Hello 001,
Hi!
I have one stupid question/ But may be You can help me?
...
example #2
I have 100k 0,2% resistor (actually 100.04k)
I parallel it with some other resistor in series with trimmer to trim overal resistance to 'pure' 100k (I can actually read It with 0,05% tolerance)
What ratio (additional resistor to trimmer values) will be right? Do I need use smallest trim as possible? How to calculate it right?
If you rearrange the formula for paralleling resistors your target resistor value is 250,1MOhm based upon on an intended value of 100k.
(100.040 * 100.000) / (100.040 - 100.000) = 250.1M
For stability reasons you should always avoid using a trimmer.
But in this particular case the current across the 250.1M path is only one 1/2500 (=250.1M/100.04k) of the current across the 100.04k.
By varying the 250.1M figure you are able to fine-trim the total resistance without even using a trimmer.
You do not even need to be able the measure the 250.1M path, just measure the complete setup.
Write down the nominal values of the resistors used for the 250.1M path to simplify following modifications!
Say you would like to integrate a trimmer with a value of 1MOhm.
If you compare 100.04k|| 250.1M with 100.04k || (250.1M + 1M) you see that it gives 100.000 - 100.000,16 = 0,16 Ohm difference, in other terms 1,6 ppm.
You would expect a flat 100k with the trimmer being in mid-position.
Therefore you would reduce that 250.1M by half the trim range of the trimmer:
100.04k || ( 250.1M - 0,5M +0,5M ) = 100k
<=> 100.04k || ( 249.6M +0,5M ) = 100k
Target value for that parallel resistor chain is now 249,6M.
Practically you would turn the trimmer to mid-position and start adjusting the parallel resistor path to 249.6M until the whole things reads out 100k flat.
With the trimmer now being roughly 1/250 of the parallel path and the parallel path responsible for a change of 1/2500 of the long-term stability of the trimmer and it's TK appears negligible.
The whole thing can be done even without the use of a computer by just having a standard resistance collection with some bigger values like 10M and 100M.
Microsoft Excel and LibreOffice Calc are convenient tools to simplify basic calculations.
This is easier with calculus (i.e., take d/dR and set dR = T), but not strictly necessary. You will need to use algebra regardless.
(If you find "quite simple" rather condescending, I don't mean that you personally have to do it, I mean that it is simple enough that a computer can, quite literally, do it. You will need a CAS system, and you will still have to write out the circuit in some mathematical form. Options: Maxima, Mathematica, MATLAB (Symbolic Math Toolbox), Xcas, Scilab, SymPy, Wolfram Alpha, etc.)
Tim
Tim, I don't agree. There is no need to use those weapons of mathematical mass destruction for such a simple problem.
Yes
It is not difficult to calculate resistor values
But what accuracy for ADDITIONAL resistors I MUST use - since its values are small in front "MAIN" resistors?
The accuracy for additional resistors does not matter as long as the steps from one resistor value to the next are small enough to allow compensation of resistor deviation from nominal value with a reasonable number of resistors involved.
example:
intended value 10k
10k (+-10%) resistor in stock measures 910 Ohm.
Put some 81 Ohm + 4,7 + 2,2 + ... series for a 10k approximation.