{"id":400206,"date":"2024-06-29T15:07:21","date_gmt":"2024-06-29T15:07:21","guid":{"rendered":"http:\/\/savepearlharbor.com\/?p=400206"},"modified":"-0001-11-30T00:00:00","modified_gmt":"2024-06-29T15:07:21","slug":"","status":"publish","type":"post","link":"https:\/\/savepearlharbor.com\/?p=400206","title":{"rendered":"A note on small-signal modeling of SEPIC CM CCM<\/span>"},"content":{"rendered":"
<\/div>\n
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\"SEPIC
SEPIC DC\/DC Converter<\/figcaption><\/figure>\n

Knowing parameters of small-signal control-to-output transfer functions makes it easier for engineers to design compensation networks of DC\/DC converters. The equations for SEPIC can be found in different works and Application Notes, but there are differences. A work has been done to solve this problem.<\/p>\n

Simplified design equations for SEPIC with Current Mode control (CM) in Continuous Conduction Mode (CCM) suitable for practical design of compensation networks are shown.<\/p>\n

Known equations<\/h2>\n

It looks like that the compensation design equations for SEPIC in Application Notes are based on the equations taken from \u201cA tutorial for small signal models in the SEPIC power stage\u201d by Joel Steenis, National Semiconductor, [2]. There are also less known works: \u201cAnalysis of the Sepic Converter\u201d by Dr. Vatch\u00e9 Vorp\u00e9rian, [1] and TI, AN-1990, \u201cCompensation for Current Mode Control SEPIC Converters\u201d, [3].<\/p>\n

Mistakes are seen in [1] and [2]. See, for example, the A2, A3, A4 definitions and later \u03c9z in [2], or the a3 and DC gain definitions in [1]. Equations in [3] are huge and it is not clear how to use them. There are also works on the researchgate web site.<\/p>\n

SEPIC CM CCM small-signal equations<\/h2>\n

The figure below, taken from TI\u2019s AN-1990 [3], shows what the small-signal control-to-output function might look like.<\/p>\n

\"SEPIC
SEPIC CM CCM, Frequency Response of the Un-Compensated System. TI AN-1990, LM3478<\/figcaption><\/figure>\n

Thanks to Christophe Basso, the small-signal control-to-output transfer function for SEPIC CM CCM was obtained. The full equation is huge, thus only simplified equations are shown for key parameters.<\/p>\n

The small signal control-to-output transfer function of SEPIC CM CCM may be written as the following:<\/p>\n

           1 + n1 \u00d7 s + n2 \u00d7 s^2 + n3 \u00d7 s^3 + ... H(s) = Adc \u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014            1 + d1 \u00d7 s + d2 \u00d7 s^2 + d3 \u00d7 s^3 + ...<\/code><\/pre>\n
The order depends on parameters taken into account. The nominator is at least 4th order and denominator is at least 5th order.<\/p>\n
It can be rewritten in more useful form as in [9]:<\/p>\n
           (1 \u2212 s \/ \u03c9rhpz) (1 + s \/ \u03c9esr) H(s) \u2248 Adc \u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014 He(s)                     (1 + s \/ wp)<\/code><\/pre>\n
Where:
Adc \u2013 the DC gain, times;
\u03c9p \u2013 the main pole frequency, rad;
\u03c9rhpz \u2013 the right-half plane zero (RHPZ) frequency, rad;
\u03c9esr \u2013 the zero caused by the output capacitance and its equivalent series resistance, rad;
He(s) \u2013 models the inductor current sampling effect as well as the slope compensation effect;<\/p>\n
The magnitude at the desired crossover frequency is:<\/p>\n
             \u221a{F\u00b2 \/ Frhpz\u00b2 + 1} \u00d7 \u221a{F\u00b2 \/ Fesr\u00b2 + 1} |H(F)| \u2248 Adc \u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014                        \u221a{F\u00b2 \/ Fp\u00b2 + 1}<\/code><\/pre>\n
All the frequencies here and later are in Hz. He(s) is ignored because it is well above the crossover frequency. If the Fesr frequency is much higher than the crossover frequency, this member may also be ignored.<\/p>\n
