Does anybody know the reason, why this was made like this?

Yes, at least partly.

Accuracy and/or precision are often part of the initial requirements for a circuit or system. As such, it is easy to see that a two-resistor voltage divider made with 1% tolerance parts will satisfy a requirement for a 5% accurate result without effort, no matter where in the resistor value range each resistor falls. Because specs are in percentages, it makes sense that part values are incremented in percentages.

Also, unless you have a meter that is 10 times more accurate than the tolerance rating of the resistors you are using AND hand-test and select parts, your initial premise is incorrect. If you series or parallel a large value resistor with a small value resistor of the same tolerance, the result will have the same tolerance. The calculated value might appear more precise on paper, but in fact it is not.

In post #1 you mention a 10.02 ohm resistor. That is a very specific value, and implies that it is discernable as different from a 10.01 or 10.03 ohm resistor. That is an accuracy of less than 1 part in a 1000. Unless you have a 4-1/2- or 5-digit ohmmeter, you cannot know that your resulting combination actually is 10.02 ohms. And no standard 1% or 0.1% tolerance parts will hold the resultant value over a temperature swing of only 10 degrees.

ak

 

Does anybody know the reason, why this was made like this?
Reminds my of the seemingly arbitrating drill bit numbering.

I told you the reason. It's a set of values chosen so that any value desired is within the tolerance limit of one of the standard values.

If you have a standard value V1, with a fractional tolerance T, then the tolerance band for resistors marked as that value runs from

V1(1-T) to V1(1+T)

For the next value up, call it V2, it will be

V2(1-T) to V2(1+T)

In order not to have a gap or overlap, you want

V1(1+T) = V2(1-T)

V2/V1 = (1+T)/(1-T)

The next value up, V3, would have this same ratio, so

V3/V1 = [(1+T)/(1-T)]^2

In general

V(k+n)/V(k) = [(1+T)/(1-T)]^n

What does n have to be in order of V(k+n)/V(k) = 10 ?

[(1+T)/(1-T)]^n = 10

and solve for n

log10([(1+T)/(1-T)]^n) = log10(10)

n·log10([(1+T)/(1-T)]) = 1

n = 1/log10((1+T)/(1-T))

At one time there were 50% tolerance parts, so for those parts you had

n = 1/log10([(1+0.5)/(1-0.5)])
n = 1/log10(1.5/0.5) = 5.68

For T = 20% (the default tolerance for a long time), this is

n = 1/log10([(1+0.2)/(1-0.2)])
n = 1/log10(1.2/0.8) = 2.096...

You need to round up to 3 otherwise you would leave a gap. The resulting values constitute the E3 sequence.

So the ideal values would be related to the next value by a factor of 10^(1/3) = 2.1544...

This would make the values in the series

10.000
21.544...
46/416...
100.000

If we round 21.544 up to 22, then the next value becomes 47.397..., which we round down to 47. So for 50% tolerance we have

V1 = 10
V2 = 22
V3 = 47
V4 = 100 (next decade)

If we next look at T = 20%, we find that we need 6 values. So this will be the E6 series.

For a variety of reasons, we want to keep the E3 series values as a subset of the E6, so we choose the values to be the geometric means of adjacent E3 values.

sqrt(10*22) = 14.832...
sqrt(22*47) = 32.156...
sqrt(47*100) = 68.557...

For reasons I've never been able to track down, they chose slightly different values. I can thing of a few possible reasons, but I don't know if any of them are the actual reason. The values they chose were

E6 = {10,15,22,33,47,68}

You can proceed with this same process for the E12 (10%) and the E24 (5%) series.

For the E48 (2%) series they chose to not require the E48 series be a subset. This is likely because doing so up to this point resulted in some gaps that were too wide to be covered by a 2% tolerance just by adding a resistor at the geometric mean.

