6.2.1. Low-Voltage Inductive Boost Converters

6.2.2. Medium-Voltage Inductive Boost Converters

$F_{\text{osc}}=6\times {10}^9/R_{\text{T}},\text{where}{R}_{\text{T}}\text{is}\text{the}\text{timing}\text{resistor.}\text{So}\text{for}\text{a}\text{20-kHz}\text{operation},R_{\text{T}}=300\text{k}\Omega \text{.}$

The $\overline{\text{FAULT}}$ pin is used to drive an external disconnect MOSFET (Fig. 6.5). During the start-up of the HV9912, the $\overline{\text{FAULT}}$ pin is held low and once the IC starts-up the pin is pulled high. This connects the LEDs in the circuit and the boost converter powers up the LEDs. In case of an output overvoltage condition or an output short circuit condition, the $\overline{\text{FAULT}}$ pin is pulled low and an external MOSFET switched off to disconnect the LEDs.

The $\overline{\text{FAULT}}$ pin is also controlled by the PWM dimming signal, so that the pin is high when the PWM dimming signal is high and vice versa. This disconnects the LEDs and makes sure that the output capacitor does not have to be charged/discharged every PWM dimming cycle. The PWM dimming input to the $\overline{\text{FAULT}}$ pin and the output of the protection circuitry are logically AND’ed to make sure that the protection circuit overrides the PWM input to the $\overline{\text{FAULT}}$ pin.

Output short circuit protection is provided by a comparator that triggers when the output current sense voltage (at the FDBK pin) is twice that of the reference voltage (at the I_REF pin). The output overvoltage protection is activated when the voltage at the OVP pin exceeds 5 V. A fault condition will exist when either of these signals are active and will result in both the GATE pin and the $\overline{\text{FAULT}}$ pin being pulled low. Once the IC goes into the fault mode, the capacitor on the COMP pin is used for timing until a reset occurs. At the end of the timing period, the HV9912 will attempt to restart. If the fault condition persists, the HV9912 will continually attempt restarts until power is removed or the fault clears.

6.2.3. Multichannel Boost Converters

6.4.1. Design Specification

6.4.2. Typical Circuit

6.4.3. Selecting the Switching Frequency (f_s)

6.4.4. Computing the Maximum Duty Cycle (D_max)

$\begin{array}{ll}{D}_{\max }\hfill & =1-\frac{{\eta }_{\min }\cdot V_{\text{in}\min }}{V_{\text{o}\max }}\hfill \\ \hfill & =0.717\hfill \end{array}$

6.4.5. Computing the Maximum Inductor Current (I_in max)

$\begin{array}{ll}{I}_{\text{in}\max }\hfill & =\frac{V_{\text{o}\max }\cdot I_{\text{o}\max }}{{\eta }_{\min }\cdot V_{\text{in}max}}\hfill \\ \hfill & =1.24\text{A}\hfill \end{array}$

6.4.6. Computing the Input Inductor Value (L1)

$\begin{array}{ll}L1\hfill & =\frac{V_{\text{in}\max }\cdot D_{\max }}{0.25\cdot I_{\text{in}\max }\cdot f_s}\hfill \\ \hfill & =254\mu \text{H}\hfill \end{array}$

$\begin{array}{ll}{P}_{\text{ind}}\hfill & =0.03\cdot V_{\text{o}\max }\cdot I_{\text{o}\max }\hfill \\ \hfill & =0.735\text{W}\hfill \end{array}$

$\begin{array}{l}\text{DCR}<\frac{0.8\cdot P_{\text{ind}}}{I_{\text{in}\max }^2}\\ \Rightarrow \text{DCR}<0.38\Omega \end{array}$

$\begin{array}{ll}{I}_{\text{sat}}\hfill & =1.2\cdot I_{\text{in}\max }\cdot \left(1+\frac{0.25}{2}\right)\hfill \\ \hfill & =1.7\text{A}\hfill \end{array}$

6.4.7. Choosing the Switching MOSFET (Q1)

$\begin{array}{ll}{V}_{\text{FET}}\hfill & =1.2\cdot V_{\text{o}\max }\hfill \\ \hfill & =84\text{V}\hfill \end{array}$

