I'm working on adding a load sharing circuit to this 18650 charger and boost converter shield:
https://www.aliexpress.com/item/32870411748.htmlThe shield consists of a lithium ion charger and both 3.3V and 5V regulators. But the USB input does not directly power the load. Instead, the load is powered by the charger and the battery, which has certain drawbacks when using it as a UPS. So I want to add a load sharing circuit that will allow the USB power source to power the load and charge the battery independently at the same time when it's plugged in, then have the circuit automatically switch to battery power when USB is not connected. The additional circuit consists of the mosfet, diode and resistor in the middle of this schematic:

I think as a practical matter this device is good for about 1A. So I'm depending on the very low RDSon of the mosfet to prevent any thermal issues so long as the battery is powering the load. But when USB is plugged in, the mosfet will be off, and the entire load will be sourced through the 1N4001 diode.
I rigged up a test circuit with parts at hand, and the closest I could get was about 760mA. The diode Vf at that current was up to a full volt, but it worked fine and the Vf stayed steady. But it was quite hot to the touch. The situation is made more prickly by the fact that the diode would be soldered to a tiny SOT23-to-SIP adapter board that's less than a centimeter square, so very little heat would be dissipated through that.
So the question is whether I can leave the diode as is, or should I change it to a 3A diode, or even a Schottky. My understanding is that the same heat would be generated in a 3A diode, but the package and leads are considerably larger, so air cooling is more effective. A Schottky of course would have a much lower Vf, and would generate less heat, but I would prefer to avoid any leakage issues if I can, and, well, I don't have one on hand.
Anyway, I just wondered if anyone here has experience running 1A through 1N400x series diodes, and if so, how did they handle that 1W.