Re: Creating a higer wattage resistor

Responding to John Fields:



I was speaking statistically. True, in your example the resultant value
would still be at the 5% tolerance value. However as unusual as finding a
resistor 5% high, finding two consecutive resistors 5% high would be a very
rare occurrence. It would be much more likely that the second resistor
would be closer to the marked value or even on the low side tending to
compensate for the 5% high side error of the first resistor.



I have requested the statistical distribution of resistors from resistor
companies but haven't been able to get it. Nevertheless I think it's
probably close to normal distribution centered around the marked value with
the tolerance set at the 3 standard deviation points. What I am saying is
that using two resistors of equal value and with the same tolerance will
move the 3 standard deviation points of the distribution down to 3.5% which
is due to the error of one resistor tending to cancel out the error of the
other.





Responding to Abstract Dissonance:



Yours is an interesting way of looking at the problem but you must remember
that actual value of the resistor is the average of all the little segments
you can divide that resistor into. Rather than having a distribution around
the marked value, all the little resistor segments have a distribution
around the actual value. It is the population of completed resistors that
are controlled to be distributed around the marked value.





"Abstract Dissonance" <Abstract.Dissonance@xxxxxxxxxxx> wrote in message
news:12f1ii3rhkg308@xxxxxxxxxxxxxxxxxxxxx

"Skeptic" <jreimer5@xxxxxxxxx> wrote in message
news:CiKHg.70288$zg.14486@xxxxxxxxxxxxxxxxxxxxxxxx
Once upon a time when surface mount parts were new, I had a design where
I needed a 2.5 ohm SMD resistor but the company didn't have any in stock
so I used four 10 ohm resistors in its place. No problem.

There is another benefit of using multiple parts to create a composite
value and that is tolerance. Using two 100 ohm, 5% resistors to make a
50 ohm load will reduce the tolerance to 3.5%. This reduction in
tolerance occurs regardless of whether the resistors are used in parallel
or series. The reduction in tolerance can be calculated by dividing the
tolerance of the individual part by the square root of the number of
parts used. (They must all be the same value and tolerance for this
formula to work.) Eg. 5% / sqrt(2) = 3.5%.


If this were true then all resistors would have 0% tolerance. Why? Because
resistance is "serial" in that you can think of a resistor as a composite
of several smaller resistors in series. It would be quite easy to get a
resistor with a tolerance that is arbitrarily small.





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