| Author |
Message |
Stewart Pinkerton
Guest
|
 Posted:
Sun Oct 09, 2005 12:33 pm Post subject:
Re: Obtaining an accurate resistor |
|
|
On Sun, 09 Oct 2005 08:48:52 GMT, "Ian Iveson"
<IanIveson.home@blueyonder.co.uk> wrote:
| Quote: | Right. I wonder whether, if I decide that no resistors must be
discarded, the final bag of 500R pairs could be more accurate than
the bag of 1k as a result of the selective pairing?
|
Only if you are able to measure them accurately, and select
complementary pairs. Otherwise, you'll gain no improvement over the
bag of 1k resistors.
| Quote: | Again, I am
trying to focus on accuracy rather than tolerance. Working this out
needs more statistics than I have in the top of my head. I feel sure
they would be probably more accurate as a result of the series
connection, however.
|
Well, you're wrong. The statitstics are very simple, and given a
Gaussian distribution of the values within that 2% tolerance band, the
vaiation will be identical for both bags, hence a random pairing will
produce 1k resistors of statistically the same variation as the bag of
1k resistors.
| Quote: | If both the 500R and 1k resistors were rated at 1/2W, then, in
service,
the resistors in the series combination will be stressed less and
so
should be more accurate in service.
|
Only if stress is an issue in the application for which they are
required.
| Quote: | That is a good point. If the 500R were 1/4W, would the probability
of drift in service be the same for the pairs as for the 1/2W 1k?
|
Yes, as the thermal cycling would be identical in each case.
| Quote: | Lastly, if you string together resistors you will add reactance to
your
circuit which, depending on what you are doing, can affect
performance.
|
Eh? If (and only if) the original resistors have identical L and C
parasitics, what you'll do is double the inductance but halve the
capacitance for a serial pair connection. The term 'add reactance' is
much too vague, and not necessarily accurate.
| Quote: | Reactance is another good point. I am a bit finnicky, so I tend to
match for both inductance and resistance. Perhaps that's more than
just finnicky, on reflection. Maybe it's a fettish, considering the
insignificance of MF resistor reactance.
|
Depends on the bandwidth. Get into a few Megs, and these things become
significant. For audio use though, I'd agree that MF parasitics are
negligible for any reasonable resistance value.
| Quote: | Anyway, with resistors from
the same batch, I have found that inductance varies in keeping with
the variation in resistance. This is not true of resistors from
different batches, and certainly not of those of different
manufacture. Different resistances of the same manufacture and
series do not have similar inductance. Based on a small sample of
less than a hundred measurements, the 500R resistors would have less
inductance, but not half. So the series connection would be worse
than the 1k singles. Considering inductance is always dominant, that
would be a case for parallel connection.
|
Indeed it would, but you'd have to work out if this has any real
significance at 100kHz. Another advantage of parallel connection is
that if one resistor should fail open-circuit, the other one will
maintain some kind of connection. Resistors hardly ever fail
short-circuit, that would need an unusual construction such as a
bifilar wirewound. If your resistors are physically small, you should
also consider that a parallel connection doubles the capacitance.
--
Stewart Pinkerton | Music is Art - Audio is Engineering
|
|
| Back to top |
|
 |
Ian Iveson
Guest
|
| Posted:
Sun Oct 09, 2005 1:48 pm Post subject:
Re: Obtaining an accurate resistor |
|
|
"Joseph Meditz" wrote
| Quote: | Suppose you bought a bag of 100 1k resistors and 200 500R
resistors.
And, without measuring them, soldered together pairs of 500R
resistors.
You would then have two bags of 1k resistors both of which will
have
the same variation. Neither set will will be of tighter tolerance
than
the other.
|
Thanks Joe
It may be true that tolerance limits are not improved. That would
depend on whether they are all measured, and whether the quoted
tolerance is valid only at the time of measuring. Unless both of
these are true, the quoted tolerance is not absolute, but is rather
a statistical measure with some confidence limit. I am quite happy
to accept that tolerance is absolute, but haven't found definitive
evidence of this. In any event, I am not looking for better
tolerance limits, I am looking for more accurate resistors.
