0
\$\begingroup\$

From the title It may seems like I need someone just do me the transfer function, but it is not.I'm looking for some directions.

I'm trying to calculate transfer function of \$ \frac{v_t}{d}|_{v_g=0} \$ of boost converter considering all resistances like inductor resistance , switch 's resistance , diode resistance and capacitor resistance (\$R_L,R_S,R_D,R_C \$) in CCM mode. This is my small signal model of the Boost converter:

schematic

simulate this circuit – Schematic created using CircuitLab

where $$X_L=sL/D'^{2}$$ $$X_C=1/sC$$ $$R_e=\frac{R+DR_S+D'R_D}{D'^2}$$ $$V_e=\frac{V_{out}-I(R_D-R_S)}{D'}$$

Here \$D=\frac{T_{ON}}{T_{ON}+T_{OFF}}\$ and \$D'=1-D\$ ,\$V_{out}\$ is output & \$V_{in}\$ input in DC and \$v_t\$ is small signal output

Now to find the transfer function of the ratio of output voltage and duty cycle, $$\frac{v_t}{R||X_C}+dI+\frac{v_t-V_e}{X_L+R_e}=0|_{v_g=0}$$ which leads to $$\frac{v_t}{d}|_{v_g=0}=\frac{R}{D'^3}.\frac{D'V_{out}-I(sL+R+R_D)}{RD'^2+(1+sCR)(sL+R+DR_S+D'R_D)}$$ This do not seems right as there should be at least two zero, but I can't find where did I do wrong. Again I thought if I replace value of \$V_{out}\$ and \$I\$ in terms of \$V_{in}\$ then it may give some good value. But as they are DC value and do not introduce any pole or zero, it did not help much.

My request is, if any one of you derive this transfer function considering all resistances, can you point me out where I may done some wrong calculation. Or any link where detail explanation for this derivation is explained. Most of the link I found use approximate transfer function and do not derive them . Some references derive transfer function using state variable, but considering all this factors , deriving from state variable very very tough for me as those references use values instead of symbols like \$R_L,R_S,R_D\$ .

FYI:

I don't give total calculation here cause it's nearly 2 page calculation and clumsy But if you need it, ask in comment . I'll add them. I go through them couple of times, hope there is no mistake.

My small signal model seems current to me as when I neglect \$R_L,R_D,R_S\$ it become the Boost converter small signal model given in Book(Fundamental of Power Electronics by Erickson & Maksimovic).

\$\endgroup\$
  • 1
    \$\begingroup\$ Its not really clear what your input and output is of your transfer function, might want to clarify. \$\endgroup\$ – Voltage Spike Jun 1 '17 at 18:18
  • \$\begingroup\$ I'm calculating a boost converter like this [allaboutcircuits.com/uploads/articles/…. \$\endgroup\$ – Anklon Jun 2 '17 at 3:11
  • \$\begingroup\$ I need a transfer function of small signal output voltage with respect to pulse width, \$\frac{v_t}{d}\$. So I took input voltage small signal \$v_g\$ zero. But I'm lost in the calculation. Need detail calculation or any link that has it. Most of the source I found just assumed resistance of switch,diode and inductor is zero. And they just write the transfer function based on that. \$\endgroup\$ – Anklon Jun 2 '17 at 3:14
  • \$\begingroup\$ in circuit ,\$v_g\$ is shown. In my equation there is \$ V_{out}\$ & \$I\$ which can be expressed in terms of \$V_g\$ . \$\endgroup\$ – Anklon Jun 5 '17 at 21:06
1
\$\begingroup\$

There are several ways to reach your goal which is to determine the control-to-output transfer function of the CCM boost converter including various losses. The easiest and most straightforward way is to use the inductor volt-seconds balance law. From there, if you determine the inductor bias during the on- and off-times including \$r_{DS(on)}\$, the diode \$V_f\$ and the rest of ohmic losses, it's not too complex. However, you will get the dc transfer function only (from which you can get the dc gain \$H_0\$ if you want). Look at the below picture to see how to do it:

enter image description here

Now, if you want to determine the dynamic response of the boost converter considering all these losses, I would suggest to study an APEC seminar given in 2013 which uses the PWM switch model. The model can be easily be rearranged to include various losses to account for semiconductor properties like diode drops or MOSFET \$r_{DS(on)}\$. The PPT for the CCM boost small-signal analysis is here:

http://cbasso.pagesperso-orange.fr/Downloads/PPTs/Chris%20Basso%20APEC%20seminar%202013.pdf

The APEC 2014 and 2015 APEC seminars carry on with small-signal analysis also. They are available from the same page:

http://cbasso.pagesperso-orange.fr/Spice.htm

and the book which describes the lossy PWM switch models is here:

https://www.amazon.com/Switch-Mode-Power-Supplies-Second-Edition/dp/0071823468/ref=pd_sim_b_5?ie=UTF8&refRID=08N1PQ31Z26697R4TGWV.

\$\endgroup\$
  • \$\begingroup\$ Another great book by the guy who pretty much invented the PWM switch model is here: amazon.com/Analytical-Techniques-Electrical-Electronic-Circuits/… \$\endgroup\$ – John D Aug 16 '17 at 16:08
  • \$\begingroup\$ Though the PWM switch model is the superior way, it's also instructive to try state-space averaging and compare the results. If you Google it you'll get the info you need. \$\endgroup\$ – John D Aug 16 '17 at 16:10
  • \$\begingroup\$ @John D, yes, state-space averaging or SSA is interesting to understand and realize that it encompasses the whole converter and not only the switching cell, unlike the PWM switch model does. Simply put, if you use SSA and later realize that you would like to consider the capacitor ESR \$r_C\$ or add any other parasitic element, you have to start all over again. Nothing like that with the PWM switch. \$\endgroup\$ – Verbal Kint Aug 16 '17 at 18:40
0
\$\begingroup\$

Not an expert here, but, have you tried expressing each element of the circuit in terms of its impedance?

By doing so, you can easily find the equivalent impedance of X_L, R_e, and V_ed, and of X_c, dl, and R. Then you'll have a circuit with a source Vg/D' and two impedances. Your output can be expressed as the a function of a voltage divider.

schematic

simulate this circuit – Schematic created using CircuitLab

Your v_t is then simply

$$ v_{out} = \frac{Z_{out}}{Z_1 + Z_{out}} v_{in} $$

And your transfer function is

$$ \frac{v_{out}}{v_{in}} = \frac{Z_{out}}{Z_1 + Z_{out}} $$

I hope this may help you.

\$\endgroup\$
  • \$\begingroup\$ I thought of that process, but instead of finding \$\frac{v_{out}}{v_{in}}\$ I need to found \$\frac{v_{out}}{d}\$ . and they don't have any relation like voltage division which you applied here. \$\endgroup\$ – Anklon Jun 5 '17 at 18:16
  • 1
    \$\begingroup\$ Do you know what $$\frac{v_{in}}{d}$$ equals? Then it would be just about multiplying $$\frac{v_{out}}{v_{in}}\frac{v_{in}}{d}$$ \$\endgroup\$ – David Jun 5 '17 at 19:20
  • \$\begingroup\$ No I don't. Finding that is as much as challenging. \$\endgroup\$ – Anklon Jun 5 '17 at 19:22
  • 1
    \$\begingroup\$ Just to be sure I understand correctly, "D" is your duty cycle, what what is "d"? \$\endgroup\$ – David Jun 5 '17 at 19:24
  • 2
    \$\begingroup\$ Oh, ok. Sorry mate, my understanding of power electronics doesn't go this far. I hope you get your answer. \$\endgroup\$ – David Jun 5 '17 at 19:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.