

Tolerance of Parallel Combination of Resistors  RF Cafe Forums

Guest Post subject: Tolerance of parallel combination of resistors
Unread postPosted: Sun Mar 13, 2005 12:23 am
I'm trying
to combine two resistors in parallel to make a non standard value
which is 2043.73 ohms. This is of course possible with the combination
of (5K6, 3K3) or the combination of (2K2, 27K). The second combination
would of course give me a closer match (in ideal situation) but
what I'm wondering now is the tolerance of the combination. Which
one would give me the better tolerance or in other words which one
is more likely to be closer to the desired value?(Consider E12,
5%values).
Could somebody tell me the general solution to
this problem & also the best practical choice?
Your help
is much appreciated An Electronics student [/list]
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Kirt Blattenberger Post subject: Unread
postPosted: Mon Mar 14, 2005 11:46 pm Offline Site Admin
User avatar
Joined: Sun Aug 03, 2003 2:02 pm Posts: 308
Location: Erie, PA Greetings:
Let's work it out.
Nominal 5600 ohms  3300 ohms = 2076.4 ohms
5% Max
Values 5880 ohms  3465 ohms = 2180.2 ohms
This represents
2180.2/2076.4 = 1.05 = +5%
5% Min Values 5320 ohms 
3135 ohms = 1972.6 ohms
This represents 1972.6/2076.4 = 0.95=
5%
So the answer is that the tolerance of the parallel combination
equals the tolerance of the individual components, as long as all
components have the same tolerance range. This holds for series
combinations, too.
 Kirt Blattenberger :smt024
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Guest Post subject: Unread
postPosted: Tue Apr 12, 2005 9:33 am
Thanks Kirt. Seeing
the problem worked out always helps. Even though your example seems
obvious, it helps to have an expert proclaim it to be so. :D
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Mike Harris Post subject: Unread
postPosted: Tue Apr 12, 2005 6:49 pm
Its nice to have an
answer, the only problem is that in this case the answer is not
correct. The reason that this particular calculation turned out
to give the same answer is only because you assumed that both resistors
were exactly 5% high, a bad assumption in real life because the
tolerance will actually distribute *randomly* with a *maximum* (three
sigma) distribution of 5%.
When you combine two random variable
you compute them as the square root of the sum of squares. So in
general when you combine resistors that have the same tolerance
in parallel the total tolerance improves by a factor of the square
root of two. Likewise, the tolerance for three resistors in series
would be found by dividing by the square root of 3, etc. etc.
Another point though is that 5% resistors don't usually distibute
randomly around their mean, but are instead bimodal. This is because
the 1% resistors have been removed from the lot. Therefore the closest
value to the specified resistor value you will ever see is the resistor
value, plus or minus 1%. So for 5% resistors the answer will come
out only a little better than 5%.
It is true though that
if you put two 1% resistors in parallel you will get an effective
tolerance of 1%/sqrt(2)=0.707%. Hope that helps.
 Mike
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Guest Post subject: Unread postPosted:
Tue Apr 12, 2005 6:52 pm
Oh, I forgot to mention. Series
resistors don't have the same tolerance as the individual resistors
either. In this case you would *multiply* the tolerance by the number
of resistors rather than dividing it. Tolerance gets worse than
the individual resistors when you combine them in series and better
when you combine them in parallel.
 Mike
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Guest Post subject: Unread postPosted: Tue Apr
12, 2005 6:54 pm
Anonymous wrote: Oh, I forgot to mention.
Series resistors don't have the same tolerance as the individual
resistors either. In this case you would *multiply* the tolerance
by the number of resistors rather than dividing it. Tolerance gets
worse than the individual resistors when you combine them in series
and better when you combine them in parallel.
 Mike
Whoops that should be multiply by *the square root* of the
number of resistors......
