SNVA866A February   2019  – January 2023 LM5155 , LM5155-Q1 , LM51551 , LM51551-Q1

 

  1.   How to design an Isolated Flyback using LM5155
  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 Current Sense Resistor Calculations
      1. 3.3.1 Current Sense Resistor and Slope Compensation Resistor Selection
      2. 3.3.2 Current Sense Resistor Filter Selection
    4. 3.4 MOSFET Selection
    5. 3.5 Diode Selection
    6. 3.6 Output Capacitor Selection
    7. 3.7 Input Capacitor Selection
    8. 3.8 UVLO Resistor Selection
    9. 3.9 Control Loop Compensation
      1. 3.9.1 Feedback Resistor Selection
      2. 3.9.2 RPULLUP Selection
      3. 3.9.3 Optocoupler Selection
      4. 3.9.4 RLED Selection
      5. 3.9.5 Crossover Frequency Selection
      6. 3.9.6 Determine Required RCOMP
      7. 3.9.7 Determine Required CCOMP
  6. 4Component Selection Summary
  7. 5Small Signal Frequency Analysis
    1. 5.1 Flyback Regulator Modulator Modeling
    2. 5.2 Compensation Modeling
  8. 6Revision History

3.5 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 19.

Equation 19. GUID-306178A3-E805-46CD-B3B3-6C043A48B9FF-low.gif

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 20. For this application, a diode with a reverse voltage rating of 40 V is selected.

The average current of the secondary side diode is estimated using Equation 20.

Equation 20. GUID-1FF28671-4795-4C67-BFF2-5677AB760AFF-low.gif

The diode must be able to conduct the value calculated in Equation 20 with some margin. For the design, the selected diode is capable of conducting 10 A of average forward current.