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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
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Kirt Blattenberger
Post subject:
Unread postPosted: Mon Mar 14, 2005 11:46 pm
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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 bi-modal. 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
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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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Kirt Blattenberger
Post subject: A common misconception about resistors...
Unread postPosted: Thu Apr 14, 2005 11:08 am
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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 bi-modal. 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 e-mail 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 sorting-KOA'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.

KOA-Speer 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 worst-case 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





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