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DMT 121 ELECTRONIC DEVICES

DMT 121 ELECTRONIC DEVICES. CHAPTER 2 DIODE APPLICATIONS. At the end of this class, students should be able to:-. Understand the concept of load-line analysis and how it is applied to diode networks. Explain the process of rectification to establish a DC level from a sinusoidal AC input.

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DMT 121 ELECTRONIC DEVICES

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  1. DMT 121ELECTRONIC DEVICES CHAPTER 2 DIODE APPLICATIONS

  2. At the end of this class, students should be able to:- Understand the concept of load-line analysis and how it is applied to diode networks. Explain the process of rectification to establish a DC level from a sinusoidal AC input.

  3. Load Line Analysis The analysis of electronic circuits can follow one of the two paths : Actual characteristic or approximate model of the device. Approximate model will be always used in the analysis VD= 0.7 V

  4. Load Line Analysis The load line plots all possible current (ID) conditions for all voltages applied to the diode (VD) in a given circuit. E / R is the maximum ID and E is the maximum VD. Where the load line and the characteristic curve intersect is the Q-point, which specifies a particular ID and VD for a given circuit. Fig. 2.1 Drawing the load line and finding the point of operation Point of operation of a circuit

  5. Load Line Analysis The intersection of load line in Fig. 2.2 can be determined by applying Kirchhoff’s voltage in the clockwise direction, which results in: Fig. 2.2 Series diode configuration ID and VD are the same for Eq. (2.1) and plotted load line in Fig. 2.2 (previous slide). Set VD = 0 then we can get ID, where Set ID = 0 then we get VD, where

  6. Example For the series diode configuration of Fig. 2.3a, employing the diode characteristics of Fig. 2.3b, determine VDQ, IDQ and VR Fig. 2.3 (a) Circuit; (b) characteristics.

  7. Solution From the result, plot the straight line across ID and VD. The resulting load line appears in Fig. 2.4. The Q points occurred at VDQ 0.78 VIDQ  18.5mA VR=IRR=IDQR=(18.5 mA)(0.5k) = 9.25 V

  8. Example For the series diode configuration of Fig. 2.13, determine VD, VR and ID. Solution:

  9. Example Repeat example 2.4 with the diode reversed Solution: Open Circuit

  10. Diode as Rectifier Rectifier: An electronic circuit that converts AC to pulsating DC. Basic function of a DC power supply is to convert an AC voltage to a smooth DC voltage.

  11. Half-Wave Rectifier The diode conducts during the positive half cycle. The diode does not conducts during the negative half cycle.

  12. Sinusoidal Input: Half Wave Rectification Fig. 2.44 Half-wave rectifier. Fig. 2.45 Conduction region (0  T/2). Fig. 2.46 Nonconduction region (T/2  T).

  13. Average Value of Half Wave Output Voltage The average value of the half-wave rectified output voltage (also known as DC voltage) is The process of removing one-half the input signal to establish a dc level is called half-wave rectification

  14. Example What is the average value of the half-wave rectified voltage? Solution: Vm/π = 15.9 V

  15. Effect of Barrier Potential(Silicon diode) Applied signal at least 0.7 for diode to turn on (Vk = 0.7V) Vi ≤ 0.7 V  diode in open circuit and Vo = 0V When conducting, Vk=0.7V ,then Vo= Vi – Vk  this cause reduction in Vo, thus reduce the resulting dc voltage level. Now Vdc  0.318 (Vm – Vk)

  16. Example Draw the output voltages of each rectifier for the indicated input voltages.

  17. Peak Inverse Voltage (PIV) PIV=peak inverse voltage and is the maximum voltage across the diode when it is not conducting/reverse bias. Can be found by applying Kirchhoff’s voltage law. The load voltage is 0V so the input voltage is across the diode at tp.

  18. Peak Inverse Voltage (PIV) Because the diode is only forward biased for one-half of the AC cycle, it is also reverse biased for one-half cycle. It is important that the reverse breakdown voltage rating of the diode be high enough to withstand the peak, reverse-biasing AC voltage. PIV=Vm OR accurately PIV (or PRV)  Vm • PIV = Peak inverse voltage • PRV = Peak reverse voltage • Vm = Peak AC voltage Diode must capable to withstand certain amount of repetitive reverse voltage

  19. Full-Wave Rectifier A full-wave rectifier allows current to flow during both the positive and negative half cycles or the full 360°. Output frequency is twice the input frequency. VDC or VAVG = 2Vm/π

  20. Full-Wave Rectification The rectification process can be improved by using more diodes in a full-wave rectifier circuit. Full-wave rectification produces a greater DC output: Half-wave: Vdc =0.318Vm =Vm/π Full-wave: Vdc =0.636Vm =2Vm/π Full Wave Rectifier Half Wave Rectifier

  21. Example Find average value of the full-wave rectified voltage?

  22. Transformer Coupling Turns ratio, n = Nsec/Npri V(sec) = nV(pri) (in RMS value) Vp(sec)=√2 x V(sec)

  23. Full-Wave Rectification • Center-Tapped Transformer Rectifier • Requires • Two diodes • Center-tapped transformer VDC=0.636(Vm)

  24. Full-Wave Center Tapped Current flow direction during both alternations. The peak output is about half of the secondary windings total voltage. Each diode is subjected to a PIV of the full secondary winding output minus one diode voltage drop PIV=2Vm(out)+0.7V

  25. PIV: Full-wave RectifierCenter-Tapped Transformer PIV can be shown by applying KVL for the reverse-biased diode. PIV across D2: 1 2 3 Substitute 4 to 2: PIV=2Vp(out) + 0.7 V 4

  26. Example • Show the voltage waveforms across each half of the secondary winding and across RL when a 100V peak sine wave is applied to the primary winding. • What minimum PIV rating must the diodes have.

