Need Help to Build a 50W Load for PSU Load Test

[Galib]

Hi,
I hope everybody is well . I Need to build a 50W load for PSU Load test . I heard that If the PSU runs with 50W Load then it is capable of running with Full Load . But sorry to say that - I have not found any 50W Load @ 5V or 6V or 12V in my market . But in my market 10W 5E Micron Resistor are available . I am thinking to make a 50W Load with these 10W 5E Micron Resistor . I want to Put 5 Micron Resistors in parallel . If I put it in parallel then it will be 50W 1E .

I have attached the Schematic and Vero board design . Please watch these Diagram and tell me - Is it OK or not ???

[shovenose]

C_hegge built himself an awesome load tester....im sure he will reply soon!

[c_hegge]

The design looks good. It's almost exactly how my PSU load tester works. However, there are two main things to keep in mind:

1. Crossloading - Some power supplies handle it better than others, but in general, you want to be drawing about the same amount of current from each of the main rails. I would suggest maybe 3A or so from each rail. That would get you a total of 61W load (15A for the 5v, 36A for the 12v and 10A for the 3.3v).

2. Heat - Resistors get HOT when used for loading, and as they do, their resistance goes up and the current draw goes down, so you may want to attach a heat sink to them.

[Galib]

Thanks both Shovenose and c_hegge for your replies .

I would suggest maybe 3A or so from each rail

But I want to draw 5A from each rail . Then what will be it's schematics ?? How many Micron Resistor have to use for each rail ?? If you give me the schematics, then It will be very helpful to me .

2. Heat - Resistors get HOT when used for loading, ................................ attach a heat sink to them.

If I use a 12V PSU Fan instead of Heat sink , then how it will be ........

[c_hegge]

OK. If you were drawing 5A from each rail, that would be about 101.5W. (25W for the 5v rail, 60W on the 12V rail and 16.5W on the 3.3v). For that, the schematic for the 5v rail would be the same as in your last post, with 5 resistors in parallel. For the 3.3v, you won't get exactly 5A with those resistors, but if you used 7 in parallel, you'd get fairly close at 4.62A (or 5.28A if you used 8).

For the 12V rail, you would need 2 for 4.8A (2.4A per resistor), BUT, the resistors you have for the other rails are only rated at 10W. 2.4A x 12V = 28.8W, which will be too much for them, so I'd recommend using 50W rated resistors.

Cooling those resistors would be difficult without a heat sink. On my load tester, I have all of the resistors mounted to heat sinks AND two 120mm fans which pump out about 500CFM of airflow combined and sound like a plane taking off, and even so, the bottom of the tester gets warm even under only moderate loads. (see attachment for a look inside my load tester)

[Galib]

Oooops !!! your Load is cost too much . I want to finish it within 100 Rupee . Each 10W Resistor costs 10 Rupee . I will buy 10 Micron Resistor at maximum . Just tell me -

1) How much Ampere Draws Per 10W Resistor in 3.3V Rail ?
2) How much Ampere Draws Per 10W Resistor in 5V Rail ? and
3) How much Ampere Draws Per 10W Resistor in 12V Rail ?

I could not find 20W Resistor in my market ...... finding 50W Resistor is Impossible !!!

[c_hegge]

Each one on the 5v rail will draw 1A or 5W (assumming they are 5 ohms). Each ine on the 3.3v rail will draw 0.66A or 2.18W (again, assuming 5 ohms). Each ine will draw 2.4A on the 12v rail, but as I said, you WILL burn them out very quickly.

It's quite easy to work out using ohms law. Basically, Voltage ÷ resistance = Current draw, and Carrent Draw x Voltage = Wattage. In your case, I'm assumming your resistors are 5 ohms, so for the 12V rail, 12 ÷ 5 = 2.4A, and 2.4A x 12V = 28.8W

[Galib]

Hahaha ...... 10W 1E Resistor are also available in the market . I mentioned there 5E because 5 Resistors in Parallel act as 1E . OK, a huge Thanks c_hegge . I got the calculation ....

In 5V rail - 10W 1E --> 5 ÷1 = 5A X 5V = (25VA X 80) ÷ 100 = 20W X 3 (Resistor in Parallel) = 60W

In 12V rail - 10W 5E --> 12 ÷5 = 2.4A X 12V = (28VA X 80) ÷ 100 = 22W X 2 (Resistor in Parallel)) = 44W

In 3.3V rail - 10W 1E --> 3.3 ÷1 = 3.3A X 3.3V = (10VA X 80) ÷ 100 = 8W X 4 (Resistor in Parallel) = 32W

So, The total Wattage is --> 60W + 44W + 32W = 136W .

And I need two PSU Fan for cooling solution and Six Switches also . And How Long ( Minutes ) have I to check attaching this load . Is 5 Minutes enough because the Resistor will produce heat too much - specially in 5V rail - it will draw 5A current !! I wish If I could add 20W resistor in 5V rail - it could much better !!

Due to heat reason - I could not attach load more than 5 minutes . Will this time be enough for testing load of PSU ??

[c_hegge]

Where did the "(___ X 80) ÷ 100" come from?

It should be:

5V rail - 10W 1E --> 5 ÷ 1 = 5A X 5V = 25W X 3 (Resistors in Parallel) = 75W

12V rail - 10W 5E --> 12 ÷ 5 = 2.4A X 12V = 28.8W X 2 (Resistors in Parallel)) = 57.6W

3.3V rail - 10W 1E --> 3.3 ÷ 1 = 3.3A X 3.3V = 10.8W X 4 (Resistors in Parallel) = 43.5W

In total, that would be 176.1W in total

Again though, you're overloading the resistors. You're going to burn them out.

I always test for 15 minutes on my tester, but if the PSU is going to blow up at all, it usually happens within 5 minutes. I'd still recommend attaching them to a heat sink as well as cooling them with a fan, since as the resistors warm up, the resistance increases and the current draw goes down, and in your case, you will probably burn them out.

[Galib]

hmmm ... Theoretically P = V X I - that means VA . But I heard that Practically Power(wattage) is the 80% of VA . So, I assumed the 80% of the VA . But I am not sure is it right or wrong !!!

OK, I will add heat sink also . But Heat sink costs much ! A medium type Heat sink costs 80 Rupees . OK - I will try to collect heat sink from old electronics goods market .