There is also a pair of double pole and zero creating a \u201cglitch\u201d due to resonances caused by the coupling capacitor and inductors. Since it looks like a \u201cglitch\u201d, there is no sense to use it in the equation for magnitude at the crossover frequency.<\/p>\n
The main resonance frequency, a first order pole, is:<\/p>\n
       (1 + D) \u00d7 Iout       (1 + D) Fp \u2248 \u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014 = \u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014      2 \u03c0 \u00d7 Cout \u00d7 Vout   2 \u03c0 \u00d7 Cout \u00d7 R<\/code><\/pre>\n
The RHPZ frequency is:<\/p>\n
                        (1 \u2212 D)\u00b2 \u00d7 R Frhpz \u2248 \u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014         2 \u03c0 \u00d7 D \u00d7 [(1 \u2212 D) \u00d7 K \u00d7 \u221a{L1 \u00d7 L2} + D \u00d7 L1]<\/code><\/pre>\n
The DC gain is:<\/p>\n
      Vout \u00d7 Gmps \u00d7 (1 \u2212 D)        Vout \u00d7 (1 \u2212 D)         R \u00d7 (1 \u2212 D) Adc \u2248 \u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014 = \u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014 = \u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014          Iout \u00d7 (1 + D)       Iout \u00d7 Rsense \u00d7 (1 + D)   Rsense \u00d7 (1 + D)<\/code><\/pre>\n
The \u201cglitch\u201d frequency is:<\/p>\n
                                1 Fglitch \u2248 \u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014           2 \u03c0 \u00d7 \u221a{Cdc \u00d7 (L1 + L2 \u2212 2 \u00d7 K \u00d7 \u221a{L1 \u00d7 L2})}<\/code><\/pre>\n
Where:
Vout \u2013 the output voltage, V;
Iout \u2013 the load current, A;
R \u2013 the load, Vout \/ Iout, Ohm;
Cout \u2013 the output capacitance, F;
Rsense \u2013 the current sense resistance, Ohm;
L1 \u2013 the first, input inductance, H;
L2 \u2013 the second, output inductance, H;
K \u2013 the coupling coefficient of the inductors;
Cdc \u2013 the coupling capacitance, F;
D \u2013 the duty cycle.<\/p>\n
The equation shows that the \u201cglitch\u201d frequency can be low when uncoupled inductors are used and should be taken into account. With tightly coupled inductors there is no glitch which is consistent with measurements [8].<\/p>\n
Now a compensation network can be designed.<\/p>\n
    \n
  1. \n
       Compute the Fp, Frhpz, Fesr, Fglitch;<\/p>\n<\/li>\n
  2. \n
       Choose a crossover frequency Fcross, between the Fp and the minimum of Frhpz and Fglitch;<\/p>\n<\/li>\n
  3. \n
       Compute the magnitude at the crossover frequency and set the compensation network gain to the 1\/gain value;<\/p>\n<\/li>\n
  4. \n
       Set the zero of the compensation network below the Fp and the pole somewhere around the Frhpz or Fesr;<\/p>\n<\/li>\n<\/ol>\n
    Example<\/h2>\n
    Let\u2019s compute the values to design a compensation network and compare the results with data from [6].<\/p>\n
    The requirements: Vin = 9\u202624 V, Vout = 12 V, Iout = 0.75 A, Fsw = 750 kHz.<\/p>\n
    <\/figcaption><\/figure>\n
    In the circuit: Cout = 32 uF, L1 = L2 = 47 uH with 0.99 coupling, Rsense = 40 mOhm from the TPS61175 datasheet.<\/p>\n
    Calculate the values for the worst case at the minimum input voltage and maximum load.<\/p>\n
    The duty cycle at the minimum input voltage is: D = 0.58.<\/p>\n
    The load resistance is:<\/p>\n
        Vout    12 V R = \u2014\u2014\u2014\u2014 = \u2014\u2014\u2014\u2014\u2014\u2014 = 16 Ohm     Iout   0.75 A<\/code><\/pre>\n
    The DC gain is:<\/p>\n
             R \u00d7 (1 \u2212 D)      16 Ohm \u00d7 (1 \u2212 0.58) Adc \u2248 \u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014 \u2248 \u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014 \u2248 106 \u2248 40.5 dB       Rsense \u00d7 (1 + D)   0.04 Ohm \u00d7 (1 + 0.58)<\/code><\/pre>\n
    The main pole frequency is:<\/p>\n
