# buck-boost converter

Discussion in 'General Electronics Discussion' started by syxuui, Jun 3, 2012.

1. ### syxuui

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0
Jun 3, 2012
hey guys, i'm now doing a negative power supply by using a buck-boost system. It works pretty well that it gives a negative 5V with a 5V DC input. Here's the problem, i'm trying to measure the output resistance of the system by treating the system as a thevenin's equivalent circuit, the thevenin's voltage source of which is -5V (as when the load resistance is really large, say, 15Mohms, the voltage at the load is -5V). I measured the current flowing through the load and voltage at the load with different load resistance and calculate output resistance of the system, and it turned out to be different for different load resistance. and i found that the output resistance increases as i decreases the load resistance value.
anyone has any idea on how this could happen??

here's a link to the intro to buck boost thing on wikipedia: (and we are doing exactly the same thing)
http://en.wikipedia.org/wiki/Buck–boost_converter

thanks

2. ### Harald KappModeratorModerator

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2,443
Nov 17, 2011
A regulated supply (linear or switched) cannot be treated as a Thevenin's equivalent circuit. A Thevenin's circuit assumes linear components. A regulated supply behaves non-linear insomuch as it will supply a constant voltage (within the accuracy limits of the regulator) as long as the current limit has not been reached. As soon as the current limit is crossed (towards higher current, of course) the behaviour depends on the characteristic of the regulator. It may go into constant current mode (thus effectively switching from ~0Ohm resistance to ~ooOhm resistance) or it may have a foldback characteristic or, or, or...

That goes for negative regulated supplies as well as for positive ones.

Harald