Question on XL4105 feedback threshold

Hello There!

I was going through XL4105 datasheet and came across the feedback pin threshold (1.25V). Is this means, when the voltage on this pin reaches to 1.25V, will it lock the output voltage? Or it means something else?

datasheet.pdf (242 KB)

It's a regulator - and wil increase the switch power if FB is below 1.25, decrease if above.

You arrange a potential divider with 2 resistors in series and connect the top resistor to the output, bottom to ground , junction to the FB pin. Choose resistor values so that 1.25v is presented to the XL4105 at your wanted voltage .......

If you haven't designed switchmode power supplies before, I suggest you give it a miss - they're tricky.

regards

Allan

If you haven't designed switchmode power supplies before, I suggest you give it a miss - they're tricky.

I would echo that. You need to use a printed circuit board, this is something you can't make on a breadboard or strip board. I have worked with specialist power supply design engineers and I have yet to have one design a printed circuit board straight off that works correctly. Normally they need a second and often a third go at the design of the PCB.

I was going through XL4105 datasheet and came across the feedback pin threshold (1.25V). Is this means, when the voltage on this pin reaches to 1.25V, will it lock the output voltage? Or it means something else?

If you post on the forum asking questions while at the same time withholding your motive for doing so don't expect to get the answer you are looking for.

There is no reason why any user should be asking that question if they intended to use the device the way it was intended to be used. All you need is a screwdriver to adjust the output voltage. You shouldn't be asking about feedback pins unless you have a hidden agenda. Do you have a hidden agenda ?

And No, it doesn't lock to the output. It's a negative feedback (I think). It's used by the feedback amplifier so it is never the output voltage.

Thanks all for your responses.

@raschemmel, Firstly my apologies if I am not expressive enough. Well, there is no hidden agenda, I am planning to make a variable power supply that's the whole idea, before jumping into it, would like to understand how its works.

Isn't there a schematic on the datasheet that shows how to wire it ?

It's right there on the feedback block on page 3 of the datasheet.

The FB pin is an input to a comparator whose other input is the 1.25V reference. If you look at the voltage divider and do the math , based on the output voltage of 5V, the voltage divider that determines the FB input would yield a voltage of 1.24V. You can see from this arrangement that the voltage divider for any output voltage should yield 1.24V . If the desired output was 12V, the resistor values would need to be chosen to yield 1.24 V on the FB pin. Most variable regulators are just that, meaning the voltage divider is actually a pot. You could use a 20k pot and adjust the output.

If you are rolling your own switching regulator I would have expected you to know all of this. If you didn't or don't, then why aren't you just buying one already made on ebay instead of making your own ?

Based on the suggestions here I will go with ready-made module only. But I still have to do some tinkering as I need to replace the on-board pot with a 10 turn pot etc. My plan is to use 5K(R2) ten turn pot with 330 ohm(R1) resister. This should give my required range of 1.25V to 20V.

5k is awfully small. Your resolution would suffer. (suck). 50k or 100k would be preferred. Like I said, it is a voltage divider so you can replace both resistors with a single pot with the wiper connecting to FB. 5K would make the adjustment rather coarse, instead of fine. There's no logical reason to use such a small value.

Hmm.. never thought of resolution, thanks for pointing it out. Also using a single pot as a voltage divider also a good idea and its makes my job simple.