SLUA560D June   2011  – March 2022 UCC28950 , UCC28950-Q1 , UCC28951 , UCC28951-Q1

 

  1.   Trademarks
  2. 1 Design Specifications
  3. 2 Functional Schematic
  4. 3 Power Budget
  5. 4 Transformer Calculations (T1)
  6. 5 QA, QB, QC, QD FET Selection
  7. 6 Selecting LS
  8. 7 Output Inductor Selection (LOUT)
  9. 8 Output Capacitance (COUT)
  10. 9 Select FETs QE and QF
  11. 10Input Capacitance (CIN)
  12. 11Setting Up the Current Sense (CS) Network (CT, RS, RRE, DA)
  13. 12Voltage Loop and Slope Compensation
  14. 13Setting Turn-on Delays to Achieve Zero Voltage Switching (ZVS)
  15. 14Turning SR FETs-off Under Light Load Conditions
  16. 15600 W FSFB Detailed Schematic and Test Data
  17. 16References
  18. 17Revision History

4 Transformer Calculations (T1)

The PSFB uses a transformer to deliver energy from the primary to the secondary. The voltage is stepped up or down through the transformers turns ratio (a1).

Equation 2. a1=NPNS=VINVOUT

Estimated FET voltage drop (VRDSON):

Equation 3. V R D S O N = 0.3   V

Select transformer turns based on 70% duty cycle (DMAX) at minimum specified input voltage. This will give some room for dropout if a PFC front end is used.

Equation 4. D M A X = 70   %
Equation 5. a 1 = V I N M I N - 2 × V R D S O N × D M A X V O U T + V R D S O N ≈ 21

Calculated typical duty cycle (DTYP) based on average input voltage.

Equation 6. D T Y P = V O U T + V R D S O N × a 1 V I N - 2 × V R D S O N ≈ 0.66

To keep the RMS current in the output capacitance to a minimum LOUT will be selected so the inductor ripple current (ΔILOUT) will be 20% of the DC output current. ΔILOUT is needed to calculate transformer peak and RMS currents

Equation 7. Δ I L O U T = P O U T × 0.2 V O U T = 10   A

Care needs to be taken in selecting a transformer with the correct amount of magnetizing inductance (LMAG). The following equations calculate the minimum magnetizing inductance of the primary of the transformer (T1) to ensure the converter operates in peak-current mode control. If LMAG is too small the magnetizing current could cause the converter to operate in voltage mode control instead of peak-current mode control. This is because the magnetizing current is too large, it will act as a PWM ramp swamping out the current sense signal across RS.

Equation 8. LMAG≥VIN×1-DTYPΔILOUT×0.5a1×fs≈2.76 mH

Figure 4-1 shows T1 primary current (IPRIMARY) and synchronous rectifiers currents, (QE (IQE) and QF (IQF)), with respect to the synchronous rectifier gate drive currents. Note that IQE and IQF are also T1’s secondary winding currents as well. Variable D is the converters duty cycle.

Figure 4-1 T1 Primary and QE and QF FET Currents

Calculate T1 secondary RMS current (ISRMS):

Equation 9. I P S = P O U T V O U T + Δ I L O U T 2 ≈ 55   A
Equation 10. I M S = P O U T V O U T - Δ I L O U T 2 ≈ 45   A
Equation 11. I M S 2 = I P S - Δ I L O U T 4 ≈ 50   A

Secondary RMS current (ISRMS1) when energy is being delivered to the secondary:

Equation 12. I S R M S 1 = D M A X 2 I P S × I M S + I P S - I M S 2 3 ≈ 29.6   A

Secondary RMS current (ISRMS2) when current is circulating through the transformer when QE and QF are both on.

Equation 13. I S R M S 2 = 1 - D M A X 2 I P S × I M S 2 + I P S - I M S 2 2 3 ≈ 20.3   A

Secondary RMS current (ISRMS3) caused by the negative current in the opposing winding during freewheeling period, please refer to Figure 4-1.

Equation 14. I S R M S 3 = Δ I L O U T 2 1 - D M A X 2 × 3 ≈ 1.1   A

Total secondary RMS current (ISRMS):

Equation 15. I S R M S = I S R M S 1 2 + I S R M S 2 2 + I S R M S 3 2 ≈ 36.0   A

Calculate T1 Primary RMS Current (IPRMS):

Equation 16. ∆ I L M A G = V I N M I N × D M A X L M A G × f S ≈ 0.47   A
Equation 17. I P P = P O U T V O U T × η + Δ I L O U T 2 1 a 1 + ∆ I L M A G ≈ 3.3   A
Equation 18. I M P = P O U T V O U T × η - Δ I L O U T 2 1 a 1 + ∆ I L M A G ≈ 2.8   A

T1 Primary RMS (IPRMS1) current when energy is being delivered to the secondary.

Equation 19. I P R M S 1 = D M A X I P P × I M P + I P P - I M P 2 3 ≈ 2.5   A

T1 Primary RMS (IPRMS2) current when the converter is free wheeling.

Equation 20. I M P 2 = I P P - Δ I L O U T 2 1 a 1 ≈ 3.0   A
Equation 21. I P R M S 2 = 1 - D M A X I P P × I M P 2 + I P P - I M P 2 2 3 ≈ 1.7   A

Total T1 primary RMS current (IPRMS)

Equation 22. I P R M S = I P R M S 1 2 + I P R M S 2 2 ≈ 3.1   A

The transformer calculations were given to Vitec a magnetic manufacturer to design a custom transformer to meet our design requirements. The transformer they designed for this application is part number 75PR8107 and the transformer has the following specifications.

Equation 23. a 1 = 21
Equation 24. L M A G = 2.8   m H

Measured leakage inductance on the Primary:

Equation 25. L L K = 4   u H

Transformer Primary DC resistance:

Equation 26. D C R P = 0.215   Ω

Transformer Secondary DC resistance:

Equation 27. D C R S = 0.58   Ω

Estimated transform losses (PT1) are twice the copper loss.

Note:

This is just an estimate and the total losses can vary based on magnetic design.

Equation 28. P T 1 ≈ 2 × I P R M S 2 × D C R P + 2 × I S R M S 2 × D C R S ≈ 7.0   W

Calculate remaining power budget:

Equation 29. P B U D G E T = P B U D G E T - P T 1 ≈ 38.1   W