SNVA994A February   2022  – March 2023 LM5157 , LM5157-Q1 , LM51571-Q1 , LM5158 , LM5158-Q1 , LM51581 , LM51581-Q1

 

  1.   1
  2.   Trademarks
  3. 1Introduction
  4. 2Example Application
  5. 3Calculations and Component Selection
    1. 3.1 Switching Frequency
    2. 3.2 Transformer Selection
      1. 3.2.1 Maximum Duty Cycle and Turns Ratio Selection
      2. 3.2.2 Primary Winding Inductance Selection
    3. 3.3 Slope Compensation Check
    4. 3.4 Diode Selection
    5. 3.5 Output Capacitor Selection
    6. 3.6 Input Capacitor Selection
    7. 3.7 UVLO Resistor Selection
    8. 3.8 Control Loop Compensation
      1. 3.8.1 Crossover Frequency (fcross) Selection
      2. 3.8.2 RCOMP Selection
      3. 3.8.3 CCOMP Selection
      4. 3.8.4 CHF Selection
  6. 4Component Selection Summary
    1. 4.1 Application Circuit
    2. 4.2 Bill of Materials
  7. 5Small Signal Frequency Analysis
    1. 5.1 Flyback Regulator Modulator Modeling
    2. 5.2 Compensation Modeling
  8. 6Revision History

3.4 Diode Selection

The diode on the secondary side must have a reverse voltage rating greater than the reflected voltage for the primary transformer winding to the secondary winding plus the secondary load voltage. The reverse voltage of the secondary diode is calculated in Equation 13. The calculation example shows the calculation for VLOAD1, this needs to be done for VLOAD2 to VLOAD4 as well.

Equation 13. V D _ r e v e r s e 1 = N S 1 N P     ×   V S U P P L Y _ m a x +   V L O A D 1 =   1.2 1     ×   16 V +   10 V =   29.2 V

Due to leakage inductance, there is a negative spike when the primary side switch is being turned off. A snubber needs to be added across the diode to help minimize this voltage spike. Even if a snubber is added, some voltage margin must be added to the value calculated in Equation 13. For this application, a diode with a reverse voltage rating of 60 V is selected.

The average current of the secondary side diode is estimated using Equation 14.

Equation 14. I D _ A V G 1 =   I L O A D 1   = 250mA

The diode must be able to conduct the value calculated in Equation 14 with some margin. For the design, the selected diode is capable of conducting 1 A of average forward current.