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I'm practicing a design of a type 3 compensator for a voltage mode buck-boost converter, so I went through the book entitled "Switch mode power supplies Spice simulation and practical design"

The topology used for calculations of the different components of the compensator is shown below (Extracted from the book)

enter image description here

The specifications are as follows:

  • Vinmin=10v (minimum input voltage).
  • Vinmax=15v (maximum input voltage).
  • Vout=-12 (Regulated output voltage).
  • Fs=100kHz (Switching frequency).
  • Fc=5kHz (Crossover frequency, chosen by the author to make the RHPZ after that frequency).
  • Iout=2A (maximum output current)
  • Rload=6ohm (DC load resistor)

I have followed all the steps given by the author but I failed to find the same values found to the compensator component. The method used to place the poles and zeros of the compensator is the manual method and not the K factor method. So the author proposed the following:

  • From the Bode plot, we can see that the required gain at 5 kHz is around -10 dB worst case.

  • To cancel the LC filter peaking, place a double zero close to the resonant frequency, 600 Hz

  • Since the zero occurs after the crossover frequency, we can place a first pole at 7 kHz

  • Place a second pole at one-half of the switching frequency, to force the gain to further roll off, 50 kHz.

  • Using the manual placement method described in Chap. 3, evaluate all the compensator elements.

  • R2 = 18.6 kohm R3 = 456 ohm C1 = 15 nF C2 = 1.3 nF C3=7 nF

    When I apply, the equation encircled in red in the figure, I come out with a different value for R2. I found R2=1.8kohm but the author finds R2=18k ohm and since C1 and C2 values depend on R2, so my compensator is not good and I do not get a sufficient phase margins.

Please can anyone ensures to me the value of R2 and does the equation encircled in red correct ?

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    \$\begingroup\$ could crop the images to their actual content, please? \$\endgroup\$ – Marcus Müller Jun 7 '20 at 21:07
  • \$\begingroup\$ @Marcus Müller I did not understand, do you mean the format of the image ? \$\endgroup\$ – learn design Jun 7 '20 at 21:11
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    \$\begingroup\$ your images are 75% white space. You can cut that off ("cropping"). \$\endgroup\$ – Marcus Müller Jun 7 '20 at 21:15
  • \$\begingroup\$ I checked the equation and it should give 16.8k and not 18.6k as mistakenly printed. These are the computed values by the macros so that you can spot where it goes wrong on your side: FC = 5.00e+003 GFC = -9.60e+000 G = 3.02e+000 PI = 3.14e+000 FZ1 = 6.00e+002 FZ2 = 6.00e+002 FP1 = 7.00e+003 FP2 = 5.00e+004 C3 = 6.98e-009 R3 = 4.56e+002 C1 = 1.58e-008 C2 = 1.36e-009 A = 6.43e+014 C = 1.87e+017 R2 = 1.68e+004. I checked and crossover is well 5 kHz with a 45° phase margin at the 10-V input. \$\endgroup\$ – Verbal Kint Jun 7 '20 at 21:18
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This is an excerpt of the book I wrote on switching power supplies. The exercise consists of stabilizing a buck-boost converter whose schematic diagram appears below:

enter image description here

In the left side of the drawing, you can see a list of variables. These variables correspond to information extracted from the open-loop response of the power stage. However, considering the negative voltage delivered by the buck-boost converter, we will observe the response after the inverting block \$E_1\$. Then, we will extract the attenuation at the selected 5-kHz crossover frequency. The phase is important and will let us compute the necessary phase boost to calibrate the type 3 compensator and place poles and zeroes. However, in this exercise, the position of these elements has been done earlier in the text. The power stage response \$H(s)\$ is here:

enter image description here

The left-side macro will now compute the components values around the op-amp to provide a gain of 9.6 dB at 5 kHz and sufficient phase boost at this frequency. The amount of phase boost will lead to the targeted phase margin, neglecting the op-amp contribution here. Owing to this approach, it is easy and fast to change the positions of the poles and zeroes and immediately see the effect on the transient response or the open-loop gain. The computed values are here:

enter image description here

The incriminated resistance \$R_2\$ is 16.8 k\$\Omega\$ and not 18.6 k\$\Omega\$ as mistakenly printed. Once these values are applied, the compensated loop gain is here:

enter image description here

It is important to note that the op-amp plays a role and can affect the expected gain and phase boost, leading to a smaller phase margin than expected. I have covered this aspect in a series of articles published on How2Power.com some time ago.

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  • \$\begingroup\$ I have correcte my script, and I got the same result. I have confused between gain dificit and the gain required at Fc.I'm about to discover that you are CBasso, All my gratitudes to you, your two books about switching power supplies are the most powerful and helpful books I have ever read about this topic. Your explanations and examples are really motivated me to move forward with SMPS design. Thanks and Thanks again for these books \$\endgroup\$ – learn design Jun 8 '20 at 19:06
  • \$\begingroup\$ Glad to read your kind comments and happy if I could modestly help you in your discovery of switching power supplies. I have recently posted new ready-made simulations files in SIMPLIS that everyone can use on their demo version. Most are also automatically compensated as in the above IsSpice example. Good luck with your designs! \$\endgroup\$ – Verbal Kint Jun 8 '20 at 20:17

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