             (1 + D)           (1 + 0.58) Fp \u2248 \u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014 \u2248 \u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014 \u2248 480 Hz      2 \u03c0 \u00d7 Cout \u00d7 R   2 \u03c0 \u00d7 33uF \u00d7 16 Ohm<\/code><\/pre>\n
    Since Resr is unknown in the circuit, let it be 20 mOhm, then the ESR zero frequency is:<\/p>\n
                   1                     1 Fesr \u2248 \u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014 \u2248 \u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014 \u2248 240 kHz        2 \u03c0 \u00d7 Cout \u00d7 Resr   2 \u03c0 \u00d7 33uF \u00d7 0.02 Ohm<\/code><\/pre>\n
    The RHPZ frequency is:<\/p>\n
                           (1 \u2212 D)\u00b2 \u00d7 R Frhpz \u2248 \u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014         2 \u03c0 \u00d7 D \u00d7 [(1 \u2212 D) \u00d7 K \u00d7 \u221a{L1 \u00d7 L2} + D \u00d7 L1]                        (1 \u2212 0.58)\u00b2 \u00d7 16 Ohm \u2248 \u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014 \u2248 16.5 kHz   2 \u03c0 \u00d7 0.58 \u00d7 [(1 \u2212 0.58) \u00d7 0.99 \u00d7 \u221a{47uH \u00d7 47uH} + 0.58 \u00d7 47uH]<\/code><\/pre>\n
    The glitch frequency is:<\/p>\n
                                    1 Fglitch \u2248 \u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014           2 \u03c0 \u00d7 \u221a{Cdc \u00d7 (L1 + L2 \u2212 2 \u00d7 K \u00d7 \u221a{L1 \u00d7 L2})}                               1 \u2248 \u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014 \u2248 164 kHz   2 \u03c0 \u00d7 \u221a{1uF \u00d7 (47uH + 47uH \u2212 2 \u00d7 0.99 \u00d7 \u221a{47uH \u00d7 47uH})}<\/code><\/pre>\n
    A crossover frequency must be in several times lower than the minimum of the RHPZ and glitch frequency. The chosen crossover frequency in [6] is 5 kHz.<\/p>\n
    The ESR zero frequency is ignored because it is too high, so the gain at 5 kHz is:<\/p>\n
                 \u221a{F\u00b2 \/ Frhpz\u00b2 + 1}       \u221a{5kHz\u00b2 \/ 16.5kHz\u00b2 + 1} |H(F)| \u2248 Adc \u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014 \u2248 106 \u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014 \u2248 10.6 \u2248 20.5 dB               \u221a{F\u00b2 \/ Fp\u00b2 + 1}          \u221a{5kHz\u00b2 \/ 480Hz\u00b2 + 1}<\/code><\/pre>\n
    Now we can compare the computed values with the measurements in Figure 2 in [6].<\/p>\n
    <\/figcaption><\/figure>\n
    A Type II compensation network can be used with any schematic.<\/p>\n
      \n
    1. \n
         The middle frequency gain around 5 kHz should be set around 1\/10.6 or \u221220.5 dB;<\/p>\n<\/li>\n
    2. \n
         The zero frequency should be around 500 Hz;<\/p>\n<\/li>\n
    3. \n
         The pole frequency should be somewhere at a frequency higher than 15 kHz.<\/p>\n<\/li>\n<\/ol>\n
      The circuit uses Type IIa compensation with built-in transconductance amplifier without explicit pole frequency. Its parameters are: the zero frequency is 1 kHz to compensate the main pole, the gain at 5 kHz is \u221223 dB.<\/p>\n
      The figure below shows the result. Since the gain of the compensation network is lower than \u221220 dB, the crossover frequency is not 5 kHz, but about 4 kHz.<\/p>\n
      <\/figcaption><\/figure>\n
      And let\u2019s look how a circuit with the same input parameters and its small-signal control-to-output transfer function would look in a design tool.<\/p>\n
      \"SEPIC,
      SEPIC, circuit design<\/figcaption><\/figure>\n
      Pay attention that C6 1uF in the original circuit has too low capacitance. The C6 capacitance should be increased to reduce circulating currents, see [10] and [11].<\/p>\n
      \"SEPIC
      SEPIC CM CCM, small-signal transfer functions, magnitude and phase<\/figcaption><\/figure>\n
      \"Power
      Power stage capacitors current sharing<\/figcaption><\/figure>\n
