Build this simple, variable output power supply using an old PC power supply. Most people would have a few old computers lying around and if not, you might be able to get one from a friend or even buy one for $5.
So, you’ve probably searched for the “how to” and found vague information. Well, that may be because there are so many manufacturers out there with various design differences.
Here, we’ll look at one of those may variations and see that this particular one is very simple to modify. Also, at the same time, we’ll try to explain some of the vague information that you may have seen.
WARNING: NEVER OPEN THE CASE WITH THE POWER PLUG CONNECTED. ONLY ATTEMPT THIS PROJECT AT YOUR OWN RISK. IF YOU’RE NOT COMFORTABLE AND FAMILIAR WITH HIGH VOLTAGES (UPWARDS OF 220VAC HOUSEHOLD CURRENT), DO NOT ATTEMPT THIS PROJECT.
Of the few PC power supplies I’ve looked at, this one turned out to be the right fit for its intended application. The basic requirement was to have a +12V power supply that is variable centered around 12V. It would be great if it can vary from 0V to 15V, but not possible within the scope of this discussion.
It turned out that the maximum voltage is around 15V (without load) and minimum of somewhere around 2.55V. The limitations for the maximum voltage, most likely, is coming from over-voltage protection (OVP) where the unit shuts down.
The minimum voltage would typically be relative to the reference voltage of the PWM controller device
|The main modification is in the feedback circuit. The picture shows where two resistors were removed (R49, R50) and R50 was replaced with a potentiometer. The green and black twisted pair of wires go to the pot mounted on the side. The pot used here is a 25Kohm “linear” pot.It turns out that this unit had feedback(FB) loops for th 5V and the 12V, so I thought it would regulate the 12V when it was loaded, but that wasn’t true. The regulation was only on the 5V.R49 was for the 5V FB loop and it was removed. Otherwise, 12V wouldn’t regulate.
|The FB pot is used as the voltage control and it’s mounted on the side. If you’re not familiar with the hookup of pots, just copy what you see in the picture. It’s not clear in the picture, but the black wire connects to both the right and the center pins.An LED was connected to indicate power. This one is powered from the 5V tap, but you can use just about any of the voltages available (+/-5, +12). See the red LED mounted with hotglue below the pot. For more information on how to connect LED, go to LED Basics page.
|Locate the power output wires coming out from the PCB. Unsolder the +5V and -5V wires. If there are others, like 3.3V, remove them also. If you want to utilise these other voltages, you still can use them. They may not regulate as accurately, though. For the remaining +12V wires, you can just use them as they are or terminate them to a banana jacks or some suitable connectors. You don’t need all of the parallel wires for the +12V and the ground. You can keep them all or reduce to about 4 and 4.
|As for the power switch, easiest thing to do is to just use what came with the PC. You can mount the switch to the side of the unit by shortening the wires, keeping track of the which wires go where. If you’re ambitious, you can redo the whole switch section, since the wires go to the 2 AC ports. See blow.
|You can possibly eliminate the extension receptacle (the one on the left in the picture above) and have the wires go straight from the switch to the PCB.
You may have read elsewhere that the power supply needs a load for it to “start up.” I’ve seen that in certain ones (really old ones), but not this one. You just have to see what you have. And if you need a load, just put one on. The load size may depend on the unit. Otherwise, if it starts up, don’t worry about it–just use it.
Also, the output current may be diminished at higher voltages than at lower. The output power is still whatever the unit is rated at. If it’s rated at 250W, then at 12V, the current is about 20A. At 15V, the output current would be about 17A. However, if you look at the label, at 12V, the max output current is 10A, or 120W. So, at 15V, you may only get 8A. These figures are not verified.
If you’re interested in a 5V supply instead, then remove R50 and replace R49 with the 10K(appx) pot. You’ll have a regulated 5V and +/- few volts variable, but not so regulated 12V.
One design variations in these old supplies is that the FB schemes are not all the same. Some regulate just the 5V. Some others have elaborate scheme that it’s hard to reverse engineer. My guess is that, if ambitious enough, you can create your own FB scheme, but that’s beyond the scope of this discussion.
Additional Info for the Advanced
These old power supplies use the TL494 PWM controllers. There are several other ones, but this seems to be more mainstream.
Looking at the partial schematic of a similar power supply below, R30 and R31 are the R49, R50 in our unit. The +12v and +5v labels go to the outputs.
|After disassembling the unit, locate the +12V output where lots of wires are bunched together, and follow the traces and identify the two resistors in question.If you were lucky enough to find that this was the FB scheme they used, simply modify as described in the discussion.If you only found the FB resistor for the 5V, then simply cut the 5V trace in the PCB between the output and the resistor and solder a jumper wire from the resistor (the side that the trace was cut) to +12V output. This should generally work unless something else is going on with the circuitry.Depending on the resistor values used in the circuit, you’ll have to calculate the appropriate value for the top resistor. Typically, you might find 4.7K ohm resistor on the bottom of the FB divider. In the schematic, it’s the R32. Then you need something like 25K pot for variable output.Calculating the resistor value for the top resistor:
R = Vout(4.7K ohms)/2.5V
where Vout is the desired output voltage; 4.7K is the bottom resistor and 2.5V is the voltage that the FB voltage is compared against. The 2.5v comes from the resistor divider, R33 & R34, where Vref is 5V and the two 4.7K resistors divide the voltage in half to 2.5V. Caution here, since this may not be standard. It may not divide by 2, necessarily.