It is certainly not true that the 500R pairs would have the same
variation, unless you mean that the 1k resistors would have more
variation than the 500R to begin with, or that by "variation" you
mean "tolerance".
My contention is that averaging always reduces variance. This
certainly improves the chance of accurate matching, but it is a bit
less certain that it reduces the deviation from the nominal value.
| Quote: | However, there are 19,900 different pairs that can be created from
the
set of 200 500Rs. This sample space is equivalent to a bag of
19,900
1k resistors. Given enough resistors you are likely to find one
that
is very very close to 1k as well as one that is very very close to
1k
+/- 1%.
|
Right. I wonder whether, if I decide that no resistors must be
discarded, the final bag of 500R pairs could be more accurate than
the bag of 1k as a result of the selective pairing? Again, I am
trying to focus on accuracy rather than tolerance. Working this out
needs more statistics than I have in the top of my head. I feel sure
they would be probably more accurate as a result of the series
connection, however.
| Quote: | If both the 500R and 1k resistors were rated at 1/2W, then, in
service,
the resistors in the series combination will be stressed less and
so
should be more accurate in service.
|
That is a good point. If the 500R were 1/4W, would the probability
of drift in service be the same for the pairs as for the 1/2W 1k?
| Quote: | Lastly, if you string together resistors you will add reactance to
your
circuit which, depending on what you are doing, can affect
performance.
|
Reactance is another good point. I am a bit finnicky, so I tend to
match for both inductance and resistance. Perhaps that's more than
just finnicky, on reflection. Maybe it's a fettish, considering the
insignificance of MF resistor reactance. Anyway, with resistors from
the same batch, I have found that inductance varies in keeping with
the variation in resistance. This is not true of resistors from
different batches, and certainly not of those of different
manufacture. Different resistances of the same manufacture and
series do not have similar inductance. Based on a small sample of
less than a hundred measurements, the 500R resistors would have less
inductance, but not half. So the series connection would be worse
than the 1k singles. Considering inductance is always dominant, that
would be a case for parallel connection.
cheers, Ian |
|
| Back to top |
|
|
Ian Iveson
Guest
|
| Posted:
Sun Oct 09, 2005 4:15 pm Post subject:
Re: Obtaining an accurate resistor |
|
|
"Ruud Broens" wrote
| Quote: | See my measurements post - if you're interested, could send you
the data, comma delimited or so,
|
Yes please Rudy!
Even though I record all the values of resistors I use, I have used
less than 100 in my life, and I have no more than 9 of any value in
stock.
Even so, my experience selecting from sets of 10 is that they seem
to be in tight clusters centred off the nominal value. That would be
consistent with a production process using blanks with a constant or
slowly changing variation from the one used to set up the trimming
machine. Or just a hysterical trimming machine.
| Quote: | for this particular batch and brand (BC 0.6W 50 ppm 1% mf)
distribution is rapidly falling off towards the extremes, so
series
connecting does sharpen the resulting peak,
|
My contention is that this is true regardless of the original
distribution, simply because averaging always reduces variance.
| Quote: | so increases the
chance that the compound value is closer to the nominal series
value.
|
Doesn't quite necessarily follow. Certainly increases the chance
that it is closer to the mean of the original distribution, but our
measurements suggest that is not the same as the nominal value. Good
for matching, but there is still an outside chance in my mind that
it may not improve the probability of getting closer to nominal. It
is possible to contrive a distribution from which more series pairs
would be further from nominal than the singles, and contrive an
argument to suggest this would worsen the chance of getting closer
to the nominal value.
Consider a distribution that is symmetrical and bell-shaped, but
shifted to one side of the nominal value, with tails still fitting
entirely within the tolerance limits. Averaging by series connection
will reduce variance, and so emphasise the bell shape. The mean
variation from nominal will not change, so on average the series
connection will not result in being closer to nominal. Fewer will be
very close to nominal, but fewer will be very far. In other words,
they will be more consistently inaccurate.
