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

2.1 Output Current Equations and Considerations

The magnetizing current waveform is identical to the typical triangular inductor current waveform that is familiar to typical buck converter operation. For the case of a coupled inductor with a single secondary winding, this magnetizing current can be written as:

Equation 6. I m = I L 1 + I L 2 × N 2 N 1

The magnetizing current equation can be expanded for the case of a multi-winding transformer but will not be discussed in this report. The AN-2292 Designing an Isolated Buck (Fly-Buck) Converter application report provides more information on this.

On a cycle-by-cycle average, the winding currents can be related to the output currents as follows:

Equation 7. I o u t 1 = I L 1
Equation 8. I o u t 2 = I L 2

Since the magnetizing current is the same as the typical buck inductor current waveform, you can calculate the peak-to-peak current ripple the same way using Equation 9:

Equation 9. Δ i m = V i n - V o u t 1 × t o n L p r i = V i n - V o u t 1 × D L p r i × f s w

This peak-to-peak magnetizing current ripple is useful for estimating the peak current through both the high-side and low-side FETs. The peak current through the primary winding is given by Equation 10:

Equation 10. 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

The peak negative current through the primary winding can be approximated by Equation 11:

Equation 11. I p r i _ n e g p k = - N 2 N 1 × I o u t 2 × 2 D 1 - D - ∆ i m 2 + I o u t 1

Equation 10 and Equation 11 are extremely important in determining whether an IC will work as a Fly-Buck™ converter in a given application. This is because the peak HS current limit of the device and the peak negative (also known as sink) current limit of a device cannot be exceeded in order for the device to properly regulate the output. This can be written mathematically as:

Equation 12. I p r i _ p o s p k ≤ I S C
Equation 13. I p r i _ n e g p k ≤ I L _ N E G

Leakage inductance and duty cycle are also critical factors in Fly-Buck™ converter operation. Real world transformers and coupled inductors have some amount of leakage inductance arising from magnetic flux that is not shared between both the coils. In practice, this leakage flux affects the power output of the secondary side by limiting the rate at which the secondary side current can ramp up. For cases of higher leakage inductance, the ramp rate of the current in the secondary side is reduced compared to that of a lower leakage case. This means that a longer amount of time is required for the same amount of energy to be transferred from primary to secondary. For a fixed frequency part, this requires that a lower maximum duty cycle be chosen to ensure that there is enough off-time for energy to be transferred from primary to secondary. By convention, the maximum duty cycle for a Fly-Buck™ converter is held to 50%, larger duty cycles would result in less off-time which would result in larger negative peak currents being reflected to the primary side. Leakage inductance should therefore be minimized to ensure the widest duty cycle range possible.