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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.

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Diodes

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  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)

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