  27. Solution 1. 2. PIV = 49.3 V

  28. Full-Wave Rectification • Bridge Rectifier • Four diodes are required • VDC = 0.636 Vm

  29. Full-Wave Bridge Rectifier The full-wave bridge rectifier takes advantage of the full output of the secondary winding. It employs 4 diodes arranged such that current flows in the direction through the load during each half of the cycle. During positive half-cycle of the input, D1 and D2 are forward-biased and conduct current. D3 and D4 are reverse-biased. During negative half-cycle of the input, D3 and D4 are forward-biased and conduct current. D1 and D2 are reverse-biased.

  30. PIV: Full-wave RectifierBridge Transformer Vp(out)=Vp(sec) – 1.4 V PIV=Vp(out) + 0.7 V

  31. Example • The transformer is specified to have a 12 Vrms secondary voltage for the standard 120 V across the primary. • Determine the peak output voltage for the bridge rectifier. • Assuming the practical model, what PIV rating is required for the diodes?

  32. Solution 1. Vp(out) = 15.6 V 2. PIV = 16.3 V

  33. Summary of Rectifier Circuits Vm = peak of the AC voltage = Vp In the center tapped transformer rectifier circuit, the peak AC voltage is the transformer secondary voltage to the tap.

  34. Power Supply Filters and Regulators In most power supply – 60 Hz ac power line voltage  constant dc voltage Pulsating dc output must be filtered to reduce the large voltage variation Small amount of fluctuation in the filter o/p voltage - ripple

  35. Power Supply Filters Filtering is the process of smoothing the ripple from the rectifier.

  36. Power Supply Filters and Regulators – Capacitor-Input Filter The capacitor input filter is widely used. A half-wave rectifier and the capacitor-input filter are shown.

  37. Power Supply Filters and Regulators Regulation is the last step in eliminating the remaining ripple and maintaining the output voltage to a specific value. Typically this regulation is performed by an integrated circuit regulator. There are many different types used based on the voltage and current requirements. A voltage regulator can furnish nearly constant output with excellent ripple rejection. 3-terminal regulators are require only external capacitors to complete the regulation portion of the circuit.

  38. Power Supply Regulators How well the regulation is performed by a regulator is measured by it’s regulation percentage. There are two types of regulation, line and load. Line regulation: how much the dc output changes for a given change in regulator’s input voltage. Load regulation: how much change occurs in the output voltage for a given range of load current values from no load (NL) to full load (FL) Line regulation Load regulation

  39. Power Supply Filters and Regulators – Capacitor-Input Filter Surge Current in the Capacitor-Input Filter: Being that the capacitor appears as a short during the initial charging, the current through the diodes can momentarily be quite high. To reduce risk of damaging the diodes, a surge current limiting resistor is placed in series with the filter and load. The min. surge Resistor values: IFSM = forward surge current rating specified on diode data sheet.

  40. Capacitor Input Filter – Ripple Voltage Ripple Voltage: the variation in the capacitor voltage due to charging and discharging is called ripple voltage Ripple voltage is undesirable: thus, the smaller the ripple, the better the filtering action The advantage of a full-wave rectifier over a half-wave is quite clear. The capacitor can more effectively reduce the ripple when the time between peaks is shorter. Figure (a) and (b) Easier to filter -shorted time between peaks. -smaller ripple.

  41. Capacitor Input Filter – Ripple Voltage • Lower ripple factor  better filter • [can be lowered by increasing the value of filter capacitor • or increasing the load resistance] • For the full-wave rectifier: Vp(rect) = unfiltered peak Ripple factor: indication of the effectiveness of the filter Vr(pp) = peak to peak ripple voltage; VDC = VAVG = average value of filter’s output voltage [half-wave rectifier]

  42. Example Determine the ripple factor for the filtered bridge rectifier with a load as indicated in the figure above.

  43. Diode Limiters (Clipper) Clippers are networks that employ diodes to “clip” away a of an input signal without distorting the remaining part of the applied waveform. Clippers used to clip-off portions of signal voltages above or below certain levels.

  44. Diode Limiter/Clipper A diode limiter is a circuit that limits (or clips) either the positive or negative part of the input voltage.

  45. Example What would you expect to see displayed on an oscilloscope connected across RL in the limiter shown in above figure.

  46. Solution

  47. Biased Limiters (Clippers) The level to which an ac voltage is limited can be adjusted by adding a bias voltage, VBIAS in series with the diode The voltage at point A must equal VBIAS + 0.7 V before the diode become forward-biased and conduct. Once the diode begins to conduct, the voltage at point A is limited to VBIAS + 0.7 V, so that all input voltage above this level is clipped off. A positive limiter

  48. Biased Limiters (Clippers) In this case, the voltage at point A must go below –VBIAS – 0.7V to forward-bias the diode and initiate limiting action as shown in the above figure. A negative limiter

  49. Modified Biased Limiters (Clippers)

  50. Example Figure above shows combining a positive limiter with a negative limiter. Determine the output voltage waveform?

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