# Derivation of Feedforward Gains for current mode control in CCM converter

I recently start to read the dissertation of Dr.Ridley, A New Small-Signal Model for Current-Mode Control and I am stuck on a basic equation.

First we have the function of the inductor current for constant frequency control : $$R_i=v_c-DT_sS_e-\frac{S_fD'T_s}{2}$$

Which is fine, I get it, then the equation below can be describe : $$D=\frac{v_{off}}{v_{on}+v_{off}}\\ D'=\frac{v_{on}}{v_{on}+v_{off}}\\ S_f=\frac{v_{off}R_i}{L}$$ Normally, if I substitute theses equations in the previous function and perturbing the ON-time voltage, I should get this : $$\frac{<\hat{i_L}>}{\hat{v_{on}}} = \frac{DS_eT_s}{V_{ap}R_i}-\frac{D^2T_s}{2L}$$ But I can't. I guess I'm missing something but I don't know what. Do you have any idea ?

EDIT : Here's what I can get : $$=-\frac{DT_sS_e}{R_i\hat{v_{on}}}-\frac{D(V_{on}+\hat{v_{on}})T_s}{2L\hat{v_{on}}}$$

$$\T_s\$$ is the period of the switching frequency, $$\D\$$ the duty cycle, $$\R_i\$$ the current sense resistor, $$\v_{on}\$$ voltage when switch is ON, $$\v_{off}\$$ voltage when the switch is OFF, $$\v_c\$$ is the control voltage. As for $$\S_f\$$ it's the current slope inductor and $$\S_e\$$ is the slope compensation of the voltage control as seen below :

• You might choose to fully explain what all the formula symbols represent because you certainly can't rely on the linked document for doing that. Nov 27, 2020 at 15:15
• Where are you stuck ? Are you able to get the first equation which has three terms down to two terms ? The last terms of both the equations seems to be direct substitution. Please explain in detail where you are stuck.
– AJN
Nov 27, 2020 at 15:31
• Ok I edited my question. Nov 27, 2020 at 19:01

There are different ways to obtain a small-signal equation from a large-signal one. The expression $$\R_i=v_c-DT_sS_e-\frac{S_fD'T_s}{2}\$$ is a large-signal one and describes the current leaving terminal c in the PWM switch model scaled by the sense resistance $$\R_i\$$. You can either perturb each of the variables and then sort out dc and ac equations, the latter being the one you want. The other option - which I always use - is partial differentiation.
The equation in question is a function of the duty ratio $$\D\$$. If you replace $$\D\$$ by the expressions you gave, you obtain: $$\I_L=\frac{v_c}{R_i}-\frac{v_{off}}{v_{on}+v_{off}}T_s\frac{S_e}{R_i}-\frac{\frac{v_{off}R_i}{L}(1-\frac{v_{off}}{v_{off}+v_{off}})}{2R_i}T_s\$$. So this function depends on $$\v_c\$$, $$\v_{on}\$$ and $$\v_{off}\$$. If you consider $$\v_c\$$ and $$\v_{off}\$$ ac-silent (their derivative is 0), then you can differentiate $$\I_L\$$ with respect to $$\v_{on}\$$ and obtain:
Factor the on- and off-voltages to reveal the duty ratio $$\D\$$ again. You see that a term remains and this is $$\(v_{on}+v_{off})\$$. If you look back at the PWM switch model, you realize that the dc on-time voltage is labeled $$\V_{ac}\$$ while the off-time voltage is labeled $$\V_{cp}\$$. Therefore, substituting these values in the on and off sum leads to: $$\V_{on}+V_{off}=V_{ac}+V_{cp}=V_{ap}\$$. It leads to:
If you carry on the exercise, you obtain the invariant gain $$\k'_f\$$ as follows:
You can now repeat the exercise for $$\k'_r\$$. If you want to simulate the entire model, there you go: