Do computer PSU need a load?

[danz409]

so i'm building a power supply from a computer PSU unit. and i remember in school that we put a huge resistor on them (cant remember the value) before powering them up for testing. i can't recall if this was just to put a load on them for accurate testing, or if it was necessarily to prevent over voltage and causing the thing from melting down, does anyone know of a constant load is necessarily on one of these things for use as a bench-top unit?

[IanB]

It's best to put some kind of load on them for proper voltage regulation. You need to put the load on the rail where the primary voltage regulator sits. On older supplies it is the 5 V rail, but on newer supplies it might be the 12 V rail. It doesn't have to be a big load, 100 mA will do.

[danz409]

what would a ideal load be? just shorting out a power resistor?

btw its just a 20 pin not even the extra 4 pin CPU power cord on the thing. its next to being a ITX unit so most likely its the 5v

[DaMaDo]

I would always just use something like this

It's pretty common in the computer modding community to use those (well more like just shorting 14-15 yourself) and I've never heard of any problems.

For computer PSU specific questions, hardforums.com is useful too.

[IanB]

I would always just use something like this

It's pretty common in the computer modding community to use those (well more like just shorting 14-15 yourself) and I've never heard of any problems.

Er...that reply has nothing to do with the original question...

[DaMaDo]

He was asking if he needed a load and I was just stating that many people run them without one without any problems.

[RCMR]

Yeah, I use them to power the battery chargers for my RC planes and have never had to use a load.

However, loading up the 5V or 3.3V rail can "boost" the output of the 12V rail (which is the one I use) a little and that's sometimes handy.

They still work either way though.

[danz409]

any recommendations for a load tho?

[onewatt]

I experimented quite a bit on ATX power supplies, and I found that no damage or hazardous condition ever occurred while operating the power supply in a NO LOAD condition.  In fact, I found that the ATX supply could regulate all DC voltage outputs perfectly without any load at all.

What I did was hook up a VARIAC that fed the ATX power supply; by varying the AC input voltage between 60 and 140Volt AC - US nominal 120Volt AC - I used my oscilloscope to monitor the PWM IC output feeding the switching transistors. I found that as I adjusted the VARIC AC voltage, the PWM would vary the pulse width fed to the switching transistors, which would regulate the DC output perfectly - without any load attached.  I would also monitor the DC voltage outputs and found them to be spot on perfectly regulated. I found that some ATX supplies used the 5Volt DC as a feedback for all voltage regulation, while others used the 12Volt DC. 

Here's the Switching frequency:

[danz409]

well this is good to hear! saves me a little work and no hot resistor in my unit wasting energy.

does the pulse width modulator have a variable resistor on it?
heck maybe i can play with that and make all the voltage settings on it variable without putting a
pot on the main power source. (did that with my last experimental one) and ran 2 amps though it with a high watt LED... cooked it

[danz409]

that was short lived. plugged it in to play with it *pop* can't find what went out. but it no longer works... my luck...
time to grab one of the other out of my massive box-o-old computer parts

[onewatt]

NO... the pulse width modulator doesn't have a pot to adjust the PW.  It's best to get the part number of the PWM IC and look up the datasheet.  As far as loading resistors go, I've used numerous ones of various wattage and types, and I can tell you that they get hot as hell.  I almost burned the skin off my finger, and several times the resistor started smoking. The only way I'd use a loading resistor after all that, no matter how huge, is screwed into a metal box with a big fan blowing on it; or limit your testing to under 30 seconds. You might want to try one of Dave's constant current loads, which is a kind of ACTIVE LOAD, where you can dial a constant current and the FET has a big heat sink.  I think building an active load is the way to go if you want a reliable and safe loading condition.

[alm]

Many switchers need a minimum load to be stable under all conditions. This should be in the datasheet if available.

As far as loading resistors go, I've used numerous ones of various wattage and types, and I can tell you that they get hot as hell.  I almost burned the skin off my finger, and several times the resistor started smoking. The only way I'd use a loading resistor after all that, no matter how huge, is screwed into a metal box with a big fan blowing on it; or limit your testing to under 30 seconds.

