A lower resistance will result in a higher wasted power which will require a higher wattage rating in the resistors. Too high a resistance and you may have issues with tolerance, and too high a resistance in parallel with the high Impedance input of the feedback pin can result in the resulting voltage being wrong.

The input is often one side of an op amp or comparator block.

It depends *why* the kit has been put together.

If you buy a lab kit of E24 values, then you get a book with 50 of each value from 10 ohms to 1 Mohm, in the full E24.

If the kit has been put together to support a set of tutorial designs (my first kit was part of the 'Phillips EE20 Electronic Engineer Kit' which had germanium transistors in it), then there's no need for the full range.

Think about what resistors are used for. It's typically to set up a ratio, at some reasonable impedance level. The impedance level can be fairly coarse. As a designer, I rarely think tighter than 'does it have to be at the 1k, or the 10k, or the 100k level?' However, I often want to nail the ratios to a few percent, and be able to set ratios over a wide range.

Resistor ranges are constant ratio (or very nearly so). To take E24, I can can set up a 2:1 ratio with 1k/2k, 1.2k/2.4k, 1.5k/3k, 1.8k/3.6k, 7.5k/15k, and maybe a few others if I think for longer (like 1.1/2.2). I don't need all those pairs in my kit, just one pair in each decade will do for 2:1 ratios. And a few more for a few others.

Because of the economies of scale, it's far cheaper to provide a tutorial kit with 20 each of a dozen different values, than 5 each of 4 dozen values. It needs less organisation, and a smaller box, as well. Whether those values are from the E6 or E24 series makes very little difference to the price, E24 is used so widely in industry. What does make a difference is tolerance, but I'd probably want to use 1% throughout for a tutorial kit anyway.

As an aside, once the book has been in use in the design lab for a month, all the 1k and 10k resistors have gone! It's possible (just sayin') that designs end up with 1.2k resistors in them because all the 1k resistors had gone when the designer came to prototype it!

## Best Answer

## Short Note

There are good reasons and a relatively long history, if you want to scoop up sporadic early attempts. It's called

rationalizationand its activity really picked up after WW II and the new-found need for manufacturing standards.Soon after WW II the National Bureau of Standards [NBS] (now NIST) engaged a multi-pronged push. You can find some of their work product in a publication called NBS Technical Note 990 (1978):

"The Selection of Preferred Metric Values for Design and Construction". (The National Bureau of Standards [NBS] is now NIST.)This activity was across many areas of manufacturing, from the number of teeth on gears to the values of resistors.

To answer your question about the

E12series, you have to go back to theE3series. There's no escaping that it. You cannot derive the values forE12without moving backward toE3. But I'll get to that, momentarily.## Charles Renard

The history of this goes back at least to Charles Renard, who proposed specific ways of arranging numbers to divide (decimal) intervals. He focused on dividing decades in 5, 10, 20, and 40 steps, where the logarithm of each step value would form an arithmetic series. His choices became known as R5, R10, R20, and R40.

Renard numbering was extended to include other special versions, such as R10/3, R20/3, and R40/3. Here, these were interpreted to mean that you would use the R10, R20, and R40 decade series approaches but would

stepacross values, taking them three at a time, for example. So R20/3 means to use R20, but to select only every 3rd term as in: \$10\cdot 10^\frac{0}{20}\approx 10\$, \$10\cdot 10^\frac{3}{20}\approx 14\$, \$10\cdot 10^\frac{6}{20}\approx 20\$, \$10\cdot 10^\frac{9}{20}\approx 28\$, \$10\cdot 10^\frac{12}{20}\approx 40\$, \$10\cdot 10^\frac{15}{20}\approx 56\$, and \$10\cdot 10^\frac{18}{20}\approx 79\$.## The Geometric E-Series

Start with the

E3series (as they did.) The idea ofis far more crucial forcoverageE3and less crucial forE12. (Still less forE24and beyond.) So you have to start atE3in order to find out why certain values are selected forE12.I'll start with the full diagram and then explain the details of each step along the way in a moment:

## Series E3

Starting with

E3, simple computation yields: \$\begin{align*}\textbf{E3}&\left\{\begin{array}{l}\lfloor 10^{1+\frac{0}{3}}+0.5\rfloor= 10\\\lfloor 10^{1+\frac{1}{3}}+0.5\rfloor= 22\\\lfloor 10^{1+\frac{2}{3}}+0.5\rfloor= 46\end{array}\right.\end{align*}\$But there's an immediate problem related to coverage. They are all even and there's no way of composing odd numbers using only even numbers.

