### Author Topic: Subtituting a 50k pot for a 10k pot?  (Read 7898 times)

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#### XOIIO

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##### Subtituting a 50k pot for a 10k pot?
« on: May 08, 2014, 07:40:37 pm »
Hey all, I have some step up/down converters which use 10 turn 10k trim pots, however I have some nice "precision" (5%) pots from an old Tektronix scope which are 50k, and would like to use those instead because they are a) easy to put in a case and access and b) have the nice double dial sort of design.

I am trying to figure out how I could substitute the 10k pot with these 50k pots and keep the same functionality. I imagine some op-amp circuit could do it, I'd like to do it with passive stuff that I might have lying around. Any tips? I'm thinking perhaps a resistor would be able to do it, though not 100% sure that might just work for increasing it.

edit: hmm, from what I see adding a resistor in parallel will decrease the value, I'll grab some and see how that works out.
« Last Edit: May 08, 2014, 07:57:32 pm by XOIIO »

#### SArepairman

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##### Re: Subtituting a 5pot for a 10k pot?
« Reply #1 on: May 08, 2014, 07:53:35 pm »
a parallel resistor will work but it will exhibit non linear behavior. you will have a curve going on. it may or may not be acceptable.

http://www.eevblog.com/files/seekPDF.pdf

somewhere in this book there are examples of trimpot modifications, I believe its in the first few chapters.

#### mariush

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##### Re: Subtituting a 10k pot for a 50k pot?
« Reply #2 on: May 08, 2014, 07:55:42 pm »
Put the 50kohm pot in parallel with a 12k resistor.

Rtotal = 1 / (1/R1 + 1/R2) = 1 / ( r1+rs)/r1r2 = r1r2/ (r1+r2)

At 50kohm , you have about 9670 ohm in total .  At 25k you have 8.2k , at 10k you have 5.45k in total, at 5k you have 3.5k in total, at 2.5k you have 2.06 kohm in total and so on.

It's not linear but if it's a 10 turn pot or something like that who cares.

#### XOIIO

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##### Re: Subtituting a 5pot for a 10k pot?
« Reply #3 on: May 08, 2014, 07:57:16 pm »
Ok wow, a 1k pot takes it from 49.8k down to .975k or so.

a parallel resistor will work but it will exhibit non linear behavior. you will have a curve going on. it may or may not be acceptable.

http://www.eevblog.com/files/seekPDF.pdf

somewhere in this book there are examples of trimpot modifications, I believe its in the first few chapters.

Damnit, you are right, it won't work at all

Put the 50kohm pot in parallel with a 12k resistor.

Rtotal = 1 / (1/R1 + 1/R2) = 1 / ( r1+rs)/r1r2 = r1r2/ (r1+r2)

At 50kohm , you have about 9670 ohm in total .  At 25k you have 8.2k , at 10k you have 5.45k in total, at 5k you have 3.5k in total, at 2.5k you have 2.06 kohm in total and so on.

It's not linear but if it's a 10 turn pot or something like that who cares.

It's for making a mini variable power supply, so it needs to be linear :/

#### XOIIO

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##### Re: Subtituting a 50k pot for a 10k pot?
« Reply #4 on: May 08, 2014, 08:02:56 pm »
This is the pot, anyone know of a 10k pot that would fit the dial on it? (would need to be the same style threading and rod for the dial and bolt that keeps it in)

on ebay and aliexpress there are 10k pots with the same style turn counters but they are at least 7 bucks each :/
« Last Edit: May 08, 2014, 08:07:37 pm by XOIIO »

#### mariush

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##### Re: Subtituting a 50k pot for a 10k pot?
« Reply #5 on: May 08, 2014, 08:11:04 pm »
It is linear in the sense that as you rotate the knob of the pot it you will get all the possible resistor values between 0 and around 9670 ohm, the only issue is that you will have to rotate the knob more to go from 8000 ohm to 9000 ohm, compared to going from 400 ohm to 1000 ohm for example.
You still get much better resolution than a simple 10k pot though.

See graph below... 0 to 50k on the x axis, the resulting resistance on y axis... 250 ohm steps ( you can do it yourself in excel/libreoffice calc.. put fixed resistor in first column, the pot on other column and put formula on third column then drag down the last two columns to have all values computed for you).

#### XOIIO

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##### Re: Subtituting a 50k pot for a 10k pot?
« Reply #6 on: May 08, 2014, 08:13:50 pm »
It is linear in the sense that as you rotate the knob of the pot it you will get all the possible resistor values between 0 and around 9670 ohm, the only issue is that you will have to rotate the knob more to go from 8000 ohm to 9000 ohm, compared to going from 400 ohm to 1000 ohm for example.
You still get much better resolution than a simple 10k pot though.

See graph below... 0 to 50k on the x axis, the resulting resistance on y axis... 250 ohm steps ( you can do it yourself in excel/libreoffice calc.. put fixed resistor in first column, the pot on other column and put formula on third column then drag down the last two columns to have all values computed for you).

Yeah I did notice that a bit on this one, I'll log it and graph it after the appointment I'm headed to, it might work out fine, I would kind of prefer tit o be a straight 10k but this is what I have on hand, and the newer ones seem to use plastic dials not aluminum.

