The Electrician
Active Member
Don't forget you need RMS current to know the heating effect. I think SCR manuals publish RMS vs. conduction angle graphs.
Here is a scope capture of the diode dissipation in the power supply I mentioned a few posts ago.
There was a 5 amp load on the supply.
The yellow trace is the voltage across the diode, the green trace is the current through the diode and the purplish trace is the math function showing the instantaneous product of the diode voltage and current.
At the bottom of the screen can be seen the average of the diode current, the RMS value of the diode current, the average diode dissipation and the duty cycle of the diode voltage.
The peak diode current is just about 20 amps, and the peak forward voltage just touches 1 volt.
These diodes would probably be ok, but there isn't any heat sinking. The manufacturer recommends a copper foil area of .8" by .8" for each lead. This power supply has essentially no foil. The diode leads are soldered into small pads which go off to the rest of the circuitry. The manufacturer also says to only have .375" lead length; this design has about 1" leads.
The RMS current doesn't really tell the whole story as to dissipation because the voltage across the diode is somewhat constant. If one were to use the RMS current to derive dissipation by use of the I^2*R formula, what would you use for R?
If the diode current isn't too outrageously high, as in this case, then it's better to use the average voltage across the diode (about .9 volts by inspection) times the average current (2.549 amps here) getting about 2.3 watts, which is quite close to the 2.32 watts calculated by the scope.