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SPECIAL DIODES

Bahan Kuliah minggu ke 5 Elektronika Dasar. SPECIAL DIODES. Jurusan Teknik Elektro UGM 2007. Zener Diodes. A zener is used in reverse breakdown mode The voltage across a zener is more or less independent of the current through it

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SPECIAL DIODES

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  1. Bahan Kuliah minggu ke 5 Elektronika Dasar SPECIAL DIODES Jurusan Teknik Elektro UGM 2007 TE-UGM-2007

  2. Zener Diodes • A zener is used in reverse breakdown mode • The voltage across a zener is more or less independent of the current through it • The function of a zener is to provide a voltage reference in a circuit TE-UGM-2007

  3. ZENER CHARACTERISTIC A zener diode is much like a normal diode. The exception being is that it is placed in the circuit in reverse bias and operates in reverse breakdown. Operation region TE-UGM-2007

  4. Some important characteristics: • Nominal Zener Voltage : • 5.1V zener, • 12V zener, etc. • Nominal Bias Current: • the Iz to get the nominal Vz • Tolerance on zener voltage, e.g. : • 12V  5%, • Maximum Power: • 1Watt zener, 5 Watt zener, etc. • Temperature coefficient: • by what % does zener voltage change as diode temp. changes 1OC • Dynamic Resistance (Rd): • Rd = V/ I TE-UGM-2007

  5. Zener Operation Region • Zeners are available with voltage breakdowns of 1.8 V to 200 V. • This curve illustrates the minimum and maximum ranges of current operation that the zener can effectively maintain it’s voltage. ∆I ∆V TE-UGM-2007

  6. Basic Zener Circuit IT IZ IL Key points: • Vin > Vz • IT = (Vin – Vz)/Rs IL+IZ TE-UGM-2007

  7. Calculation: Find R Suppose a 5.1 Volt zener is connected to a 12 Volt supply through a resistor. The zener requires a 15 mA bias, and the load is 510 Ohms. Find the required resistor value. • load current: IL = 5.1V / 510 = 10 mA • total current: IT = IL + IZ= (10 + 15) = 25 mA • drop across R: VR = 12V – 5.1V = 6.9 V • R value : R = VR / IT = 6.9 V / 25 mA = 276 Ohms • Select standard value resistor: R = 270 Ohms TE-UGM-2007

  8. DATA SHEET TE-UGM-2007

  9. Zener Diode Applications Regulation In this simple illustration of zener regulation circuit, the zener diode will “adjust” it’s impedance based on varying input voltages and loads (RL) to be able to maintain it’s designated zener voltage. Zener current will increase or decrease directly with voltage input changes. The zener current will increase or decrease inversely with varying loads. Again, the zener has a finite range of operation. TE-UGM-2007

  10. Calculation: Find PMAX • A 10 V zener has 20 mA of bias current. The load resistor across the zener is 20 Ohms. What power rating should the zener have? Remember: if the load is removed, all current is in the zener. • Find total current: • IT= IBIAS + ILOAD = 20mA + 50mA = 70 mA • Find power in zener (Pz) at a current (Iz) = 70 mA: • Pz = Vz Iz = 10V  70ma = 700 mW • Double value for reliability: • Use a zener rated for 1.5 Watts or higher TE-UGM-2007

  11. Troubleshooting Although precise power supplies typically use IC type regulators, zener diodes can be used alone as a voltage regulator. A properly functioning zener will work to maintain the output voltage within certain limits despite changes in load. TE-UGM-2007

  12. Zener Limiting Zener diodes can be used as limiters. The difference to consider for a zener limiter is a it’s zener breakdown characteristics. TE-UGM-2007

  13. Voltage Surge Protectors • Fast, high-voltage transients, called “spikes”, on AC power lines can damage electronic equipment. • Back-to-back zeners can clip off the spikes. TE-UGM-2007

  14. Varactor Diodes A reverse-biased PN junction makes a voltage-controlled capacitor TE-UGM-2007

  15. Varactor Capacitance Fig 3.12 Capacitance range: from 50 pF to 500 pF TE-UGM-2007

  16. Varactor Diodes A varactor diode is best explained as a variable capacitor. Think of the depletion region a variable dielectric. The diode is placed in reverse bias. The dielectric is “adjusted” by bias changes. TE-UGM-2007

  17. Calculation: C & fR • If the varactor of figure 3.12 is biased at VR =5 V. • Find the capacitance from the graph. • Find the resonant frequency with a 253 uH inductor. • From the graph, C = 100 pF. • Resonant frequency fR = 1/(2LC) = 1.0 MHz TE-UGM-2007

  18. Varactor Tuner Similar tuners are used in TVs, cell-phones, etc. TE-UGM-2007

  19. Varactor Diodes The varactor diode can be useful in filter circuits as the adjustable component. TE-UGM-2007

  20. The PIN Diode Usable at high-frequencies TE-UGM-2007

  21. PIN DIODE The pin diode is also used in mostly microwave frequency applications. It’s variable forward series resistance characteristic is used for attenuation, modulation, and switching. In reverse bias exhibits a nearly constant capacitance. TE-UGM-2007

