varying resistance precisely!!!!!! - Page 1
Forum section: electronics / beginners

author: saipavan
author_rank: Contributor
timestamp: April 04, 2014, 03:55:30 am
content: for one of my project i require a variable resistor with precise control.unfortunately a 10 turn pot is not available inm my locality.i have a bunch of normal varying resistors(10k).is it possile to vary the resistance precisely using multiple variable resistors?If so,please suggest ,how can i do that..thanks........

author: FrankBuss
author_rank: Unknown
timestamp: April 04, 2014, 04:14:18 am
content: How precise do you need it? If you have 10 potentiometers, 10k each, and you need a 100k variable resistor, you could connect them all in series and it would be effectively a 10 turn pot.

author: pickle9000
author_rank: Super Contributor
timestamp: April 04, 2014, 04:26:51 am
content: or 100k 10k 1k (coarse, med, fine)The fewer the better

author: sleemanj
author_rank: Super Contributor
timestamp: April 04, 2014, 04:32:18 am
content: Quote from: saipavan on April 04, 2014, 03:55:30 amWhat is your locality?Do you need a panel pot, or a trimpot?3296 style trimpots and WXD3 panel mount potentiometers can be bought on ebay for a few dollars.

author: mariush
author_rank: Super Contributor
timestamp: April 04, 2014, 05:00:29 am
content: Why did you have to create another thread?It's about this right?https://www.eevblog.com/forum/beginners/controlling-an-lm317/msg417819/#msg417819If you didnt understand something, why dont you ask again?There's two simple rules you can take advantage of:Resistors in series : Total resistance = R 1 + R 2 + ... R nResistors in parallel : 1 / Total Resistance = 1 / R1 + 1 / R2 + 1 / R3For two resistors, the last formula becomes 1/R = 1/R1 + 1/R2 = (R1 + R2)/R1xR2 so then R = R1xR2 / (R1+R2)Now let's imagine you have a 1kohm resistor and that 10 kohm potentiometer. What happens when you put them in parallel?When the pot is at 0, the potentiometer acts as a jumper, shorting the 1kohm resistor, so R = 0 ohm.When the pot is at 100 ohm: R = 100 x 1000 / 1100 = 90.9 ohmWhen the pot is at 1k: R = 1000 x 1000 / 2000 = 500 ohmWhen the pot is at 2k: R = 2000 x 1000 / 3000 = 666 ohmWhen the pot is at 3k: R = 3000 x 1000 / 4000 = 750 ohm[..]When the pot is at 5k then R = 820 ohm , at 7k R will be 875 ohm , at 10k R will be 910 ohmSo just by putting a 1kohm resistor in parallel with a 10kohm potentiometer, you have a 0-910 ohm potentiometer. Now by turning the potentiometer you'll get a much finer resistance change.Now you can put this in series with a second 10k potentiometer and you basically have the rough adjust (change from a resistance value to another) and when you adjust the second potentiometer you can fine tune the resistance.For example, if you put both potentiometers at the middle of the rotation, you'd get 5000 ohm (rought adjust) + 820 ohm ( the 1kohm and 5000 ohm of pot in parallel gives you 820 ohm) so now you have a 5820 ohm resistor.Want an even finer adjustment? Do the math for a 470 ohm fixed resistor and see what happens.edit: corrected the error, thanks for mentioning it.

author: pa2ees
author_rank: Contributor
timestamp: April 04, 2014, 05:15:24 am
content: I agree with Mariush...However, just as a correction:Quote from: mariush on April 04, 2014, 05:00:29 amwhen the pot is a 0, it's a short, so R = 0 ohms.You could put 3 pots in parallel. If you used all the same values, you'd get 1/3 the max value of an individual pot, but a significant increase in fine tuning. However, the tuning might get a little messy.Try some of these suggestions out, and let us know which solution you liked the best.- Erik

author: saipavan
author_rank: Contributor
timestamp: April 04, 2014, 05:24:02 am
content: I will do the fine tuning with a resistor connected in parallel ,and another pot for coarse adjustment.Thanks for the help !!!!