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

[JKowalski]

Personally I would just go with the series resistor. Least hassle, and relatively efficient. 60%? I'd have to calculate how efficient the regulator is but I don't think it would be much of an improvement.


1n400's are schottkys so you would need many more of them to get the same voltage drop  A resistor is a much more elegant solution, even if you need a couple in parallel to handle the power.

[Thomeeque]

1n400's are schottkys so you would need many more of them to get the same voltage drop   A resistor is a much more elegant solution, even if you need a couple in parallel to handle the power.

(source)

If I read it correctly, it should drop even more then I've stated before - slightly above 0.8V at 0.3A?

T.

[trad3mark]

WINRAR!

we have it!

3 x 1N4004 in series gives me a voltage of... 6.5! Nice!

time to get going on the final run of putting this onto board. Cheers for the epic help all! EXCITED TO THE MAX!

[maarten]

Your 7805 regulator is internally protected against overheating, so it will shut off automatically and work again when it has cooled down; so there is no risk as to the heat for your circuit. However, the heat dissipation will also be depending upon the temperature of the environment; this will mean that in a hotter environment (heating, lights...) your regulator may shut down quite fast and often. If you know for certain that this won't be the case, you might go for the regulator. Otherwise I would go for either PRR's suggestion, and buy a 2 watt resistor - which will cost you only maybe 20 or 50 cents, or use the diodes; in the latter case make sure that you get them oriented in the right way, I think Thomeeques suggestion was perfectly valid: give it another try...

Maarten

[PRR]

The heater IS a resistor. So a "voltage divider" just needs one more resistor.

Yes, nearly a whole Watt (no matter how you do it). If you only build transistor pedals, you won't have any >1W resistors.

As Tomas says: use a bunch. I have used thirteen 39-ohm resistors (because a box of 39-ohm resistors fell off a truck in front of me!). In this case, five 47 ohm in parallel will work.

The four diodes should work (use 1N400x parts, not small-signal diodes). I've run a 3.3V CPU from a 5V power supply that way.

Nothing "wrong" with using a 42-transistor chip to do the job of a resistor. But it is elegant to use simple solutions for simple problems. But I guess these days, everybody has LM7805, nobody does big resistors.

> you wouldn't want to be holding it for too long

Yes, that's about right for a TO-220 package with 0.8W in open air. You can hold it, but you'd rather let go soon.

And in this case, it gets as hot as it gets in about a minute.

These parts "can" run MUCH hotter. Hot enough to boil spit. Or give you a blister. LM7805 is rated for 125 deg C, like 250 deg F.

When you close it up in a box, without free air around it, it will run hotter. You also have tube-heat adding to the oven. Try that for a few minutes, whip it open, and touch-test. If spit on your finger doesn't sizzle, it's probably OK.

> I'd have to calculate how efficient the regulator

Assuming identical voltage to the heater, and a perfect linear regulator, the efficiency is identical.

This regulator is "imperfect" because it pulls 5mA extra to power itself. 9V*5mA= 0.045W extra. Compared to the essential 0.81W needed to drop 9.0V to 6.3V at 0.3A, hardly worth mentioning.

> WINRAR!

Assuming the input is pretty solid 9V: The diodes do the same as resistor or regulator, except maybe you have them on-hand. They too will run pretty hot, 1/4W each. We know the 1N400x will take 0.9W "forever", so they are well within their safe zone.

Conceptually, if the "9V" varies, the 7805 will attempt to hold a fixed voltage, the resistor holds a semi-constant fraction, and the diode-string actually makes dip/rise worse. Since R.G. and a billion others offer good solid 9V supplies at popular price, that's maybe not a big deal.

[JKowalski]

1n400's are schottkys so you would need many more of them to get the same voltage drop   A resistor is a much more elegant solution, even if you need a couple in parallel to handle the power.

If I read it correctly, it should drop even more then I've stated before - slightly above 0.8V at 0.3A?

T.

Huh, I always thought the 1n400x family were Schottky rectifiers.  Mistake, mistake.