I need to sink some T-18 package zeners. This package is an axial epoxy cylinder, looks like a typical diode, except for the diameter. The zeners are rated at 5W dissapation. They do get hot when not sinked.
I mistakenly ordered from Digikey some TO-18 heatsinks, which obviously do not fit. I don't see any specific listing for this type.
A trick: Drill out a standoff for the diode to (just) slip into, using silicone grease. That can be inserted (more silicone grease) in the round "loop" of a slip-on TO-220 heatsink (i had to add one to 2 layers of thin copper sheet to build up to the inner diameter of that loop).
Be creative.
In article , snipped-for-privacy@nowhere.com mentioned...
Well, if you look inside the PCs SMPS, you might find a couple 3A or so rectifiers that look like your zener, but they are soldered to a heatsink. Or should I say just one lead is soldered to the heatsink. They insert a copper or more likely steel heatsink into the board, and solder one lead of the diode to it, and the other lead to the board. Some boards use the very large copper pads as the heatsinks. Most of the heat from these 3A diodes comes out thru the heavy leads.
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Excellent point! Better than the "surround with metal" scheme i mentioned. So, to maximize heatsinking, a copper block with a hole for the lead should be sweat-soldered on each end; adding fins to each block and a fan will help some more. Just remember, that if the diode got too hot *without* a heatsink, that it is being overstressed, and a different part and/or parallel units should be used.
epoxy
zeners
obviously do
All good, except your last point. Unlike rectifiers, zeners dissipate a substantial amount of power. To approach the five watt rating, the zener will get very hot, unless it's sinked.
In article , snipped-for-privacy@nowhere.com mentioned...
You mean that rectifiers don't dissipate a substantial amount of power? Especially those on a heatsink? Gimme a Break! DUH!
-- @@F@r@o@m@@O@r@a@n@g@e@@C@o@u@n@t@y@,@@C@a@l@,@@w@h@e@r@e@@ ###Got a Question about ELECTRONICS? Check HERE First:### http://users.pandora.be/educypedia/electronics/databank.htm My email address is whitelisted. *All* email sent to it goes directly to the trash unless you add NOSPAM in the Subject: line with other stuff. alondra101 hotmail.com Don't be ripped off by the big book dealers. Go to the URL that will give you a choice and save you money(up to half). http://www.everybookstore.com You'll be glad you did! Just when you thought you had all this figured out, the gov't changed it: http://physics.nist.gov/cuu/Units/binary.html@@t@h@e@@a@f@f@l@u@e@n@t@@m@e@e@t@@t@h@e@@E@f@f@l@u@e@n@t@@
heatsink.
and
of
Correct. Rectification diodes do not dissipate a significant percentage of the power that passes through them. Efficiency of a rectification diode is in the vicinity of 99%. In theory, a rectification diode could perform rectification without any dissipation; it's simply the device physics that does not permit this.
By contrast, a zener diode cannot perform the function of voltage regulation except by dissipation of power. The zener operates in a nonswitching region of V/R curve.
While in theory they might do this, in practice they don't. In practice, a rectifier drops around 0.7V; so, it dissipates 0.7V times the rms current through it (or 1.4V, for a bridge rectifier), and there's no way around that. Which is why power rectifiers come in heat-sinkable packages.
regulation
region
AFAICT from looking at catalogs, it's easier to find regular diodes in high-power cases than it is to find Zeners.
This is probably because generally you *don't* try to drop a whole lot of power in a Zener; rather, if you have a lot of power to drop, you use the Zener at low current to control something like a power transistor, that is dirt cheap and can easily dissipate a lot of power.
By the way, you didn't tell us what problem you're trying to solve, but this solution might be appropriate in your situation. It might be cheaper and quicker than soldering custom heat sinks to a bunch of diodes.
of
is
that
aThis is all correct. My objection to the other poster's remarks is that the efficiency is highly dependent upon the operating regime. The number derived from the above can be as little as a few watts for a 400 watt rectifier circuit.
By contrast, a typical zener diode circuit dissipates about half the power passing through it; the rest is consumed by the dropping resistor.
True.
this
It's a low voltage supply for a couple of comparators. I have a 75V supply rail to work with. The 7800, 7900 series regulators cannot take this directly. The first stage is to drop to 15V through a 1500 ohm resistor and the zener. The zener dissipation is on the order of 1.5 watts. Since the construction is perf board, the zener isn't heatsinked to pads; hence the need for some kind of sink. The suggestion I like the best is to solder some copper sheet metal to the zener leads. The result is passed through 7805, 7905, 7812, 7912.
zener.
How do you get 1.5W? I get 75v-15v = 60v drop across the 1.5k resistor =
40mA; therefore max of 40mA through the 15v Zener; therefore Pd(max) = 40mA
I'm assuming that since you're building this on perfboard you're only making small quantities, so the guiding concern is more what parts you have on hand rather than what would be cheapest in large quantities. So, whatever works, go for it.
40mA
consuming
about
Mistake in-head calc :).
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