Resistor values and tolerance for checking multimeters

[001]

Hi! My stupid question

At my homebrew "lab" I have few used meters:
Instek GDM-8245 50000 counts ± (0.1% + 4) for 500 Ohm range and ± (0.1% + 2) for others
Fluke 87V  6000 counts ± (0.2 % + 1) for 600k range and  ± (0.6 % + 1) for all above
Fluke 117  6000 counts ± (0.9 % + 2) for 600 Ohm range and  ± (0.9 % + 1) for all above

Can I "check" all of them with NOS wirewound 0.25% 1ppm 1/2W resistors ? (Since 0.25% for 0-300k resistor values is lower than 0.2% for 600k multimeter RANGE for example)
What values I must use? 1/2 of range (I mean 300Ohm/3k/30k/300k for 6000 Count meter) or similar to range nominal (I mean 560Ohm/5k6/56k/560k for 6000 Count meter)?

[Edwin G. Pettis]

The normal minimum accuracy ratio for calibration is 5:1, preferential is 10:1, however if your requirements are not too high on absolute accuracy you can get away with a bit lower ratio, say 3:1 or 4:1, anything less than that will not provide enough accuracy to get your meters close to stated accuracy.  Since the best accuracy is ±0.1%, ±.025% or ±.02% resistors would work there, TCR <5 PPM/°C.  In the case of these meters, mid-range values should suffice to check accuracy.  You can certainly use resistors close to the full range value but make sure they are slightly lower than full scale.

[Cerebus]

That's a 'me too' to everything Edwin said. Those measurement ratios he's talking about are called "test uncertainty ratios" (TUR) by the metrology crowd; I mention it as it might be a useful search term for you.

Can I "check" all of them with NOS wirewound 0.25% 1ppm 1/2W resistors ? (Since 0.25% for 0-300k resistor values is lower than 0.2% for 600k multimeter RANGE for example)

I think from the way you've put that, that perhaps there is a slight misunderstanding of how those accuracy figures are normally presented and what they mean.

In figures like "± (0.6 % + 1)" the first figure is for the uncertainty in the reading, the second the uncertainty in the range. You may also see figures such as "± (0.6 % + 0.06%)" or "± (200ppm + 100ppm)". You can read "± (0.6 % + 0.06%)" as "± (0.6% of reading + 0.06% of range)". They typically represent the effects of gain (and linearity) errors and offset errors respectively.

Let's have a concrete example, drawn from the figures you've presented. Let's pick

Fluke 87V  6000 counts ± (0.2 % + 1) for 600k range and  ± (0.6 % + 1) for all above

and a resistor under test of 300k. Let's assume that the resistor is perfect, an exact 300k. On the 600k range the error would be "± (0.2 % + 1 count)" with 6000 counts representing 600k so 1 count 100 ohms, so that calculates out as "± (600 ohms + 100 ohms)" or simply "± 700 ohms". So with that 'perfect' 300k resistor our meter could read anything from 299,300 to 300,700 ohms.

To meet a 10:1 TUR for that range with a 300k resistor you'd need a resistor whose true value is 300k ±70 ohms, or 300k ± 0.023%. A marginal 4:1 TUR would call for a 300k ± 0.0583% resistor.

Note by the way, as Edwin hints, some meters have calibration routines that rely on testing against a 'near full range' resistor. My Keithley 197 specifies a 190 ohm calibration resistor for the 200 ohm range, and so on.

[001]

Thank You!
Awesome answers 

Can You talk me also: if I use TWO serial soldered resistors what is resultated tolerance?
For example 10k[± 0.5 %] + 100k[± 0.25 %] = 110k[± ?? %]

And what about overall tolerance if I solder it parallel? 9.(09)k[± ?? %]

[e61_phil]

Thank You!
Awesome answers 

Can You talk me also: if I use TWO serial soldered resistors what is resultated tolerance?
For example 10k[± 0.5 %] + 100k[± 0.25 %] = 110k[± ?? %]

And what about overall tolerance if I solder it parallel? 9.(09)k[± ?? %]

You can simply caculate the worst case error of such a circuit.

[Alex Nikitin]

Thank You!
Awesome answers 

Can You talk me also: if I use TWO serial soldered resistors what is resultated tolerance?
For example 10k[± 0.5 %] + 100k[± 0.25 %] = 110k[± ?? %]

And what about overall tolerance if I solder it parallel? 9.(09)k[± ?? %]

Now you are just been lazy, sorry to say. 

