I used an old atx from an old computer to make my own cheap bench power supply and it works fine.
I wanted to add an adjustable voltage output, so i followed this schematic.
For R1 i have a 220 ohm resistor and for R2 i have a 10k ohm potentiometer.
So the adjusting of the voltage works however the first half rotation of the potentiometer is very very sensitive and then it reaches about 11 volts and the rest of the turn of the potentiometer does not change the output voltage.
Is there another resistor value needed for R1 maybe to have a less sensitive output and change the output trueout the whole turn of the potentiometer instead of only using the first half?
This is not a low dropout regulator. It needs about 2volt across to work properly.
You're lucky if you are getting 11volt with 12volt input.
Pot value should be 1892ohm for 12volt (theoretical) output.
If you're using a 10k pot, then that 12volt is reached at a ~19% turn of the pot.
The rest of the pot does not work (as you have discovered).
Leo..
Bringamosa:
Thanks Wawa, so bottomline is, this pot is not usable for this project? Putting a resistor in parallel with the pot will not work i guess?
No, you need a 2k pot for smooth control (assuming R1 is ~220ohm).
There is always 1.25volt across R1, so the ratio R1:pot must be 1.25:10.75 for 12volt (theoretical) output.
Leo..
If you are salvaging pots from an old radio, check them with your DMM before you use them to make sure that they are "linear taper" pots. Some pots used in radios for volume control (for example) are "audio (exponential) taper".
groundFungus:
If you are salvaging pots from an old radio, check them with your DMM before you use them to make sure that they are "linear taper" pots. Some pots used in radios for volume control (for example) are "audio (exponential) taper".
How do i check if they are linear? Hookup the DMM and then? turn the knob and see how resistance picks up?
Connect between the wiper (usually center terminal) and one end. Turn the pot from one extreme to the center. A linear pot will read 1/2 of the total pot resistance (5K for a 10K pot). An audio taper pot will be very different than 1/2 the total.
You also need to be cautious on how much current you use in the lower voltage range. The regulator has to do something with all that voltage it's dropping, so it sheds it as heat. Having a heatsink on it will help.
Well, the radio has one 100k pot and 2 40k pots so to bad. Will leave the variable output out. 5 bucks for item and shipment while not sure how often I'll be using something else then 3.3 5 or 12 volts is not worth ordering
A pot usually has a code on it.
A number (the resistance) and a letter.
A = audio = log
B = lin
C = reverse log
I gave you the formula/ratio to calculate the pot/resistor.
If you want to use that 10k pot for a 12volt range, then the resistor should be 10k/10.75*1.25= 1.16k
A 1k2 fixed resistor is close enough.
~220-240 ohm is usually used for the fixed resistor, because this regulator needs some load on it's output for stability. You might have to use some permanent >=5mA load from output to ground. Make sure you also use the recommended bypass caps on input and output (see datasheet).
Leo..
Leo thanks once again for your response and patience with noobs like me.
Code on the pot is B10K. That's all it gives. So, if i'm going to go with a 1k2 resistor i'll have to put some load on it and see how well it all works.
Nothing else to be found written on the pot so a google search will give me several datasheets but it looks like i can not find any of them telling me something about using caps. Will google on a bit maybe i'll find something eventually.