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] Top 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 Top Profile
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 Top 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 Top 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 Top
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...... Top
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 Top 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
Top Profile 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! Top
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
