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?
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.
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!!
:?: :?:
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!!
:?: :?:
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.
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
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.