Converter In-Rush Thermistors etc, UPDATE -4
UPDATE on 17 Nov, Update 2 on 22 Nov, Update 3 on 16 Jan 14, Update 4 on 19 Oct 14. --------------- In a recent thread there was mention of how you can wreck your gizmo (in my case a converter) ...
Forum Discussion
I've read the thread up to here, and thought it might help to make some brief comments about the NTC resistance and its effect on the charging of a capacitor bank. It's been posted here that a charged capacitor has energy stored equal to 1/2 CV**2. It's also been posted that the thermistor needs to dissipate energy of 1/2 CV**2. Both are correct. The total energy required to charge a capacitor through a resistor (the resistor is the thermistor in our case) is CV**2 (not 1/2 CV**2). Half of that CV**2 energy is dissipated in the resistor/thermistor, and half gets through and is stored in the capacitor.
It doesn't matter how much resistance you are charging the capacitor through - the total energy lost due to that resistance (energy dissipated in that resistance) will be the same.
However, the resistance does have a great effect on the power dissipated, i.e., on the rate at which the 1/2 CV**2 energy is dissipated. The energy dissipated in the resistor appears as heat. The heat is transferred to the surroundings. As the resistance of the thermistor goes up, the current goes down and the capacitor charges more slowly. The same amount of energy is dissipated in the resistor, but it gets dissipated more slowly, and the resistor/thermistor has longer to transfer its heat to the surroundings, so it doesn't get as hot. That's why the resistance calculations produce a "minimum" resistance. If you use more resistance, you just slow things down - decreasing the maximum inrush current and allowing more time for heat to be dissipated. If you go below the minimum, you risk damaging the thermistor as the energy is dissipated too quickly for the device to rid itself of that heat without damage.
It doesn't matter how much resistance you are charging the capacitor through - the total energy lost due to that resistance (energy dissipated in that resistance) will be the same.
However, the resistance does have a great effect on the power dissipated, i.e., on the rate at which the 1/2 CV**2 energy is dissipated. The energy dissipated in the resistor appears as heat. The heat is transferred to the surroundings. As the resistance of the thermistor goes up, the current goes down and the capacitor charges more slowly. The same amount of energy is dissipated in the resistor, but it gets dissipated more slowly, and the resistor/thermistor has longer to transfer its heat to the surroundings, so it doesn't get as hot. That's why the resistance calculations produce a "minimum" resistance. If you use more resistance, you just slow things down - decreasing the maximum inrush current and allowing more time for heat to be dissipated. If you go below the minimum, you risk damaging the thermistor as the energy is dissipated too quickly for the device to rid itself of that heat without damage.
