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Calculating the negative input impedance of a converter

Ishtiyaq Makda
Prodigy 10 points


Dear all,

I have designed a Differential Mode EMI filter for a regulated isolated dc-dc boost converter and now in a phase of designing a parallel resonance damping branch for a filter. I was reading an article on EE Times from Robert Kollman (Senior Applications Manager and Distinguished Member of Technical Staff at Texas Instruments) on the following link:

http://www.eetimes.com/document.asp?doc_id=1273268&piddl_msgorder=thrd#msgs

While calculating negative input impedance of the converter he did it by using simple ohm's law and said in the description that:

Negative input impedance
This voltage-current line has a slope that essentially defines the dynamic impedance of the power supply. The slope of this line is the negative of the input voltage divided by the input current. That is, with Pin = V • I, we have V = Pin/I; and so dV/dI = –Pin/I2 or dV/dI ˜ –V/I.

This approximation is a bit of an over-simplification because the control loop impacts the frequency response of the input impedance. But many times, this simplification is good enough when current-mode control is involved.

Can someone please explain me that in the presence of Current-mode control, how this simple assumption is good enough?

Thank you very much.

Regards.