Yes it is feasible, to use the center tap connection on your transformer for lower output voltages. Actually a good idea....That way the heat dissipated by the regulator will be cut in half.

How to detect if it is in high range? Many ways to do this...but the simplest would be to add, in parallel with the input filter caps (C1, C2), a LED with its current limiting resistor, and a 16 volt zener in series. That way, in the low range, the voltage will not be enough to breakdown the zener +led voltages and it will be dark. In the high range, there will be sufficient voltage to light it up.

That project is WRONG.
It uses 240 ohm and 5k ohm resistors which are for the more expensive LM138. The LM338 needs 120 ohms and 2.5k ohms.

Thank you both!

audioguru: I suspected something was wrong when the resistances didn't match the LM338 datasheet, but I figured I was just ignoring something.

As for the LEDs, I had already thought of using a zener diode to detect the high voltage, but what about an LED to detect low voltage? The only circuit I could think of would involve a JFET turning off the "low" LED when the zener breaks down, but by the time that the JFET switches, the "low" LED could already by overvolted...

EDIT:

Also, when the supply is switched to the low voltage, what happens when R2 is at 2.5k?

You should always question a design when it looks like it was drawn with an etcher-sketch...

Your 24V transformer will have a peak voltage of 34V and the bridge rectifier and ripple on the filter capacitor will reduce it to 29V.
But for a 30V output the LM338 needs an input of at least 33V.

You need a comparator with a voltage reference to detect a voltage setting that is above or below 12.5V then it switches the transformer with a relay.

An LM10 is an opamp with an adjustable voltage reference that can be used as the comparator.

I'm getting ready to build the same thing.
I have 19VDC coming in from an existing PSU and I need:
12V - 2.5A avg (3A Peak) AND 5V - 0.5A avg (1A Peak)

This is the basic design I started with


So I'm going to run that into a LM338(5A) w/heatsink to bring it down to 12V
That 12V will go to a 12V load AND an LM7805T to give me my 5V

So on the LM338 I'll put a 120 ohm and 1.2k ohm resistor.
________________________

I'm confused on a few things.
1) All the diagrams are showing me .1uf cap across the +/gnd on the input side and 1uf on the output side
Do I need to increase these values for the 3-4Amp load?

2) Do my resistors need to be a certain wattage or is 1/4w fine?
I mean the load itself isn't going through it so I don't see why it would matter

Thanks,

Kenny

For voltage selection detection you could use a two color LED and add a zener in series with one of the LED connections. That way you will get one color with a low voltage and a second color when both light with the high voltage.

Well I want both the 5V and 12V at the same time.

I modified the diagram I found with what I was going to do.
Does this look right?

Your 7805 regulator is missing an important capacitor from its output to ground.
If it is not beside the LM338 then it will also need its own input capacitor.

ahh ok, same specs as the other two are fine right?
And 1/4watt resistors are fine?

Sorry these must sound like stupid questions to you, I'm trying to learn

-Kenny

You have your 5V load disconnected from ground and a series capacitor instead.
The 5V regulator does not have a capacitor from its output (+5V) to ground.

Calculate the power in the resistors to make sure:
1) 1.25V squared divided by 120 ohms= 0.013W.
2) (1.25V/120 ohms) squared x 1200 ohms= 0.13W.

I don't know if you can cool the regulators enough.
The peak power in the LM338 is 28W.
The peak power in the 7805 is 7W.

hmm alright, what if I used a regulator to go from 19>15 then another 15>12 then another 12>5