Now consider the same distribution, skewed in the opposite direction
to the shift. Averaging will in this case move the *mode*
(distribution peak) away from the nominal value. You could then say
that the most common series value is less accurate than the most
common single value. Variance would still be reduced, and the mean
variation from nominal would still be the same I think (not sure
about this though).
| Quote: | For tube circuits, the voltage rating is much improved with
series connecting, eg. a 0.6 W resistor can handle some 300 V
without adverse effects, smaller resistors are usually 200 V tops.
50 ppm/K, 2ppm/Vr resistors only cost about 1 eurocent, in bulk,
|
Right. I must admit that I don't understand resistor voltage
ratings. The only way to get 1W through 1Mohm for example, is to put
1kV across it. So what does a 1W, 1Mohm, 500V spec mean? A resistor
that never gets very hot, I suppose.
| Quote: | so easy to just find out by gettin' some .
|
Next opportunity, I'll pick up a whole roll of surplus resistors,
just for the hell of it.
| Quote: | nb increasing the change of tighter tolerance is not the same
as equals tighter tolerance, so multiple series connecting
seems overkill to no avail.
|
To most of the world, overkill is what we seem to be here for!
thanks
cheers, Ian |
|
| Back to top |
|
|
Stewart Pinkerton
Guest
|
| Posted:
Sun Oct 09, 2005 5:32 pm Post subject:
Re: Obtaining an accurate resistor |
|
|
On Sun, 09 Oct 2005 14:13:46 GMT, "Ian Iveson"
<IanIveson.home@blueyonder.co.uk> wrote:
| Quote: |
"Stewart Pinkerton" <patent3@dircon.co.uk> wrote in message
news:kp2ik1lbcuj7u47f9kdgqoc8d4n2csup6e@4ax.com...
On Sun, 09 Oct 2005 08:48:52 GMT, "Ian Iveson"
IanIveson.home@blueyonder.co.uk> wrote:
Right. I wonder whether, if I decide that no resistors must be
discarded, the final bag of 500R pairs could be more accurate than
the bag of 1k as a result of the selective pairing?
Only if you are able to measure them accurately, and select
complementary pairs. Otherwise, you'll gain no improvement over
the
bag of 1k resistors.
Again, I am
trying to focus on accuracy rather than tolerance. Working this
out
needs more statistics than I have in the top of my head. I feel
sure
they would be probably more accurate as a result of the series
connection, however.
Well, you're wrong. The statitstics are very simple, and given a
Gaussian distribution of the values within that 2% tolerance band,
the
vaiation will be identical for both bags, hence a random pairing
will
produce 1k resistors of statistically the same variation as the
bag of
1k resistors.
No I don't think I am wrong, and I think you are confusing mean with
variance. The mean would stay the same, but the variance would
reduce.
|
No, the mean *and* the variance would stay the same. The mean is
simply the central value, and is independent of the number of items.
| Quote: | For a normal distribution, doesn't variance decrease in
proportion to root n, where n is the number of resistors in the
chain?
|
No, since they still lie within a 2% tolerance band.
| Quote: | Stewart, you have inspired the Ultimate Solution (cue phil...),
thanks.
If I get lots of resistors of different manufacture, I should get
close to a normal distribution centred on the nominal value. If I
then reduce variance by series connection, I get more accurate
resistors.
|
Alternatively, get some 910 ohm resistors, and put them in series with
some others between 82 and 110 ohms to get a value within 0.1% of
1kohm without much difficulty.
--
Stewart Pinkerton | Music is Art - Audio is Engineering |
|
| Back to top |
|
|
Ian Iveson
Guest
|
| Posted:
Sun Oct 09, 2005 7:13 pm Post subject:
Re: Obtaining an accurate resistor |
|
|
"Stewart Pinkerton" <patent3@dircon.co.uk> wrote in message
news:kp2ik1lbcuj7u47f9kdgqoc8d4n2csup6e@4ax.com...
| Quote: | On Sun, 09 Oct 2005 08:48:52 GMT, "Ian Iveson"
IanIveson.home@blueyonder.co.uk> wrote:
Right. I wonder whether, if I decide that no resistors must be
discarded, the final bag of 500R pairs could be more accurate than
the bag of 1k as a result of the selective pairing?
