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
Also to consider is the thermal time constant of the thermistor. The 20A, 5 ohm device has a time constant of 194s. That sounds like the device will hold 5 ohms for quite a while. The converter will be on after a fraction of a second.
If the converter is connected to a (low) battery, the device is now consuming power: P = I^2 * R = 13A^2 * 5 ohm = 845W (that's huge!)
We know that's not possible. Due to the thermistor's voltage drop, the converter will operate at reduced current and somewhat lower power dissipation. Still, the problem is that the thermistor is too large. It's thermal time constant is too big. That's the reason it's getting destroyed.
Parallex converter solved this problem by connecting the thermistor to the diode bridge heatsink. The device absolutely needs a heatsink.
Sal
If the converter is connected to a (low) battery, the device is now consuming power: P = I^2 * R = 13A^2 * 5 ohm = 845W (that's huge!)
We know that's not possible. Due to the thermistor's voltage drop, the converter will operate at reduced current and somewhat lower power dissipation. Still, the problem is that the thermistor is too large. It's thermal time constant is too big. That's the reason it's getting destroyed.
Parallex converter solved this problem by connecting the thermistor to the diode bridge heatsink. The device absolutely needs a heatsink.
Sal
DryCamper11 wrote:ken white wrote:
Unless the 2 Ohm Thermistor can handle 2.5 times more energy than the 5 Ohm Thermistor, it will have less excess energy capacity given the same set of conditions...
It will also allow over twice the surge current to flow, which is what it should be limiting...
And it will have 6.25 times the power being dissipated for the 2.5X increase in current flow. These numbers aren't exact. The faster current flow charges the caps faster, and reduces the current flow sooner. Still, the thermistor is failing due to heating, and that heating is a result of inrush current flow and is proportional to that current squared. Reducing current flow and spreading it out over time are both effective means of reducing the heating that is blowing up the thermistor. Increasing R does both.