$\begin{array}{ll}{I}_{\text{FET}}\hfill & \approx I_{\text{in}\max }\cdot \sqrt{D_{\max }}\hfill \\ \hfill & =1.05\text{A}\hfill \end{array}$

6.4.8. Choosing the Switching Diode (D1)

6.4.9. Choosing the Output Capacitor (C_o)

$\begin{array}{ll}∆v_{\text{p}-\text{p}}\hfill & =∆I_{\text{o}}\cdot R_{\text{LED}}\hfill \\ \hfill & =0.63\text{V}\hfill \end{array}$

$I_{\text{o}\text{max}}=C_{\text{o}}\cdot \frac{∆v_{\text{p}-\text{p}}}{D_{\text{max}}\cdot T_{\text{s}}}$

$\begin{array}{ll}{C}_{\text{o}}\hfill & =\frac{I_{\text{o}\max }\cdot D_{\max }}{∆v_{\text{p}-\text{p}}\cdot f_s}\hfill \\ \hfill & =1.99\mu \text{F}\hfill \end{array}$

$\begin{array}{ll}{I}_{\text{rms}}\hfill & =\sqrt{D_{\max }\cdot I_{\text{o}\max }^2+(1-D_{\max })\cdot {(I_{\text{in}\text{max}}-I_{\text{o}\text{max}})}^2}\hfill \\ \hfill & =0.56\text{A}\hfill \end{array}$

6.4.10. Choosing the Disconnect MOSFET (Q2)

$\begin{array}{ll}{R}_{\text{on,25}°\text{C}}\hfill & =\frac{0.01\cdot V_{\text{o}\max }}{I_{\text{o}\text{max}}\cdot 1.4}\hfill \\ \hfill & =1.43\Omega \hfill \end{array}$

6.4.11. Choosing the Input Capacitors (C1 and C2)

$C_{\text{in}}\ge \frac{1}{{(2\cdot \pi \cdot f_{\text{LC}})}^2\cdot L_{\text{source}}}=3.95\mu \text{F}$

$\begin{array}{l}{R}_{\text{eq}}={(1-D_{\text{max}})}^2\cdot R_{\text{LED}}\hfill \\ R_{\text{eq}}={(1-0.717)}^2\cdot 18\hfill \\ R_{\text{eq}}=1.4416\Omega \hfill \end{array}$

$R_{\text{source,max}}=1.44\Omega$

6.4.12. Choosing the Timing Resistor (R_T)

$\frac{1}{f_{\text{s}}}\approx R_{\text{T}}\cdot 18\text{pF}$

6.4.13. Choosing the Two Current Sense Resistors (R1 and R2)

$\begin{array}{ll}R2\hfill & =\frac{0.15\text{W}}{I_{\text{o}\max }^2}\hfill \\ \hfill & =1.22\Omega \hfill \end{array}$

$\begin{array}{ll}R1\hfill & =\frac{0.25}{1.125\cdot I_{\text{in}\max }}\hfill \\ \hfill & =0.18\Omega \hfill \end{array}$

$\begin{array}{ll}{P}_{R1}\hfill & =I_{\text{FET}}^2\cdot R1\hfill \\ \hfill & =0.2\text{W}\hfill \end{array}$

6.4.14. Selecting the Current Reference Resistors (R3 and R4)

$R3+R4=\frac{1.25\text{V}}{50\mu \text{A}}=25\text{k}\Omega$

$\frac{1.25\text{V}}{R3+R4}\cdot R4=I_{\text{o}\text{max}}\cdot R2$

$\begin{array}{l}{R}_{R2}=8.68\text{k}\Omega ,1/8\text{W},1\%\\ R_{R1}=16.32\text{k}\Omega ,1/8\text{W},1\%\end{array}$

6.4.15. Programming the Slope Compensation (R_slope and R7)

$\begin{array}{ll}DS\hfill & =\frac{V_{\text{o}\max }-V_{\text{in}\min }}{L}\hfill \\ \hfill & =0.145\text{A/}\mu \text{s}\hfill \end{array}$

$R_{\text{slope}}=\frac{10\cdot R7\cdot f_{\text{s}}}{DS(\text{A/}\mu \text{s})\cdot {10}^6\cdot R1}$

