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 : $$ R_i<i_L>=v_c-DT_sS_e-\frac{S_fD'T_s}{2} $$

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 : $$ <i_L>=-\frac{DT_sS_e}{R_i\hat{v_{on}}}-\frac{D(V_{on}+\hat{v_{on}})T_s}{2L\hat{v_{on}}} $$

\$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 :

enter image description here

  • 1
    \$\begingroup\$ 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. \$\endgroup\$
    – Andy aka
    Nov 27, 2020 at 15:15
  • \$\begingroup\$ 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. \$\endgroup\$
    – AJN
    Nov 27, 2020 at 15:31
  • \$\begingroup\$ Ok I edited my question. \$\endgroup\$
    – Gontran
    Nov 27, 2020 at 19:01

1 Answer 1


There are different ways to obtain a small-signal equation from a large-signal one. The expression \$R_i<i_L>=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:

enter image description here

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:

enter image description here

If you carry on the exercise, you obtain the invariant gain \$k'_f\$ as follows:

enter image description here

You can now repeat the exercise for \$k'_r\$. If you want to simulate the entire model, there you go:

enter image description here

As you can see from the above, Ridley's model cannot find its dc operating point and is therefore constrained to ac simulations only. If you want to explore transient simulations with a current-mode model, I would recommend Vatché Vorpérian's version which predicts sub-harmonic oscillations with only two current sources. It is unbeatable in terms of simplicity. It has been extensively described in the book I published (with an original auto-toggling CCM-DCM model) but also through many APEC seminars (2013 and 2014) I taught.


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