power supply design -12v, 3a
Hi Mengghee !
Don´t worry about the kind of the load. It is a DC power supply with output protection (short circuit, thermal, etc). Of course you should read the LM350 datasheet (National Semiconductors) to design properly your power supply. But it is very simple. With the components I listed, I think the only difficult is to know how to connect the resistor and the potentiometer.
Before that, the transformer secondary wires connected to the rectifier bridge (tilde terminals), the positive terminal connected to the positive of big capacitor and the negative (groud/earth0 connected to the negative of the same big capacitor (the filter) . Basic unregulated power supply. It should provide a 24 to 28Vdc with ripple.
Ok, the LM350T (TO220 package) has 3 pins: looking at it front view: pin 1 is the adjust, pin 2 is the output voltage (regulated) and pin 3 (the rightmost) is the input (from the positive of the filter capacitor).
Connect the resistor between Voutput and Adjust pins (2 and 1) and the potentiometer (the center terminal and one of the other terminal) between the pin Adjust (2 of the LM350T) and the ground (the negative terminal of the big capacitor). The remaining capacitor (smaller) you just connect between output pin
(positive of the power supply) and ground (negative of the power supply).
Sorry for not drawing the schematic, but I think the directions are enough. If you have any doubt, ask again. Remember this power supply will not give you output voltages below 1.25V (because it is used as reference voltage at the LM350). But the construction is very simple and all components are easy to find and buy.
Special attention to the heatsink. LM350T has a thermal resistance junction to case of 3K/W and the maximum junction temperature is 125C. So at 1.25V output and 24V input and 3A current, you have (24-1.25)V x 3A = 68W (impossible to dissipate)
The LM350T with a 4K/W heatsink (a very common size) and the thermal compound could dissipate only 125C/ (3+4+1)K/W = 15W.
One alternative is to use "K" package (LM350K) which has a lower thermal resistance (1.2K/W) together with a larger heatsink (maybe 2K/W) to obtain 125/(1.2+2+1) = 30W power dissipated (or 10V difference between input and output at 3A current), below this output voltage, your output current must be below 3A. This is another disadvantage: you are restricted to some voltage and currents, or in the other words, you cannot use it for the full range of voltages with 3A consumption.