SNVA790A October   2020  – July 2022 LMR36520

 

  1.   Abstract
  2.   Trademarks
  3. 1Introduction
  4. 2 Fly-Buck Converter Device Operation
    1. 2.1 Output Current Equations and Considerations
  5. 3LMR36520 Fly-Buck Converter Design
    1. 3.1 Coupled Inductor
    2. 3.2 Primary Output Capacitor
    3. 3.3 Rectifying Diode
    4. 3.4 Secondary Output Capacitor
    5. 3.5 Preload Resistor
    6. 3.6 Zener Diode
    7. 3.7 Snubber Circuit
  6. 4Experimental Results
    1. 4.1 Steady State
    2. 4.2 Secondary Output Voltage
    3. 4.3 Load Transient
    4. 4.4 Start-up
    5. 4.5 Output Current
  7. 5Conclusion
  8. 6References
  9. 7Revision History

3.7 Snubber Circuit

When the switch node voltage transitions from a high voltage to a low voltage or vice versa, large transient spikes can be reflected from the secondary to the primary side, resulting in spikes on the switch node voltage waveform and primary winding current waveform. These spikes are caused by the resonant LC tank formed by the leakage inductance of the secondary winding and the junction capacitance of the rectifier diode. These spikes can be large enough to exceed the electrical ratings of the converter and result in instability or damage to the device.

To mitigate the risk of these spikes, a simple RC snubber circuit placed in parallel with the secondary rectifying diode can be used. The following are steps to calculate the snubber components are outlined:

Estimate the resonant frequency of the LC tank using the following:

Equation 41. f t a n k = 1 2 π × L l e a k a g e × C j u n c t i o n

Given a typical leakage inductance of 1% for the 22-μH inductance results in Lleakage = 0.22 μH. From the MURA160 data sheet, the junction capacitance during the maximum reverse voltage applied to the diode is 5 pF typical. The approximate ftank = 151 MHz.

Calculate the RC ratio such that the frequency at which the pole is located is as far from the ftank frequency as possible to result in maximum attenuation:

Equation 42. f s n u b = 1 2 π × R s n u b × C s n u b

Selecting Rsnub = 200 Ω and Csnub = 100 pF results in fsnub = 1125 Hz, which is approximately five decades away from the ftank.

Determine the best value of Csnub to minimize snubber power loss by using the snubber power equation:

Equation 43. P s n u b = C s n u b V D 2 f s w
Equation 44. P s n u b = 200   p F × 34   V 2 × 400000   H z
Equation 45. P s n u b = 92.4   m W

This power will be dissipated by the snubber resistor, so the power rating of Rsnub must be greater than Psnub.