Proper Power Supply - Different Voltages

[row-she]

Hello there,

I'd like to build a power supply that puts out different voltages (18v, 12v, 9v, 5v) for my experimental board.
I was thinking of using a 18vdc wall wart and voltage regulators. Found this schematic from AMZ...

http://www.muzique.com/images2/power9.gif

... which is almost exactly what I need.

- I have a 18vdc / 2000ma Wall wart
- I'd like to use the L7800 series regulators (L7812cv, L7809cv, L7805cv)

Couple of questions here:

-the 12v out needs maybe 600ma (900ma tops)
-the 9v and 5v each maybe 300ma

-> should I use heatsinks?? not quite shure about the power dissipation...
-> whould you regulate 18 to 12 then 12 to 9 then 9 to 5 (doesn't make much sense I just realize ... anyways...)
-> it's probably better to regulate 18 to 12, 18 to 9 and 18 to 5, right?

-> also I there are no values in the schematic (which is totally understandable - not everything's got to be open source, right?..)
    I have a pretty good idea what to put there (the typical 100u, 47u, maybe I'd throw in a couple of 47ns..)
    BUT... any recommendations?

-> also - do I really need the diodes between IN and OUT of the 7800s ?? I don't really get it...

... so... that's it for now... THANKS IN ADVANCE FOR YOUR ANSWERS!!

Cheers,
Andy

[balkanizeyou]

900ma is dangerously close to the maximum of the 7812 chip (it can provide 1A according to a datasheet) and when regulating from 18V it will have to dissipate (18-12)*0,9=5,4W, so you absolutely have to use a heatsink. Same goes with the 9v and 5v regulators.

-> whould you regulate 18 to 12 then 12 to 9 then 9 to 5 (doesn't make much sense I just realize ... anyways...)

Yeah, that's not a good idea - your 7812 regulator would have to provide roughly 900+300+300mA=1,5A which it simply cannot do. So, as you said - you want to regulate 18 to 5V, 18 to 9V and 18 to 12V.

And yes, you want those diodes in there, they form a protection against the situation, in which the output voltage is higher than the input voltage (it can happen for example with a big capacitance at the output with shorted input) - usually when Vout>Vin the regulator gets damaged, but the diode conducts the excess voltage from the capacitor to the input pin.

[TejfolvonDanone]

-> should I use heatsinks?? not quite shure about the power dissipation...

According to this datasheet the thermal resistance between junction and air is 65 °C/W. So for 1 W discipation it's junction's temperature rises 65 °C.
The maximum discipations are input to output voltage difference of the regulator * output current:
12V -> 6V*0.9A = 5.4W
9V -> 9V*0.3A = 2.7W
5V -> 13V*0.3A = 3.9W
The maximum junction tremperature of the regulator is 125 °C. All of them would overheat without a heatsink.
Also 1A is the maximum current for the regulator so you might want to add a transistor to increase the output current capability. Fig 13 in this datasheet: https://www.fairchildsemi.com/datasheets/LM/LM7805.pdf

-> also - do I really need the diodes between IN and OUT of the 7800s ?? I don't really get it...

The diodes are for protection. You don't really NEED them but it's always good to have some protection.

-> also I there are no values in the schematic (which is totally understandable - not everything's got to be open source, right?..)
    I have a pretty good idea what to put there (the typical 100u, 47u, maybe I'd throw in a couple of 47ns..)
    BUT... any recommendations?

100uF electrolytic and a 100nF ceramic cap for every regulator's out and 100uF +100nF for the input.

[Mark Hammer]

As a more general strategy, I think the thing to remember about power supplies is that they are general-purpose devices.  You don't necessarily know what you will be using them with, know how far you will need to stretch its performance, or have absolute assurances about the performance and current demands of the things you will be using it with.

So the best course of action is to design/build around the worst-case scenario and have as many safeguards as you can manage.

[row-she]

Hey,

thanks for the advice! Appreciate it. I have some 7,2K/W heatsinks laying here, this
should be enough for all the different 7800s, I guess?

Just one more thing,
should I put them on with a little bit of heatsink-creme (sorry don't know the english word)
or use one of those isolation-discs? I heard in in some situations you can't make a electrical connection
with the chassis or heatsink because the dissipation plate is internally connected to one of the terminals...

So the best course of action is to design/build around the worst-case scenario and have as many safeguards as you can manage.

thanks, that sounds like something I should keep in mind - I much to often tend to walk down quick-and-dirty-lane 

Cheers

[TejfolvonDanone]

thanks for the advice! Appreciate it. I have some 7,2K/W heatsinks laying here, this
should be enough for all the different 7800s, I guess?

If you'd like to calculate the heat discipation you could use the thermal-electric analogy. This means that heat "circuits" can modelled with electrical components and circuits. You have to substitute
Temperatue - voltage
Power (discipation) - current
Thermal resistance - resistance.
So in this case you have a 5.4W "current" source feeding 2 resistors in series: the junction to case resistors and the case to air (=heatsink).
So the junction's tepmerature will rise above the ambient temperature 5.4 W* (5 K/W + 7.2K/W) = 65.88 K. If you say the ambient tepmerature is about 50°C (in the enclosure this is reasonable) the junction will be around 115°C which isn't really far from the maximum tepmerature of the junction. That's almost the worst case scenario.

should I put them on with a little bit of heatsink-creme (sorry don't know the english word)
or use one of those isolation-discs? I heard in in some situations you can't make a electrical connection
with the chassis or heatsink because the dissipation plate is internally connected to one of the terminals...

You should always put some paste on it because it helps discipate the heat.
If you don't put an insulator between the IC and the heatsink you just connect the heatsink electrically to the given leg of the IC. On the 78xx series it's the ground. The question is if you connect several 78xx's ground leg together is it a problem? Because in your case the ground is shared you connect it together anyways so it isn't a problem here.