SLVAF95 april   2023 TPS7H5001-SP

 

  1.   Abstract
  2.   Trademarks
  3.   Description
  4.   Features
  5.   Applications
  6. 1System Overview
    1. 1.1 Block Diagram
    2. 1.2 Design Considerations
    3. 1.3 System Design Theory
      1. 1.3.1 Switching Frequency
      2. 1.3.2 Transformer
      3. 1.3.3 RCD and Diode Clamp
      4. 1.3.4 Output Diode and MOSFET
      5. 1.3.5 Output Filter and Capacitance
      6. 1.3.6 Compensation
      7. 1.3.7 Controller Passives
  7. 2Test Results
    1. 2.1 Testing and Results
      1. 2.1.1 Test Setup
      2. 2.1.2 Test Results
        1. 2.1.2.1 Efficiency
        2. 2.1.2.2 Frequency Response
        3. 2.1.2.3 Thermal Characteristics
        4. 2.1.2.4 Output Voltage Ripple
        5. 2.1.2.5 Load Step
        6. 2.1.2.6 Start-Up
        7. 2.1.2.7 Shutdown
        8. 2.1.2.8 Component Stresses
  8. 3Design Files
    1. 3.1 Schematics
    2. 3.2 Bill of Materials
    3. 3.3 Assembly Drawings
  9. 4Related Documentation

1.3.2 Transformer

The transformer of the design consists of two major values, turns ratio and primary side inductance. Equation 2 provides the turns ratio as a function of duty cycle. If the turns ratio is put in the maximum duty cycle of the converter, the calculations provide a maximum turns ratio. The TPS7H5001-SP design targeted a duty cycle of 33%. This allows for room in both directions for the duty cycle to change due to transients as well as keep it above the minimum on-time. Equation 2 and Equation 3 calculates the turns ratio of the transformer.

Equation 2. N p s = V i n × D ( V o u t + V D i o d e ) × ( 1 - D )
Equation 3. N p s = 28   V × 0 . 33 ( 5   V + 0 . 5   V ) × ( 1 - 0 . 33 ) = 2 . 5

Often the turns ratio slightly changes in the design due to how the transformer is manufactured. In this case, an axillary winding was planned, but not used. That factor plus the need to have a whole value of turns caused the turns ratio to be 8:3. Using a turns ratio of 3:1 is recommended, if the axillary winding is not needed, because while the duty cycle changes, there is less of a parasitic spike on the switch node since there are less turns on the primary side.

The primary inductance of the transformer is found from picking an appropriate ripple current. A higher inductance often means reduced current ripple, thus lower EMI and noise, but a higher inductance also increases physical size and limits the bandwidth of the design. A lower inductance does the opposite, increasing current ripple, lowering EMI, lowering noise, decreasing physical size, and increasing the limited bandwidth of the design. The percent ripple current can be anywhere from 20% to 80% depending on the design. Use Equation 4 and Equation 5 to find the primary inductance from the percentage ripple current.

Equation 4. L P R I = V i n 2 × D 2 V o u t × I o u t × f o s c × % R i p p l e
Equation 5. ( 28   V   ) 2 × 0 . 33 2 5   V × 10   A × 500   k H z × 0 . 4 = 8.54   μ H

There are quite a few physical limitations when making transformers, so often this inductance changes slightly. For the TPS7H5001-SP design, us a primary inductance of µH. This corresponds to a percent ripple of around 0.38. The peak and primary currents of the transformer are also generally useful for figuring out the physical structure of the transformer, as listed in Equation 6 through Equation 13.

Equation 6. I R i p p l e = V o u t × I o u t × % R i p p l e V i n × D
Equation 7. I R i p p l e = 5   V × 10   A × 0 . 38 28   V × 0 . 33 = 2 . 06   A
Equation 8. I P r i P e a k = V o u t × I o u t V i n × D × η + I R i p p l e 2
Equation 9. I P r i P e a k = 5   V × 10   A 28   V × 0 . 33 × 0 . 8 + 2 . 06 2 = 7 . 79   A
Equation 10. IPriRMS=D×(Vout×IoutVin×D)2+IRipple23
Equation 11. I P r i R M S = 0 . 33 × ( 5   V × 10   A 28   V × 0 . 33 ) 2 + ( 2 . 06   A ) 2 3 = 3 . 33   A
Equation 12. ISecRMS=(1-D)×Iout2+(IRipple×Nps)23
Equation 13. I S e c R M S = 0 . 67 × ( 10   A ) 2 + ( 2 . 06   A × 2 . 67 ) 2 3 = 8 . 78   A