Pot Power Discussion

richard nineteenfortyone richard1941 at gmail.com
Sat Mar 4 20:28:59 MST 2017


I want to emphasize that for the pot you showed in the picture, the
temperature rise is not uniform.   10 watts over the entire winding is no
problem.  10 watts in the top 1% of the winding is probably lethal.   In
reliability engineering, the reliability of resistors is dependent on the
temperature rise, which is dependent on the power dissipation.

On Sun, Feb 26, 2017 at 10:28 AM, Richard J. Nelson <rjnelsoncf at cox.net>
wrote:

> Hello TomC,
>
>
>
> And, as always, the challenge is to make the tradeoffs of cost vs.
> performance.  A high quality (durable) pot is quite expensive.  My most
> useful test power pot is a 5 ohm, 500 watt, Omite Model R that would cost
> me $480 (Newark 2014 prices) to replace*(1)*.
>
>
>
>
>
>
>
> The practical physical considerations aside, the question is the
> calculation (methods and tools) of the values (current and power) involved
> with a  simple "user variable" resistor.  I use several “tools” in my
> studies of Calculating Pot Power that includes an HP-48 and Excel
> Spreadsheet with embedded VBA code - thanks to Namir’s optimization skills
> as presented at past HHCs.
>
>
>
> What is especially surprising to me is the lack of a good explanation of
> the problem in text books.  I am sure this is an old issue, and it should
> be documented somewhere, and is probably in one of the manufacturers old
> application notes.  *Anyone reading this that can provide a reference
> (see the two questions below) would be most helpful.  *
>
>
>
> *Q1*  How does the dissipated power change with pot rotation?  It turns
> out that the power always peaks in a narrow (25%) range.
>
>
>
> *Q2*  What is the optimum Load/Pot ratio?  Most texts are concerned with
> (limited to?) linearity and suggest that the load resistor not be any lower
> than 10 times the value of the pot.  If you follow this rule of thumb you
> would NEVER use a pot for power control because if its (power wasting)
> inefficiency.  When the power is plotted the load may actually be 1/10th
> (0.1) the pot value and the dissipated (wasted) pot power is still very
> reasonable.  In fact, a ratio as low as 0.05 (optimum?) is still below the
> sharp upward turn of the dissipated power-rotation curve.  Of course the
> maximum current rating of the pot must also be observed.
>
>
>
> The advantages of using a pot as a power control is simple and complete.
> Many modern elaborate and expensive electronics circuits for LED dimming do
> not offer the good control down to zero that a pot will.  I ran into this
> issue with microscope USB lighting and getting the “perfect” exposure using
> a video camera.
>
>
>
> Even using the internet is a research challenge on the pot power subject.
> I have “discovered” several interesting relationships involved and I think
> I now know the answers to the above two questions.  *These should be in
> the text books.*
>
>
>
> *What are your experiences?*
>
>
>
> *(1)*  I will email my description of (mounting) this pot (4 pp article
> titled *Five Ohm, 500 Watt Ohmite Rheostat*) on request.  It has a very
> useful “oldie but goodie”  Nomograph for calculating I, E, R, & P as
> Appendix A.
>
>
>
> X < > Y,
>
>
>
> Richard
>
> P.S.  It seems that one of the major pot design weaknesses is the wiper
> terminal current path as you noted.  Quality (expensive) pots have lots of
> detailed data for their specifications, but these start in the $5+ range in
> cost.  The variety of pots in the market place is truly astounding and
> selecting the right one for the job is a real challenge.
>
>
>
> *Other Mounted High Power Rheostat (pot) Examples*
>
>
>
>
>
>
>
> 25 ohms, 50 watts         50 ohms, 150 watts             100 ohms, 100
> watts
>
>
>
> -----Original Message-----
> From: hhc-bounces at lists.brouhaha.com [mailto:hhc-bounces at lists.
