You will find the maximum continuous current (1A), and the forward voltage (Vf) at some temperature and 1A. This allows the power dissipation to be calculated as:
P = V*I
This is an approximation. The temperature coefficient of Vf allows a more accurate determination of power at other temperatures. The diode self heats due to the dissipation, and the dissipation changes as it heats because Vf changes. For this, the thermal resistance of the diode is needed to determine the temperature rise, but the overall calculation is not straight forward, simply because the temperature depends on the dissipation and the dissipation depends on temperature. Thermal resistance is °C/W, for the thermal path from junction to ambient air at some temperature. The maximum allowed junction temperature is also required. If the temperature coefficient is ignored this calculation is simple. It can be ignored because dissipation actually reduces with increasing temperature (Vf reduces). Also to simplify things just do the calculations for the maximum allowed junction temperature, and keep everything below that. Note that the thermal resistance is in a certain situation, with the length of the connecting wires in free air and the pads on the PCB providing most of the dissipation of heat. If the air flow is restricted or the wires shortened, the thermal resistance might increase.
That is why the diode is rated in amps, and a derating curve is used to determine the actual rated current. The derating curve for forward current takes ambient temperature and self heating into account. The current rating is reduced at ambient temperatures above 75°C. Use another curve to determine the voltage Vf at 1A is 0.95V. With a current of 1A and ambient of 75°C and Vf of 0.95V, the dissipation in the diode is 1A * 0.95V = 0.95W, so the temperature rise is 50°C * 0.95W =47.5°C. Using the maximum junction temperature, 175°-47.5° = 127.5 degrees. This would be the absolute highest ambient temperature allowed. However it has been reduced to only 75°C to allow for:
1) short 9mm leads which increases the thermal resistance. Note other components nearby can increase the ambient too.
2) half wave current waveform (as a diode rectifier) which has a high peak current. The current of the single half cycle has to double to get the same load current. This is because the diode junction temperature substantially follows the current waveform at 50/60Hz. This basically doubles the temperature rise for the same average (rms) power or current.
The derating does not represent worse case situation, but acts as a prompt to the designer using the data sheet not to take stuff for granted. The derating is based on that half wave current and the thermal time constant of the junction, allowing for an extra watt of heating in the half wave current situation. It does not allow for capacitor input type of filter which has high peak current pulses, because this invokes the maximum repetitive surge rating, and the time involved becomes important.
Personally I would always derate diodes used in power rectifiers, so they do not get too hot to the touch (case temperature not >50°C). This is a simple way to deal with unknown peak currents, thermal time constants, unknown forward (circuit) resistance and all that stuff. There are also problems with soldered joints and insulation external to the diode if the leads get up near the maximum junction temperature.
Incidentally this data sheet specifies the power dissipation (Pd) as 3W and the thermal resistance as 50°C/W, so that would cause a temperature rise of 150°C (of the junction), and it follows the maximum junction temperature is 175°C at an ambient of 25°C.
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