At least one of these numbers must change, but they cannot change

10for obvious reasons.To change just one, the only possibilities are: \$\begin{align*}\textbf{E3}_1&\left\{\begin{array}{l}10\\\textbf{23}\\46\end{array}\right.\end{align*}\$, or else, \$\begin{align*}\textbf{E3}_2&\left\{\begin{array}{l}10\\22\\\textbf{47}\end{array}\right.\end{align*}\$.

But \$\textbf{E3}_1\$ still has a problem related to coverage. The difference between 46 and 23 is itself just 23. And this combined value is a number already in the sequence. (This means that you can't reach a new integer by putting two 23 Ohm resistors in series -- the 46 Ohm resistor suffices. So the coverage is poorer.) In contrast, \$\textbf{E3}_2\$ doesn't have that problem, as the differences and sums provide useful values not already in the sequence.

Rationalized, it is: \$\begin{align*}\textbf{E3}&\left\{\begin{array}{l}10\\22\\\textbf{47}\end{array}\right.\end{align*}\$

## Series E6

The next step is to examine

E6. First and foremost,E6must preserve the values that were determined forE3. That's a given that cannot be avoided. Accepting that requirement, the computed (andE3retained) values forE6are \$\begin{align*}\textbf{E6}&\left\{\begin{array}{l}10\\\lfloor 10^{1+\frac{1}{6}}+0.5\rfloor= 15\\22\\\lfloor 10^{1+\frac{3}{6}}+0.5\rfloor= 32\\\textbf{47}\\\lfloor 10^{1+\frac{5}{6}}+0.5\rfloor= 68\end{array}\right.\end{align*}\$But a coverage problem shows up, again. The difference between 32 and 22 is 10 and this is one of the values already in the sequence. Also, 47 minus 32 is 15. So there are at least two problems to solve. And 32 is involved in both. Changing it to 33 solves both these problems at once and provides the needed coverage.

Rationalized, it is: \$\begin{align*}\textbf{E6}&\left\{\begin{array}{l}10\\15\\22\\\textbf{33}\\\textbf{47}\\68\end{array}\right.\end{align*}\$

## Series E12

E12must preserve the values that were determined forE6, of course. The computed (andE6retained) values forE12are: \$\begin{align*}\textbf{E12}&\left\{\begin{array}{l}10\\\lfloor 10^{1+\frac{1}{12}}+0.5\rfloor= 12\\15\\\lfloor 10^{1+\frac{3}{12}}+0.5\rfloor= 18\\22\\\lfloor 10^{1+\frac{5}{12}}+0.5\rfloor= 26\\\textbf{33}\\\lfloor 10^{1+\frac{7}{12}}+0.5\rfloor= 38\\\textbf{47}\\\lfloor 10^{1+\frac{9}{12}}+0.5\rfloor= 56\\68\\\lfloor 10^{1+\frac{11}{12}}+0.5\rfloor= 83\end{array}\right.\end{align*}\$More coverage problems, of course. 83 minus 68 is 15 and 15 is already in the sequence. Adjusting that to 82 solves this issue. But 26 has a prior span of 4 and a following span of 7; and 38 has a prior span of 5 and a following span of 9. These spans should, roughly speaking, be monotonically increasing. This situation is quite serious and the only options really are to adjust 26 to the next nearest upward alternative of 27 and to adjust 38 to its nearest upward alternative of 39.

Rationalized, it is: \$\begin{align*}\textbf{E12}&\left\{\begin{array}{l}10\\12\\15\\18\\22\\\textbf{27}\\\textbf{33}\\\textbf{39}\\\textbf{47}\\56\\68\\\textbf{82}\end{array}\right.\end{align*}\$

## Notes

coverageas possible.(The above is exactly true when using the theoretical values rather than the preferred values. But since the preferred values have been

adjustedby the rationalizing process, there will be some deviation.)P.S. Links to my prior writing on this topic are found here and here.