Cheers

Alex

[Cerebus]

Thank You!
Awesome answers 

Can You talk me also: if I use TWO serial soldered resistors what is resultated tolerance?
For example 10k[± 0.5 %] + 100k[± 0.25 %] = 110k[± ?? %]

And what about overall tolerance if I solder it parallel? 9.(09)k[± ?? %]

Now you are just been lazy, sorry to say. 

Cheers

Alex

Perhaps, perhaps not. It's easy to assume that maths that is 'obvious' to thee and me is also 'obvious' to others. Remembering back to school (long enough ago that we used clay tablets) percentages and fractions were two of the basics that some people just didn't 'get'. If you don't 'get' percentages then it's quite easy to assume that some special formula is needed, especially when you're already using a formula to calculate resistances in parallel. So it may not be laziness, it might be the need of a little remedial maths* (for which, obviously, this isn't the place).

* We're none of us immune. I can discuss closed loops over residue number fields, or the various kinds of countable and uncountable infinities  until your eyes glaze over, but I can't do simple integrals without a cheat sheet handy.

[Alex Nikitin]

Thank You!
Awesome answers 

Can You talk me also: if I use TWO serial soldered resistors what is resultated tolerance?
For example 10k[± 0.5 %] + 100k[± 0.25 %] = 110k[± ?? %]

And what about overall tolerance if I solder it parallel? 9.(09)k[± ?? %]

Now you are just been lazy, sorry to say. 

Cheers

Alex

Perhaps, perhaps not. It's easy to assume that maths that is 'obvious' to thee and me is also 'obvious' to others. Remembering back to school (long enough ago that we used clay tablets) percentages and fractions were two of the basics that some people just didn't 'get'. If you don't 'get' percentages then it's quite easy to assume that some special formula is needed, especially when you're already using a formula to calculate resistances in parallel. So it may not be laziness, it might be the need of a little remedial maths* (for which, obviously, this isn't the place).

* We're none of us immune. I can discuss closed loops over residue number fields, or the various kinds of countable and uncountable infinities  until your eyes glaze over, but I can't do simple integrals without a cheat sheet handy.

Well, I rarely take such a negative view, however in this case we have a question on the basic math, which only requires a bit of research on the net even if you forgot that bit (unlike the original question, which was about metrology methods and completely relevant in this section).

Cheers

Alex

[Cerebus]

Well, I rarely take such a negative view, however in this case we have a question on the basic math, which only requires a bit of research on the net even if you forgot that bit (unlike the original question, which was about metrology methods and completely relevant in this section).

That suggests that I've failed to make myself clear, my bad.

My suggestion is that some people develop a blind spot, around fractions and percentages, where they fail to see that it's just basic arithmetic. My other half goes into a blind panic when presented with irrational fractions, or fractions that don't have a power of 2 on the bottom, but if presented the same calculation as a series of multiplications and divisions can do it in her head. I'm sure that there must be variants on the word dyslexia but applied to fractions and percentages.

Add to that, when starting out, some formulae for things electronic, even basics like resistors in parallel, can seem like magical incantations because of a failure on those teaching (or writing books) to provide adequate explanation of their derivation. I certainly remember when it felt like that to me for some things - I got handed R_e = ²⁵/_{Ic(in mA)} without any decent explanation, let alone any mention of the word transconductance or our old friend V_T = ^kT/_q.

Combine the two and you have a perfect recipe for confusion. To you, to me, it's obvious that it's just basic maths, but some people may need a hand to see that before they realise that they are already equipped to  tackle the problem. Not seeing the forest for the trees and all that.

Of course I may be being too charitable and perhaps the OP really is a lazy sod but I'm prepared to give him the benefit of the doubt.

I do agree that this is one the OP ought to be able to work out for himself, he perhaps just needed that pointed out first.

So, 001, Alex and I both think you can probably work that out for yourself, even if you have to think for a few minutes first. If you get stuck, you can always ask for help.

[Alex Nikitin]

Well, I rarely take such a negative view, however in this case we have a question on the basic math, which only requires a bit of research on the net even if you forgot that bit (unlike the original question, which was about metrology methods and completely relevant in this section).

That suggests that I've failed to make myself clear, my bad.

My suggestion is that some people develop a blind spot, around fractions and percentages, where they fail to see that it's just basic arithmetic.

I understand, and if the TS would make at least an attempt at some calculations, like he did in the first post, I would have no objections  .

Cheers

Alex

[001]

2 Mr Nikitin
Why so negative? 

[Cerebus]

2 Mr Nikitin
Why so negative? 

The poor bugger lives in the Midlands, so he hasn't seen the sun since 1976, that's bound to make a man a bit grumpy. 