Only if you are able to measure them accurately, and select
complementary pairs. Otherwise, you'll gain no improvement over
the
bag of 1k resistors.
Again, I am
trying to focus on accuracy rather than tolerance. Working this
out
needs more statistics than I have in the top of my head. I feel
sure
they would be probably more accurate as a result of the series
connection, however.
Well, you're wrong. The statitstics are very simple, and given a
Gaussian distribution of the values within that 2% tolerance band,
the
vaiation will be identical for both bags, hence a random pairing
will
produce 1k resistors of statistically the same variation as the
bag of
1k resistors.
|
No I don't think I am wrong, and I think you are confusing mean with
variance. The mean would stay the same, but the variance would
reduce. For a normal distribution, doesn't variance decrease in
proportion to root n, where n is the number of resistors in the
chain?
Stewart, you have inspired the Ultimate Solution (cue phil...),
thanks.
If I get lots of resistors of different manufacture, I should get
close to a normal distribution centred on the nominal value. If I
then reduce variance by series connection, I get more accurate
resistors.
thanks, Ian
| Quote: |
If both the 500R and 1k resistors were rated at 1/2W, then, in
service,
the resistors in the series combination will be stressed less
and
so
should be more accurate in service.
Only if stress is an issue in the application for which they are
required.
That is a good point. If the 500R were 1/4W, would the probability
of drift in service be the same for the pairs as for the 1/2W 1k?
Yes, as the thermal cycling would be identical in each case.
Lastly, if you string together resistors you will add reactance
to
your
circuit which, depending on what you are doing, can affect
performance.
Eh? If (and only if) the original resistors have identical L and C
parasitics, what you'll do is double the inductance but halve the
capacitance for a serial pair connection. The term 'add reactance'
is
much too vague, and not necessarily accurate.
Reactance is another good point. I am a bit finnicky, so I tend to
match for both inductance and resistance. Perhaps that's more than
just finnicky, on reflection. Maybe it's a fettish, considering
the
insignificance of MF resistor reactance.
Depends on the bandwidth. Get into a few Megs, and these things
become
significant. For audio use though, I'd agree that MF parasitics
are
negligible for any reasonable resistance value.
Anyway, with resistors from
the same batch, I have found that inductance varies in keeping
with
the variation in resistance. This is not true of resistors from
different batches, and certainly not of those of different
manufacture. Different resistances of the same manufacture and
series do not have similar inductance. Based on a small sample of
less than a hundred measurements, the 500R resistors would have
less
inductance, but not half. So the series connection would be worse
than the 1k singles. Considering inductance is always dominant,
that
would be a case for parallel connection.
Indeed it would, but you'd have to work out if this has any real
significance at 100kHz. Another advantage of parallel connection
is
that if one resistor should fail open-circuit, the other one will
maintain some kind of connection. Resistors hardly ever fail
short-circuit, that would need an unusual construction such as a
bifilar wirewound. If your resistors are physically small, you
should
also consider that a parallel connection doubles the capacitance.
--
Stewart Pinkerton | Music is Art - Audio is Engineering |
|
|
| Back to top |
|
|
Ian Iveson
Guest
|
|
| Back to top |
|
|
Ian Iveson
Guest
|
| Posted:
Sun Oct 09, 2005 9:52 pm Post subject:
Re: Obtaining an accurate resistor |
|
|
pps
| Quote: | If I get lots of resistors of different manufacture, I should get
close to a normal distribution centred on the nominal value. If I
then reduce variance by series connection, I get more accurate
resistors.
|
....probably. |
|
| Back to top |
|
|
Ruud Broens
Guest
|
| Posted:
Sun Oct 09, 2005 11:37 pm Post subject:
Re: Obtaining an accurate resistor |
|
|
"Ian Iveson" <IanIveson.home@blueyonder.co.uk> wrote in message
news:gZ62f.84949$iW5.43451@fe3.news.blueyonder.co.uk...
