Converter In-Rush Thermistors etc, UPDATE -4

Salvo wrote:
How can you keep getting it so wrong? The 2 ohm device does not dissipate way more energy. During turn-on, the 2 ohm and 5 ohm device dissipate EQUAL amount of energy. Further, the energy dissipated is 1/2 * C * V^2. The energy consumed by any thermistor in the OP's converter is equal to 31 J. The part fails at 300 J. Turn-on is not the problem. I've covered that a long time ago.

Arguing high inrush current can compromise the power switch is a bunch of baloney. This surge is for such a short time duration, it has no energy behind it. If you're concerned about surge currents then take a look at the surge when turning on the A/C. That is orders of magnitude worse.

Regarding Ken. I don't see his posts any more. I'm done with his childish nonsense.

Sal

Not exactly true.

The thermister is always consuming energy (power) and the amount of energy it consumes is based on its hot and cold values.

The total energy stored in a capacitor is valid during the initial charge period and the thermister must be able to handle this impulse energy when cold - its specification.

But when the thermister is hot, or at a very low resistance, it will dissipate much more energy than when cold.

Oh, and I told you about the capacitor specification and you stated it was BS...

I so enjoy you're circular reasoning....

EDIT:

Salvo wrote:
That's right, the energy of the capacitor (1/2 CV^2) has nothing to do with this. Why mention it?

The energy absorbed by the thermistor is I^2 * R * t. Where I is a exponential function of the RC time constant.

Sal

ken white wrote:


The energy stored in a capacitor is what the thermistor must absorb, and while the energy stored is determined by the (CV^2)/2 equation, the initial movement of charge (current) is the issue.


:h