Posted on 02/03/2024 5:48:43 PM PST by Paul R.
I've built power supplies using full wave rectifiers in the past, and a couple circuits with single diode half wave rectification for non-critical supplies, but in all cases the power requirements were low and the diodes I had on hand several times overrated for the current and voltage / power involved.
This time though, I want to add a half power function to a 1500 watt 120 volt quartz heater. (Over-simplification - see below comments. But ~half of the 1500 watts is the design target.)*
So... I know a diode in series with the quartz element and no power supply capacitor** will essentially cut the power in half, as current is flowing ~1/2 the time. (Area under the curve.) So, is a silicone diode rated at 10 amps & 1000 reverse volts (I have a few in my parts box) going to have enough current headroom?
1500 watts at 120 volts means the quartz element draws ~ 12.5 amps. Half that is 6.25 amps, but since the element will be operating at a lower temperature, current drawn and power emitted will be a bit higher. That will be slightly mitigated by the diode's forward voltage drop. I'm pretty sure the current won't be over 8 amps, but, it also seems to me that 8 amps actual is pushing a 10 amp diode pretty hard. Safest bet might be to parallel 2x 10a diodes, or, order some 20 or 25 amp diodes? (I'd prefer to use the parts on hands.)
Note that the switch & diodes will be in a steel "handy box", most likely, so I can give the diodes significant but not "great" heat sinking.
From experience I know the 1000v reverse voltage rating is more than enough. I've never even blown a 600 volt diode in a 120 volt AC rectification application. (Yes, I know, one has to be careful with DC ratings vs. where the peak of a dirty sine wave may go.)
Comments by those with some experience are welcome / requested!
No, I just keep Schottky “diodes”.
Which are P-N
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