The power stage of the UCC25640x EVM [3] is considered to demonstrate the Type 3 compensator [1.12.2.3] design which is shown in Figure 1-27. Lets consider 10kHz as a cross over frequency (f_c) for the loop gain.

1. From Figure 1-22, the open loop gain $$\left(G_{plant}(s)=\frac{V_{out}(s)}{FBreplica(s)}\right)$$ is close to -25dB at 10kHz.
2. So $$G_c(s)$$ should be 25dB at the cross over frequency.
3. Assuming $$f_L\ll f_z\ll f_c\ll f_{p2}\ll f_{p1}$$ in [1.12.2.3], $$G_c(s)$$ can be approximated as $$G_o\cdot \frac{f_c}{f_z}$$ . For a given phase lead ( $$\theta$$ ), cross over frequency ( $$f_c$$ ), f_z, f_p2 can be found out using following equations [Chapter 9.5 in Reference 9]: $$f_c=\sqrt{f_z\cdot f_{p2}}$$ , $$f_z=f_c\sqrt{\frac{1-\sin (\theta )}{1+\sin (\theta )}}$$ , $$f_{p2}=f_c\sqrt{\frac{1+\sin (\theta )}{1-\sin (\theta )}}$$ . So, $$G_c(s)\cong G_o\cdot \frac{f_c}{f_z}=G_o\cdot \frac{\sqrt{f_z\cdot f_{p2}}}{f_z}=G_o\cdot \sqrt{\frac{f_{p2}}{f_z}}$$ .
4. For a phase lead of 52^o, f_z and f_p2 should be 3.4kHz and 29kHz respectively.
5. Since f_z, f_p2 are found out, G_o can be obtained using following expression: $$G_o\cdot \sqrt{\frac{f_{p2}}{f_z}}=17.78\Rightarrow G_o=6.126$$ (25dB=17.78).
6. f_p1 is a high frequency pole which is used to eliminate the high frequency noise. It is recommended to place this pole close to ESR of the output capacitor. Here f_p1 is chosen as 479kHz.
7. f_L should be chosen such that controller should be able to regulate the output voltage when the converter operates in the burst mode. So, f_L should be less than the burst mode frequency. In this design, f_L is considered as 88Hz.
8. Rup and Rlow can be found out using following expressions: $$\frac{V_o-V_{ref}}{R_{up}}=I_{ref}+\frac{V_{ref}}{R_{low}}$$ where V_o is output voltage and V_ref, I_ref are reference voltage and bias current through the reference pin of the shunt regulator. To make V_o independent of the I_ref, the I_ref should be much lower than $$\frac{V_o-V_{ref}}{R_{up}}$$ . So, $$\frac{V_o-V_{ref}}{R_{up}}=\frac{V_{ref}}{R_{low}}$$ . In the EVM, TLVH431 is considered for which reference voltage is given as 1.24V. For this design, $$\frac{V_o-V_{ref}}{R_{up}}$$ is considered as 73uA. So R_up obtained as 147kOhm. And from $$\frac{V_o-V_{ref}}{R_{up}}=\frac{V_{ref}}{R_{low}}$$ , R_low obtained as 16.98kohm.
9. Consider C_f as 10pF. So, R_v can be obtained as $${\omega }_{p1}=\frac{1}{R_v\cdot C_f}\Rightarrow f_{p1}=\frac{1}{2\cdot \pi \cdot R_v\cdot C_f}\Rightarrow R_v=\frac{1}{2\cdot \pi \cdot 479kHz\cdot C_f}\Rightarrow R_v=33.2kohm$$
10. R_LED can be obtained as $$G_o=\frac{R_{fb}CTR}{R_{LED}}\left(1+\frac{R_v}{R_{up}}\right)\Rightarrow R_{LED}=\frac{R_{fb}CTR}{G_o}\left(1+\frac{R_v}{R_{up}}\right)\Rightarrow R_{LED}=\frac{100\cdot {10}^3\cdot 0.2}{6.126}\left(1+\frac{33.2k}{147k}\right)\Rightarrow R_{LED}=4kohm$$
11. C_v can be obtained as $${\omega }_L=\frac{1}{\left(R_v+R_{up}\right)\cdot C_v}\Rightarrow C_v=\frac{1}{2\cdot \pi \cdot f_L\cdot \left(R_v+R_{up}\right)}\Rightarrow C_v=\frac{1}{2\cdot \pi \cdot 88\cdot \left(33.2k+147k\right)}\Rightarrow C_v=10nF$$
12. C_p, R_p are obtained as $${\omega }_z=\frac{1}{\left(R_{LED}+R_p\right)\cdot C_p}\text{ , }{\omega }_{p2}=\frac{1}{R_p\cdot C_p}\Rightarrow f_z=\frac{1}{2\cdot \pi \cdot \left(R_{LED}+R_p\right)\cdot C_p}{\text{ , f}}_{p2}=\frac{1}{2\cdot \pi \cdot R_p\cdot C_p}$$ $$\Rightarrow 3.4kHz=\frac{1}{2\cdot \pi \cdot \left(R_{LED}+R_p\right)\cdot C_p}\text{ , 29kHz}=\frac{1}{2\cdot \pi \cdot R_p\cdot C_p}\Rightarrow C_p=10nF\text{ , }{R}_p=540ohm$$ .
13. R_bias is used to bias the shunt regulator. R_bias is obtained as $$R_{bias}=\frac{V_{opto}}{I_{bias}}=\frac{1V}{1mA}=1kohm$$ .

Figure 1-27 Type 3 Compensator