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!
Diode voltage drop is only .7 volts which will be pretty much insignificant to the formula power(P) = voltage(E) times current(I). P=E^2/R
I don’t believe that running 2 diodes in parallel is a great idea since the slightest change in characteristics is going to cause one diode to do most of the work.
These days higher amp diodes are not that expensive. I see some 15 A for under 2 dollars.
Since you don’t know the full characteristics of the element then the easiest way to get the actual current would be to set up the circuit and measure the current with a meter. Otherwise just over design and go for worst case and then some.
Ah, something like emitter resistors on a transistor audio amplifier with multiple parallel transistors on the output?
Here, I figure (equalized + headroom) maybe 5A worst case(!) per diode, so, 5x5x0.2 = 5 watts — hmmm, that handybox be gettin’ pretty warm!
OTOH, the aluminum “curve” sign some driver took out beside my property has been laying in the ditch for over 4 months. I even dragged it out where the highway dept. would not miss it. They put up a new one and left the old one by the ditch. And have been working literally “right there” since. Pretty nice sheet of 1/8”(?) aluminum for a heat sink there...
When in doubt, build it stout. Diodes, capacitors and resistors are cheap. Make it three times bigger than you need.
I’d have to series several Schottky diodes to get the reverse voltage rating, no?
That paints a familiar picture. We used to open a dewar of liquid nitrogen in the elevator, hit the next floor and then hide. (Intel)
Hahaha! Well, if I’m willing to wait a few days (and find something else(s) to get up to free shipping), I just found:
I’ll never need to buy silicon diodes in this power range again!
I’m looking at voltage regulators and / or constant current sources for some automotive projects, so, getting up to the free shipping may not be too tough.
I’m still curious what the general engineering rule of thumb is...
Pretty nice sheet of 1/8”(?) aluminum for a heat sink there...
***********************************
Oh Yeah - Grab that sucker, blast it clean. You got some dandy heat sink material there.
Thank You for the link. Interesting....
Like this?
https://www.digikey.com/en/products/detail/wolfspeed,-inc./C6D20065D1/19250272
Very nice.
But for what I want to do(!), the cost - benefit ratio looks like it came from Biden.
Precisely! Same idea.
:-)
The steel post is in good shape too. Several possible uses...
Re: voltage drop...
The Schottky diode has a relatively small voltage drop, usually between 0.15 to 0.45 V.
Decision time.
Just get a Variac then you can get all the control you want.
Hook that old console TV up to a 100A 3 Phase Wye supply and find one of those TV evangelist channels and wait for the moment that the preacher says the magic phrase:
“Now Folks, be sure to first send in your check for $500 and then put your hands underneath the boot on the back of your TV screen and FEEL THE POWER !!!”
I don’t breadboard much any more, but I usually double the expected rating.
I have a warehouse full of components, but usually can’t find what I’m looking for. Recently, I ordered 100 10k resistors and 2N3904 switches, only to find hundreds on a shelf about a week after the new one arrived.
Maybe you should walk around my shop and just pick up what you need...
I’d like to see a smith chart...
I harvested lots of Sprague "orange drop" capacitors from discarded TV sets as a teen. A great source of parts for the parts box. Resistors, caps, chokes. All neatly sorted for future use. One of the "future" uses was charging them up using a Ford Model T spark coil. Place the charged cap in a small paper bag. Leave it out on the desk and wait for the curious and unsuspecting to dump the part into a bare hand.
LED’s blow pretty well too.
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