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ZENER DIODES

Chia sẻ: Van Kent Kent | Ngày: | Loại File: PDF | Số trang:19

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VOLTAGE REGULATION A voltage regulator circuit automatically maintains the output voltage of a power supply constant, regardless of - a change in the load - a change in the source voltage ZENER DIODES The simplest of all voltage regulators is the zener diode voltage regulator. A zener diode is a special diode that p is optimized for operation in the g breakdown region. ZENER DIODE CIRCUIT The zener diode is typically connected reverse biased, in parallel with the load. Resistor Rs limits current to zener. zener ZENER DIODE CHARACTERISTICS In the forward region, the zener diode acts like a regular silicon diode, with a 0.7 volt drop...

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  1. 3.1 ZENER DIODES
  2. VOLTAGE REGULATION A voltage regulator circuit automatically maintains the output voltage of a power supply constant, regardless of - a change in the load - a change in the source voltage
  3. ZENER DIODES The simplest of all voltage regulators is the zener diode voltage regulator. A zener diode is a special diode that p is optimized for operation in the g breakdown region.
  4. ZENER DIODE CIRCUIT The zener diode is typically connected reverse biased, in parallel with the load. Resistor Rs limits current to zener. zener
  5. ZENER DIODE CHARACTERISTICS In the forward region, the zener diode acts like a regular silicon diode, with a 0.7 volt drop when it conducts.
  6. ZENER DIODE CHARACTERISTICS In the reverse bias region, region a reverse leakage current flows until the breakdown voltage is reached. At this point, the reverse current, called zener current Iz, increases sharply. sharply
  7. ZENER DIODE CHARACTERISTICS Voltage after breakdown is also b kd i l called zener voltage Vz. Vz Vz remains nearly a constant, even constant though current Iz varies considerably.
  8. ZENER DIODE RATINGS A zener spec sheet typically provides - the maximum power rating Pz - the nominal zener voltage Vz at test current IzT - the maximum DC zener current IzM Example: - 1N752 has a power rating of 500mW, a h ti f 500 W nominal zener voltage of 5.6V at a test current of 20 mA, a maximum zener mA current of 80 mA.
  9. ZENER DIODE CIRCUIT EXAMPLE In a zener diode voltage regulator circuit, VDC(IN) = 20V Vz = 9V, Rs = 82Ω, RL = 100 20V, V 9V R 82 100Ω Q: Calculate IL. A: IL = VL/RL = VZ/RL = 9V/100 Ω = 90 mA 82Ω 20V 100Ω 9V
  10. ZENER DIODE EXAMPLE (CONT’D) Q: Calculate IT. A: IT = VRs/Rs = [VDC(IN) - Vz]/Rs = [20 - 9]/82 Ω = 134 mA 82Ω 20V 9V 100Ω
  11. ZENER DIODE EXAMPLE (CONT’D) Q: Calculate Iz A: Iz = IT - IL = 134mA - 90mA = 44mA 82Ω 20V 9V 100Ω
  12. ZENER DIODE EXAMPLE (CONT’D) Q: If zener Pz = 500mW, is Iz safe ? A: IzM = Pz / Vz = 500 mW / 9V = 55 mA Since circuit Iz = 44 mA (< 55mA), zener is safe
  13. ZENER DIODE EXAMPLE (CONT’D) Q: Suppose load changes from 100 Ω to 200 Ω. Q pp g Find new IL, Iz, IT. Is Iz safe ? A: IL = VL/RL = VZ/RL = 9V/200 Ω = 45 mA 20V 9V 200Ω
  14. ZENER DIODE EXAMPLE (CONT’D) A: IT = VRs/Rs = [VDC(IN) - Vz]/Rs = [20 - 9]/82 Ω = 134 mA (no change) Iz = IT - IL = 134mA - 45mA = 89mA Since IzM = 55mA, zener is not safe. 20V 9V 200Ω
  15. ZENER DIODE EXAMPLE ( (CONT’D) ) Q: Suppose load changes from 100 Ω to 80 Ω. Find new IL, Iz, IT. Is Iz safe ? A: IL = VL/RL = VZ/RL = 9V/80 Ω = 112 5 mA 112.5 20V 9V 80Ω
  16. ZENER DIODE EXAMPLE (CONT D) (CONT’D) A: IT = VRs/Rs = [VDC(IN) - Vz]/Rs = [20 - 9]/82 Ω = 134 mA (no change) Iz = IT - IL = 134mA - 112 5mA = 21.5 mA 112.5mA 21 5 Since IzM = 55 mA, zener is safe. 20V 9V 80Ω
  17. ZENER DIODE EXAMPLE (CONT’D) Q: Suppose zener regulator circuit has no load connected, is Iz safe ? A: IT = VRs/Rs = [VDC(IN) - Vz]/Rs = [20 - 9]/82 Ω = 134 mA (no change) Iz = IT = 134 mA this is very Unsafe 20V 9V
  18. ZENER DIODE EXAMPLE ( (CONT’D) ) Q: In the previous example, suppose VDC(IN) changes from 20V to 30V. Is Iz safe ? A: IL = 9V/100 Ω = 90 mA IT = [30-9]/82 Ω = 256 mA Iz = IT - IL = 256 mA - 90 mA = 166 mA Very Unsafe 82Ω 30V 9V 100Ω
  19. ZENER DIODE EXAMPLE ( (CONT’D) ) Q Q: Suppose VDC(IN) changes from 20V to 18V. pp ( ) g Is Iz safe ? A: IL = 9V/100 Ω = 90 mA IT = [18-9]/82 Ω = 110 mA Iz = IT - IL = 110 mA - 90 mA = 20 mA Safe 82Ω 18V 9V 100Ω
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