SNVA940 November   2021 LM5157 , LM5157-Q1 , LM51571-Q1 , LM5158 , LM5158-Q1 , LM51581 , LM51581-Q1

 

  1.   Trademarks
  2. 1LM5157 Boost Design Example
  3. 2Calculations and Component Selection
    1. 2.1  Switching Frequency
    2. 2.2  Inductor Calculation
    3. 2.3  Slope Compensation Check
    4. 2.4  Inductor Selection
    5. 2.5  Diode Selection
    6. 2.6  Output Capacitor Selection
    7. 2.7  Input Capacitor Selection
    8. 2.8  UVLO Resistor Selection
    9. 2.9  Soft-Start Capacitor Selection
    10. 2.10 Feedback Resistor Selection
    11. 2.11 Control Loop Compensation
      1. 2.11.1 Crossover Frequency (fcross) Selection
      2. 2.11.2 RCOMP Selection
      3. 2.11.3 CCOMP Selection
      4. 2.11.4 CHF Selection
    12. 2.12 Power Loss and Efficiency Estimation
  4. 3Implementation Results
  5. 4Small Signal Frequency Analysis
    1. 4.1 Boost Regulator Modulator Modeling
    2. 4.2 Compensation Modeling
    3. 4.3 Open Loop Modeling

2.3 Slope Compensation Check

According to the theory of the peak current mode control, the slope of the compensation ramp must be greater than half of the sensed inductor current falling slope to prevent subharmonic oscillation at high duty cycle. Therefore, the following inequality in Equation 7 will be satisfied.

Equation 7. 0.5 × V LOAD + V F - V SUPPLY L M × A CS × Margin < 500 mV × f sw  

where

  • ACS is the equivalent current sensing gain.
  • 500mV is the slope compensation peak voltage.
  • VF is the forward voltage of the diode.

Typically 82% of the sensed inductor current falling slope is an optimal value of the slope compensation, which reflects to a margin of 1.6. If the inequality is failed, the inductance value of LM must be increased so that the falling slope can be smaller. If the LM inductance value is changed the peak current must be re-calculated and the device selection must be re-examined. In this example, the inequality is verified in Equation 8, Equation 9, and Equation 10.

Equation 8. 0.5 × V LOAD + V F - V SUPPLY L M × A CS × Margin = 0.5 × 12 V + 0.5 V - 3 V 1.5 μH × 0.095 × 1.6 = 0.481 × 10 6
Equation 9. 500 mV × f sw = 500 mV × 2.1 MHz = 1.05 × 10 6
Equation 10. 0.481 × 10 6 < 1.05 × 10 6