1 / 246

Diodes

Diodes. 1. Diode. Class of non-linear circuits having non-linear v-i Characteristics Uses Generation of : DC voltage from the ac power supply Different wave (square wave, pulse) form generation Protection Circuits Digital logic & memory circuits. Creating a Diode.

anakin
Télécharger la présentation

Diodes

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. Diodes 1

  2. Diode • Class of non-linear circuits • having non-linear v-i Characteristics • Uses • Generation of : • DC voltage from the ac power supply • Different wave (square wave, pulse) form generation • Protection Circuits • Digital logic & memory circuits

  3. Creating a Diode • A diode allows current to flow in one direction but not the other. • When you put N-type and P-type silicon together gives a diode its unique properties.

  4. Diode Equivalent circuit in the reverse direction Equivalent circuit in the forward direction.

  5. Operation Reverse Bias • -ve voltage is applied to Anode • Current through diode = 0 (cut off operation) • Diode act as open circuit Forward Bias +ve voltage applied to Anode • Current flows through diode • voltage Drop is zero (Turned on) • Diode is short circuit

  6. The two modes of operation of ideal diodes Reverse biased Reverse Voltage 10 V Forward biased Forward Current 10 mA

  7. Ex 3.2

  8. Rectifier circuit Input waveform Equivalent circuit when vi0 Output waveform. Equivalent circuit when vi≤ 0 Waveform across diode

  9. Exercise 3-3

  10. Battery Charger

  11. Diodes are ideal , Find the value of I and V Figure 3.6 Circuits for Example 3.2.

  12. Example 3.2. Assumption Both Diodes are conducting

  13. Assumption Both Diodes are conducting Node A Node B Not Possible Thus assumption of both diode conducting is wrong

  14. Example 3.2(b). Assumption # 2 Diodes 1 is not conducting Diodes 2 is conducting Assumption is correct

  15. Figure E3.4 Diodes are ideal , Find the value of I and V

  16. Figure E3.4 Diodes are ideal , Find the value of I and V I= 2mA V= 0V I= 0A V= 5V I= 0A V= -5V I= 2mA V= 0V

  17. Figure E3.4 Diodes are ideal , Find the value of I and V I= 3mA V= 3V I= 4mA V= 1V

  18. Diodes are ideal , Find the value of I and V Figure P3.2

  19. Figure P3.2Diodes are ideal , Find the value of I and V Diode is conducting I = 0.6 mA V = -3V Diode is cut-off I = 0 mA V = 3V Diode is conducting I = 0.6 mA V = 3V Diode is cut-off I = 0 mA V = -3V

  20. Problem 3-3 Diodes are ideal , Find the value of I and V D1 Cut-Off & D2 Conducting I = 3mA D1 Cut-Off & D2 Conducting I = 1mA , V=1 V

  21. Figure P3.4 In ideal diodes circuits, v1is a 1-kHz, 10V peak sine wave. Sketch the waveform ofvo

  22. In ideal diodes circuits, v1is a 1-kHz, 10V peak sine wave. Sketch the waveform ofvo Vp+ = 10V Vp- = 0V f = 1 K-Hz Vp+ = 0V Vp- = - 10V f = 1 K-Hz Vo = 0V

  23. Figure P3.4 In ideal diodes circuits, v1s a 1-kHz, 10V peak sine wave. Sketch the waveform ofvo Vp+ = 10V Vp- = -10V f = 1 K-Hz Vp+ = 10V Vp- = 0V f = 1 K-Hz Vp+ = 10V Vp- = 0V f = 1 K-Hz

  24. Figure P3.4 In ideal diodes circuits, v1s a 1-kHz, 10V peak sine wave. Sketch the waveform ofvo

  25. Figure P3.4 In ideal diodes circuits, v1s a 1-kHz, 10V peak sine wave. Sketch the waveform ofvo Vp+ = 0V Vp- = -10V f = 1 K-Hz V0 = 0V Vp+ = 10V Vp- = -5V f = 1 K-Hz

  26. Figure P3.4 In ideal diodes circuits, v1s a 1-kHz, 10V peak sine wave. Sketch the waveform ofvo Vp+ = 10V Vp- = -5V f = 1 K-Hz

  27. Problem 3-4(k) For Vi >0 V D1 is cutoff D2 is conducting vo=1V For Vi < 0 V is conducting D2 is cutoff vo=vi+1V - 9 V

  28. Problem 3-4(k)

  29. Figure P3.6 X = A . B X = A + B

  30. Problem 3-4 (c) vo=zero

  31. Problem 3-4(f) Vi is a 1kHz 10-V peak sine wave. +ve Half Cycle with 10 V peak at 1 KHz

  32. Problem 3-4(h) vo=zero

  33. Problem 3.5 vi is 10 V peak sine wave and I = 100 mA current source. B is battery of 4.5 V . Sketch and label the iB 100 mA 4.5 v

  34. Solution P3-5 100 mA vo 4.5 v

  35. Problem 3-5 100 mA 4.5 v

  36. Problem 3-5 10 4.5 100 mA

  37. REVERSE POLARITY PROTECTOR

  38. REVERSE POLARITY PROTECTOR • The diode in this circuit protects a radio or a recorder etc... In the event that the battery or power source is connected the wrong way round, the diode does not allow current to flow.

  39. I1 2 I3 Problem 3-9 I1 2 I3 D1& D2 Conducting I1=1mA I3=0.5 mA I2=0.5 mA V= 0 V D1=off, D2=On I1= I3=0.66 mA V = -1.7 V

  40. Problem 3-10 D conducting I=0.225 mA V=4.5V D is not conducting I=0A V=-2V

  41. Problem 3-16 V REDGREEN 3V On Off D1 conducts 0 V Off Off -3 V Off On D2 conducts

  42. Quiz No 3 DE 28 EE -A Sketch vO if vi is 8 sin  Find out the conduction angle for the diode & fraction of the cycle the diode is conducting

  43. 8V I1 I2 I Solution Quiz No 3 vi/2 10-10-07

  44. Sketch vO if vi is 10 sin  Find out the conduction angle for the diode & fraction of the cycle the diode is conducts +12 V D1 never conducts Vi<5V D2 is cut-off, Vo=5V Vi>5V D2 is conducts D1 D2 5 22-10-07

  45. Quiz No 3 DE 27 CE -B Sketch vO if vi is 10 sin  Find out the conduction angle for the diode & fraction of the cycle the diode is conducts D1 never conducts Vi<5V D2 is cut-off, Vo=Vi Vi>5V D2 is conducts

  46. Problem • Assume the diodes are ideal, sketch vo if the input is 10sin (9) • Find out the conduction angles for Diode D1 & D2 (4) and the fraction of the cycle these diodes conduct. (2)

More Related