SNVA991 October   2022 LM5123-Q1

 

  1.   How to Design a Boost Converter Using LM5123
  2.   Trademarks
  3. 1Design Example
  4. 2Calculations and Component Selection
    1. 2.1  Switching Frequency
    2. 2.2  Initial Inductor Calculation
    3. 2.3  Current Sense Resistor Selection
    4. 2.4  Inductor Selection
    5. 2.5  Output Capacitor Selection
    6. 2.6  Input Capacitor Selection
    7. 2.7  Feedback Resistor Selection
    8. 2.8  UVLO Resistor Selection
    9. 2.9  Soft-Start Capacitor Selection
    10. 2.10 Control Loop Compensation
      1. 2.10.1 Crossover Frequency (fcross) Selection
      2. 2.10.2 RCOMP Selection
      3. 2.10.3 CCOMP Selection
      4. 2.10.4 CHF Selection
    11. 2.11 MOSFET selection
  5. 3Implementation Results
  6. 4Small Signal Frequency Modeling
    1. 4.1 Boost Regulator Modulator Modeling
    2. 4.2 Compensation Modeling
    3. 4.3 Open Loop Modeling
  7. 5Resources

4.1 Boost Regulator Modulator Modeling

Table 4-1 includes equations model the plant (control-to-output) of a peak current mode boost regulator in continuous conduction mode.

Figure 4-1 Modulator transfer function
Table 4-1 Power Plant Equations
Simplified FormulaComprehensive Formula
Modulator Equations
Modulator Transfer Function v ^ L O A D v ^ C O M P = A M 1 + s ω Z e s r ∙ 1 - s ω Z r h p 1 + s ω P l f v ^ L O A D v ^ C O M P = A M 1 + s ω Z e s r ∙ 1 - s ω Z r h p 1 + s ω P l f ∙ 1 + s Q ∙ ω n + s 2 ω n
Modulator DC Gain A M = R L O A D ∙ D ' 2 ∙ R C S ∙ A C S ((1)) A M = R L O A D ∙ D ' K D ∙ R C S ∙ A C S ((1))((3))
RHP Zero ω Z r h p = R L O A D ∙ D ' 2 L m
ESR Zero ω Z e s r =   1 C O U T ∙ R E S R
Low Frequency Pole ω P l f =   2 C O U T ∙ R L O A D ω P l f =   K D C O U T ∙ R L O A D ((3))
Sub-Harmonic Double PoleNot Considered ω n = π ∙ f S W
Quality FactorNot Considered Q = 1 π ∙ D ' 1 + s e s n - 1 2
Slope CompensationNot Considered s e = V S L ∙ f S W ((1))
Sensed Rising Inductor SlopeNot Considered s n = V S U P P L Y ∙ R C S ∙ A C S L M ((2))
Not Considered K E X =   R C S ∙ A C S ∙ D ∙ D ' 2 ∙ L M ∙ f S W ((2))
Not Considered K M = 1 1 2 - D ∙ R C S ∙ A C S L M ∙ f S W + V S L ∙ A C S V L O A D ((1))((2))
Not Considered K D = R L O A D ∙ D ' 2 R C S ∙ A C S ∙ 1 K M + K E X D ' ((2))((3))
(1) VSL is the peak voltage of the internal slope compensation 45 mV
(2) ACS is the gain of the current sense amplifier, 10 V/V.
(3) KD is approximately equal to 2. This estimation simplifies the equation while still providing accurate high frequency modeling