      Conclusion<\/h2>\n
      Equations are able to simplify design of compensation networks.<\/p>\n
      Equations of small-signal control-to-output transfer functions of DC\/DC converters can be found, for example, in [9], [12], [13], [14]. But keep in mind that the equations are simplified and the curves are different from real ones, especially in high frequency range.<\/p>\n
      Measurements for SEPIC in different modes can be found here: \u201cPractical control measurements for the Sepic converter\u201d, [8]. Unfortunately, some details are missed there.<\/p>\n
      Christophe Basso published his new book \u201cTransfer Functions of Switching Converters\u201d recently about small-signal modeling, [14]. You can also find a lot of other useful information on his site [12].<\/p>\n
      Such tools like \u201cPower Stage Designer Tool\u201d, \u201cLTpowerCAD\u201d, \u201cCircuit Calculator\u201d are intended to simplify work for engineers and offer equation-based design.<\/p>\n
      References<\/h2>\n
        \n
      1. \n
           Dr. Vatch\u00e9 Vorp\u00e9rian, \u201cAnalysis of the Sepic Converter\u201d. Ridley Engineering. Power Systems Design Europe, November 2006. Link<\/a>.<\/p>\n<\/li>\n
      2. \n
           Joel Steenis, National Semiconductor, \u201cA tutorial for small signal models in the SEPIC power stage\u201d. Link<\/a>.<\/p>\n<\/li>\n
      3. \n
           Texas Instruments, AN-1990 Compensation for Current Mode Control SEPIC Converters. Link<\/a>.<\/p>\n<\/li>\n
      4. \n
           Texas Instruments, AN-1484, \u201cDesigning A SEPIC Converter\u201d. Link<\/a>.<\/p>\n<\/li>\n
      5. \n
           Texas Instruments, TPS61175 datasheet. Link<\/a>.<\/p>\n<\/li>\n
      6. \n
           Texas Instruments, SLVA384, \u201cDesigning a Split-Rail SEPIC With the TPS61175\u201d. Link<\/a>.<\/p>\n<\/li>\n
      7. \n
           Texas Instruments, Markus Zehendner, Matthias Ulmann, \u201cPower Stage Topology Reference Guide\u201d. Link<\/a>.<\/p>\n<\/li>\n
      8. \n
           Ridley Engineering site. Link<\/a>.<\/p>\n<\/li>\n
      9. \n
           Power Stage Designer User Guide, Chapter 11. Link<\/a>.<\/p>\n<\/li>\n
      10. \n
           Power Tip 32: Beware of circulating currents in a SEPIC coupled-inductor \u2013 Part 1. Link<\/a>.<\/p>\n<\/li>\n
      11. \n
           Power Tip 33: Beware of circulating currents in a SEPIC coupled-inductor \u2013 Part 2. Link<\/a>.<\/p>\n<\/li>\n
      12. \n
           Christophe Basso, personal site. Link<\/a>.<\/p>\n<\/li>\n
      13. \n
           Christophe Basso, \u201cSwitch-Mode Power Supplies, Second Edition: SPICE Simulations and Practical Designs\u201d, 2nd Edition. Link<\/a>.<\/p>\n<\/li>\n
      14. \n
           Christophe Basso, \u201cTransfer Functions of Switching Converters\u201d. Stairway Press. Link<\/a>.<\/p>\n<\/li>\n
      15. \n
           Power Stage Designer Tool. Link<\/a>.<\/p>\n<\/li>\n
      16. \n
           LTpowerCAD. Link<\/a>.<\/p>\n<\/li>\n
      17. \n
           «Circuit Calculator», an electronics circuit design tool for Android. Link<\/a>.<\/p>\n<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n
        <\/div>\n

         \u0441\u0441\u044b\u043b\u043a\u0430 \u043d\u0430 \u043e\u0440\u0438\u0433\u0438\u043d\u0430\u043b \u0441\u0442\u0430\u0442\u044c\u0438          https:\/\/habr.com\/ru\/articles\/574828\/<\/a><\/p>\n","protected":false},"excerpt":{"rendered":"
        <\/div>\n
        \n
        \n
        \n
        \n
        SEPIC DC\/DC Converter<\/figcaption><\/figure>\n
        Knowing parameters of small-signal control-to-output transfer functions makes it easier for engineers to design compensation networks of DC\/DC converters. The equations for SEPIC can be found in different works and Application Notes, but there are differences. A work has been done to solve this problem.<\/p>\n