  22. Schottky Diodes • Not a PN junction • Fast, but reverse breakdown voltage less than 50 V TE-UGM-2007

  23. Schottky diode The Schottky diode’s significant characteristic is it’s fast switching speed. This is useful for high frequencies and digital applications. It is not a typical diode in the fact that it does not have a p-n junction, instead it consists of a heavily doped n-material and metal bound together. TE-UGM-2007

  24. TUNNEL Diode The tunnel diode has negative resistance. It will actually conduct well with low forward bias. With further increases in bias it reaches the negative resistance range where current will actually go down. This is achieved by heavily doped p and n materials that creates a very thin depletion region. TE-UGM-2007

  25. TUNNEL DIODE CHARACTERISTIC The step-recovery diode is also used for fast switching applications. TE-UGM-2007

  26. LASER DIODE The laser diode (light amplification by stimulated emission of radiation) produces a monochromatic (single color) light. Laser diodes in conjunction with photodiodes are used to retrieve data from compact discs. TE-UGM-2007

  27. Laser is an abbreviation of …... • Light Amplification by Stimulated Emission of Radiation • “Stimulated emission”antonym of “spontaneous emission” • optical transition stimulated by the effect of electric field of light waveon the contrary usually emission occur spontaneously without help of electric field TE-UGM-2007

  28. What is the difference between LED and LD? • LED is light emitting diode • LD is laser diode • Diode is a semiconductor device which has an effect of rectification • Both LED and LD are semiconductor diode with a forward bias. Both emit light • LED emits light by spontaneous emission mechanism, while LD has an optical cavity which enables multiplication of photon by stimulated emission TE-UGM-2007

  29. Explain how the light is transmitted through optical fiber. cladding CROSS SECTION N=1.46 N=1.48 Light Ray Entering Core from Air core Light is propagated by Total internal reflection • Light is transmitted along the core by total reflection mechanism at the boundary with the cladding layer TE-UGM-2007

  30. LEDs: Light Emitting Diodes • Brightness proportional to current • Colors: red, white, blue, green, orange, yellow • Drop across an LED is about 1.5 Volts TE-UGM-2007

  31. Optical Diodes The light-emitting diode (LED) emits photons as visible light. It’s purpose is for indication and other intelligible displays. Various impurities are added during the doping process to vary the color output. TE-UGM-2007

  32. Calculation: Power in an LED • How much power does an LED consume if it requires 25 mA and has a forward drop of 2.0 Volts? • P = V  I = 2V  .025A = 50 mW TE-UGM-2007

  33. THE 7-SEGMENT DISPLAY • Bright, but consumes a lot of power • Typically multiplexed to conserve power TE-UGM-2007

  34. 7 SEGMENT DISPLAY The seven segment display is an example of LEDs use for display of decimal digits. TE-UGM-2007

  35. Power in a 7-Segment Display • How much power would a 4-digit 7-segment LED display consume if each LED required 10 mA and had a forward drop of 1.5 Volts? • Power in one LED: • PLED = V  I = 2V  .01A = 20 mW • Assume all segments are lit, then: • Power in a Digit: • PD = 7  PLED = 7  20mW = 140 mW • Total Power: • PT = 4  PD = 4  140 mW = 560 mW • That’s over half a Watt! TE-UGM-2007

  36. Multiplexing to Reduce Power • Suppose a 4-digit display requires 400 mW if all segments are lit. If the display is multiplexed so that each digit is lit in a continuous sequence (1,2,3,4,1,2,3,4...) how much power would the display use? • Since each digit is on for only 25% of the time, • P = 0.25  400 mW = 100 mW TE-UGM-2007

  37. Troubleshooting Although precise power supplies typically use IC type regulators, zener diodes can be used alone as a voltage regulator. A properly functioning zener will work to maintain the output voltage within certain limits despite changes in load. TE-UGM-2007

  38. DIODA FOTO TE-UGM-2007

  39. Fig. Description Harga • A Photo Transistor $0.70 $0.55 • A Photo Transistor 0.40 0.30 • B Photo Transistor 0.45 0.40 • C Photo Transistor 0.40 0.30 • A Photo Diode 0.75 0.65 • A Photo Diode 0.50 0.40 • H Photo Diode 0.50 0.40 • D Photo Darlington/Motorola 2.95 2.50 • D Photo Darlington 2.25 1.95 • D Photo Transistor GE 1.85 1.65 • D Photo Transistor GE 1.85 1.65 • E Photo Diode 1.20 0.90 • D Photo Transistor/Motorola 1.90 1.60 • F Photo Darlington 2.50 1.95 • F Photo Diode 1.60 1.40 • G Solar Cell 2.10 1.95 TE-UGM-2007

  40. FOTODIODA INFRA MERAH TE-UGM-2007

  41. FOTODIODA InGaAs TE-UGM-2007

  42. FOTODIODA ULTRAVIOLET TE-UGM-2007

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  46. photodiode The photodiode is used to vary current by the amount of light that strikes it. It is placed in the circuit in reverse bias. As with most diodes when in reverse bias, no current flows when in reverse bias, but when light strikes the exposed junction through a tiny window, reverse current increases proportional to light intensity. TE-UGM-2007

  47. Fundamentals of photodiode - - - - + + + + ｐ-type ｎ-type Depletion layer • Illuminate the pn junction • Electrons and holes are generated by an excitation across the gap • Generated electrons and holes are separated and drift to electrodes by diffusion potential TE-UGM-2007

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