: "Ruud Broens" wrote
:
: > See my measurements post - if you're interested, could send you
: > the data, comma delimited or so,
:
: Yes please Rudy!
consider it done :-)
: Even though I record all the values of resistors I use, I have used
: less than 100 in my life, and I have no more than 9 of any value in
: stock.
shame on you ;-)
:
: Even so, my experience selecting from sets of 10 is that they seem
: to be in tight clusters centred off the nominal value. That would be
: consistent with a production process using blanks with a constant or
: slowly changing variation from the one used to set up the trimming
: machine. Or just a hysterical trimming machine.
heh. emotions in power tools - no thanks, dangerous enuf already :-)
:
: > for this particular batch and brand (BC 0.6W 50 ppm 1% mf)
: > distribution is rapidly falling off towards the extremes, so
: > series connecting does sharpen the resulting peak,
:
: My contention is that this is true regardless of the original
: distribution, simply because averaging always reduces variance.
:
yep, the watering down effect
: > so increases the
: > chance that the compound value is closer to the nominal series
: > value.
:
: Doesn't quite necessarily follow. Certainly increases the chance
: that it is closer to the mean of the original distribution, but our
: measurements suggest that is not the same as the nominal value. Good
: for matching, but there is still an outside chance in my mind that
: it may not improve the probability of getting closer to nominal. It
: is possible to contrive a distribution from which more series pairs
: would be further from nominal than the singles, and contrive an
: argument to suggest this would worsen the chance of getting closer
: to the nominal value.
well, i'd expect a feedback process to target the nominal value
probably set up more accurately than my DVM - it has a basic
0.15 % accuracy, that i assume is a systematic error, repeatability
is much better, so could be all measurements were 0.15 % too low
:
: Consider a distribution that is symmetrical and bell-shaped, but
: shifted to one side of the nominal value, with tails still fitting
: entirely within the tolerance limits. Averaging by series connection
: will reduce variance, and so emphasise the bell shape. The mean
: variation from nominal will not change, so on average the series
: connection will not result in being closer to nominal. Fewer will be
: very close to nominal, but fewer will be very far. In other words,
: they will be more consistently inaccurate.
yes, i had that inprecisely formulated, thanks
: Now consider the same distribution, skewed in the opposite direction
: to the shift. Averaging will in this case move the *mode*
: (distribution peak) away from the nominal value. You could then say
: that the most common series value is less accurate than the most
: common single value. Variance would still be reduced, and the mean
: variation from nominal would still be the same I think (not sure
: about this though).
ok, needin' some coffee ... report back l8er maybe ;-)
:
: > For tube circuits, the voltage rating is much improved with
: > series connecting, eg. a 0.6 W resistor can handle some 300 V
: > without adverse effects, smaller resistors are usually 200 V tops.
: > 50 ppm/K, 2ppm/Vr resistors only cost about 1 eurocent, in bulk,
:
: Right. I must admit that I don't understand resistor voltage
: ratings. The only way to get 1W through 1Mohm for example, is to put
: 1kV across it. So what does a 1W, 1Mohm, 500V spec mean? A resistor
: that never gets very hot, I suppose.
:
: > so easy to just find out by gettin' some .
:
: Next opportunity, I'll pick up a whole roll of surplus resistors,
: just for the hell of it.
:
: > nb increasing the change of tighter tolerance is not the same
: > as equals tighter tolerance, so multiple series connecting
: > seems overkill to no avail.
:
: To most of the world, overkill is what we seem to be here for!
:
: thanks
:
: cheers, Ian
:
: |
|
| Back to top |
|
|
Joseph Meditz
Guest
|
| Posted:
Mon Oct 10, 2005 1:40 am Post subject:
Re: Obtaining an accurate resistor |
|
|
| Quote: | Suppose you bought a bag of 100 1k resistors and 200 500R
resistors.
And, without measuring them, soldered together pairs of 500R
resistors.