$\begin{array}{ll}{R}_{\text{slope}}\hfill & =\frac{10\cdot 1k\cdot 200k}{0.2682\cdot {10}^6\cdot 0.15}\hfill \\ \hfill & =76.62\text{k}\Omega \hfill \end{array}$

$\begin{array}{l}R7=510,1/8\text{W},1/\%\\ R_{\text{slope}}=39k,1/8\text{W},1\%\end{array}$

6.4.16. Setting the Inductor Current Limit (R5 and R6)

$V_{\text{CLIM}}\ge 1.35\cdot I_{\text{in}\max }\cdot R1+\frac{4.5\cdot R7}{R_{\text{slope}}}$

$\begin{array}{lll}{V}_{\text{CLIM}}\hfill & =\hfill & 1.35\cdot 1.24\cdot 0.18+\frac{4.5\cdot 510}{39k}\hfill \\ \hfill & =\hfill & 0.36\hspace{0.17em}\text{V}\hfill \end{array}$

$\begin{array}{l}R5=20k,1/8\text{W},1\%\\ R6=8.06k,1/8\text{W},1\%\end{array}$

6.4.17. Capacitors at V_DD and REF pins

6.4.18. Setting the Overvoltage Trip Point (R8 and R9)

$\begin{array}{ll}{V}_{\text{open}}\hfill & =1.2\cdot V_{\text{o}\max }\hfill \\ \hfill & =84\text{V}\hfill \end{array}$

$\begin{array}{ll}R8\hfill & =\frac{{(V_{\text{open}}-5\text{V})}^2}{0.1}\hfill \\ \hfill & =64\text{k}\Omega \hfill \end{array}$

$\begin{array}{ll}R9\hfill & =\frac{R8}{(V_{\text{open}}-5\text{V})}\cdot 5\text{V}\hfill \\ \hfill & =3.95\text{k}\Omega \hfill \end{array}$

$\begin{array}{l}R8=68k,1/8\text{W},1\%\\ R9=3.9k,1/8\text{W},1\%\end{array}$

6.4.19. Designing the Compensation Network

$\text{Loop}\text{gain}=R_{\text{s}}\cdot G_{\text{m}}\cdot Z_{\text{c}}(s)\cdot \frac{1}{15}\cdot \frac{1}{R_{\text{cs}}}\cdot G_{\text{ps}}(s)$

$G_{\text{ps}}(s)=\frac{(1-D_{\max })}{2}\cdot \frac{1-s\cdot \frac{L1}{{(1-D_{\max })}^2\cdot R_{\text{LED}}}}{1+s\cdot \frac{R_{\text{LED}}\cdot C_{\text{o}}}{2}}$

$G_{\text{ps}}(s)=\frac{0.283}{2}\cdot \frac{1-s\cdot \frac{330\cdot {10}^{-6}}{{(0.283)}^2\cdot 18}}{1+s\cdot \frac{18\cdot 2\cdot {10}^{-6}}{2}}$

$G_{\text{ps}}(s)=0.1415\cdot \frac{1-s\cdot 2.28912\cdot {10}^{-4}}{1+s\cdot 1.8\cdot {10}^{-5}}$

$G_{\text{ps}}(s)=0.1415\cdot \frac{1-i\cdot 2.8766}{1+i\cdot 0.226188}$

$\begin{array}{l}{\left.\left|G_{\text{ps}}(s)\right|\right|}_{f_{\text{c}}=2\text{kHz}}=A_{\text{ps}}=0.40996\\ {\left.\angle G_{\text{ps}}(s)\right|}_{f_{\text{c}}=2\text{kHz}}={\phi }_{\text{ps}}=-83.57°\end{array}$

$\begin{array}{ll}{\phi }_{\text{boost}}\hfill & ={\phi }_{\text{m}}-{\phi }_{\text{ps}}-90°\hfill \\ \hfill & =38.57°\hfill \end{array}$

$\begin{array}{l}{\phi }_{\text{boost}}\le 0°\Rightarrow \text{Type-I}\text{controller}\\ 0°\le {\phi }_{\text{boost}}\le 90°\Rightarrow \text{Type-II}\text{controller}\\ 90°\le {\phi }_{\text{boost}}\le 180°\Rightarrow \text{Type-III}\text{controller}\end{array}$

