Other Parts Discussed in Thread: LM76003
Tool/software:
Hello, I am trying to design a LM76003-based buck converter circuit that responds gracefully to a short-circuit on the output. I see that the LM76003 has overcurrent protection self-protection, which I would like to rely upon.
My thinking is to simply size the other components in the circuit such that the LM76003 will self-protect before any other components get destroyed. It seems to me the main component that could get destroyed by overcurrent is the inductor. I'm having trouble determining how to size the inductor though.
I see the I_HS_LIMIT (Short-circuit, high-side current limit) is specified as min 4.35A, typ 5.5A, max 6.8A. I understand that I should use an inductor with saturation current above this 6.8A value. But what about the current rating of the inductor? I imagine that in a short-circuit condition, even if the current spikes up to the 6.8A value, the hiccup current-protection mode will lead to a fairly low duty cycle on the inductor, meaning perhaps it doesn't need a current rating ≥6.8A.
Even if the above is true, and I could get away with a smaller current rating on the inductor, another failure mode I'd like to be aware of is an overcurrent condition that doesn't quite rise to the level of a short-circuit. Say a load is connected which draws 4A. 4A is higher than the chip's labeled current "limit" of 3.5A, but would the LM76003 attempt to drive the circuit at 4A, since that's lower than I_HS_LIMIT? If so, it seems I should select an inductor which can take such current continuously.