        Simplified design equations for SEPIC with Current Mode control (CM) in Continuous Conduction Mode (CCM) suitable for practical design of compensation networks are shown.<\/p>\n
        Known equations<\/h2>\n
        It looks like that the compensation design equations for SEPIC in Application Notes are based on the equations taken from \u201cA tutorial for small signal models in the SEPIC power stage\u201d by Joel Steenis, National Semiconductor, [2]. There are also less known works: \u201cAnalysis of the Sepic Converter\u201d by Dr. Vatch\u00e9 Vorp\u00e9rian, [1] and TI, AN-1990, \u201cCompensation for Current Mode Control SEPIC Converters\u201d, [3].<\/p>\n
        Mistakes are seen in [1] and [2]. See, for example, the A2, A3, A4 definitions and later \u03c9z in [2], or the a3 and DC gain definitions in [1]. Equations in [3] are huge and it is not clear how to use them. There are also works on the researchgate web site.<\/p>\n
        SEPIC CM CCM small-signal equations<\/h2>\n
        The figure below, taken from TI\u2019s AN-1990 [3], shows what the small-signal control-to-output function might look like.<\/p>\n
        SEPIC CM CCM, Frequency Response of the Un-Compensated System. TI AN-1990, LM3478<\/figcaption><\/figure>\n
        Thanks to Christophe Basso, the small-signal control-to-output transfer function for SEPIC CM CCM was obtained. The full equation is huge, thus only simplified equations are shown for key parameters.<\/p>\n
        The small signal control-to-output transfer function of SEPIC CM CCM may be written as the following:<\/p>\n
                   1 + n1 \u00d7 s + n2 \u00d7 s^2 + n3 \u00d7 s^3 + ... H(s) = Adc \u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014            1 + d1 \u00d7 s + d2 \u00d7 s^2 + d3 \u00d7 s^3 + ...<\/code><\/pre>\n
        The order depends on parameters taken into account. The nominator is at least 4th order and denominator is at least 5th order.<\/p>\n
        It can be rewritten in more useful form as in [9]:<\/p>\n
                   (1 \u2212 s \/ \u03c9rhpz) (1 + s \/ \u03c9esr) H(s) \u2248 Adc \u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014 He(s)                     (1 + s \/ wp)<\/code><\/pre>\n
        Where:
        Adc \u2013 the DC gain, times;
        \u03c9p \u2013 the main pole frequency, rad;
        \u03c9rhpz \u2013 the right-half plane zero (RHPZ) frequency, rad;
        \u03c9esr \u2013 the zero caused by the output capacitance and its equivalent series resistance, rad;
        He(s) \u2013 models the inductor current sampling effect as well as the slope compensation effect;<\/p>\n
        The magnitude at the desired crossover frequency is:<\/p>\n
                     \u221a{F\u00b2 \/ Frhpz\u00b2 + 1} \u00d7 \u221a{F\u00b2 \/ Fesr\u00b2 + 1} |H(F)| \u2248 Adc \u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014                        \u221a{F\u00b2 \/ Fp\u00b2 + 1}<\/code><\/pre>\n
        All the frequencies here and later are in Hz. He(s) is ignored because it is well above the crossover frequency. If the Fesr frequency is much higher than the crossover frequency, this member may also be ignored.<\/p>\n
        There is also a pair of double pole and zero creating a \u201cglitch\u201d due to resonances caused by the coupling capacitor and inductors. Since it looks like a \u201cglitch\u201d, there is no sense to use it in the equation for magnitude at the crossover frequency.<\/p>\n
        The main resonance frequency, a first order pole, is:<\/p>\n
               (1 + D) \u00d7 Iout       (1 + D) Fp \u2248 \u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014 = \u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014      2 \u03c0 \u00d7 Cout \u00d7 Vout   2 \u03c0 \u00d7 Cout \u00d7 R<\/code><\/pre>\n
        The RHPZ frequency is:<\/p>\n