Or you could take the EE approach and calculate the dissipation based on the expected voltage and figure out if it's below the power rating of the resistor. Power resistors can get quite hot under normal load, so just avoid touching them.

[BravoV]

Or you could take the EE approach and calculate the dissipation based on the expected voltage and figure out if it's below the power rating of the resistor. Power resistors can get quite hot under normal load, so just avoid touching them.

+1

With just these very basic formulas like -> Watt = I^2 x R  , R = V / I , you don't need to go through trial & error that might possibly catch fire or even burn down your house.

[onewatt]

I'm not a EE, but I did do the wattage calculations, and I can tell you from experience that just because a resistor falls within the calculated wattage, using P=IE, does not calculate the heat.  I was using a wattage many times what I calculated that I needed, but the heat dissipated was dangerous; that's why they make flame retardant resistors.

But, I'd like to know how to calculate the thermal heat generated by a high power resistor.  Maybe you can explain how to calculate that, so I can improve my power resistor selection.  How big a heat sink would I need to safely dissipate the heat?

For example: An ATX 5Volt and 1Amp, I would use a 5Ohm resistor at 5Watts.  All I can say is good luck that connecting a 5Ohm 5Watt resistor to an ATX power supply that doesn't burn your house down.

[onewatt]

Many switchers need a minimum load to be stable under all conditions. This should be in the datasheet if available.

Thiat might be true if we were talking about any switcher -  but we're not.  We're talking about an ATX Computer Switcher that must follow the ATX power supply design specification.

[danz409]

all an all. i think i found a solution that makes things far easier AND makes it so i can have a variable supply.

http://engineeringshock.com/lm338-adjustable-voltage-regulator-module-p-110.html

now i just need to find a transformer that puts out ~36v hopefully i have one in my parts stash.
also, couldn't find what amps this thing can push but i did notice that in the tags for the youtube video it has 5A so i'm assuming 5 amps

[alm]

I was using a wattage many times what I calculated that I needed, but the heat dissipated was dangerous; that's why they make flame retardant resistors.

How did you determine that the dissipation was dangerous? A power resistor might be designed to operate up to >= 150 °C ambient, and the resistor itself may reach temperatures close to 300 °C. A random datasheet for a 5 W wirewound resistor shows that it's expected to generate a 170 °C temperature rise at its rated power. The resistor is designed to handle this, the only way this may be dangerous if it's touching inflammable materials or skin. The exception to this is power resistors designed to be mounted on a heat sink, typically this includes the larger (eg. 50 W) power resistors. These will have mounting lugs and the requirement of a heat sink will be clearly stated in the data sheet.

Thiat might be true if we were talking about any switcher -  but we're not.  We're talking about an ATX Computer Switcher that must follow the ATX power supply design specification.

Indeed. According to the ATX12V power supply design guide, referred to by the ATX specification v2.2, the typical minimum current for a 250 W power supply (page 15) would be 1 A on each of the two 12 V rails, 0.3 A on the 5 V rail and 0.5 A on the 3.3 V rail. Regarding no-load operation, it is stated that it should not be hazardous or cause damage. It's not required to remain operating.

[Rerouter]

danz just be aware that the devices loose there current capability at higher voltage differentials, e.g. 35V in but 10V out will likely mean bugger all current before you hit the thermal limit of the thing (25V x 1A = 25W way too high for a TO-220 package without a serious heatsink)

this is why some designs switch taps, or have a switching pre-regulator,

[danz409]

danz just be aware that the devices loose there current capability at higher voltage differentials, e.g. 35V in but 10V out will likely mean bugger all current before you hit the thermal limit of the thing (25V x 1A = 25W way too high for a TO-220 package without a serious heatsink)

this is why some designs switch taps, or have a switching pre-regulator,

all i ask is 2amps @ 7v to drive some high end LEDs and anything more is a bonus.

[RCMR]

Might I state the obvious and suggest that an incandescent bulb of a suitable voltage and wattage on the 12V bus might be the cheapest, easiest and most convenient load for dissipations of 5W or more?

[IanB]

Here's a conversion I did:

https://www.eevblog.com/forum/index.php?topic=4541.msg60384#msg60384

The big white thing strapped to the heat sink is a 10 ohm resistor I attached to the 5 V rail as a dummy load. I also did the bulb trick and used an incandescent indicator lamp to provide some additional load.