> brouhaha.com] On Behalf Of Tom C
> Sent: Sunday, February 26, 2017 6:41 AM
> To: hhc at lists.brouhaha.com
> Subject: Re: HHC Digest, Vol 94, Issue 12
>
>
>
> Richard:
>
>
>
> Certainly, a most interesting and practical challenge.  I've dealt with
> this issue many times.
>
>
>
> Another practical factor that is rarely considered is the actual contact
> resistance of the wiper on the potentiometer.  This is a practical
> limitation that may to show itself in the initial design, but will 'rear
> its ugly head' after some time in the field.
>
>
>
> Thanks!
>
> TomC
>
> --------------------------------------------
>
> On Fri, 2/24/17, hhc-request at lists.brouhaha.com <
> hhc-request at lists.brouhaha.com> wrote:
>
>
>
> Subject: HHC Digest, Vol 94, Issue 12
>
> To: hhc at lists.brouhaha.com
>
> Date: Friday, February 24, 2017, 2:00 PM
>
>  Send HHC mailing list submissions to
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>  Today's Topics:
>
>     1. Re: Help with a "simple" ohms law  problem (Monte Dalrymple)
>
>    2. RE: Help with a "simple" ohms law  problem (Richard J. Nelson)
>
>
>
>  ----------------------------------------------------------------------
>
>  Message: 1
>
> Date: Thu, 23 Feb 2017 22:19:18 -0800
>
> From: "Monte Dalrymple" <monted at systemyde.com>
>
> To: "Handheld Computing Conference discussion list"
>
>     <hhc at lists.brouhaha.com>
>
> Subject: Re: Help with a "simple" ohms law problem
>
> Message-ID: <7AE2348BA32E4E1AAA89D046CFC9B56E at jssmv21>
>
>  Hmm...  I didn't see the word "peak" anywhere in your  original email
>
>
>
>
>
>  ----- Original Message -----
>
> From: "Richard J. Nelson" <rjnelsoncf at cox.net>
>
> To: "'Handheld Computing Conference discussion list'"
>
>  <hhc at lists.brouhaha.com>
>
> Sent: Thursday, February 23, 2017 9:52 PM
>
> Subject: RE: Help with a "simple" ohms law problem
>
>
>
>  Hello Richard,
>
>
>
>
>
>  All data you need is provided.  See my orange response  text below.
>
>
>
>
>
>
>
>
>
>  From: hhc-bounces at lists.brouhaha.com
>
> [mailto:hhc-bounces at lists.brouhaha.com <hhc-bounces at lists.brouhaha.com>]
>
>  On Behalf Of richard nineteenfortyone
>
> Sent: Thursday, February 23, 2017 8:54 PM
>
> To: Handheld Computing Conference discussion list <hhc at lists.brouhaha.com>
>
> Subject: Re: Help with a "simple" ohms law problem
>
>
>
>
>
>  Tell me what R1 or R2 is, and I will tell you the  power.   I can do that
> as  well using this equation I derived.
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>  You only give R1+R2, which is insufficient data.  What  are we supposed
> to  guess for the pot setting?
>
>  ??
>
>  That is the critical element, and you are asked to determine  the peak
> power  dissipated.  You will arrive at the pot setting in the  process.
>
>
>
>
>
>  X < > Y,
>
>
>
>
>
>  Richard
>
>
>
>
>
>  On Thu, Feb 23, 2017 at 4:32 PM, Richard J. Nelson <rjnelsoncf at cox.net
>
>  <mailto:rjnelsoncf at cox.net <rjnelsoncf at cox.net>>
>
> > wrote:
>
>  Hello HHCers,
>
>
>
>
>
>  I ran across a simple voltage divider problem that seems to  be not well
> understood/discussed in all of the texts and references I  have consulted,
> the internet included.  I know we have some  technically/mathematically
> savvy  folks that attend HHCs so I am asking for inputs - except  from RH,
> NS, or  RP.
>
>
>
>
>
>  The problem is three resistors and calculating the power in  two of them,
> R1  + R2.
