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Energy Handling Capability
In operations where a single thermistor is used to limit the inrush current this calculation is not necessary, but when you start to use 2 or more thermistors then care must be taken so that energy does not exceed its rating.
Ex ) Lets say we use 3 MS22 12103 in series, with the initial tolerance being +/- 25%, and the rated energy(ETH) = 250joules.
Worst case: 1st NTC = 120Ω*1.25 = 150Ω 2nd NTC = 120Ω nominal 3rd NTC = 120Ω *0.75 = 90Ω
Voltage drop 1 = 311.08V on the 150Ω Voltage drop 2 = 186.64V on the 120Ω Voltage drop 3 = 248.86V on the 90Ω
The total energy(ET) = 655 joules = (Ip)^2*(R)* time.
Since the time that the parts are exposed is equal, then the only varying factor is their resistance. Hence the energy divided across three thermistors is a ration of their resistance to total resistance as shown:
E = (ET) (R1/RT) E1= (150/360) * 655 = 272.92 joules E3 = (120/360) * 655 = 218 joules E2 = (90/360) * 655 = 163.75 joules
As you can see E1>ETH hence this part will have a much shorter life than the other two thermistors. In order to use this solution with the part number mentioned the tolerance of the thermistor needs to be tighter. To find the tolerance, the condition below needs to be met using the following:
(RH/(RH +2 RL))ET<= ETH
where
RH= (1+(x/100))R x = the tolerance desired RL= (1-(x/100))R x = the tolerance desired
Thus:
((1+(x/100))R) /(((1+(x/100))R)+2(1-(x/100))R) ≤ (ETH/ ET) = (1+(x/100)) / ((1+(x/100)) + (2-(2*(x/100)))) ≤ (ETH/ ET) = (1+(x/100)) / (3-(x/100)) ≤ (ETH/ ET) = (1+(x/100)) / (3-(x/100)) ≤ (250/655) 655(1+(x/100)) ≤ 250 (3 - (x/100)) 655 + 655(x/100) ≤ 750 – 250(x/100) 905(x/100) ≤ 95 R ≤ 95/905 R ≤ 0.105 x/100 ≤ 10.57% Thus the tolerance has to be ± 10% Nom.
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