[danz409]

cool. ill do the same. just found another power supply. still can't seem what happend with the other. but. meh. pays to have a billion of the things laying around.

i got a resistor laying around ill just mount that to the heat sink myself to prevent any over heating,

[onewatt]

I think if you're compelled to use a dummy load, the method IanB used looks terrific.  Nice build IanB.  I like the chunky cement power resistor and heat sink. 

How did you determine that the dissipation was dangerous? A power resistor might be designed to operate up to >= 150 °C ambient, and the resistor itself may reach temperatures close to 300 °C.

I think a temperature of 300 °C is dangerous. I think it's dangerous because you might be a dopey electronics tech like me and put the resistor dummy load on the end of the ATX power cable and lay it a bunch of papers or a wood desk.  When you start to see the paper smoke and the wood starts to smoke and burn, you'll understand why these power resistors can be unwittingly dangerous.  We're all use to using 1/4 watt resistors and they don't usually get hot plugged into a perfboard, but power resistors are a breed unto themselves. If you're unfamiliar with working with power resistors, you might reflexively reach for the resistor to move it, and then you'll touch the resistor and burn yourself because you were unaware of the resistor's temperature.

At one time I was going to get a Non-contact Infrared Thermometer that could measure the resistor's temperature from a safe distance with a targeting laser that you point at the spot you want to measure.  I would let the resistor sit in a ceramic dish while dangling from the end of an ATX power cable.

I'm still waiting for an EE to calculate the thermal properties of a power resistor and show me how to calculate the heat generated of various wattage resistors run at the same current and voltage.  What specs do I need to use and what's the math?  All I can think of doing is using a thermocouple temperature sensor and feed the data into a logging DMM and then upload the temp data into an excel spread sheet and graph the data.  To bad I don't have a logging DMM     or a thermocouple temperature probe

   

[IanB]

I'm still waiting for an EE to calculate the thermal properties of a power resistor and show me how to calculate the heat generated of various wattage resistors run at the same current and voltage.  What specs do I need to use and what's the math?  All I can think of doing is using a thermocouple temperature sensor and feed the data into a logging DMM and then upload the temp data into an excel spread sheet and graph the data.

Well, the answer to this is both simple and not so simple. Firstly, the heat generated is simple. The resistive power law says heat is current times voltage, which is the same as voltage squared over resistance, which is the same as current squared times resistance.

So if I put my 10 ohm resistor across a 5 V power supply, the heat generation will be (5 x 5) / 10 = 2.5 watts.

What is not so simple is what temperature the resistor must reach to dissipate 2.5 W into the surroundings. This will depend quite a lot on the size of the resistor body, the air flow around the resistor, and the temperature of the surroundings. Or in my case any heat sink the resistor is touching. Probably the most reliable way to proceed is to connect the resistor to a regulated power supply and feed the expected worst case current through it, then measure the surface temperature with an IR temperature probe. Experimental data always trumps calculations, especially when some parameters are uncertain.

The temperature of a hot resistor is given by the heat transfer law:

(heat transfer) = (heat transfer coefficient) x (surface area) x (temperature difference)

In this equation the heat transfer coefficient is the thing we generally don't know very well. There may be some estimates given on the resistor data sheet to assist, but these are only estimates. It will all depend on how and where the resistor is installed.

Here's an example calculation.

Let's suppose our big ceramic resistor is 50 mm long and 10 mm on a side. Then its surface area will be 50 x 10 x 4 = 2000 mm2.

Further, let's presume the heat transfer coefficient will be 10 W/m2/degC.

Lastly, let's assume the ambient air temperature is 25 C.

Then to dissipate 2.5 W we have:

2.5 W = (10 W/m2/degC) x (2000e-6 m2) x (T - 25 C)

Solving for T we get:

T = 2.5 / 10 / 2000e-6 + 25 = 150 C

Quite toasty 

It's important to stress here that the value of 10 W/m2/degC for the heat transfer coefficient is an estimate. We really don't know it with any accuracy unless we measure it for our specific build. But it's clear that even a piddling 2.5 W of heat dissipation has the potential to make things quite hot.