You would then have two bags of 1k resistors both of which will
have
the same variation. Neither set will will be of tighter tolerance
than
the other.
|
| Quote: | Thanks Joe
It may be true that tolerance limits are not improved. That would
depend on whether they are all measured, and whether the quoted
tolerance is valid only at the time of measuring. Unless both of
these are true, the quoted tolerance is not absolute, but is rather
a statistical measure with some confidence limit. I am quite happy
to accept that tolerance is absolute, but haven't found definitive
evidence of this. In any event, I am not looking for better
tolerance limits, I am looking for more accurate resistors.
|
| Quote: | It is certainly not true that the 500R pairs would have the same
variation, unless you mean that the 1k resistors would have more
variation than the 500R to begin with, or that by "variation" you
mean "tolerance".
|
To be precise, I mean the variance of the distribution. If you have
two independent distributions and add them together, the mean of the
sum is the sum of the means and the variance of the sum is the sum of
the variances. The variance is a measure of the spread of values. The
spread for the 1k resistors is +/- 10R, and the spread of the 500R
resistors is +/- 5R. So, when you connect two 500Rs together, i.e.,
sum their resistances, the spread of the sum is now +/- 10R.
| Quote: | My contention is that averaging always reduces variance. This
certainly improves the chance of accurate matching, but it is a bit
less certain that it reduces the deviation from the nominal value.
|
If I understand you, the you are actually referring to the variance of
the estimate of the mean.
A set of a very large number of randomly distributed resistors has a
mean value. Let's call this the true mean. Because of the large
numbers involved, you can't ever know the true mean, you can only
estimate it from a bag of resistors. The larger the bag of resistors,
the closer your estimate of the mean will get to the true mean.
Joe |
|
| Back to top |
|
|
Ian Iveson
Guest
|
| Posted:
Mon Oct 10, 2005 4:42 am Post subject:
Re: Obtaining an accurate resistor |
|
|
whoa there joe
apologies for no capitals or anything that requires a shift key.
knackered keyboard.
| Quote: | To be precise, I mean the variance of the distribution. If you
have
two independent distributions and add them together, the mean of
the
sum is the sum of the means and the variance of the sum is the sum
of
the variances. The variance is a measure of the spread of values.
The
spread for the 1k resistors is +/- 10R, and the spread of the 500R
resistors is +/- 5R. So, when you connect two 500Rs together,
i.e.,
sum their resistances, the spread of the sum is now +/- 10R.
you say you mean variance, but then you confuse variance with |
tolerance limits. they are very different measures. variance is the
sum of squares of differences from the mean of all members of the
population.
i got the formula wrong and suggested the ratio is 1/rootN. that is
for the standard deviation. the variance of the sample mean is
actually 1/N times the variance of the population. check last para
of
http://www.gps.caltech.edu/~tapio/acm118/Handouts/variance_estimation.pdf
it is from this relationship that you get...
| Quote: | If I understand you, the you are actually referring to the
variance of
the estimate of the mean.
A set of a very large number of randomly distributed resistors
has a
mean value. Let's call this the true mean. Because of the large
numbers involved, you can't ever know the true mean, you can only
estimate it from a bag of resistors. The larger the bag of
resistors,
the closer your estimate of the mean will get to the true mean.
that last sentence is very tricky. on average, the mean of a sample |
will be the same as the population mean regardless of sample size.
perhaps you mean that the variance of your estimates will reduce.
then i am right with you. if you take many bags of n resistors and
find the mean of each bag, then the variance of the bag means would
be 1nth of the variance of the population. Hence, if you connect the
resistors in each bag in series, which effectively averages them,
the resulting resistances would show a variance of 1nth of the
variance of the population.
if i string n resistors together to make a number of compound
resistors, then the standard deviation of the compound values would
be 1/rootn that of the singles.
standard deviation is the most widely accepted measure of variation
within a population.
note this is variation from the mean, not from the nominal resistor
value. if i take resistors from various manufactures, however, the
mean of the population would be very likely to be normal *and* the
mean would be very probably the same as the nominal value.
in other words, accuracy would, on average, improve by a factor of
root n, where n is the number of resistors in series.
thanks, that's a precise enough theory to partly test with rudy's
data. it will take me a while though...and its all from one
manufacture so i can only illustrate the effect of reducing
variation from the mean, rather than from the nominal value.
woo hoo. thanks
cheers, ian |
|
| Back to top |
|
|
Ian Iveson
Guest
|
| Posted:
Tue Oct 11, 2005 4:42 am Post subject:
Re: Obtaining an accurate resistor |
|
|
"Simon G Best" <s.g.best@btopenworld.com> wrote in message
news:diatfi$d9q$1@nwrdmz01.dmz.ncs.ea.ibs-infra.bt.com...
| Quote: | Hello!
