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
You are dealing with a thermal transient during turn-on. The pulse of the transient is much smaller than the thermal time constant of the thermistor. The thermistor has no time to cool down during the transient. That means your argument is not correct. Resistance has no effect on how hot the device gets during the transient.
Resistance does have an effect on steady state use. The greater the resistance the hotter the device gets. The 5 ohm device will always be hotter than the 2 ohm device. And we know that heat kills.
Just to clear up another misconception, the surge current calculation used in the tutorial is not correct. They use
I_surge = V_max / R_thermistor
The ac has an appreciable source impedance as does the converter input filter. The ac will not be at 170V when pulling 30A or whatever. The Honda gen that the OP is using also has bandwidth limitations. I doubt it can correct for the increased load (or in this case surge current) within 4 ms. The ac will have a significant voltage drop. That means the surge current is a lot less that these calculations show.
Sal
Resistance does have an effect on steady state use. The greater the resistance the hotter the device gets. The 5 ohm device will always be hotter than the 2 ohm device. And we know that heat kills.
Just to clear up another misconception, the surge current calculation used in the tutorial is not correct. They use
I_surge = V_max / R_thermistor
The ac has an appreciable source impedance as does the converter input filter. The ac will not be at 170V when pulling 30A or whatever. The Honda gen that the OP is using also has bandwidth limitations. I doubt it can correct for the increased load (or in this case surge current) within 4 ms. The ac will have a significant voltage drop. That means the surge current is a lot less that these calculations show.
Sal
DryCamper11 wrote:
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.
