SLUAAN0 December   2022 TPS62933F

 

  1.   Designing a Simple and Low-Cost Flybuck Solution With the TPS62933F
  2.   Trademarks
  3. 1Introduction
  4. 2Flybuck Converter Device Overview
    1. 2.1 Operation Description
    2. 2.2 Equations for Maximum Output Current
  5. 3Design Flybuck With TPS62933F
    1. 3.1 Primary Voltage and Turns Ratio
    2. 3.2 Feedback Resistor
    3. 3.3 Rectifier Diode
    4. 3.4 Primary Inductance
    5. 3.5 Primary Turns
    6. 3.6 Input and Output Capacitor
    7. 3.7 Pre-Load
    8. 3.8 Factors Affecting Voltage Regulation
    9. 3.9 Avoiding Low-Side Sink Current Limit
  6. 4Experimental Results
  7. 5Conclusion
  8. 6References

3.4 Primary Inductance

For the TPS62933F, the minimum high-side current is 4.2A, and therefore the maximum magnetizing current ripple that can be tolerated is given by Equation 18:

Equation 18. ∆ i m = 2 × I H S _ L I M I T m i n - I O U T 1 + N 2 N 1 I O U T 2 + N 3 N 1 I O U T 3 = 4.48 A

Using Equation 18 for the maximum magnetizing current ripple calculation, the minimum primary inductance is given by Equation 19.

Equation 19. L p r i ( m i n ) = V I N ( m a x ) - V O U T 1 ∆ i m × f S W V O U T 1 V I N ( m a x ) = 1.77 μ H

Too much high Δim may not be good for the efficiency and output voltage ripple. Assuming the 40% ripple of the rate output current of the device, then the optimized minimum primary inductance is given by Equation 20:

Equation 20. L p r i = V I N ( m a x ) - V O U T 1 0.4 × 3 × f S W V O U T 1 V I N ( m a x ) = 6.6 μ H

We chose 6.8 μH for the primary inductor, therefore Δim(max) = 1.16 A. When all the outputs are with full loading, the positive and negative primary winding peak current is the highest and check the peak value not be over the high side FET over current limit (see Equation 21).

Equation 21. i S W _ p o s p k ( m a x ) = I O U T 1 + N 2 N 1 I O U T 2 + ∆ i m ( m a x ) 2 = 2.54 A

The primary output side is not always loaded, and without the primary load, the negative current is more obvious. So consider the worst case, IOUT1 = 0 A, and the secondary output side with full loading, check the negative primary winding peak current, as given by Equation 22.

Equation 22. i S W _ n e g p k ( m a x ) = - N 2 N 1 I O U T 2 + N 3 N 1 I O U T 3 1 + D 1 - D - ∆ i m 2 + I O U T 1 = - 1.2 A

For the TPS62933F, the minimum high-side source current limit, ILIMHS(min), is 4.2 A and the minimum low-side sink current limit, ILIMLSSOC(min), is 1.2 A. Therefore the positive and negative primary winding peak current can meet the current requirements with the normal leakage of a transformer.