Converter In-Rush Thermistors etc, UPDATE -4

Salvo wrote:
You are dealing with a thermal transient during turn-on.

Agreed.

The pulse of the transient is much smaller than the thermal time constant of the thermistor.

Again, agreed, assuming we are talking about the bulk thermistor time constant - but we aren't. I thought about discussing this, but didn't think it was worth getting into that detail. An NTC thermistor is typically made of a sintered semiconductor material pressed into a disk. The active semiconductor regions in the thermistor are actually quite small.

The thermistor has no time to cool down during the transient.

It's not the entire thermistor that cools down. It is the tiny active region(s) that must transfer heat from that region into the adjacent material. That transfer has a time constant that is much much less than the bulk time constant of the entire device. You can damage the thermistor without producing significant heating of the entire device if enough current is passed through it in a short time.

That means your argument is not correct. Resistance has no effect on how hot the device gets during the transient.

We disagree. The resistance of the thermistor controls how much heat (energy=1/2 CV**2) is dissipated in the semiconductor active region(s) of the thermistor per unit time. Given the small size of the semiconductor region, the total energy dissipated as heat has sufficient time to diffuse out of that region before the region is damaged even though the entire device does not have time to cool significantly.

Resistance does have an effect on steady state use. The greater the resistance the hotter the device gets.

No, the greater the resistance, the longer the time it takes to dissipate the 1/2 CV**2 energy (during the initial charge cycle of the capacitor). Power is energy divided by time, and the energy remains constant, while increasing resistance increases the time over which that energy must be dissipated.

In steady state, we are no longer interested in the initial (zero power) resistance - we are interested in the steady state resistance, which is close to zero. We aren't talking about steady state design here and in the initial charge cycle, it is the zero power resistance (5 ohm vs. 2 ohm) that is critical. That's why the calculations all include C = the capacitance and produce a minimum resistance - the resistance needed to ensure that the energy dissipated in the device is spread over a sufficient period that heat at the semiconductor junctions of the thermistor can escape those junctions without excessively heating the semiconductor material to the point it is damaged. More resistance is OK. Less is not. That's why the number calculated is a "minimum."

The 5 ohm device will always be hotter than the 2 ohm device.

The 5 ohm device will dissipate the same energy in the initial charge cycle as a 2 ohm device, but it will be spread out over time. The temperature of the active region(s) of the device will be lower as the heat in that region can move out of the active region before it is damaged. We're not talking about bulk heating of the entire device, but localized heating.

In steady state (which is not what we are concerned about), the temperature of the device is a function of the steady state resistance. I agree that if the steady state resistance of the 5 ohm thermistor is greater, then its temperature will be higher. However, provided the device is rated for the steady state current that it actually sees, it will happily survive that higher temperature. The problem encountered here is not with steady state failure. It is with surge/turn on failure, and that is where the zero power resistance (2 ohms vs. 5 ohms) is critical. To put it another way: the maximum temperature of the regions that are sensitive to heat damage (heat produced by current flow through those regions) will be less for the 5 ohm device at surge/turn than for a 2 ohm device because current flow is less.