Forum Discussion
Salvo
Nov 20, 2013Explorer
I didn't misinterpret the 194s spec. I didn't say the thermistor will hold 5 ohms for 194 s, or what ever. You are putting words in my mouth.
The topic I was responding to is in regards to what's the best thermistor to install. There are really two issues.
1. Thermal time constant. There's a coupling between the thermal time constant and the part's ability to change it's base resistance. In other words, a part with a small time constant will be able to reduce its resistance faster. In this application, you want a fast part. It's not the turn-on stress that's killing the thermistor, but the duration of time the thermistor's resistance is still high when the converter is in operation.
2. Thermistor resistance. As I've shown, choosing a high resistance part is a bad idea. You want a resistance that protects the diode from excessive surge currents. You don't choose a resistance that's 12 times what you actually need. That's a recipe for disaster. And it come in the form of exploding thermistors.
Sal
The topic I was responding to is in regards to what's the best thermistor to install. There are really two issues.
1. Thermal time constant. There's a coupling between the thermal time constant and the part's ability to change it's base resistance. In other words, a part with a small time constant will be able to reduce its resistance faster. In this application, you want a fast part. It's not the turn-on stress that's killing the thermistor, but the duration of time the thermistor's resistance is still high when the converter is in operation.
2. Thermistor resistance. As I've shown, choosing a high resistance part is a bad idea. You want a resistance that protects the diode from excessive surge currents. You don't choose a resistance that's 12 times what you actually need. That's a recipe for disaster. And it come in the form of exploding thermistors.
Sal
ktmrfs wrote:Salvo wrote:
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.
SalDryCamper11 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.
sorry salvo you aren't intrepeting a NTC thermal time constant spec correctly. it is the time it takes for the thermistor to COOL down in free air with no forced cooling when the full rated current is REMOVED from the device and no current is flowing through it. In other words, the time that it takes a NTC Inrush Current Limiter to recover to usually 50% of its initial resistance. It's an indication of how long it will take before the thermistor will "behave" normally when it needs to limit current again.
The time constant to heat up is going to be highly dependent on the current waveform and is going to be an integral of power over time using the thermistor heat rise in Degrees/watt. Highly dependent on the application, but very quick in most cases, nowhere near 194 seconds, more like seconds or less.
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