What is the practical limit/best practice for the average rectified output current of a diode in AC applications?

Question

Background: As part of a little project I'm looking to add a diode to a hair dryer to cut the output in half. This is the circuit:

The potential diode in question is at the bottom, and the simulated current through it is in the center graph: 2.184 A peak (edit: 3.1 A). By my math that would mean right about 0.7 A average (edit: 0.99). Technically, this would fall below the 1.0 A spec for average rectified output current of a 1N4004 diode. However this sheet by Vishay states

IFAV decreases with an increasing value of the reverse voltage during the interval of no current flow.

In my case the reverse voltage is 120V sinusoidal; potentially 240V if used in another country. Unfortunately the 1N4004 datasheet doesn't seem to indicate the reverse voltage for the IFAV spec.

So for my education, ultimately my questions are:

1) Practically, should a 1N4004 diode be able to handle 0.7A (0.99) IFAV with 120-240V reverse voltage for an indefinite amount of time? Or is that too close to the spec, and reducing life of the part? More generally, how close is too close to the IFAV spec?

2) If practically possible, would best practice be to use the 1N4004, or a 1N5404 (3A IFAV)? A 1N5404 would be harder for me to fit but is probably doable.

Answer 1

Disclaimer: you're modifying a UL (or CE) labeled device; if you burn down your house or electrocute your dogs it'll be on you. If your insurance company is particularly asinine, then if a Mad Bomber tosses a Molitov cocktail into your bathroom window and burns down your house, they may use the presence of your un-plugged but modified hair dryer as an excuse not to pay. Because -- insurance companies.

From the sheet that you link to:

In the absolute maximum ratings, one or both of the following are given:

• The maximum permissible average rectified output current for zero diode voltage (reverse).
• The maximum permissible average rectified output current for the maximum value of VRRM during the time interval at which no current is flowing.

So, find a diode that specifies the second, and use it.

In practical terms, the diode heats up mostly proportionally to the average rectified current, because the drop is fairly constant, and power = current * voltage. I believe that the reason the number depends on the reverse voltage is because the part's reverse breakdown voltage will drop with increasing temperature, and higher current = higher dissipation = higher temperature.

Note, too, that the given ratings assume a certain ambient temperature, and a "typical" installation with free air circulation. If you use your 3A rated diode, but put it into the same narrow, no-airflow space, its junction temperature may get just as high or higher than the "1A" diode.