                                (1 \u2212 D)\u00b2 \u00d7 R Frhpz \u2248 \u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014         2 \u03c0 \u00d7 D \u00d7 [(1 \u2212 D) \u00d7 K \u00d7 \u221a{L1 \u00d7 L2} + D \u00d7 L1]<\/code><\/pre>\n
        The DC gain is:<\/p>\n
              Vout \u00d7 Gmps \u00d7 (1 \u2212 D)        Vout \u00d7 (1 \u2212 D)         R \u00d7 (1 \u2212 D) Adc \u2248 \u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014 = \u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014 = \u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014          Iout \u00d7 (1 + D)       Iout \u00d7 Rsense \u00d7 (1 + D)   Rsense \u00d7 (1 + D)<\/code><\/pre>\n
        The \u201cglitch\u201d frequency is:<\/p>\n
                                        1 Fglitch \u2248 \u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014           2 \u03c0 \u00d7 \u221a{Cdc \u00d7 (L1 + L2 \u2212 2 \u00d7 K \u00d7 \u221a{L1 \u00d7 L2})}<\/code><\/pre>\n
        Where:
        Vout \u2013 the output voltage, V;
        Iout \u2013 the load current, A;
        R \u2013 the load, Vout \/ Iout, Ohm;
        Cout \u2013 the output capacitance, F;
        Rsense \u2013 the current sense resistance, Ohm;
        L1 \u2013 the first, input inductance, H;
        L2 \u2013 the second, output inductance, H;
        K \u2013 the coupling coefficient of the inductors;
        Cdc \u2013 the coupling capacitance, F;
        D \u2013 the duty cycle.<\/p>\n
        The equation shows that the \u201cglitch\u201d frequency can be low when uncoupled inductors are used and should be taken into account. With tightly coupled inductors there is no glitch which is consistent with measurements [8].<\/p>\n
        Now a compensation network can be designed.<\/p>\n
          \n
        1. \n
             Compute the Fp, Frhpz, Fesr, Fglitch;<\/p>\n<\/li>\n
        2. \n
             Choose a crossover frequency Fcross, between the Fp and the minimum of Frhpz and Fglitch;<\/p>\n<\/li>\n
        3. \n
             Compute the magnitude at the crossover frequency and set the compensation network gain to the 1\/gain value;<\/p>\n<\/li>\n
        4. \n
             Set the zero of the compensation network below the Fp and the pole somewhere around the Frhpz or Fesr;<\/p>\n<\/li>\n<\/ol>\n
          Example<\/h2>\n
          Let\u2019s compute the values to design a compensation network and compare the results with data from [6].<\/p>\n
          The requirements: Vin = 9\u202624 V, Vout = 12 V, Iout = 0.75 A, Fsw = 750 kHz.<\/p>\n
          <\/figcaption><\/figure>\n
          In the circuit: Cout = 32 uF, L1 = L2 = 47 uH with 0.99 coupling, Rsense = 40 mOhm from the TPS61175 datasheet.<\/p>\n
          Calculate the values for the worst case at the minimum input voltage and maximum load.<\/p>\n
          The duty cycle at the minimum input voltage is: D = 0.58.<\/p>\n
          The load resistance is:<\/p>\n
              Vout    12 V R = \u2014\u2014\u2014\u2014 = \u2014\u2014\u2014\u2014\u2014\u2014 = 16 Ohm     Iout   0.75 A<\/code><\/pre>\n
          The DC gain is:<\/p>\n
                   R \u00d7 (1 \u2212 D)      16 Ohm \u00d7 (1 \u2212 0.58) Adc \u2248 \u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014 \u2248 \u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014 \u2248 106 \u2248 40.5 dB       Rsense \u00d7 (1 + D)   0.04 Ohm \u00d7 (1 + 0.58)<\/code><\/pre>\n
          The main pole frequency is:<\/p>\n
                   (1 + D)           (1 + 0.58) Fp \u2248 \u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014 \u2248 \u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014 \u2248 480 Hz      2 \u03c0 \u00d7 Cout \u00d7 R   2 \u03c0 \u00d7 33uF \u00d7 16 Ohm<\/code><\/pre>\n
          Since Resr is unknown in the circuit, let it be 20 mOhm, then the ESR zero frequency is:<\/p>\n
