Questions about a multi out dc supply

[strungout]

Oi.

So, I'm thinking about (and gathering some info on) building a regulated dc 'brick'. I want to have a +/-18v out to power a 10 band EQ ( http://www.diystompboxes.com/smfforum/index.php?topic=103893.0 ) and a few +9v outs, maybe 2 or 3, and I'd be using 78xx's.

This http://www.cgs.synth.net/modules/cgs14_psu.html fits part of the bill.

I have a 36vct (18-0-18) 2A transformer. This is RMS, so the actual voltage I'll get to feed the regulators, after the diode bridge will be (18v x 1.414) - 1.1 = 23.9v (I'd be using an RS201 2A Diode bridge I have. The datasheet says it's max drop is 1.1v), correct?

Now, that shouldn't heat up the +/-18v regulators too much, but a 9v regulator will have to work hard, no? Or could I just take the regulated +18v and knock it down to +9v with resistors to power the 9v outs? Or yet again just go with a separate transformer?

Any help appreciated. Now I'll go play around with the EQ some more...

[Seljer]

It'll work, its below the maximum voltage rating, you'll just have to plop on a bigger heatsink and possibly derate the current capacity.


Doing the math: the voltage drop from 24V to 9V is 15V

Looking at the datasheet, for a TO220 package the thermal resistance from semiconductor junction to case is 5°C/W, from junction to air is 35°C . From a quick glance at the selection on mouser, a heatsink for a TO220 device can have a thermal resistance of anywhere between 5°C to 25°C depending on how big a of a chunk of metal it is. So lets say you get something thats not huge and has a rating of about 15°C/W (when you include the thermal paste and isolater pad and everything).

The combined thermal resistance from the junction to air is 15+5=20°C/W. Maximum temperature for the junction is usually around 120°C. Say the ambient temperature is 28°C (on a hot stage). You've got a total temperature difference of 120-28=92°C.

92°C / 20°C/W = 4.6watts

4.6watts / 15V drop over the regulator = 307mA running through it without it overheating.


If you're dropping it from 24V you're going to be turning more power than your 9v load is using into heat anyhow. Putting in resistors or zener diodes or even other 780x devices (so yes, re-regulating the 18V output is an option) will enable you to take a portion of the thermal load off the final 7809 regulator and distribute it among the other components, but it's still going to be heating up something.

[R.G.]

So, I'm thinking about (and gathering some info on) building a regulated dc 'brick'. I want to have a +/-18v out to power a 10 band EQ ( http://www.diystompboxes.com/smfforum/index.php?topic=103893.0 ) and a few +9v outs, maybe 2 or 3, and I'd be using 78xx's.

Just as a bit of kibitzing, if you can't get separate grounds on your 9V outputs, there's no real improvement to be gained (other than possibly heat and/or current splitting) to be had by separate regulators.

I have a 36vct (18-0-18) 2A transformer. This is RMS, so the actual voltage I'll get to feed the regulators, after the diode bridge will be (18v x 1.414) - 1.1 = 23.9v (I'd be using an RS201 2A Diode bridge I have. The datasheet says it's max drop is 1.1v), correct?

Yes, you'll see 18V*1.414 = 25.452 peaks before diode drops; there are two diode drops in a full wave bridge, and the data sheet does say it's 1.1V per element (i.e. diode) at 2A peak. The problem is that the forward peak current in a full wave rectifier is ill-defined. The data sheet shows as much as 1.2V at 10A; I'd use that. So call it 24.25V peak under no-load and nominal line.

Nominal line?? Yes, the AC power line changes. There's a primary voltage spec on that transformer too. The AC power line commonly goes to 125Vac, and smart designers allow for +/- 10% line variations at least.

So your output DC could be 26.68 at high line, no load.

When you load it, the voltage sags due to the winding resistances and ripple run-down on the DC filter caps. Not a big deal because you have so much voltage to start with, but worth remembering for your next power supply design.

If you do an 18V output, positive or negative, from a bipolar ~25V power supply, you get 26.68V - 18V = 8.68V across the regulators, and a dissipation of 8.68W per amp of current pulled through the power supply. As Seljer said, you calculate the heat, the ambient temperture, the max junction temperature (in this case at internal cutoff, which 7800s do) and out pops a maximum heat sink thermal resistance.