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Kirt Blattenberger
Post subject: Unread postPosted: Tue Apr 12, 2005 9:47 pm
Offline Site Admin User avatar
Joined: Sun Aug
03, 2003 2:02 pm Posts: 308 Location: Erie, PA Greetings:
I have to defend my original assertion. I assumed worst case
(at the extreme of the rated 5% tolerance) to simplify the explanation
of how the total tolerance of a parallel combination of two resistors
is related to the tolerance of the individual resistors. That is
the question asked by the poster concerning the combination of two
resistors.
The parallel case has been worked out. Here is
the series case:
Nominal 5600 ohms + 3300 ohms = 8900
ohms
5% Max Values 5880 ohms + 3465 ohms = 9340 ohms
This represents 9340/8900 = 1.05 = +5%
5% Min Values
5320 ohms + 3135 ohms = 8455 ohms
This represents 8455/8900
= 0.95= 5%
So as you can see, it holds for the series case,
too.
 Kirt Blattenberger :smt024
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Profile
Kirt Blattenberger Post subject: A common
misconception about resistors... Unread postPosted: Thu Apr 14,
2005 11:08 am Offline Site Admin User avatar
Joined:
Sun Aug 03, 2003 2:02 pm Posts: 308 Location: Erie, PA
Greetings:
Mike Harris wrote: Another point though is
that 5% resistors don't usually distibute randomly around their
mean, but are instead bimodal. This is because the 1% resistors
have been removed from the lot. Therefore the closest value to the
specified resistor value you will ever see is the resistor value,
plus or minus 1%. So for 5% resistors the answer will come out only
a little better than 5%.
I knew that the above statement
was not accurate, but could not locate information on the Web to
back it up. It is a very common misconception, however, and Mike
is not alone in his assumption (it seems reasonable). So, I wrote
to a very prominent resistor manufacturer, KOA Speer, and asked
the question. I also wrote to Dale, but have not yet received a
response.
My email dialog with KOA Speer follows.
Greetings:
Is it true that the way resistors are separated
into tolerance ranges for sale is by measuring each one and putting
them in bins according to whether they fall within 0% to 1%, 2%
to 5%, 6% to 10%, etc? If so does it mean that the probability of
finding a value within 1% of nominal in a batch of 5% resistors
is nearly zero, and that the value distribution for 5% resistors
is typically bimodal?
Thank you for your time.
Sincerely,
Kirt Blattenberger
KOA Speer representative's response.
Good morning:
Please note that there is no sortingKOA's
parts are made to the actual tolerance.
I hope this information
helps you. If you have any more questions, please let us know
Best regards Dawn McGriff
So, the myth can end here.
:D
 Kirt Blattenberger :smt024
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Guest Post subject: Tolerances of resistors Unread
postPosted: Thu Apr 14, 2005 5:20 pm
The observation that
(certain) resistor manufacturers sorted their product for tight
tolerances came before the Deming quality revolution. I've seen
it, but won't name company names because the guilty have changed
their ways.
KOASpeer is a Japanese company, and the Japanese
took up Deming well before we in the US did. It's not surprising
that they don't screen, as they should have their processes under
statistical control.
For military designs, I was taught to
do worstcase analysis  what Kirt did, above, and quite correctly.
For commercial designs, connecting many resistors does tighten the
standard deviation (assuming a truncated Gaussian distribution of
values)  but not the worst case. So if you're willing to test and
throw out some units, it might be (slightly) cheaper to use two
resistors. At current prices for 1% resistors, it'll be hard to
convince me that the test and yield impacts of multiple resistors
is at all worth it!
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guest2 Post subject:
Tolerance of Resistors Unread postPosted: Fri Apr 15, 2005 7:17
am
Perhaps capacitors are different, but a few months back
I spoke with someone at ATC about this very subject and he told
me that they do indeed pick out the tight tolerance parts. Additionally,
Coilcraft inductors are manufactured to a tighter tolerance, not
picked So it seems that different manufacturers approach this
in different ways
Posted 11/12/2012