                         1                     1 Fesr \u2248 \u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014 \u2248 \u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014 \u2248 240 kHz        2 \u03c0 \u00d7 Cout \u00d7 Resr   2 \u03c0 \u00d7 33uF \u00d7 0.02 Ohm<\/code><\/pre>\n
          The RHPZ frequency is:<\/p>\n
                                 (1 \u2212 D)\u00b2 \u00d7 R Frhpz \u2248 \u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014         2 \u03c0 \u00d7 D \u00d7 [(1 \u2212 D) \u00d7 K \u00d7 \u221a{L1 \u00d7 L2} + D \u00d7 L1]                        (1 \u2212 0.58)\u00b2 \u00d7 16 Ohm \u2248 \u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014 \u2248 16.5 kHz   2 \u03c0 \u00d7 0.58 \u00d7 [(1 \u2212 0.58) \u00d7 0.99 \u00d7 \u221a{47uH \u00d7 47uH} + 0.58 \u00d7 47uH]<\/code><\/pre>\n
          The glitch frequency is:<\/p>\n
                                          1 Fglitch \u2248 \u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014           2 \u03c0 \u00d7 \u221a{Cdc \u00d7 (L1 + L2 \u2212 2 \u00d7 K \u00d7 \u221a{L1 \u00d7 L2})}                               1 \u2248 \u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014 \u2248 164 kHz   2 \u03c0 \u00d7 \u221a{1uF \u00d7 (47uH + 47uH \u2212 2 \u00d7 0.99 \u00d7 \u221a{47uH \u00d7 47uH})}<\/code><\/pre>\n
          A crossover frequency must be in several times lower than the minimum of the RHPZ and glitch frequency. The chosen crossover frequency in [6] is 5 kHz.<\/p>\n
          The ESR zero frequency is ignored because it is too high, so the gain at 5 kHz is:<\/p>\n
                       \u221a{F\u00b2 \/ Frhpz\u00b2 + 1}       \u221a{5kHz\u00b2 \/ 16.5kHz\u00b2 + 1} |H(F)| \u2248 Adc \u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014 \u2248 106 \u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014 \u2248 10.6 \u2248 20.5 dB               \u221a{F\u00b2 \/ Fp\u00b2 + 1}          \u221a{5kHz\u00b2 \/ 480Hz\u00b2 + 1}<\/code><\/pre>\n
          Now we can compare the computed values with the measurements in Figure 2 in [6].<\/p>\n
          <\/figcaption><\/figure>\n
          A Type II compensation network can be used with any schematic.<\/p>\n
            \n
          1. \n
               The middle frequency gain around 5 kHz should be set around 1\/10.6 or \u221220.5 dB;<\/p>\n<\/li>\n
          2. \n
               The zero frequency should be around 500 Hz;<\/p>\n<\/li>\n
          3. \n
               The pole frequency should be somewhere at a frequency higher than 15 kHz.<\/p>\n<\/li>\n<\/ol>\n
            The circuit uses Type IIa compensation with built-in transconductance amplifier without explicit pole frequency. Its parameters are: the zero frequency is 1 kHz to compensate the main pole, the gain at 5 kHz is \u221223 dB.<\/p>\n
            The figure below shows the result. Since the gain of the compensation network is lower than \u221220 dB, the crossover frequency is not 5 kHz, but about 4 kHz.<\/p>\n
            <\/figcaption><\/figure>\n
            And let\u2019s look how a circuit with the same input parameters and its small-signal control-to-output transfer function would look in a design tool.<\/p>\n
            SEPIC, circuit design<\/figcaption><\/figure>\n
            Pay attention that C6 1uF in the original circuit has too low capacitance. The C6 capacitance should be increased to reduce circulating currents, see [10] and [11].<\/p>\n
            SEPIC CM CCM, small-signal transfer functions, magnitude and phase<\/figcaption><\/figure>\n
            Power stage capacitors current sharing<\/figcaption><\/figure>\n
            Conclusion<\/h2>\n
            Equations are able to simplify design of compensation networks.<\/p>\n
