Type 3 compensator design for Buck-Boost converter

Question

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)

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 ?

Answer 1

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:

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:

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:

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:

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.