Table 6.1

Compensation networks

Type	Circuit Diagram	Transfer Function
I		$Z_{\text{c}}(s)=\frac{1}{s{C}_{\text{c}}}$
II		$Z_{\text{c}}(s)=\frac{1}{s(C_{\text{c}}+C_{\text{z}})}\cdot \frac{1+s\cdot R_{\text{z}}\cdot C_{\text{z}}}{1+s\cdot \frac{C_{\text{z}}\cdot C_{\text{c}}}{C_{\text{z}}+C_{\text{c}}}\cdot R_{\text{z}}}$

$\begin{array}{ll}K\hfill & =\tan (45°+\frac{{\phi }_{\text{boost}}}{2})\hfill \\ \hfill & =2.077\hfill \end{array}$

$\begin{array}{ll}{\omega }_{\text{z}}\hfill & =\frac{1}{R_{\text{z}}\cdot C_{\text{z}}}=\frac{2\cdot \pi \cdot f_{\text{c}}}{K}\hfill \\ \hfill & =\mathrm{6,050}\text{rad}/\text{s}\hfill \end{array}$

$\begin{array}{ll}{\omega }_{\text{p}}\hfill & =\frac{C_{\text{z}}+C_{\text{p}}}{C_{\text{z}}\cdot C_{\text{p}}\cdot R_{\text{z}}}=(2\cdot \pi \cdot f_{\text{c}})\cdot K\hfill \\ \hfill & =\mathrm{26,100}\text{rad}/\text{s}\hfill \end{array}$

$R_{\text{s}}\cdot G_{\text{m}}\cdot \left(\frac{1}{2\cdot \pi \cdot f_{\text{c}}\cdot (C_{\text{z}}+C_{\text{c}})}\cdot \frac{\sqrt{1+K^2}}{\sqrt{1+{(1/K)}^2}}\right)\cdot \frac{1}{15}\cdot \frac{1}{R_{\text{cs}}}\cdot A_{\text{ps}}=1$

$R_{\text{s}}\cdot G_{\text{m}}\cdot \left(\frac{1}{2\cdot \pi \cdot f_{\text{c}}\cdot }\cdot \frac{\sqrt{1+K^2}}{\sqrt{1+{(1/K)}^2}}\right)\cdot \frac{1}{15}\cdot \frac{1}{R_{\text{cs}}}\cdot A_{\text{ps}}=(C_{\text{z}}+C_{\text{c}})$

$C_{\text{z}}+C_{\text{c}}=41\text{nF}$

$\begin{array}{ll}{C}_{\text{c}}\hfill & =(C_{\text{z}}+C_{\text{c}})\cdot \frac{{\omega }_{\text{z}}}{{\omega }_{\text{p}}}\hfill \\ \hfill & =9.5\text{nF}\hfill \end{array}$

$C_{\text{z}}=31.5\text{nF}$

$\begin{array}{ll}{R}_{\text{z}}\hfill & =\frac{1}{{\omega }_{\text{z}}\cdot C_{\text{z}}}\hfill \\ \hfill & =5.247\text{k}\Omega \hfill \end{array}$

$\begin{array}{l}{C}_{\text{c}}=10\text{nF},50\text{V},C0G\text{capacitor}\\ C_{\text{z}}=33\text{nF},50\text{V},C0G\text{capacitor}\\ R_{\text{z}}=5.1k,1/8\text{W},1\%\text{resistor}\end{array}$

If the MPS boost controller MP4013B were used instead, calculations would have to be repeated with G_m = 370 μA/V. Also because the current sense reference voltages are fixed, new values of R_cs (R1) and R_s (R2) would also need to be substituted into the equations. However, a first approximation for the scaling will be approximately 370/550, giving $C_{\text{z}}+C_{\text{c}}=27\text{nF}$.