>
>
>
>
>
>
>
>
>
>
>
>
>
>  If you think that you understand this type of problem I  would love to
> hear  from you.  As a "test" I provide the following  values.  R1 + R2 =
> 2K.  Vin =
>
> 7 volts,  RL = 2K.  What is the power dissipated  by the pot - R1+R2?
>
>
>
>
>
>  If you calculated 0.026366174 watts you are correct and
>
> please contact
>
>  (help) me for a discussion.
>
>
>
>
>
>
>
>
>
>
>
>
>
>  Thanks.
>
>
>
>
>
>  X < > Y,
>
>
>
>
>
>  Richard
>
>
>
>
>
>
>
>  _______________________________________________
>
> HHC mailing list
>
> HHC at lists.brouhaha.com
>
> <mailto:HHC at lists.brouhaha.com <HHC at lists.brouhaha.com>>
>
> http://lists.brouhaha.com/mailman/listinfo/hhc
>
>
>
>
>
>
>
>  --
>
>
>
>  NOT sent from that galldurned iPhone with its crippled
>
> bluetooth and
>
>  authoritarian restrictions.
>
>
>
>
>
>
>
>  -----------------------------------------------------------
> ---------------------
>
>
>
>  > _______________________________________________
>
> > HHC mailing list
>
> > HHC at lists.brouhaha.com
>
> > http://lists.brouhaha.com/mailman/listinfo/hhc
>
>
>
>
>
>
>
>
>
>  ------------------------------
>
>  Message: 2
>
> Date: Fri, 24 Feb 2017 09:25:33 -0700
>
> From: "Richard J. Nelson" <rjnelsoncf at cox.net>
>
> To: "'Handheld Computing Conference discussion list'"
>
>     <hhc at lists.brouhaha.com>
>
> Subject: RE: Help with a "simple" ohms law problem
>
> Message-ID: <000001d28eba$9f997020$decc5060$@cox.net>
>
> Content-Type: text/plain;
>
> charset="us-ascii"
>
>  Hello Monte,
>
>  A component has a power rating.  If the power varies -
>
> a pot varies - you
>
> must consider that the user may set it at the position where
>
> the power is
>
> highest.  You need to know that point.  I describe
>
> this point/peak as a
>
> percentage of maximum.
>
>
>
>  What the classical texts  do not teach is that the
>
> power of a pot will vary
>
> with its setting, thus the word peak.  This
>
> relationship is the heart of the
>
> problem - which is ignored by most (I am looking for the
>
> exception) texts.
>
> This is a "simple" problem that gets very messy in its
>
> analysis..
>
>  The classical texts teach that you should not load the pot
>
> less than ten
>
> times its value.  They also teach that pots are not a
>
> "practical" way for
>
> power control.  Perhaps that is why this issue is swept
>
> under the carpet.
>
> My analysis shows that you may effectively load the pot to
>
> 1/10th its value
>
> and still not get into the steep part of the power/cost
>
> curve.
>
>
>
>  This is a highly calculation intensive "problem."
>
>
>
>  X < > Y,
>
>  Richard
>
>
>
>  -----Original Message-----
>
> From: hhc-bounces at lists.brouhaha.com
>
> [mailto:hhc-bounces at lists.brouhaha.com <hhc-bounces at lists.brouhaha.com>]
>
> On Behalf Of Monte Dalrymple
>
> Sent: Thursday, February 23, 2017 11:19 PM
>
> To: Handheld Computing Conference discussion list <hhc at lists.brouhaha.com>
>
> Subject: Re: Help with a "simple" ohms law problem
>
>  Hmm...  I didn't see the word "peak" anywhere in your
>
> original email
>
>
>
>
>
>  ----- Original Message -----
>
>  From: "Richard J. Nelson" <rjnelsoncf at cox.net>
>
> To: "'Handheld Computing Conference discussion list'"
>
>  <hhc at lists.brouhaha.com>
>
> Sent: Thursday, February 23, 2017 9:52 PM
>
> Subject: RE: Help with a "simple" ohms law problem
>
>
>
>  Hello Richard,
>
>
>
>
>
>  All data you need is provided.  See my orange response
>
> text below.