            Equations of small-signal control-to-output transfer functions of DC\/DC converters can be found, for example, in [9], [12], [13], [14]. But keep in mind that the equations are simplified and the curves are different from real ones, especially in high frequency range.<\/p>\n
            Measurements for SEPIC in different modes can be found here: \u201cPractical control measurements for the Sepic converter\u201d, [8]. Unfortunately, some details are missed there.<\/p>\n
            Christophe Basso published his new book \u201cTransfer Functions of Switching Converters\u201d recently about small-signal modeling, [14]. You can also find a lot of other useful information on his site [12].<\/p>\n
            Such tools like \u201cPower Stage Designer Tool\u201d, \u201cLTpowerCAD\u201d, \u201cCircuit Calculator\u201d are intended to simplify work for engineers and offer equation-based design.<\/p>\n
            References<\/h2>\n
              \n
            1. \n
                 Dr. Vatch\u00e9 Vorp\u00e9rian, \u201cAnalysis of the Sepic Converter\u201d. Ridley Engineering. Power Systems Design Europe, November 2006. Link<\/a>.<\/p>\n<\/li>\n
            2. \n
                 Joel Steenis, National Semiconductor, \u201cA tutorial for small signal models in the SEPIC power stage\u201d. Link<\/a>.<\/p>\n<\/li>\n
            3. \n
                 Texas Instruments, AN-1990 Compensation for Current Mode Control SEPIC Converters. Link<\/a>.<\/p>\n<\/li>\n
            4. \n
                 Texas Instruments, AN-1484, \u201cDesigning A SEPIC Converter\u201d. Link<\/a>.<\/p>\n<\/li>\n
            5. \n
                 Texas Instruments, TPS61175 datasheet. Link<\/a>.<\/p>\n<\/li>\n
            6. \n
                 Texas Instruments, SLVA384, \u201cDesigning a Split-Rail SEPIC With the TPS61175\u201d. Link<\/a>.<\/p>\n<\/li>\n
            7. \n
                 Texas Instruments, Markus Zehendner, Matthias Ulmann, \u201cPower Stage Topology Reference Guide\u201d. Link<\/a>.<\/p>\n<\/li>\n
            8. \n
                 Ridley Engineering site. Link<\/a>.<\/p>\n<\/li>\n
            9. \n
                 Power Stage Designer User Guide, Chapter 11. Link<\/a>.<\/p>\n<\/li>\n
            10. \n
                 Power Tip 32: Beware of circulating currents in a SEPIC coupled-inductor \u2013 Part 1. Link<\/a>.<\/p>\n<\/li>\n
            11. \n
                 Power Tip 33: Beware of circulating currents in a SEPIC coupled-inductor \u2013 Part 2. Link<\/a>.<\/p>\n<\/li>\n
            12. \n
                 Christophe Basso, personal site. Link<\/a>.<\/p>\n<\/li>\n
            13. \n
                 Christophe Basso,<\/p>\n<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[],"tags":[],"class_list":["post-400206","post","type-post","status-publish","format-standard","hentry"],"_links":{"self":[{"href":"https:\/\/savepearlharbor.com\/index.php?rest_route=\/wp\/v2\/posts\/400206","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/savepearlharbor.com\/index.php?rest_route=\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/savepearlharbor.com\/index.php?rest_route=\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/savepearlharbor.com\/index.php?rest_route=\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/savepearlharbor.com\/index.php?rest_route=%2Fwp%2Fv2%2Fcomments&post=400206"}],"version-history":[{"count":0,"href":"https:\/\/savepearlharbor.com\/index.php?rest_route=\/wp\/v2\/posts\/400206\/revisions"}],"wp:attachment":[{"href":"https:\/\/savepearlharbor.com\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=400206"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/savepearlharbor.com\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=400206"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/savepearlharbor.com\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=400206"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}