They also, for reasons I don't know, allowed some gaps. To avoid gaps, the number of values in the series should be at least 58. They chose to go with 48. This could be because they wanted to continue the trend of doubling the number of standard values in each succeeding series, but I don't know that for sure.

 

I don't deny the sincerity and application of the author, but in these situations it is much smarter to design your circuit so that you don't need to work to that level of precision. Heavens above, the PCB traces will be a bigger source of error.

 

Of course you still will have the tolerance of the resistors themselves, like 1%,
but that is he max – if you are lucky, some of the tolerances will cancel each other.

Last not least, if you do have a really good Ohm meter, you can measure the actual result after each step
and can use that as the input to the next iteration.

Here are flaws in your methodology:

1) Tolerances don't cancel. They add.
2) A resistance of 10.02Ω is begging for 0.1% accuracy. You need an ohmmeter that is better than that and most are not.
The resolution and accuracy of a good ohmmeter at that range is about 0.1Ω.
3) The resistance of the test leads in the ohmmeter alone is going to be of that order.
4) The resistance in your circuit connections is going to exceed 0.1Ω.

Thanks to @RobertPink for pointing this out.

 

1) Tolerances don't cancel. They add.

Not the case.

If I take 100 100 Ω, 1% resistors and put them in series, I do NOT end up with a 10 kΩ resistance with a 100% tolerance.

IF the distribution of resistors were independent of each other and gaussian with zero mean error, then they would tend to cancel out. In that case, putting those hundred resistors in series would yield a 10 kΩ resistor with a 0.1% tolerance. This is how ensemble assemblies work, including using an ensemble of atomic clocks to get universal time to a higher degree of accuracy and precision than any one clock can achieve on it's own.

But resistor value distributions are neither independent nor gaussian. They are quite far from it. If the resistors come from the same lot then there will tend to be a systematic bias in the mean, which by itself is enough to defeat building a significantly better effective resistance using an ensemble. Plus, sorting practices wreak havoc on the distributions from a statistical standpoint.

 

If you want a 10.02 ohm resistor, measure some 10 ohm resistors until you find one (if you've got a meter good enough).

My pet hate: circuits that use E96 resistors to drive LEDs .

 

If you want a 10.02 ohm resistor, measure some 10 ohm resistors until you find one (if you've got a meter good enough).

My pet hate: circuits that use E96 resistors to drive LEDs .

It doesn't really matter. One company I worked for stocked all 1% resistors when 5% or 10% would be just as fine.
It was cheaper to stock all 1% than to stock 1%, 5%, and 10% resistors.

 

Thank you WBahn!

Good Info.

BTW, I worked for 7 years developing electronic analog circuits for a well known medical company in my last life, unfortunately I got distracted with product management and now consider myself a paper pusher...
(I sometimes miss the old days).

I know very well that for a mass produced system you want to to be as tolerance independent, cheap and "no adjustments needed" as possible.

That does not mean that in your garage you can dream of the most precise piece of hand-made electronics....

As I said before, there are people who still fiddle with audio amplifier with discrete transistors, when a well designed "cheap" Op-Amp will give you some nV/SQRT(Hz) noise, that you probably will never achieve. (In the end you ARE designing a circuit that some super smart guys have already designed onto a chip).

I am happy that my post started such a lively discussion!

 

As I said before, there are people who still fiddle with audio amplifier with discrete transistors, when a well designed "cheap" Op-Amp will give you some nV/SQRT(Hz) noise, that you probably will never achieve. (In the end you ARE designing a circuit that some super smart guys have already designed onto a chip).

I know it's just one example. . .
Most of the really quiet audio preamps are discrete-transistor. Designing one to beat an off-the-shelf op-amp for noise isn't particularly difficult, due to the difficulties of getting the circuit on silicon:- no capacitors of any size, difficulty in making pnp transistors on the same chip as npn, and the substrate diodes. (correct me if I'm wrong, semiconductor physicists).
You can also make an op-amp that precisely suits your circuit for current noise, voltage noise, power supply voltage etc.

