zener heatsinks?

[Robert Morein]

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.

 

[Robert Baer]

Robert Morein wrote:

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.

 

[Watson A.Name - "Watt Sun]

In article <n7-dnSdmLoiRj3qiRVn-gg@comcast.com>, nowhere@nowhere.com
mentioned...

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.

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.


--
@@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:###
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goes directly to the trash unless you add NOSPAM in the
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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@@

 

[Robert Baer]

"Watson A.Name - Watt Sun, Dark Remover" wrote:

In article <n7-dnSdmLoiRj3qiRVn-gg@comcast.com>, nowhere@nowhere.com
mentioned...
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.

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.

----------- SNIPped spamad ---------------------

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.

 

[Robert Morein]

"Robert Baer" <robertbaer@earthlink.net> wrote in message
news:3FE7A7D6.38FD5144@earthlink.net...

"Watson A.Name - Watt Sun, Dark Remover" wrote:

In article <n7-dnSdmLoiRj3qiRVn-gg@comcast.com>, nowhere@nowhere.com
mentioned...
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.

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.

----------- SNIPped spamad ---------------------

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.

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.

 

[Watson A.Name - "Watt Sun]

In article <NqadnWhTYtR_xHWiRVn-sw@comcast.com>, nowhere@nowhere.com
mentioned...

"Robert Baer" <robertbaer@earthlink.net> wrote in message
news:3FE7A7D6.38FD5144@earthlink.net...
"Watson A.Name - Watt Sun, Dark Remover" wrote:

In article <n7-dnSdmLoiRj3qiRVn-gg@comcast.com>, nowhere@nowhere.com
mentioned...
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.

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.

----------- SNIPped spamad ---------------------

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.

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.

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 <at> 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@@

 

[Robert Morein]

"Watson A.Name - Watt Sun, Dark Remover" <alondra101@hotmail.com> wrote in
message news:MPG.1a520fd95da131d0989a79@news.dslextreme.com...

In article <NqadnWhTYtR_xHWiRVn-sw@comcast.com>, nowhere@nowhere.com
mentioned...

"Robert Baer" <robertbaer@earthlink.net> wrote in message
news:3FE7A7D6.38FD5144@earthlink.net...
"Watson A.Name - Watt Sun, Dark Remover" wrote:

In article <n7-dnSdmLoiRj3qiRVn-gg@comcast.com>, nowhere@nowhere.com
mentioned...
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.

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.

----------- SNIPped spamad ---------------------

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.

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.

You mean that rectifiers don't dissipate a substantial amount of
power? Especially those on a heatsink? Gimme a Break! DUH!

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.

 

[Walter Harley]

"Robert Morein" <nowhere@nowhere.com> wrote in message
news:FJ6dnaRLbbPhT3WiRVn-hQ@comcast.com...

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.

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.

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.

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.

 

[Robert Morein]

"Walter Harley" <walterh@cafewalterNOSPAM.com> wrote in message
news:vuhm0mklmm9ne9@corp.supernews.com...

"Robert Morein" <nowhere@nowhere.com> wrote in message
news:FJ6dnaRLbbPhT3WiRVn-hQ@comcast.com...
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.

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.

This 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.

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.

AFAICT from looking at catalogs, it's easier to find regular diodes in
high-power cases than it is to find Zeners.

True.

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.

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.

 

[Walter Harley]

"Robert Morein" <nowhere@nowhere.com> wrote in message
news:6LWdnbIbd-uRGnSiRVn-iQ@comcast.com...

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.

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
* 15V = 0.6W, but less than that if the load (regulators etc.) are consuming
any of it (which they must, if only in their own quiescent current of about
5mA each).

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.

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.

 

[Robert Morein]

"Walter Harley" <walterh@cafewalterNOSPAM.com> wrote in message
news:vujrbo8qqd5d86@corp.supernews.com...

"Robert Morein" <nowhere@nowhere.com> wrote in message
news:6LWdnbIbd-uRGnSiRVn-iQ@comcast.com...
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.

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
* 15V = 0.6W, but less than that if the load (regulators etc.) are
consuming
any of it (which they must, if only in their own quiescent current of
about
5mA each).

Mistake in-head calc .