Note that the amount of current out of the power supply is CRUCIAL to the amount of heating. Knowing how much current comes out is necessary to being able to pick a heat sink.

For 9V outputs, you have other problems, chief among them being the really high voltage you're putting in.  For 9V out, a 7809 only needs 11V in. If you feed it 26.68V, it will dissipate 26.68-9 = 17.6W per amp of current. This is getting right on up there where TO-220s are barely able to keep up on any kind of heat sink.

It's better to waste the power in resistors. You can use a series resistor so that at max current to the load(s), the resistor eats all but 2V of the incoming power supply. So if you must have 1A out, the resistor needs to drop 26.68 -9 -2 = 15.68V, and then R = 15.68/1A = 15.68 ohms and dissipates 15W. The resistor gets HOT too, but power resistors at the 10 and 25W level are more reliable dissipating that much power than TO-220s.

Obviously, if you only need 100ma, not 1A, the power wasted drops by 10:1, too.

Gotta know or at least estimate how much current is needed.

[strungout]

Baby steps. Baby steps...

RG - If you do an 18V output, positive or negative, from a bipolar ~25V power supply, you get 26.68V - 18V = 8.68V across the regulators, and a dissipation of 8.68W per amp of current pulled through the power supply.(...)

Here, you're talking about the current draw of my EQ? Meaning the higher the current draw of the effect I plug in, the higher the heat dissipation ability is needed, ie, higher rated heatsink?
I measured the current draw from my EQ at 41.3mA @ 17v (Actually, 16.92v. Single supply at this moment). So, I'm looking at about 43.7mA @ 18v.
If I understand correctly, the heatsink will need to be able to dissipate 8.68w per amp, so, with a current draw of 43.7mA it's 8.68 x 0.0437 = 0.379w for the pull of the EQ? But then, I have to decide how much current max I want to be able to pull from the supply (for the total outlets I wanna put in) and decide on a heatsink based on that?

Seljer - If you're dropping it from 24V you're going to be turning more power than your 9v load is using into heat anyhow. Putting in resistors or zener diodes or even other 780x devices (so yes, re-regulating the 18V output is an option) will enable you to take a portion of the thermal load off the final 7809 regulator and distribute it among the other components, but it's still going to be heating up something

RG - For 9V outputs, you have other problems, chief among them being the really high voltage you're putting in.  For 9V out, a 7809 only needs 11V in. If you feed it 26.68V, it will dissipate 26.68-9 = 17.6W per amp of current. This is getting right on up there where TO-220s are barely able to keep up on any kind of heat sink.

It's better to waste the power in resistors. You can use a series resistor so that at max current to the load(s), the resistor eats all but 2V of the incoming power supply. So if you must have 1A out, the resistor needs to drop 26.68 -9 -2 = 15.68V, and then R = 15.68/1A = 15.68 ohms and dissipates 15W. The resistor gets HOT too, but power resistors at the 10 and 25W level are more reliable dissipating that much power than TO-220s.

Obviously, if you only need 100ma, not 1A, the power wasted drops by 10:1, too.

I think I'd rather build for the most amps possible to be available, tho I doubt I'll ever plug in fx that are gonna draw as much, at the same time. The pedal with the highest current draw I ever had is my RP355 multifx that has a 1.5A adapter. The manual doesn't say how much it actually draws tho.

So, the simplest option seems to be to add a 15R 15W resistor. I won't need the 78/7909 regulators since I'm feeding the output from the 7818's output.

I had bought some heatsink a few years back when I wanted to build an lm3886 amp, but for 78/7915 (using this same 36ct 2A transformer). I'm not sure if they are suitable because I don't remember if the guy chose em for me of I did... I can only guess at what the c/W rating was since they aren't marked. They look exactly like this one from mouser http://www.digikey.com/product-detail/en/593002B03400G/HS336-ND/1216343 and the size is pretty much the same, mine's 29x25x12. The one from digikey is rated at 13.4 c/w.

I need to get out of this chair for a while