$\begin{array}{l}{C}_{\text{c}}=6.8\text{nF},50\text{V},C0G\text{capacitor}\\ C_{\text{z}}=22\text{nF},50\text{V},C0G\text{capacitor}\\ R_{\text{z}}=7.5k,1/8\text{W},1\%\text{resistor}\end{array}$

6.4.20. Output Clamping Circuit

6.5.1. Design Specification

6.5.2. Typical Circuit

6.5.3. Selecting the Switching Frequency (f_s)

6.5.4. Computing the Maximum Inductor Current (I_in max)

$\begin{array}{ll}{I}_{\text{in}\max }\hfill & =\frac{V_{\text{o}\max }\cdot I_{\text{o}\max }}{{\eta }_{\min }\cdot V_{\text{in}\min }}\hfill \\ \hfill & =0.915\text{A}\hfill \end{array}$

6.5.5. Computing the Input Inductor Value (L1)

$\begin{array}{ll}L1\cdot i_{\text{Lpk}}\cdot \left(\frac{1}{V_{\text{in}\text{min}}}+\frac{1}{V_{\text{o}\text{max}}-V_{\text{in}\text{min}}}\right)\hfill & =\frac{0.95}{f_{\text{s}}}\hfill \\ \hfill & =4.75\mu \text{s}\hfill \end{array}$

$\begin{array}{ll}{I}_{\text{in}\text{max}}\hfill & =\frac{1}{2}\cdot i_{\text{Lpk}}\cdot \frac{4.75\mu \text{s}}{5\mu \text{s}}\hfill \\ \hfill & =0.475\cdot i_{\text{Lpk}}\hfill \end{array}$

$\begin{array}{ll}{i}_{\text{Lpk}}\hfill & =\frac{I_{\text{in}\max }}{0.475}\hfill \\ \hfill & \approx 1.93\text{A}\hfill \end{array}$

$\begin{array}{ll}L1\hfill & =\frac{0.95}{200k}\cdot \frac{9\text{V}\cdot (70\text{V}-9\text{V})}{70\text{V}\cdot 1.93\text{A}}\hfill \\ \hfill & =19.3\mu \text{H}\hfill \end{array}$

$\begin{array}{ll}L{1}_{\text{nom}}\hfill & =\frac{L1}{1.2}\hfill \\ \hfill & =16.08\mu \text{H}\hfill \end{array}$

$\begin{array}{ll}{I}_{\text{Lrms}}\hfill & =i_{\text{Lpk}}\cdot \sqrt{\frac{0.9}{3}}\hfill \\ \hfill & =1.057\text{A}\hfill \end{array}$

6.5.6. Computing the On- and Off-Times of the Converter

$\begin{array}{ll}{t}_{\text{on}\_\text{sw}}\hfill & =\frac{L{1}_{\text{nom}}\cdot i_{\text{Lpk}}}{V_{\text{in}\min }}\hfill \\ \hfill & =3.22\mu \text{s}\hfill \end{array}$

$\begin{array}{ll}{t}_{\text{on}\_\text{diode}}\hfill & =\frac{L{1}_{\text{nom}}\cdot i_{\text{Lpk}}}{V_{\text{o}\max }-V_{\text{in}\min }}\hfill \\ \hfill & =467\text{ns}\hfill \end{array}$

$\begin{array}{ll}{D}_{\max }\hfill & =t_{\text{on}\_\text{sw}}\cdot f_{\text{s}}\hfill \\ \hfill & =0.644\hfill \end{array}$

$\begin{array}{ll}D1\hfill & =t_{\text{on}\_\text{diode}}\cdot f_{\text{s}}\hfill \\ \hfill & =0.0934\hfill \end{array}$

6.5.7. Choosing the Switching MOSFET (Q1)

$\begin{array}{ll}{V}_{\text{FET}}\hfill & =1.2\cdot V_{\text{o}\max }\hfill \\ \hfill & =84\text{V}\hfill \end{array}$

$\begin{array}{ll}{I}_{\text{FET}}\hfill & \approx i_{\text{Lpk}}\cdot \sqrt{\frac{D_{\max }}{3}}\hfill \\ \hfill & =0.895\text{A}\hfill \end{array}$

