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Tolerance of parallel combination of ressitors

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supernova

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I'm trying to combine two resistors in parallel to make a non standard value which is 2043.73 ohms. This is of course possible with the combination of (5K6, 3K3) or the combination of (2K2, 27K). The second combination would of course give me a closer match (in ideal situation) but what I'm wondering now is the tolerance of the combination. Which one would give me the better tolerance or in other words which one is more likely to be closer to the desired value?(Consider E12, 5%values).

Could somebody tell me the general solution to this problem & also the best practical choice?

Your help is much appreciated
:wink:
 
The combination that gives the closest match to your target value is the best choice, ignoring secondary effects like temperature coefficient.
 
When connecting resistors in parallel the only thing that goes down is the resistance. The wattage is now the sum of the individual wattages and so is the tolerance. Course you could get luckly. One could be less than the marked value, and the one could be higher and the net result is exactly what you want.
 
k7elp60 said:
The wattage is now the sum of the individual wattages and so is the tolerance.

Wattage adds up only when resistance of each resistor is the same.

Tolerance of the combination is the same as the individual resistors if both have the same tolerance %.

e.g. a 1K 1W 5% and 10K 1W 5% in parallel gives:

0.909K 5%, wattage slightly larger than 1W but definitely not 2W.
 
This is getting very confusing now.
Can somebody else join & help.
Reference to a web article or book would be much useful.
I think neither of these two are my answer!!
:?: :?:
 
supernova said:
This is getting very confusing now.
Can somebody else join & help.
Reference to a web article or book would be much useful.
I think neither of these two are my answer!!
:?: :?:

Get your pen and paper out and do the sums!.

Pick your two resistor values you want to put in parallel, then work out their maximum and minimum values based on their tolerances. Then work out the parallel value for the two maximum values and the two minimum values. Compare those with the calculated exact value, then work out the tolerance to the maximum and minimum ones.

Simple maths!.
 
I would suggest adding 10 ohms in series with your 2k2//27k combination, to get a bit closer. Also use metal film (1%) resistors for better accuracy (of course, dependant on required wattage).

Or try 2k2//33k//220k = 2043.34 ohms
Or even better 2k2//33k//470k//470k//5M6//56M = 2043.7350 ohms
 
eblc1388 wrote:
Wattage adds up only when resistance of each resistor is the same.
Well the wattage rating of a resistor depends upon it's construction and physical size. The amount of power it is actually dissipating depends on the current flow thru the resistor and the voltage drop accross it.
 
Dr.Em is correct, just use a multimeter and find the best one which will give the closest value for u, the total power will be same anyway, cuz it depends on the total curret flow, and the voltage will be the same across parallel resistors. So y are u worried about the wattage as the power rating is not going to decrease anyway.
 
eblc1388 said:
Wattage adds up only when resistance of each resistor is the same.

Sorry, my mistake, should be:

"Wattage adds up only when each resistor is the same."
 
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