There are plenty of resistor calculators on the in internet, but all I found stop short at 2 resistors.
While large resistor values can be assembled to a good tolerance relatively easy by adding values, small resistor values need to be designed by paralleling existing values.
And there is some pesky math there. (Yes, math and physics run the world).
It might be logical to many of the users in this forum, but for beginners it is sometimes hard to grasp how to get to a 10.02 Ω resistor for instance.
(All of this is of course based on the formula Rres= 1/(1/R1 + 1/R2 + 1/R3….), basically we are adding
smaller and smaller 1/R3, 1/R4 … until we get close enough to 1/Rres.)
Here is the procedure:
Start with the closest value you can get, in this case you might choose a 12 Ω resistor or
10 Ω + 1 Ω. (You can use one if the above mentioned internet calculators to get the best starting value).
Remember: You always need to start with a higher value than the desired value as paralleling resistor will always lower
the resulting value.
OK, here we go…
Let’s say we start with 11 Ω, we are almost 10% off the goal.
Take the inverse of the desired resistor 1/10.02 Ω = 0.099800…
{
Subtract the inverse of the currently used resistor (11 Ω) = 0.090909
0.099800  0.090909 = 0.008891308 invert it > 112 Ω (omit the minus)
choose the next higher available value > 120 Ω
Invert and add to the inverted first chosen resistor 1/120 + 0.090909 = 0.099242
Invert and we are at 10.07 Ω a 0.5% error.
{
Not good enough? Repeat:
Take the inverse of the desired resistor 1/10.02 = 0.099800
{
Subtract the inverse of the currently resistor (now 10.07 Ω) = 0.092242 from the
desired resistor again.
Result:  0.0005579 invert it > 1792 Ω choose the next higher value > 1800 Ω
Invert add to the inverted first chosen resistor 1/1800 + 0.092242 = 0.099798
Invert and we are at 10.0202 Ω a 0.002% error.
}
Repeat until you are satisfied. BTW, when you are at the last step, you might want to try also the next LOWER available value. Sometimes instead of an error of + 0.002, you might get to – 0.001 for instance.
In this example we now have 4 resistors 10+1 in series, with 120 and 1800 in parallel,
not too bad for a mathematical 0.002% difference to the desired value.
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.
I am attaching an Excel sheet with the calculation in formulas, you only need to add values to column D.
For those who don’t trust Excel sheets, there is also a screenshot with the formulas used.
I hope this is useful for some.
While large resistor values can be assembled to a good tolerance relatively easy by adding values, small resistor values need to be designed by paralleling existing values.
And there is some pesky math there. (Yes, math and physics run the world).
It might be logical to many of the users in this forum, but for beginners it is sometimes hard to grasp how to get to a 10.02 Ω resistor for instance.
(All of this is of course based on the formula Rres= 1/(1/R1 + 1/R2 + 1/R3….), basically we are adding
smaller and smaller 1/R3, 1/R4 … until we get close enough to 1/Rres.)
Here is the procedure:
Start with the closest value you can get, in this case you might choose a 12 Ω resistor or
10 Ω + 1 Ω. (You can use one if the above mentioned internet calculators to get the best starting value).
Remember: You always need to start with a higher value than the desired value as paralleling resistor will always lower
the resulting value.
OK, here we go…
Let’s say we start with 11 Ω, we are almost 10% off the goal.
Take the inverse of the desired resistor 1/10.02 Ω = 0.099800…
{
Subtract the inverse of the currently used resistor (11 Ω) = 0.090909
0.099800  0.090909 = 0.008891308 invert it > 112 Ω (omit the minus)
choose the next higher available value > 120 Ω
Invert and add to the inverted first chosen resistor 1/120 + 0.090909 = 0.099242
Invert and we are at 10.07 Ω a 0.5% error.
{
Not good enough? Repeat:
Take the inverse of the desired resistor 1/10.02 = 0.099800
{
Subtract the inverse of the currently resistor (now 10.07 Ω) = 0.092242 from the
desired resistor again.
Result:  0.0005579 invert it > 1792 Ω choose the next higher value > 1800 Ω
Invert add to the inverted first chosen resistor 1/1800 + 0.092242 = 0.099798
Invert and we are at 10.0202 Ω a 0.002% error.
}
Repeat until you are satisfied. BTW, when you are at the last step, you might want to try also the next LOWER available value. Sometimes instead of an error of + 0.002, you might get to – 0.001 for instance.
In this example we now have 4 resistors 10+1 in series, with 120 and 1800 in parallel,
not too bad for a mathematical 0.002% difference to the desired value.
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.
I am attaching an Excel sheet with the calculation in formulas, you only need to add values to column D.
For those who don’t trust Excel sheets, there is also a screenshot with the formulas used.
I hope this is useful for some.
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