Using the wall(s) of an enclosure as a heatsink?

[trad3mark]

Hey all,
I'm working on a pedal at the moment that uses a 5V Regulator with two diodes to power the filament of a 12AX7. Great. Everything works as it should! Anyway, as we all know, voltage regulators get hot. Now, because of the pedal, the enclosure is a very very tight squeeze, and realistically, there just isn't room for a heatsink unless it was in a bigger box. So, i read a while ago about someone using the wall of the enclosure (cos it's a normal Hammond diecast box like...) as a heatsink. It makes sense to me personally. However, is it a suitable solution? Can anyone shed some light on this, or better yet, if they've done it, how's about a pic?
cheers all,
tm

[sean k]

I did it using an amp in an old pot and used the wall, because it was a pressure pot and thicker than normal, to dissapate heat from a TDA2003 chip. I drilled holes around the chip, in the body of the pot, on the assumption that by drilling holes I would increase the surface area for heat to dissapate. Funnily enough the amp works fine and doesn't seem to heat up at all.

[trad3mark]

hahaha! that's awesome using a kitchen pot as an amp! Nice!

As far as the heatsink thing goes though, makes me think it's very doable to just mount it to the hammond...

[petemoore]

 Basically just easier to see hot how it'll get, you can calculate it to some degree, ie sometimes you'll have the idea [such as a read informing you] that there's gonna be a lot of heat, but with the cooling fins it won't build up etc.
  I did that using a 6v regulator for an x7's heater, raco got warm.
 

[maarten]

The metal back side of a 7805 regulator is internally connected to ground (as is the middle pin). Now if you if you are raising the voltage to 6.3 volts by connecting the middle pin by the 2 diodes to the ground, you will that the voltage will drop to 5 volts again if you are using your enclosure as a heatsink - unless the enclosure is not connected to ground. Usually it is, to protect the circuit from outside impulses such as radio. In that case the regulator's backside will short out the diodes. There are isolation sets for this, which will electronically isolate and yet conduct the heat to the enclosure. I think you can use the same set as is used for transistors of the TO 220 packaging type.
Maarten

[trad3mark]

Cheers for that maarten. Helpful, although not good news.

Might be easier to just get a 12V reg and have 12V going into the tube instead of 6.4V. I think if i used a 12V reg, it wouldn't get as hot as the 5V with the 2 diodes. I might be able to avoid a heat sink all together if i used the 12V reg, right??

[trad3mark]

actually.... If i was thinking about mounting it to the wall to act as a heatsink, would the tube still work with 5V going to the 9th pin?

if not, is there a way i can go from 9V dc to 12V dc using a ka7812?

[zyxwyvu]

maarten brought up the most important issue here: isolating the part from the enclosure. You really do not want them connected electrically unless they really are both at ground. This is generally not the case, and is not in your case. You can buy mica insulators very cheaply online that are thermally but not electrically conductive.

actually.... If i was thinking about mounting it to the wall to act as a heatsink, would the tube still work with 5V going to the 9th pin?

if not, is there a way i can go from 9V dc to 12V dc using a ka7812?

Why not just use an adjustable regulator, like the LM317 to get 6.3V exactly? The regulator would have to dissipate less than a watt, assuming you have one tube. Keep in mind the metal tab on these is connected to the output, not ground.

[PRR]

Why don't you use a resistor?

I mean, what is a tube's heater? A precision millivolt nanosecond device? No, it is a resistor, as sophisticated and fussy as an electric heater. Give it a little more or less, it runs a little hotter or less-hot. That does not affect the electrons: oxide cathode make plenty of electrons, and the number which actually flow is almost entirely up to the Grid.

If you need to work the tube at MAX current, such as a 70 Watt guitar amp's EL34s, you want to be above 6.1V. If you work 24/7 and NEED it to always work, you want to stay below 6.5V. But for a stompbox... bah, 5.5V will work, 7V isn't going to burn-out in a decade of gigs.