#### scientist

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##### Re: Subtituting a 50k pot for a 10k pot?
« Reply #7 on: May 08, 2014, 08:29:38 pm »
This is the pot, anyone know of a 10k pot that would fit the dial on it? (would need to be the same style threading and rod for the dial and bolt that keeps it in)

on ebay and aliexpress there are 10k pots with the same style turn counters but they are at least 7 bucks each :/

Is $7 really that much? #### XOIIO • Super Contributor • Posts: 1592 • Country: ##### Re: Subtituting a 50k pot for a 10k pot? « Reply #8 on: May 08, 2014, 09:54:51 pm » This is the pot, anyone know of a 10k pot that would fit the dial on it? (would need to be the same style threading and rod for the dial and bolt that keeps it in) on ebay and aliexpress there are 10k pots with the same style turn counters but they are at least 7 bucks each :/ Is$7 really that much?

Not really, it's just having to wait for it and it being cheaper quality than the metal dials, when I have a potential solution right here.

Shipping stuff from ebay usually takes a good 15 days:/

#### XOIIO

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##### Re: Subtituting a 50k pot for a 10k pot?
« Reply #9 on: May 08, 2014, 10:09:23 pm »
I'd need to hook it to a motor for a nice fine steady reading, instead of by hand, but it doesn't work to well, it plummets pretty fast.

#### SArepairman

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##### Re: Subtituting a 50k pot for a 10k pot?
« Reply #10 on: May 09, 2014, 07:02:37 am »
you can get metal duodials on ebay for 12$http://www.ebay.com/sch/i.html?_odkw=helipot&_sop=15&_osacat=0&_from=R40&_trksid=p2045573.m570.l1313.TR3.TRC1.A0.H0.Xduodial&_nkw=duodial&_sacat=0 sometimes you can find a cheaper helipot that comes with the duo-dial (ive seen em go for like 10$ w/shipping), thought you wil have to wait there are no 10k ones around now

here is a nice 10k pot multiturn, helipot brand
http://www.ebay.com/itm/HELIPOT-R10K-L-25-NSPP-R10KL25-/310848076771?pt=BI_Control_Systems_PLCs&hash=item485ffd3be3

nice and cheap but no duo dial. im pretty sure if you wait one will come up. unless you want chinese for half the price..
« Last Edit: May 09, 2014, 07:06:50 am by SArepairman »

#### XOIIO

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##### Re: Subtituting a 50k pot for a 10k pot?
« Reply #11 on: May 09, 2014, 08:25:17 am »
you can get metal duodials on ebay for 12$http://www.ebay.com/sch/i.html?_odkw=helipot&_sop=15&_osacat=0&_from=R40&_trksid=p2045573.m570.l1313.TR3.TRC1.A0.H0.Xduodial&_nkw=duodial&_sacat=0 sometimes you can find a cheaper helipot that comes with the duo-dial (ive seen em go for like 10$ w/shipping), thought you wil have to wait there are no 10k ones around now

here is a nice 10k pot multiturn, helipot brand
http://www.ebay.com/itm/HELIPOT-R10K-L-25-NSPP-R10KL25-/310848076771?pt=BI_Control_Systems_PLCs&hash=item485ffd3be3

nice and cheap but no duo dial. im pretty sure if you wait one will come up. unless you want chinese for half the price..

Yeah, I do have the duodials though, it just depends if they fit that one. For now I may just hit up the local store and grab a regular old 10k pot. It is just going to be a cheap little variable power supply that takes up less space than my proper one.

#### SArepairman

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##### Re: Subtituting a 50k pot for a 10k pot?
« Reply #12 on: May 09, 2014, 08:47:12 am »
you can get metal duodials on ebay for 12$http://www.ebay.com/sch/i.html?_odkw=helipot&_sop=15&_osacat=0&_from=R40&_trksid=p2045573.m570.l1313.TR3.TRC1.A0.H0.Xduodial&_nkw=duodial&_sacat=0 sometimes you can find a cheaper helipot that comes with the duo-dial (ive seen em go for like 10$ w/shipping), thought you wil have to wait there are no 10k ones around now

here is a nice 10k pot multiturn, helipot brand
http://www.ebay.com/itm/HELIPOT-R10K-L-25-NSPP-R10KL25-/310848076771?pt=BI_Control_Systems_PLCs&hash=item485ffd3be3

nice and cheap but no duo dial. im pretty sure if you wait one will come up. unless you want chinese for half the price..

Yeah, I do have the duodials though, it just depends if they fit that one. For now I may just hit up the local store and grab a regular old 10k pot. It is just going to be a cheap little variable power supply that takes up less space than my proper one.

don;'t forget the cheap option of putting a 10k pot in series with a smaller pot for fine adjustment.

if yoou don't mind multiple knobs it could be alot cheaper, lots of test equipment does this

i have even seen "coarse" adjust switches (rotary) with a fine adjust pot, this method is more stable (as your switch connected resistors will have likely less drift then a coarse adjust trimpot).
you could really ghetto-cheap it up using coarse adjust 10 position dip switch + single turn 1k ohm pot if your application allows it.

« Last Edit: May 09, 2014, 08:54:30 am by SArepairman »

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