0 Members and 1 Guest are viewing this topic.
I get 458mV.LVR03 so P = 3WR = 0.07V = (3*0.07)0.5 = 0.458V
Quote from: Hero999 on July 16, 2015, 08:01:33 pmI get 458mV.LVR03 so P = 3WR = 0.07V = (3*0.07)0.5 = 0.458VHow did you get that Hero999? 3x.07=0.21v which divided by 2 makes 105mV.
Quote from: theleakydiode on July 16, 2015, 08:07:48 pmQuote from: Hero999 on July 16, 2015, 08:01:33 pmI get 458mV.LVR03 so P = 3WR = 0.07V = (3*0.07)0.5 = 0.458VHow did you get that Hero999? 3x.07=0.21v which divided by 2 makes 105mV.That is not divided by 2 it is to the exponent of 1/2 (aka square root).
It seems like a low value, does this make them a bad choice for peak current sensing in a flyback converter? As the inductor peak current will be much higher than the average current.
max voltage drop, not max voltage. thats an absurdely low max voltage
It's interesting they've stating the maximum voltage rating, just by the amount of power the resistor can safely dissipate.Most resistors can withstand momentary voltage surges which will exceed the power rating, so long as the limiting element voltage isn't exceeded. For example a metal film 1W 100R resistor may only be able to take 10V continuously but it can take brief 100V pulses, as long as the average power dissipated doesn't exceed 1W. However a 1kV pulse would cause it to arc over and be damaged.