6.5.8. Choosing the Switching Diode (D1)

6.5.9. Choosing the Output Capacitor (C_o)

$\begin{array}{ll}∆v_{\text{p}-\text{p}}\hfill & =∆I_{\text{o}}\cdot R_{\text{LED}}\hfill \\ \hfill & =0.55\text{V}\hfill \end{array}$

$I_{\text{o}\text{max}}=C_{\text{o}}\cdot \frac{∆v_{\text{p}-\text{p}}}{D_{\max }\cdot T_{\text{s}}}$

$\begin{array}{ll}{C}_{\text{o}}\hfill & =\frac{I_{\text{o}\max }\cdot D_{\max }}{∆v_{\text{p}-\text{p}}\cdot f_{\text{s}}}\hfill \\ \hfill & =0.585\mu \text{F}\hfill \end{array}$

$\begin{array}{ll}{I}_{\text{rms}}\hfill & =\sqrt{(1-\text{D}1)\cdot I_{\text{o}\text{max}}^2+\frac{\text{D}1}{3}\cdot {\left(i_{\text{Lpk}}-I_{\text{o}\max }\right)}^2}\hfill \\ \hfill & =0.34\text{A}\hfill \end{array}$

6.5.10. Choose the Disconnect MOSFET (Q2)

$\begin{array}{ll}{R}_{\text{on,25}°\text{C}}\hfill & =\frac{0.01\cdot V_{\text{o}\max }}{I_{\text{o}\max }\cdot 1.4}\hfill \\ \hfill & =5\Omega \hfill \end{array}$

6.5.11. Choosing the Input Capacitors (C1 and C2)

$C_{\text{in}}\ge \frac{1}{{(2\cdot \pi \cdot f_{\text{LC}})}^2\cdot L_{\text{source}}}=3.95\mu \text{F}$

$M=\frac{V_{\text{o}\text{max}}}{V_{\text{in}\text{min}}}=\frac{70}{9}=7.778$

$R_{\text{source},\max }=\frac{M-1}{M^2\cdot (M-2)}\cdot R_{\text{LED}}=1.404\Omega$

6.5.12. Calculating the Timing Resistor (R_T)

$\frac{1}{f_{\text{s}}}\approx R_{\text{T}}\cdot 18\text{pF}$

6.5.13. Calculating the Two Current Sense Resistors (R1 and R2)

$\begin{array}{ll}R2\hfill & =\frac{0.4\text{V}}{I_{\text{o}\max }}\hfill \\ \hfill & =4\Omega \hfill \end{array}$

$\begin{array}{ll}R1\hfill & =\frac{0.25}{i_{\text{Lpk}}}\hfill \\ \hfill & =0.12\Omega \hfill \end{array}$

$\begin{array}{ll}{P}_{R1}\hfill & =I_{\text{FET}}^2\cdot R1\hfill \\ \hfill & =0.096\text{W}\hfill \end{array}$

6.5.14. Selecting the Current Reference Resistors (R3 and R4)

$R3+R4=\frac{1.25\text{V}}{50\mu \text{A}}\le 25\text{k}\Omega$

$\frac{1.25\text{V}}{R3+R4}\cdot R4=I_{\text{o}\max }\cdot R2=0.1\cdot 3.9=0.39\text{V}$

$\begin{array}{l}R3=19.1\text{k}\Omega ,1/8\text{W},1\%\\ R4=8.66\text{k}\Omega ,1/8\text{W},1\%\end{array}$

6.5.15. Setting the Inductor Current Limit (R5 and R6)

$V_{\text{CLIM}}\ge 1.2\cdot i_{\text{Lpk}}\cdot R1$

$\begin{array}{ll}{V}_{\text{CLIM}}\hfill & =1.2\cdot 1.93\cdot 0.12\hfill \\ \hfill & \ge 0.278\text{V}\hfill \end{array}$

$\begin{array}{l}R5=20k,1/8\text{W},1\%\\ R6=6.04k,1/8\text{W},1\%\end{array}$

6.5.16. Capacitors at V_DD and REF Pins

6.5.17. Setting the Overvoltage Trip Point (R8 and R9)

$\begin{array}{ll}{V}_{\text{open}}\hfill & =1.15\cdot V_{\text{o}\max }\hfill \\ \hfill & =80.5\text{V}\hfill \end{array}$