I've read through the threads, and thought I'd respond :-)
I'll state now, up-front, that I'm taking 'tolerance' to mean
'certainly no more than [whatever the tolerance is] away from the
nominal value' (and that any resistors which fall outside their
stated tolerances are therefore duff, and to be regarded as
defective).
|
Fair enough. Still some cloud around Phil's contention that service
life is included in the quoted tolerance, which of course would mean
there would be tails on the final distribution regardless of
rejection at the factory. Does seem unlikely though so I will assume
as you have done.
| Quote: |
Ian Iveson wrote:
I need two accurate resistors.
[...]
The kind of resistors I want are only available at 1% tolerance.
I am only building one example of the circuit. I don't care about
anything except accuracy in both cases.
For each resistor, R1 and R2, am I better off:
a) Buying one resistor of the right nominal value.
For R1, it could be as much as 1% above or below the nominal value
(obviously). For R2, you could end up with one of the pair being
as much as 2.020202...% above the other (if one's 1% up, and the
other's 1% down).
|
OK. same as for singles.
| Quote: |
b) Buying X resistors each of 1/X th of the nominal value and
connecting them all in series.
That'll tend to 'home in' on the mean value of those resistors.
The bigger X is, the better (but you'll still occasionally be
unlucky). However, the mean value of a bunch of resistors is not
the same as the nominal value. Indeed, the mean doesn't even have
to be particularly close to the nominal for the resistors to meet
the tolerance requirements (the smaller the range of actual
values, the further from nominal the mean can be).
|
Yes. From the few we have measured. For example, Rudy's set of 159
56k 1% can be seen here
http://www.ivesonaudio.pwp.blueyonder.co.uk/rudysdata.gif
It just about conforms to a worst case scenario: it is offset in one
direction, and skewed in the other. Hence the mode will move away
from the nominal value. The mean will of course stay the same, and
the distribution variance will decrease, resulting in a set of more
accurately inaccurate strings.
| Quote: | So, for R1, this isn't much good. But, for the pair of R2s, this
will tend to be good, as long as each R2 in a pair is taken from
the same, well-mixed bunch of resistors. But it's still possible
to be unlucky, and get R2s in a pair that differ by up to about 2%
(from each other).
|
Yes. So accurate matching may be more likely, but if for a one-off
circuit that needs to be right, there is no substitute for selection
and trimming.
| Quote: |
c) Buying Y resistors of the correct nominal value and selecting
the best one.
That's much better than (b) for R1, but there's still no guarantee
of a close match. All Y resistors might be, say, 0.3% to 0.7%
below nominal.
|
OK
| Quote: |
d) Buying X * Y resistors, each of 1/X th of the nominal value,
and selecting the best series combination of X resistors.
That won't work for R1, as you'll just end up with Y networks
tending to be of the mean, rather than nominal, value. You might,
just possibly, be (very) lucky, but you're much more likely to
find a better match with (c).
|
Yes
| Quote: | For the R2s, assuming the X*Y resistors are in a well-mixed batch,
this is the best method, yet. It's like (b), but with lots of
networks to choose from. With Y such networks, you'd have Y*(Y -
1)/2 possible pairs :-)
|
Woohoo!
| Quote: | If I decide to produce the circuit in quantity, should that make
a difference to which I choose?
