what potentiometer should I go with to my power supply

[xReM1x]

http://www.lion.co.il/p/2017538#.VbghM_lVhjY
http://www.lion.co.il/p/2706034#.VbggpvlVhjY
http://www.lion.co.il/p/3525218#.VbggmflVhjY

i dont care about resistence, i use it to as a potentiometer. (0v to watever).

[TheMG]

Without knowing any details about your circuit it is impossible to tell which potentiometer will suit your application.

Post a schematic.

[xReM1x]

Without knowing any details about your circuit it is impossible to tell which potentiometer will suit your application.

Post a schematic.

thats the thing, any potentiometer will work just fine. it used as a voltage divider from a voltage refrenece the control the voltage.
every potentiometer will work for this, it just a matter of quality of the potentiometer.

[SL4P]

So they question is not which pot, but..
How much current don you want to drop in the divider, and
How much current do you want draw off the wiper?

As you note, any pot will do, but with varying success.

[xReM1x]

So they question is not which pot, but..
How much current don you want to drop in the divider, and
How much current do you want draw off the wiper?

As you note, any pot will do, but with varying success.

the voltage refrenece is 2.5V.
I would like to be variable from 0V-2.5V. the closer it is to 0V and 2.5V the better.
the wiper will go directly to an op amp. so no current at all. (expect like 1nA  )

[SL4P]

Well, before someone does the work for you...
Consider your pot directly spanning the 2V5 rail and 0V
The wiper as you know will give you full range swing between 0 and 2V5

So, let's say you want to give away 1mA in the divider, you can use ohms law to identify the required resistance across the two ends of the pot.  we'll leave that to you.

Assuming as you say, the input impedance of the op amp is negligible, you may discount that for this simple calculation.

One other thing you could add is a small cap to smooth the voltage from the wiper into the op amp.
e.g .01uF from the wiper to ground will flatten out most noise of the wiper travelling across the pot's carbon track.

[xReM1x]

Well, before someone does the work for you...
Consider your pot directly spanning the 2V5 rail and 0V
The wiper as you know will give you full range swing between 0 and 2V5

So, let's say you want to give away 1mA in the divider, you can use ohms law to identify the required resistance across the two ends of the pot.  we'll leave that to you.

Assuming as you say, the input impedance of the op amp is negligible, you may discount that for this simple calculation.

One other thing you could add is a small cap to smooth the voltage from the wiper into the op amp.
e.g .01uF from the wiper to ground will flatten out most noise of the wiper travelling across the pot's carbon track.

As for the cap, I alerady have 1uF at the wiper to ground.

So, let's say you want to give away 1mA in the divider, you can use ohms law to identify the required resistance across the two ends of the pot.  we'll leave that to you.

Assuming as you say, the input impedance of the op amp is negligible, you may discount that for this simple calculation.

why do I need to calculate it? or I dont? as I said it goes to an op amp. no needie to calculate anything.
so what pot will give me the most close voltage to 0V and the most close voltage to 2.5V? the vishay one look nice.

[SL4P]

The calculation is needed to determine what pot you need to use (or close enough)...
For example, if you are going to use a 50K pot across 2V5, that will identify the current flowing through the pot - which is important on several levels - as well as not burning out the pot !

Of course, you will use a preferred 'stock' value for the pot, which will determine the exact current & power dissipated in the pot at a given voltage.

Also, remember when you do select a specific pot, to buy a LINEAR taper (profile) version of the pot!

[xReM1x]

The calculation is needed to determine what pot you need to use (or close enough)...
For example, if you are going to use a 50K pot across 2V5, that will identify the current flowing through the pot - which is important on several levels - as well as not burning out the pot !

Of course, you will use a preferred 'stock' value for the pot, which will determine the exact current & power dissipated in the pot at a given voltage.

Also, remember when you do select a specific pot, to buy a LINEAR taper (profile) version of the pot!

I can get any pot value I want.
even if I choose 100R pot, I = V/R
2.5/100= 0.025A
I2R: 0.06W.
all the pots will go down to 1R so they are all basically the same and they will all fit. what I'm asking is what pot should I choose as for quality, precision, reabilty and such. leave the resistence and the calculation for me.
oh and yes ofcourse linear. (they all linear)

[codeboy2k]

I don't know the other brands.  The prices are so close, I would just say go with Vishay, it's well known and respected.

why do I need to calculate it? or I dont? as I said it goes to an op amp. no needie to calculate anything

This is arrogant thinking.  You should always do calculations to determine if the value you picked is right for your system.

As for which value to use, I would pick a higher resistance, rather than a lower one, since you're buffering it with an op amp then you don't need a low impedance source. The higher resistance will give you smoother control because you will have more resistance per each degree of rotation. For example, a 10 turn pot has 3600 degrees of rotation, so a 100 ohm pot has 100/3600 ohms per degree, or 27.8 millohms per degree. That's so small it's going to be hard to control and going to be jumpy.  On the other hand, a 20k pot has 5.6 ohms per degree, which you can control more precisely and get a smoother response from your pot.

You want to go high, but not too high, especially for a reference voltage, as the higher resistances have more Johnson noise (thermal noise).  For example, going up to 100k might seem like a good idea because it wastes very little power, but it could actually be bad for your system if it introduces noise that affects your system.  I would aim to keep the resistance as high as you can go (to minimize wasted power and get a better control) but to still keep the thermal noise below 1 LSB of your system. So determine what 1 LSB is for your system and then use that value and the thermal noise formula to determine your maximum resistance you can tolerate.



Where:
k = boltzmann's constant (1.374 x 10^-23 J/deg K
T = Absolute temperature (deg K)
R = Resistance in Ohms
B = System Bandwidth (in Hz)

You want to keep E_N below 1 LSB. Also remember that your temperature inside the box might be around 40 to 50 C , or 313 to 323K. 

as an example: a 16-bit system in 50kHz BW can easily use a 20k pot, but a 24-bit system would need to use no more than a 100 ohm pot.

There are many reasons why you need to calculate it. I've shown you at least two of them here.

[dom0]

You want to go high, but not too high, especially for a reference voltage, as the higher resistances have more Johnson noise (thermal noise).  For example, going up to 100k might seem like a good idea because it wastes very little power, but it could actually be bad for your system if it introduces noise that affects your system.

Not an issue with pots, since low bandwidths are no issue.

Very high values are still bad, though, due to leakage currents affecting output and noise pick-up (which usually is many orders of magnitude stronger than intrinsic noise).

In most cases anything between, say, 10 k? (to avoid high dissipation and other issues with low resistances) and 100 k? fits the bill.