I am trying to figure out what would be the closed loop transfer function of the LT1619 controller. This is the first step I want to make in order to design the feedback loop.

What I already know about the subject

-I know to work with poles and zeroes, right half plane zero, etc.
-I understand how current mode control works (but not at a very deep level)
-I know to draw the asymptotes of impedances

What I know about my problem
-The circuit: power stage+controller setup enter image description here -The transfer function is \$T(s)=G_{cv}(s)K(s)A(s)\$ where \$K(s)\$ is the transfer function of the voltage divider formed by \$R_3\$ and \$R_4\$, a function which I think doesn't influence the loop since I didn't add any capacitors, \$A(s)\$ which is the compensator function and \$G_{cv}(s)\$ is the control to output transfer function, influenced by the power stage. I know \$A(s)\$ (with the help of this document) and \$K(s)\$, but I don't know \$G_{cv}(s)\$. This document gives some examples, but I don't know how to work with them to obtain an answer. My circuit is different than what is shown there.

The actual question is how do I obtain the control to output transfer function for the given controller (\$\frac{V_o}{V_c}\$)? I wonder how does an engineer figure out the frequency response of the chip he/she chooses as I believe that is a very important detail.

  • \$\begingroup\$ Vc is a node inside the control loop and not an input (as far as I know). \$\endgroup\$ – Andy aka Sep 1 '17 at 16:36
  • \$\begingroup\$ Vc Is the compensation pin, one example is here ti.com/lit/an/slva503a/slva503a.pdf (internal to the control loop it's a control voltage for the pwm stage) \$\endgroup\$ – sstobbe Sep 1 '17 at 20:35

What you want is the open-loop transfer function of this boost converter to think about a possible compensation strategy. The closed-loop transfer function is really when the converter operates with the loop physically closed and you can't measure it with a switching converter (you can reconstruct it). Why? Because this is a control system fighting perturbation like \$I_{out}\$ or \$V_{in}\$ while the setpoint (or the input) is \$V_{ref}\$ but is fixed inside the IC. So the true closed-loop transfer function would be that linking \$V_{out}\$ to \$V_{ref}\$ but nobody does that. See the small sketch below:

enter image description here

The plant is your boost converter power stage and the left-side triangle is the error amplifier. You clearly see how the perturbations via their open-loop coefficients affect the output. By closing the loop, you will reduced their contribution by \$\frac{1}{1+T(s)}\$ in which \$T(s)\$ is the loop gain: \$T(s)=G(s)H(s)\$. \$G(s)\$ represents the compensator transfer function (that you shape via the elements connected to your pin 3) while \$H(s)\$ represents the power stage dynamic response you need to determine before attempting to compensate.

How can you determine that power stage response? This is good because it is the subject of my next APEC 2018 seminar : )

  1. Use a small-signal transfer function obtained from the literature. Capture it in a mathematical solver and get the Bode plot. A good source for the transfer function is here.
  2. Build you hardware prototype using the LT part and extract the power stage response. It can be tricky here but as pin 3 is an OTA output, I believe (to be confirmed) that you can override it with a dc+ac source and plot \$V_{out}\$ to \$V_c\$. If not:
  3. Use a SPICE model to extract the dynamic response of the boost converter. You can use LTspice and apply the PWM-CM model to obtain the power stage and then further enrich the schematic with an OTA to compensate your loop:

enter image description here

You will find the model in my page and all the calculations around the OTA compensation in a seminar I gave in 2010. You must add parasitics such as the output capacitor ESR \$r_C\$ and the inductor ESR \$r_L\$ to show their effect on the response. Basically, you're all set!

  • \$\begingroup\$ You mean any open-loop transfer function will do? I'm thinking about control to output, line to output and output to control voltage. \$\endgroup\$ – Daniel Tork Sep 2 '17 at 8:13
  • \$\begingroup\$ Yes, the datasheet confirms that it's the output of the OTA amplifier as far as I can tell. \$\endgroup\$ – Daniel Tork Sep 2 '17 at 8:15
  • \$\begingroup\$ If it wasn't, then we probably wouldn't be dealing with a current source and as a consequence the compensator frequency response would be different than what is shown here:google.ro/… \$\endgroup\$ – Daniel Tork Sep 2 '17 at 8:18
  • \$\begingroup\$ Yes, you need to shape the compensator with poles and zeros so that the overall open-loop gain \$T(s)\$ crosses over at the chosen frequency \$f_c\$ with sufficient phase and gain margins. But first, you need the power stage dynamic response, the control-to-output transfer function. Except \$Z_{in}(s)\$ all the other transfer functions share a common denominator \$D(s)\$ so the poles are the same. The easiest way is to use a quick simulation to get there and the LTspice example is a way to do it. The PPT I linked gives all the details around the OTA compensation, check it out. \$\endgroup\$ – Verbal Kint Sep 2 '17 at 9:13
  • \$\begingroup\$ Thanks a lot for PPT, I read it. I found a method of extracting the open loop power stage gain using LTspice, but it seems it can only be used as some reference. The method is this: linear.com/solutions/4672 \$\endgroup\$ – Daniel Tork Sep 3 '17 at 11:05

Dr. Ray Ridley is one of the pioneers of analyzing current-mode control. The best way to analyze this is the PWM switch model, which you can read about on his website: Ridley Engineering

It's too complex a topic for a simple answer here, but you can find what you need at the link above or by Googling "PWM switch model" or looking for Chris Basso's site.

Another great reference is Dr. Vatche Vorperian's book: Fast Analytic Techniques


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