Electronics > Beginners

Current question in switchmode boost converter

<< < (2/2)

Thanks all ... Sounds like such spikes in current shouldn't be a concern for my resistors. Works for me. I will change their values though for smaller ones, as recommended.



1.6 A peak current is high (too high for my taste) for this kind of converter. Datasheet specified maximum switch current is 1.5 A. I suggest using something else (like LM2585) for this purpose, to get decent efficiency. Besides, by using proper regulator with error amplifier and PWM-modulator you'll get smaller ripple as a bonus :) There is a application circuit in LM2585 datasheet, page 20, with just the voltages you'll need.

Also, make sure that your inductor does not saturate, otherwise there is a risk that switching device will be destroyed. Current rating of the inductor should be as high as your other constraints will allow, since reducing the DC resistance of the inductor will increase efficiency.


The bottom line is the two resistors R1 and R2 do not conduct any of that peak current, they are there simply to give feedback of the output voltage for the chip. The chip needs to know if the output voltage is too high, too low or just right so that it can adjust it's operation to control that voltage That is all that these resistors do.

If the output voltage is high enough then the chip will stop switching until the output voltage drops again. (In reality it does not just stop switching or start switching it can vary its switching a bit instead of turning it on or off. But you get the idea.)

So when you have no load the output voltage stays right up where it should be 24V in your case and the chip stops switching, so that peak current no longer exists in fact there is no current except for the leakage current in your feedback resistors which is wasted (So note that the higher these resistor values the lower your leakage current just make sure they are low enough to make things work alright).

In theory this chip could have been designed so that we could connect the output straight back into the chip and it could read the voltage directly, some chips do this, but their voltages may not be adjustable. We use the "resistive divider" as feedback. As far as the chip is concerned 1.25V on pin 5 is just right, more than that is too high and less than that is too low. The voltage we get out of our resistive divider is

Vresdiv = (Voutput * R1)/(R1+R2)

So if we know that Vresdiv should be 1.25V for our desired Voutput we can choose R1 and R2 to satisfy this formula.

So what would happen if we did not feedback the output voltage? Lets say we just disconnected R2 or R2 somehow broke out of our circuit, say we soldered the resistor in badly. Then Pin 5 on our chip would be pulled down to ground always. So our chip would believe that the output voltage was always too low it would keep boosting the voltage higher and higher until either some protection kicked in or the chip was killed by high voltage on Pin 1. In fact using an external switch (Transistor, MOSFET, IGBT) and a high voltage inductor (car ignition coil) this chip could produce a very high voltage indeed thousands of volts if you like.

*NB* you can not use the external switch schematic in the data sheet to produce high voltages

This is definitely getting into some good meaty explanations. Thanks AchMed99 and Feanor!


--- Quote from: Rhythmtech on October 11, 2010, 02:08:23 pm ---This is definitely getting into some good meaty explanations. Thanks AchMed99 and Feanor!

--- End quote ---
I'm glad to see the question about divider resistors has opened the door to feedback explanation.
Feedback is one of the most useful concepts for understanding things. Once you catch it, you find that it's everywhere, from natural systems to the most complex control, to the way you can drive a car staying between the lines on the road.


[0] Message Index

[*] Previous page

There was an error while thanking
Go to full version