How do I learn RPL and solve this problem with it?

09262017, 02:29 PM
Post: #28




RE: How do I learn RPL and solve this problem with it?
(09262017 01:02 AM)brickviking Wrote: I'm no good at the mathematics behind all this, I've been a bit lost. However, I realised that there's also a realworld value of each resistor, not just a theoretic L<R<U where L and U are the lower/upper 5% bounds. As soon as you add a resistor which ALSO has that same 5% tolerance, you have to hope that for an exact match, your chosen resistors aren't both at the lower edges of their tolerance bands (or higher edges) affecting how close their real (not nominal) value will be to the required theoretic value. Will the two resistors that are low still fall within 5% of the combined value? Ditto for two high resistors in parallel or series. You have a point there. The error in each resistor doesn't really matter much, and here's why: He wants to replace R (with a certain error, let's say 1%) with 2 parallel resistors that also come with a certain error of 1%. 1/R=1/R1+1/R2, therefore let's say both R1 and R2 are at the lower bound of error (the real R1'=0.99*R1, and R2'=0.99*R2) it turns out that: 1/R' = 1/R1'+1/R2' = 1/0.99*(1/R1+1/R2) = 1/(0.99*R) The same will happen if you pick the upper bounds, (just with 1.01 instead of 0.99). In other words, 2 resistors in parallel with a certain tolerance will cause the resulting resistance to be within the same tolerance. However, if the selected pair has an error you'll have that bias + the original tolerance. For example if the selected R1 and R2 produce a value that is 1% higher, you'll have a maximum error of 2% and a minimum of 0%, biased in the same direction. So the user needs to pick the pair that produces the smallest error and accept that little bias, otherwise you'll be forced to use resistors with a smaller tolerance than the original requirement, so the result stays within bounds. But that should be up to the designer, the problem at hand should find all 4 or 5 combinations and present them to the designer. The designer in the end will pick the closest pair, or whatever pair he can find in stock. 

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