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

3.1 Coupled Inductor

Once the above input and output parameters are known, the first thing to do is to select an appropriately-sized coupled inductor or transformer by first finding the turns ratio required to generate the desired secondary output voltage. From there, the critical inductance can be calculated the same way as for a typical non-isolated buck converter.

Two windings will be needed to establish the two output voltages. The secondary output voltage can be described by applying Kirchoff’s voltage law around the secondary loop:

Equation 14. V o u t 2 = V o u t 1 × N 2 N 1 - V f
Equation 15. 3.3   V = 5   V × N 2 N 1 - V f

By targeting a diode forward voltage drop (Vf) of 1 V, the ratio of N2/N1 comes to be 0.86. The closest whole number ratio is 1:1, therefore, a coupled inductor is selected. This will result in the secondary output voltage being slightly greater than 3.3 V. To address this, a series connected Zener diode and resistor can be placed in parallel with the secondary load to clamp the output to 3.3 V.

Because the magnetizing inductance current waveform is the same as that of the inductor current waveform for a typical inductor, the primary winding inductance can be calculated using the same methodology.

Equation 16. L p r i = V L ∆ t ∆ i

The Δi term in Equation 16 represents the peak-to-peak magnetizing current ripple and is typically set around 30% to 40% of the magnetizing current supplied by the converter which is 1 A in this case.

Setting Δi = 0.4 A then yields:

Equation 17. L p r i = 36   V - 5   V 5   V 36   V 0.4 × 400000   H z = 26.9   μ H

This Δi term can be adjusted to be higher or lower based on the needs of the application. The selected Δi term can work as long as the primary winding current does not exceed either the high-side current limit or the negative current limit. The calculated Lpri in this case is not a typical nominal inductance value so a 22-μH coupled inductor is selected instead resulting in the following peak-to-peak magnetizing current ripple:

Equation 18. Δ i m = V i n - V o u t 1 × D L p r i × f s w = 36   V - 5   V × 5   V 36   V 22   u H × 400000   H z = 0.489   A

Now to ensure this magnetizing ripple current does not violate our peak current conditions:

Equation 19. I p r i _ p o s p k = I o u t 1 + N 2 N 1 × I o u t 2 + ∆ i m 2 = 0.5   A + 1 1 × 0.5   A + 0.489   A 2 = 1.244   A
Equation 20. I p r i _ n e g p k = - 1 × 0.5 * 2 × 0.5 1 - 0.5 - 0.489   A 2 + 0.5   A = - 0.744   A

The minimum LMR36520 HS current limit is 2.4 A, while the negative current limit is -1.7 A according to the data sheet. Slight variations occur between chip-to-chip which is why it is important to use the minimum HS current limit when evaluating this condition to ensure that the limit is not violated under worst case conditions.

The selected 22-μH coupled inductor is, therefore, adequate to ensure that the peak current limits are not violated. This inductor should also have a saturation current rating that is at least as large as the maximum short circuit current limit of the device. Here, the maximum short circuit current limit is used to ensure the limit is not violated under worst case conditions as previously discussed.