[Alex Nikitin]

2 Mr Nikitin
Why so negative? 

The poor bugger lives in the Midlands, so he hasn't seen the sun since 1976, that's bound to make a man a bit grumpy. 

I live in the sunny North West England, thank you very much! And at my age I have to be grumpy otherwise what's the point?!



Cheers

Alex

[Cerebus]

I live in the sunny North West England, thank you very much! And at my age I have to be grumpy otherwise what's the point?!

Sorry, I clearly misremembered.

Once you reach the age where:
     if you chase young women they don't giggle, they scream,
     when the beer doesn't taste as good as it used to, and the hangover lasts twice as long,
     you make a "ohh" noise when you bend down to pick up a dropped resistor from the floor,
there's only two pleasures left, being a curmudgeon and muttering about "T'youth of today! When I were a lad...".

[001]

I`m a dummy a little. sorry

Since DMM acuracy at DC volts far better when DC ampers is it good idea to measure drop across external shunt?
1Ohm 0.01% for example?

[azer]

Can You talk me also: if I use TWO serial soldered resistors what is resultated tolerance?
For example 10k[± 0.5 %] + 100k[± 0.25 %] = 110k[± ?? %]

And what about overall tolerance if I solder it parallel? 9.(09)k[± ?? %]

This is called error propagation. Have a look at http://ipl.physics.harvard.edu/wp-uploads/2013/03/PS3_Error_Propagation_sp13.pdf
For the series case it is sqrt((10k*0.005)^2 + (100k*0.0025)^2)/110k=0.0023 so 110k +-0.23%
For the parallel case it is  sqrt(0.005^2+0.0025^2+0.0023^2)=0.006 so 9.09k +- 0.6%

[001]

2 azer

Awesome link!
Thanx a lot! 

[Alex Nikitin]

Can You talk me also: if I use TWO serial soldered resistors what is resultated tolerance?
For example 10k[± 0.5 %] + 100k[± 0.25 %] = 110k[± ?? %]

And what about overall tolerance if I solder it parallel? 9.(09)k[± ?? %]

This is called error propagation. Have a look at http://ipl.physics.harvard.edu/wp-uploads/2013/03/PS3_Error_Propagation_sp13.pdf
For the series case it is sqrt((10k*0.005)^2 + (100k*0.0025)^2)/110k=0.0023 so 110k +-0.23%
For the parallel case it is  sqrt(0.005^2+0.0025^2+0.0023^2)=0.006 so 9.09k +- 0.6%

Unfortunately, that calculation is for measurement uncertainties and not correct for component tolerances. e61_phil answer earlier was correct. To calculate the tolerance of a network with several resistors you have to calculate maximum and minimum possible values and then calculate deviations from the nominal value, in percent. I.e. maximum possible value for the series case would be 10.05K + 100.25K = 110.3K, and the minimum possible value is 9.95K+99.75K = 109.7K, the deviation is +/-0.3K from 110K or about +/-0.273%

Cheers

Alex

[Cerebus]

Unfortunately, that calculation is for measurement uncertainties and not correct for component tolerances.

Philosophically, there's an interesting question because they are both tolerances and (rather poorly specified) uncertainties.

I doubt that you could say with any conviction that the uncertainties were independent (uncorrelated) and so meet the conditions for using root-sum-of-squares to combine their values. And furthermore I've never seen a resistor data sheet that said how many standard deviations the tolerance interval represents.

However you can take comfort in the fact that good quality resistor's values will follow something like a Gaussian distribution and so offer better overall combined tolerance than you'll get from a worst case calculation using the tolerance bands, but you're a fool if you rely on getting anything better than worst case figures.

[Alex Nikitin]

Unfortunately, that calculation is for measurement uncertainties and not correct for component tolerances.

Philosophically, there's an interesting question because they are both tolerances and (rather poorly specified) uncertainties.

I doubt that you could say with any conviction that the uncertainties were independent (uncorrelated) and so meet the conditions for using root-sum-of-squares to combine their values. And furthermore I've never seen a resistor data sheet that said how many standard deviations the tolerance interval represents.

However you can take comfort in the fact that good quality resistor's values will follow something like a Gaussian distribution and so offer better overall combined tolerance than you'll get from a worst case calculation using the tolerance bands, but you're a fool if you rely on getting anything better than worst case figures.