Well, for a single R1, (c) is the best of the options, even though
it's not guaranteed to work very well in practice (but would still
work better than the other options). Options (b) and (d) are just
lousy. So, for producing the circuit in quantity, you could either
settle for 1% plain and simple with option (a), or you could,
perhaps, automate the measuring and selecting process for option
(c).
|
Yes. I'm getting worried about this agreeing thing.
| Quote: | For pairs of R2s, it's either option (b) or (d), depending on
whether or not you want to do the measuring. With (d) you have to
measure and select, but with (b) you can just get on with it (as
long as the batches of resistors are well-mixed, of course). With
(b), though, there might be quality control issues, as it's always
possible for bad matches to occasionally occur by chance. But,
for both options, have a large X.
|
Yes. sigh...
But what about if I get many smallish batches from different makers
and mix them. Then they should be gaussian and with a mean of the
nominal value, considering the nominal value is the only target
value.
I have a bridge with a parallel port. Always wondered how to use it.
Might be interesting to find out. But my mind is already wandering
to valve matching, sepp, pse, and randomly selected massively
parallel mixed pentode and triode. RSMPMPT. Catchy eh?
| Quote: | Personally, though, I don't much like these kinds of approaches.
I'd much rather make the need for calibration a feature and a
selling point, complete with exciting panel meters and the like
:-)
|
And a bluetooth-enabled microprocessor-controlled "total precision
management" calibration system with optional manual override and a
"reset defaults" option.
Thanks Simon
cheers, Ian |
|
| Back to top |
|
|
Guest
|
| Posted:
Wed Oct 12, 2005 11:00 am Post subject:
Re: Obtaining an accurate resistor |
|
|
Ian Iveson wrote:
[quote]whoa there joe
apologies for no capitals or anything that requires a shift key.
knackered keyboard.
To be precise, I mean the variance of the distribution. If you
have
two independent distributions and add them together, the mean of
the
sum is the sum of the means and the variance of the sum is the sum
of
the variances. The variance is a measure of the spread of values.
The
spread for the 1k resistors is +/- 10R, and the spread of the 500R
resistors is +/- 5R. So, when you connect two 500Rs together,
i.e.,
sum their resistances, the spread of the sum is now +/- 10R.
you say you mean variance, but then you confuse variance with
tolerance limits. they are very different measures. variance is the
sum of squares of differences from the mean of all members of the
population.
i got the formula wrong and suggested the ratio is 1/rootN. that is
for the standard deviation. the variance of the sample mean is
actually 1/N times the variance of the population. check last para
of
http://www.gps.caltech.edu/~tapio/acm118/Handouts/variance_estimation.pdf
it is from this relationship that you get...
If I understand you, the you are actually referring to the
variance of
the estimate of the mean.
A set of a very large number of randomly distributed resistors
has a
mean value. Let's call this the true mean. Because of the large
numbers involved, you can't ever know the true mean, you can only
estimate it from a bag of resistors. The larger the bag of
resistors,
the closer your estimate of the mean will get to the true mean.
that last sentence is very tricky. on average, the mean of a sample
will be the same as the population mean regardless of sample size.
perhaps you mean that the variance of your estimates will reduce.
then i am right with you. if you take many bags of n resistors and
find the mean of each bag, then the variance of the bag means would
be 1nth of the variance of the population. Hence, if you connect the
resistors in each bag in series, which effectively averages them,
the resulting resistances would show a variance of 1nth of the
variance of the population.
if i string n resistors together to make a number of compound
resistors, then the standard deviation of the compound values would
be 1/rootn that of the singles.
standard deviation is the most widely accepted measure of variation
within a population.
note this is variation from the mean, not from the nominal resistor
value. if i take resistors from various manufactures, however, the
mean of the population would be very likely to be normal *and* the
mean would be very probably the same as the nominal value.
in other words, accuracy would, on average, improve by a factor of
root n, where n is the number of resistors in series.
thanks, that's a precise enough theory to partly test with rudy's
data. it will take me a while though...and its all from one
manufacture so i can only illustrate the effect of reducing
variation from the mean, rather than from the nominal value.
woo hoo. thanks
cheers, ian[/quote] |
|
| Back to top |
|
|
|
|
|
|
Profile
Log in to check your private messages
Log in