Re: E96 Series Computation
- From: The Phantom <phantom@xxxxxxx>
- Date: Wed, 13 Apr 2005 11:25:14 -0700
On Wed, 13 Apr 2005 12:51:57 GMT, Fred Bloggs <nospam@xxxxxxxxxx> wrote:
>
>
>The Phantom wrote:
>> On Thu, 17 Mar 2005 15:36:38 GMT, Fred Bloggs <nospam@xxxxxxxxxx> wrote:
>>
>>
>>>Robert Monsen wrote:
>>>
>>>
>>>>Ok, caching is a usually a good strategy.
>>>>
>>>>Sadly, I don't have a basic interpreter, so I guess I'm stuck with the
>>>>clumsy table lookup. It's ok, since I already have all the values typed
>>>>(actually cut and pasted) in here:
>>>>
>>>>http://home.comcast.net/~rcmonsen/resistors.html
>>>>
>>>>You have a cute routine, though. Curve fitting is fun. Somebody had alot
>>>>of fun doing it, I think. That "if 919, output 920" is hilarious.
>>>>
>>>>I'm hoping Bloggs posts the simpler scheme he claims to know.
>>>>
>>>
>>>If I said I have the answer then that means I have the answer.
>>>
>>>For the **E96** series the basic calculation is something like this,
>>>where Log's are base 10, and R is input value normalized to 100<=R<1000,
>>>IROUND is round-to-nearest-integer function:
>>>
>>>X=1.5*LOG10(R) % compute and scale log base 10 of R
>>>F=FRACT(X) % fractional part of X
>>>Y=INT(X) % integer part of X
>>>RSTD=IROUND(10^((Y+ IROUND(64*F)/64)/1.5)) % standard value output
>>>
>>>In words, for normalized R in range 10 to <1000 to be converted to
>>>standard value with hand calculator:
>>>
>>>1) compute logarithm base 10 and multiply by 1.5
>>>2) note integer portion and subtract off
>>>3) multiply remaining fraction by 64
>>>4) mentally round that result up/down to nearest integer and then divide
>>>by 64
>>>5) add back in original integer portion subtracted in step 2)
>>>6) divide the above by 1.5
>>>7) take antilog ( raise 10 to this power)
>>>8) round result to nearest integer
>>>
>>>It turns out that 920 is a genuine error in the original tabulation of
>>>E192, not a part of E96 and lesser precision series, and 919 should be
>>>the entry instead for E192, unless there is an ancient multi-point
>>>smoothing interpolation I am overlooking. This and a bit more logic to
>>>filter the RSTD calculated produces the correct results, and there is a
>>>very simple theoretical basis for this that I am certain follows the
>>>original thinking and computation. The original table was not generated
>>>using the above formula- the formula is a luxury not available to the
>>>original number crunchers and is an analog of their computation. The
>>>formulation for the other series, which is similar, and of any other
>>>series to be created with specified tolerance makes for an elegant
>>>generalization of which the E96 formula is just one instance.
>>>
>>>Some numbers from pi 31415927 say to spot check 314,141,415,159,592,927:
>>>
>>>314 X=3.7453945 F=0.7453945 Y=3 RSTD=316
>>>141 X=3.2238287 F=0.2238287 Y=3 RSTD=140
>>>415 X=3.9270721 F=0.9270721 Y=3 RSTD=412
>>>159 X=3.3020957 F=0.3020957 Y=3 RSTD=158
>>>592 X=4.1584826 F=0.1584826 Y=4 RSTD=590
>>>927 X=4.4506196 F=0.4506196 Y=4 RSTD=931
>>>
>>>These results are right on. The formula has been checked in other ways
>>>so that 1) standard values are always returned and 2) deviations from
>>>absolute closest standard value to arbitrary input R are hopelessly lost
>>>as noise compared to tolerance.
>>
>>
>> I guess I don't understand what your algorithm is supposed to do. Assume for example that
>> the design of a filter required an exact value of 835.3 ohms. Your algorithm says the
>> nearest standard value (in the E92 series) is 825 ohms. But wouldn't 845 ohms be closer?
>>
>
>Statistically, no, it is a +/- 1.2% split- the maximum error due to
>tolerance is 2.2% for either standard value selection and the nominal
>value advantage of the 845 over the 825 is only 0.07% or a fraction 1/30
>of that. You can't make a 0.01% tolerance final value out of 1% parts-
>choose another resistor line. If you want to delude yourself into
>thinking you have the closest value then you would extend the algorithm
>to compute:
>RSTD1=IROUND(10^((Y+ FLOOR(64*F)/64)/1.5)) and
>RSTD2=IROUND(10^((Y+ CEILING(64*F)/64)/1.5))
>Choose the one which minimizes |RSTD1,2-R|.
>Happy now?
You said in another post:
"I did post it under "E96 series computation", E192 and E48 left as
exercise for the student:"
It looks like the changes for E48 and E192 would be:
X=1.5*LOG10(R) % compute and scale log base 10 of R
F=FRACT(X) % fractional part of X
Y=INT(X) % integer part of X
RSTD=IROUND(10^((Y+ IROUND(32*F)/32)/1.5)) % standard value output
and
X=1.5*LOG10(R) % compute and scale log base 10 of R
F=FRACT(X) % fractional part of X
Y=INT(X) % integer part of X
RSTD=IROUND(10^((Y+ IROUND(128*F)/128)/1.5)) % standard value output
It would seem that a version for the E24 series would be:
X=1.5*LOG10(R) % compute and scale log base 10 of R
F=FRACT(X) % fractional part of X
Y=INT(X) % integer part of X
RSTD=IROUND(10^((Y+ IROUND(16*F)/16)/1.5)) % standard value output
But this version, when given an input value of 361 returns 348, which isn't a standard
value in the E24 series.
Do you have another algorithm which works for the E24 and E12 series?
.
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