Hmm, tolerances are normally not a result of a natural distribution of values, components are sorted and tested according to these limits. It is perfectly possible for a batch of resistors to be all near the top allowable limit, or even worse, a 1% tolerance component can have all values between +/-0.5% from the nominal value removed, because these were sorted out for a better grade!

Cheers

Alex

[001]

So intersting   

Is it my stupid question or sciencific holywar?

[CatalinaWOW]

The various demands for precision (4:1 up through 10:1) are rules of thumb (good ones) for getting the uncertainties in your measurement device to be "negligible".  If you really want to do it right you have to dive into some fairly serious math.  Define your own definition of "negligible".  See what all of the errors add up to.  Compute the probability of exceeding your definition.  Decide if you can live with that.

Rules of thumb are used for a variety of reasons ranging from "can't do the math" to "the information needed to do the math isn't available".  The former is common, but correctable.  The latter is also common and much harder to deal with.  For example many people will do a "worst case analysis" based on tolerance values.  But tolerance values do not really mean that the component will not be outside that range, they mean that there is some probability that the part is within that range.  Each part type will have a probability distribution, and that distribution will have tails outside the tolerance range.  This is true for parts that are 100% tested and binned and doubly so for parts generated under some statistical control process.  Even the simplest information about these distributions is often difficult or impossible to get.  Things like what is the total probability that a part will be outside the range?  Is it 1% or 0.01%.  Distributions are harder to obtain.  Totally solving the problem may mean testing hundreds of items and generating your own statistics.  This only makes sense if you have a really important problem or voltnut OCD.

[Cerebus]

Hmm, tolerances are normally not a result of a natural distribution of values, components are sorted and tested according to these limits. It is perfectly possible for a batch of resistors to be all near the top allowable limit, or even worse, a 1% tolerance component can have all values between +/-0.5% from the nominal value removed, because these were sorted out for a better grade!

Hence my qualification of "good quality" resistors. If someone is any good at making resistors, say Welwyn or Vishay, they do actually aim to manufacture a, say, 10k 0.1% resistor and the resultant product will have a Gaussian distribution around some mean pretty close to 10k and a 3 sigma level at 0.1% out. Some of those resistor ranges don't have the wider tolerances available in that range's package style so they don't have a continuous spread of values that they could be binned into. I.e. a range will have only E96 values, but only 0.1% tolerance available in that style so there isn't a set of 1% bins to put the outliers in.

Once they've got the technology and experience in place from making precision resistors it tends to naturally transfer to the more run of the mill items. I've got a batch of Vishay VR37 200ppm 1/2W 10M 5% resistors that actually measure out to much better than 1%. That range also includes 10% and 1% tolerances (in the same case style) so if they were picking out the 1 percenters from the spread then I wouldn't see any in my nominally 5% batch. I've seen this more than once at the better end of the market.

Whereas your Chinesium 10k may well, as you say, get binned into 9k1 10%, 10k 5% (with no resistors between 9k9 and 10k1), 10k 1% (all the resistors missing from the 5% batch) and so on.

[Edwin G. Pettis]

What actual value spread you find in a batch of resistors depends on what the manufacturer does with the end product, they may provide the product as it comes off the assembly line with whatever normal variance of values are within the tolerance, some may pick out the tighter tolerances into bins, and depending on the process, you may actually get a relatively small spread of values within a given tolerance, depending on the width of the tolerance.  It all depends on the manufacturing processes and variables within.  Commodity resistors made in large batches may have a smaller distribution of values than smaller batches, it all depends on how they are made.  Setup costs really determine what is available, technically a resistor in most types can be made in any value possible, standardized values such as the E96 set were made to limit setup costs, any value that doesn't fit into one of those standard sets requires a special setup and is going to cost a lot more to manufacture even though it is the same resistor.  You'll notice when you ask Vishay (for example) for a special non-standard value (in a standard package), they'll quote you a lot higher price, well some of that is due to the way they make their resistors.

Most precision wire wound resistors are hand made as making them by automated machine is difficult and usually only for very large orders, particularly for tighter tolerances, loose tolerances roughly >0.5% can be machine made, particularly power resistors.  PWW resistors that are hand made do not have a particular distribution as it depends on the skill of the calibrator and how much time is spent calibrating the resistor, most manufacturers don't want to spend any more time than necessary calibrating a resistor, time is expensive.  You may get resistors distributed all over the tolerance band or sometimes in a much smaller tolerance band.  Most PWW manufacturers are only concerned that the resistors are within tolerance and not any particular distribution pattern.

Some of this has already been discussed in other threads, there is no simple rules here, it mostly depends on the manufacturer and how they are set up to handle resistors.