$\begin{array}{ll}R8\hfill & =\frac{{(V_{\text{open}}-5\text{V})}^2}{0.1}\hfill \\ \hfill & =57\text{k}\Omega \hfill \end{array}$

$\begin{array}{ll}R9\hfill & =\frac{R8}{(V_{\text{open}}-5\text{V})}\cdot 5\text{V}\hfill \\ \hfill & =3.78\text{k}\Omega \hfill \end{array}$

$\begin{array}{l}R8=56k,1/8\text{W},1\%\\ R9=3.6k,1/8\text{W},1\%\end{array}$

6.5.18. Designing the Compensation Network

$\text{Loop}\text{gain}=R_{\text{s}}\cdot G_{\text{m}}\cdot Z_{\text{c}}(s)\cdot \frac{1}{15}\cdot \frac{1}{R_{\text{cs}}}\cdot G_{\text{ps}}(s)$

$M=\frac{V_{\text{o}\text{max}}\cdot I_{\text{o}\text{max}}}{V_{\text{o}\text{max}}\cdot I_{\text{o}\text{max}}-0.5\cdot L{1}_{\text{nom}}\cdot i_{\text{Lpk}}^2\cdot f_{\text{s}}}$

$M=\frac{70\cdot 0.1}{70\cdot 0.1-0.5\cdot 15\cdot {10}^{-6}\cdot {1.93}^2\cdot 200\cdot {10}^3}$

$M=\frac{7}{1.41265}=4.9552$

$G_{\text{r}}=\frac{M-1}{2\cdot M-1}=\frac{3.95522}{8.9104}=0.4439$

$G_{\text{ps}}(s)=2\cdot \frac{I_{\text{o}\text{max}}}{i_{\text{Lpk}}}\cdot \frac{G_{\text{r}}}{1+s\cdot R_{\text{LED}}\cdot C_{\text{o}}\cdot G_{\text{r}}}$

$G_{\text{ps}}(s)=2\cdot \frac{0.1}{1.93}\cdot \frac{0.4439}{1+s\cdot 55\cdot 2\cdot {10}^{-6}\cdot 0.4439}=\frac{0.4439}{1+s\cdot 48.829\cdot {10}^{-6}}$

$G_{\text{ps}}(s)=\frac{0.046}{1+s\cdot 0.6136}$

$\begin{array}{l}{\left.\left|G_{\text{ps}}(s)\right|\right|}_{f\text{c}=2\text{kHz}}=A_{\text{ps}}=0.039\\ {\left.\angle G_{\text{ps}}(s)\right|}_{f\text{c}=2\text{kHz}}={\phi }_{\text{ps}}=-31.5°\end{array}$

$\begin{array}{ll}{\phi }_{\text{boost}}\hfill & ={\phi }_{\text{m}}-{\phi }_{\text{ps}}-90°\hfill \\ \hfill & =45°+31.5°-90°\hfill \\ \hfill & =-13.5°\hfill \end{array}$

$\begin{array}{l}{\phi }_{\text{boost}}\le 0°\Rightarrow \text{Type-I}\text{controller}\\ 0°\le {\phi }_{\text{boost}}\le 90°\Rightarrow \text{Type-II}\text{controller}\\ 90°\le {\phi }_{\text{boost}}\le 180°\Rightarrow \text{Type-III}\text{controller}\end{array}$

$R2\cdot G_{\text{m}}\cdot \left(\frac{1}{2\cdot \pi \cdot f_{\text{c}}\cdot C_{\text{c}}}\cdot \right)\cdot \frac{1}{15}\cdot \frac{1}{R1}\cdot A_{\text{ps}}=1$

$C_{\text{c}}=R2\cdot G_{\text{m}}\cdot \left(\frac{1}{2\cdot \pi \cdot f_{\text{c}}}\cdot \right)\cdot \frac{1}{15}\cdot \frac{1}{R1}\cdot A_{\text{ps}}$

$C_{\text{c}}=3.9\cdot 550\cdot {10}^{-6}\cdot \left(\frac{1}{12566}\cdot \right)\cdot \frac{1}{15}\cdot \frac{1}{0.12}\cdot 0.039=3.698\text{nF}$