Also: a heater is a poor resistor: when cold it sucks a big current. Adding some dumb fixed resistance in series reduces the start-up shock.

Heaters themselves certainly don't need "clean" power: they work fine on raw AC. Yes, heater leads near delicate audio paths can radiate hum/buzz; but presumably your 9V is DC clean enough for the rest of the pedal, so probably clean enough to run to the heater.

You got 9V? You can't get 9V to 12V any trivial way. So you want to start from 9V and arrive at 6.3V with 0.3A of current. (9.0V-6.3V)/0.3A= 10 ohms near-enuff. Heat is (9.0V-6.3V)*0.3A= 0.81W, use 2W part.

Dissipation is exactly the same as the reg (actually like 0.05W less for reg-brain power). No embarassing tab-to-case shorts. No screw nor washer. No trimmer-diodes or bypass caps.

[trad3mark]

that would make things much easier PRR.

So i just put a 10R between 9V and the 9th pin? That's it? I was thinking this morning, a voltage divider would be an easy and effective solution considering i've a good few trimmers around, but a resistor.... that's easier again.

[served]

I have the same thing in my mind. But I am going to use the isolation and connect the regulator to the box. I just cant see better way. And I cant see that it is wrong. It will work, why it should not.

[trad3mark]

going to go with a super simple voltage divider. With my 9.4V input, and by using a 150R and a 330R, you can get 6.463V. pretty spot on imo.

[JKowalski]

going to go with a super simple voltage divider. With my 9.4V input, and by using a 150R and a 330R, you can get 6.463V. pretty spot on imo.

Problem is, once you hook up that filament to your divider you will definitely NOT have 6.463 anymore. The filaments are something like 20 ohms load, so that 20 ohms load would be in parallel with the 330R, giving you around 18 ohms instead. A voltage divider with 150 ohms and 18 ohms would output  around 1V, not 6.463 volts. Plus the divider would waste power.


To get the required stability, your voltage divider would need to be made out of, say, a 1.5 ohm and a 3.3 ohm resistor instead. That wastes alot of power, but it will get you the right voltage with the heaters as a load. Overall, not a good solution.

[trad3mark]

hmmm... back to square one.

See, i don't have any 2W resistors. I think the one's i have are the normal 1/4W ones. (the kinda beige coloured 4 band ones). at my disposal i have a few 5V regs and a 12V reg. what about 5V to the fillament?

[trad3mark]

just a little update, i went back to where i was before, with the 5V reg and the two diodes. I played for 10mins, with gain on full etc, really pushing the thing like. Then i checked with my finger how hot it was. It's just the back of it that's hot like. It was pretty hot to the touch. Didn't burn me or anything, but you wouldn't want to be holding it for too long like. I then kept playing for another 10 mins and it didn't seem to be any hotter.

So... new question(s). Is that the hottest it's going to get, and if so..... would i get away with just leaving it as is??? what other possible solutions do i have?

[Thomeeque]

hmmm... back to square one.

See, i don't have any 2W resistors. I think the one's i have are the normal 1/4W ones.

Use more of them in some combination (depends on what values you have got) to spread power between them - eg. if you'd have four 39Ω/0.25W in parallel, they would make 9.75Ω/1W. T.

[Thomeeque]

..or try 4-5 silicon diodes in series, each will drop around 0.6V..

[earthtonesaudio]

..or try 4-5 silicon diodes in series, each will drop around 0.6V..

...and have to pass 300mA.

[trad3mark]

..or try 4-5 silicon diodes in series, each will drop around 0.6V..

yah this didn't work.

ok, so realistically, what are my options here, apart from leaving it as is with the 5V reg running hot (ish) at 6.4V. what if i only used one diode or ran it at 5v?

[Thomeeque]

..or try 4-5 silicon diodes in series, each will drop around 0.6V..

...and have to pass 300mA.

Rather probably more to stand initial heating..? I'd go with 1N400x's..

yah this didn't work.

Really? But it should, give it another try.. T.