>
>
>
>
>
>
>
>
>
>  From: hhc-bounces at lists.brouhaha.com
>
> [mailto:hhc-bounces at lists.brouhaha.com <hhc-bounces at lists.brouhaha.com>]
>
>  On Behalf Of richard nineteenfortyone
>
> Sent: Thursday, February 23, 2017 8:54 PM
>
> To: Handheld Computing Conference discussion list <hhc at lists.brouhaha.com>
>
> Subject: Re: Help with a "simple" ohms law problem
>
>
>
>
>
>  Tell me what R1 or R2 is, and I will tell you the
>
> power.   I can do that as
>
>  well using this equation I derived.
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>  You only give R1+R2, which is insufficient data.  What
>
> are we supposed to
>
>  guess for the pot setting?
>
>  ??
>
>  That is the critical element, and you are asked to determine
>
> the peak power
>
>  dissipated.  You will arrive at the pot setting in the
>
> process.
>
>
>
>
>
>  X < > Y,
>
>
>
>
>
>  Richard
>
>
>
>
>
>  On Thu, Feb 23, 2017 at 4:32 PM, Richard J. Nelson <rjnelsoncf at cox.net
>
>  <mailto:rjnelsoncf at cox.net <rjnelsoncf at cox.net>>
>
> > wrote:
>
>  Hello HHCers,
>
>
>
>
>
>  I ran across a simple voltage divider problem that seems to
>
> be not well
>
>  understood/discussed in all of the texts and references I
>
> have consulted,
>
>  the internet included.  I know we have some
>
> technically/mathematically savvy
>
>  folks that attend HHCs so I am asking for inputs - except
>
> from RH, NS, or
>
>  RP.
>
>
>
>
>
>  The problem is three resistors and calculating the power in
>
> two of them, R1
>
>  + R2.
>
>
>
>
>
>
>
>
>
>
>
>
>
>  If you think that you understand this type of problem I
>
> would love to hear
>
>  from you.  As a "test" I provide the following
>
> values.  R1 + R2 = 2K.  Vin =
>
>  7 volts,  RL = 2K.  What is the power dissipated
>
> by the pot - R1+R2?
>
>
>
>
>
>  If you calculated 0.026366174 watts you are correct and
>
> please contact
>
>  (help) me for a discussion.
>
>
>
>
>
>
>
>
>
>
>
>
>
>  Thanks.
>
>
>
>
>
>  X < > Y,
>
>
>
>
>
>  Richard
>
>
>
>
>
>
>
>  _______________________________________________
>
> HHC mailing list
>
> HHC at lists.brouhaha.com
>
> <mailto:HHC at lists.brouhaha.com <HHC at lists.brouhaha.com>>
>
> http://lists.brouhaha.com/mailman/listinfo/hhc
>
>
>
>
>
>
>
>  --
>
>
>
>  NOT sent from that galldurned iPhone with its crippled
>
> bluetooth and
>
>  authoritarian restrictions.
>
>
>
>
>
>
>
>  -----------------------------------------------------------
> -----------------
>
> ----
>
>
>
>  > _______________________________________________
>
> > HHC mailing list
>
> > HHC at lists.brouhaha.com
>
> > http://lists.brouhaha.com/mailman/listinfo/hhc
>
>
>
>
>
>  _______________________________________________
>
> HHC mailing list
>
> HHC at lists.brouhaha.com
>
> http://lists.brouhaha.com/mailman/listinfo/hhc
>
>
>
>
>
>  ------------------------------
>
>  _______________________________________________
>
> HHC mailing list
>
> HHC at lists.brouhaha.com
>
> http://lists.brouhaha.com/mailman/listinfo/hhc
>
>  End of HHC Digest, Vol 94, Issue 12
>
> ***********************************
>
> _______________________________________________
>
> HHC mailing list
>
> HHC at lists.brouhaha.com
>
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>
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-- 
NOT sent from that galldurned iPhone with its crippled bluetooth and
authoritarian restrictions.
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