by using a diode or a group of diodes. Half wave rectifiers use one diode, while a full wave rectifier uses multiple diodes.

4. HALF-WAVE RECTIFICATION A half wave rectifier is defined as a type of rectifier that only allows one half-cycle of an AC voltage waveform to pass, blocking the other half-cycle. Half-wave rectifiers are used to convert AC voltage to DC voltage, and only require a single diode to construct.

5. Positive Half-Wave Rectifier

6. Working During positive half cycle of supply voltage, diode is said to be forward biased and acts as a short circuit and current flows through 𝑅 𝐿 . Therefore 𝑉 = +𝑉 𝑖𝑛 During negative half cycle of supply voltage, diode is said to be reverse biased and acts as an open circuit, no current flows through 𝑅 𝐿 . Therefore 𝑉 = 0.

7. Input and output waveform The graph above actually shows a positive half wave rectifier. This is a half-wave rectifier which only allows the positive half-cycles through the diode, and blocks the negative half-cycle.

8. When the diode is forward-biased T he peak output voltage is Where VPi = 1.414Vi, where Vi is the rms level of the sinusoidal input voltage to the rectifier circuit . The diode peak forward current is

9. During the negative half-cycle of the input the reverse-biased diode offers a very high resistance. So there is only a very small reverse current (IR), giving an output The peak reverse voltage, or peak inverse voltage (PIV), applied to the diode is voltage The average and rms values of the half-wave rectifier waveform can be determined by Vo( ave ) = 0.318 Vpo Vo( rms ) = 0.5 Vpo .

10. Example A diode with Vp = 0.7 V is connected as a half-wave rectifier. The load resistance is 500 Ω and the ( rms ) ac input is 22 V. Determine the peak output voltage, the peak load current, and the diode peak reverse voltage.

11. Solution

12. Negative Half Wave rectifier

13. Waveforms

14. Practice problems A half-wave rectifier circuit has a 15V ac input and a 330 Ω load resistance. Calculate the peak output voltage, the peak load current, and the diode maximum reverse voltage. A half-wave rectifier produces a 40mA peak load current through a 1.2 k Ω resistor. If the diode is silicon, calculate the rms input voltage and the diode PIV.

15. Solution 20.51 V, 62.2 mA, 21.2 V 33.95 V, 48.7 V

16. Full Wave Rectifier A full wave rectifier is defined as a type of rectifier that converts both halves of each cycle of an alternating wave (AC signal) into a pulsating DC signal. We can further classify full wave rectifiers into:  Centre-tapped Full Wave Rectifier  Full Wave Bridge Rectifier

17. Circuit diagram

18. Working During positive half cycle of supply voltage Diode D1 will be forward biased and hence conduct, while D2 will be reverse biased and will not conduct. As D1 conducts, current flows through the path D1-𝑅 𝐿 -centre tap. Therefore 𝑉 = +𝑉 𝑖𝑛 . During negative half cycle of supply Diode D1 will be reverse biased, while D2 will be forward biased and hence conduct. As D2 conducts, current flows through the path D2-R 𝐿 -Centre tap. Thus current keeps on flowing through 𝑅 𝐿 .

19. Waveforms

20. Negative full wave rectifier

21. Full wave Bridge Rectifier

22. Working During the positive half-cycle of input voltage, diodes D1, and D4 are in series with RL. Load current I L flows from the positive input terminal through D1 to R L and then through R L and D4 back to the negative input terminal. Direction of the load current through R L is from top to bottom. During this time D2 and D3 will be reverse Biased.

23. The bridge rectifier has two forward-biased diodes in series with the supply voltage and the load. Because each diode has a forward voltage drop (V F ), the peak output voltage is The average value of the FWR is The rms value of the FWR is

24. Problem Determine the peak output voltage and current for the bridge rectifier circuit shown in Fig. below. when V i = 30 V, R L = 300 Ω , and the diodes have V F = 0.7 V.

25. Solution

26. Other form of Bridge rectifier

27. Advantages of full wave rectifier The rectification efficiency of full wave rectifier is double than the half wave rectifier because it converts both the cycles of AC to DC. Ripple factor is less in full wave rectifier so waveform is smooth. The ripple frequency is also double so they are easy to filter out. The full wave rectifier produces high DC output voltage and current so the output power is higher. The full wave rectifier have better transformer utilization

28. Filter circuit Filter circuit are used to remove the ac component present in the output obtained from rectifier. Filters are the circuits which converts pulsating dc to pure dc. Capacitor and inductor are used in filter circuits. Filter circuit are connected between rectifier and load. Most popular used filters are capacitor and RC filter.

29. Capacitor Filter circuit In capacitor filter circuit, capacitor is connected parallel with the output of the rectifier in a linear power supply. Capacitor increases the dc voltage and decreases the ripple voltage components of the output.

30. Half wave rectifier with capacitor filter

31. Working During positive half cycle of input, diode is forward biased and capacitor charges to a peak value V m . When input starts decreasing below its peak value, capacitor will be fully charged and holds the charge until the input ac supply to the rectifier reaches the negative half cycle. During negative half cycle of input, diode gets reverse biased and blocks the current through it. During this non-conduction period, Vs is less than Vc so the capacitor discharges through 𝑅𝐿 until Vc is less than Vs. This cycle repeats producing a very smooth dc voltage. Time required by the capacitor is very small to charge while its discharging time is very large, ripple in the output gets reduced.

32. Full wave rectifier with Capacitor filter

33. Analysis

34. The capacitor value is calculated as, The repetitive current (IFRM) can be determined from The average forward current for each diode is half of the load current.

35. The reverse voltage across D3 is Capacitor discharge time is equal to half the input waveform time period.

36. Problem The full-wave rectifier dc power supply as shown in figure below is to supply 20 V to a 500 Ω load. The peak-to-peak ripple voltage is not to exceed 10% of the average output voltage, and the ac input frequency is 60 Hz. Accurately calculate the required reservoir capacitor value.

37. Solution

38. Transformer Selection The transformer specification for a full-wave bridge rectifier power supply is determined similarly to that for a half-wave circuit, with some exceptions. Two diode voltage drops are involved in calculating the secondary rms voltage. For full-wave rectifiers with capacitor filters, power supply transformers manufacturers recommend Which gives

39. As with all transformers, the primary current is

40. RC ∏ FILTER

41. The ripple voltage that appears across the capacitor in a rectifier power supply can be _attenuated by the use of an additional resistor and capacitor, which together function as an ac voltage divider. The combination of C1, R1, and C2 is referred-to-as π filter, because of the π-shaped arrangement of the circuit components. Assuming a constant output load current, the capacitor continues to charge and discharge, producing a sawtooth (ripple) waveform across C1 regardless of the presence of the additional components

42. The sawtooth waveform is composed of a fundamental ac voltage (same frequency as the ripple) and a number of smaller-amplitude, higher-frequency harmonic components. Due to their higher frequencies, the harmonic components are more severely attenuated than the fundamental frequency component by the voltage division across R1 and C2. This combined with the smaller input amplitude of the harmonics means that the waveform developed across C2 (the filter output) is essentially an attenuated version of the sinusoidal fundamental component. The peak value of the fundamental component of the sawtooth waveform can shown to be

43. The ac voltage developed across C2 is the filter ac output and is given by

44. LC ∏ Filter