Also there's tremendous satisfaction in assembling seven 3p transistors to make an op-amp that's quieter than one costing a fiver.

 

Have a look at this

https://www.passdiy.com/pdf/diyopamp.pdf
https://www.bursonaudio.com/about-us/discrete-circuits/

 

I don’t think I’D be using small power MOSFETs for audio. They tend to be really noisy.
My 7 transistors are constant current source, long tailed pair (2), current mirror (2), voltage amplifier and a constant current load for the voltage amplifier.
JFET input for high impedance source, bipolar for low impedance source.
For really low noise, a single transistor input with a resistive load - quieter, but at the expense of gain and distortion.

 

Hi Wolframore

The dyiopamp article is a great read, thanks.

The Bursonaudio's self made Op-amps will cost you, but then
you get one where " Each transistor is chemically optimized for its application " .

Marketing got a hold of that one..

 

Yeah you gotta ignore the marketing, they make it sound like there are no advantages to being on the same die, like thermal coupling and the silicon die can be more consistent than transistors made from two different dies. But there’s a market for discrete opamps. That Burson page also gets class A and AB mixed up. And class D is actually pretty good as they are way beyond the Nyquist frequency required to perform beyond human hearing.

 

Just how many NE5532s did that signal go through before it got onto the .wav file that you're listening to on your fancy discrete-only hi-fi?

 

If you have a good quality ohmmeter you can make your own precision resistor very cheaply. Depending upon how high a resistance you're after and how much current is expected to be carried through it you can simply use bus wire or ni-chrome wire and wrap your own resistor. Knowing the "ohms per inch" value you just wrap till you achieve the necessary resistance. You can make your own extreme high precision resistor this way. For example, if you need a 10.378 ohm wire you can use a standard el-cheapo 10 ohm carbon resistor and wrap it with a number of turns of simple 28 awg bus wire (configured in series), until it reads exactly 10.378 ohm. For high resistance use Ni-chrome otherwise your finished resistor will resemble a ball of yarn. Depending upon the current and power drop you may need to use lower gauge (thicker) wire. You can also wrap it in a parallel configuration (each end of the wrap to one end of the resistor) to get values under the donor resistor's value. For extremely low resistive values (essentially current shunts), you can simply use a length of bus wire by itself.

 

If you have a good quality ohmmeter you can make your own precision resistor very cheaply. Depending upon how high a resistance you're after and how much current is expected to be carried through it you can simply use bus wire or ni-chrome wire and wrap your own resistor. Knowing the "ohms per inch" value you just wrap till you achieve the necessary resistance. You can make your own extreme high precision resistor this way. For example, if you need a 10.378 ohm wire you can use a standard el-cheapo 10 ohm carbon resistor and wrap it with a number of turns of simple 28 awg bus wire (configured in series), until it reads exactly 10.378 ohm. For high resistance use Ni-chrome otherwise your finished resistor will resemble a ball of yarn. Depending upon the current and power drop you may need to use lower gauge (thicker) wire. You can also wrap it in a parallel configuration (each end of the wrap to one end of the resistor) to get values under the donor resistor's value. For extremely low resistive values (essentially current shunts), you can simply use a length of bus wire by itself.

So what happens to your precision 10.378 Ω resistor constructed this way when the temperature of that cheap 10 Ω resistor, which likely has a temperature coefficient larger than -200 ppm/°C, changes by 10 °C? The resistance drops by 20 mΩ and your effort to get a resistance accurate to the milliohm has been wasted. Even a change of 1 °C will change it by 2 mΩ.

Let's say that you wanted to parallel 28 AWG nichrome wire, which has about 4 Ω/ft of resistance, with this 10 Ω resistor (and let's say that you happened to find one that is exactly 10 Ω) to get 9.95 Ω. How long would the piece of nichrome wire be that you are wrapping around this resistor? You'd need 1990 Ω, or about 500 ft of nichrome wire.

 

Or get some manganin