Do computer PSU need a load? - Page 1
Forum section: electronics / beginners

author: danz409
author_rank: Contributor
timestamp: January 08, 2012, 01:48:15 am
content: 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?

author: IanB
author_rank: Super Contributor
timestamp: January 08, 2012, 01:55:39 am
content: 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.

author: danz409
author_rank: Contributor
timestamp: January 08, 2012, 02:11:52 am
content: 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

author: DaMaDo
author_rank: Contributor
timestamp: January 08, 2012, 03:26:30 am
content: I would always just use something likethisIt'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.

author: IanB
author_rank: Super Contributor
timestamp: January 08, 2012, 03:29:19 am
content: Quote from: DaMaDo on January 08, 2012, 03:26:30 amEr...that reply has nothing to do with the original question...

author: DaMaDo
author_rank: Contributor
timestamp: January 08, 2012, 03:30:58 am
content: He was asking if he needed a load and I was just stating that many people run them without one without any problems.

author: RCMR
author_rank: Frequent Contributor
timestamp: January 08, 2012, 03:36:55 am
content: 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.

author: danz409
author_rank: Contributor
timestamp: January 08, 2012, 04:42:38 am
content: any recommendations for a load tho?

author: onewatt
author_rank: Contributor
timestamp: January 08, 2012, 04:51:53 am
content: 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:

author: danz409
author_rank: Contributor
timestamp: January 08, 2012, 05:01:11 am
content: 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 apot 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

author: danz409
author_rank: Contributor
timestamp: January 08, 2012, 05:19:25 am
content: 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

author: onewatt
author_rank: Contributor
timestamp: January 08, 2012, 05:35:29 am
content: 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.

author: Unknown
author_rank: Unknown
timestamp: January 08, 2012, 05:46:50 am
content: Many switchers need a minimum load to be stable under all conditions. This should be in the datasheet if available.Quote from: onewatt on January 08, 2012, 05:35:29 amOr 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.

author: BravoV
author_rank: Super Contributor
timestamp: January 08, 2012, 06:07:11 am
content: Quote from: alm on January 08, 2012, 05:46:50 am+1With 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.

author: onewatt
author_rank: Contributor
timestamp: January 08, 2012, 06:09:40 am
content: 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.

author: onewatt
author_rank: Contributor
timestamp: January 08, 2012, 06:22:27 am
content: QuoteThiat 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.

author: danz409
author_rank: Contributor
timestamp: January 08, 2012, 06:34:10 am
content: 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.htmlnow 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

author: Unknown
author_rank: Unknown
timestamp: January 08, 2012, 06:44:15 am
content: Quote from: onewatt on January 08, 2012, 06:09:40 amHow 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.Quote from: onewatt on January 08, 2012, 06:22:27 amIndeed. According to theATX12V 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.

author: Rerouter
author_rank: Super Contributor
timestamp: January 08, 2012, 06:52:19 am
content: 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,

author: danz409
author_rank: Contributor
timestamp: January 08, 2012, 06:55:14 am
content: Quote from: Rerouter on January 08, 2012, 06:52:19 amall i ask is 2amps @ 7v to drive some high end LEDs and anything more is a bonus.

author: RCMR
author_rank: Frequent Contributor
timestamp: January 08, 2012, 06:57:38 am
content: 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?

author: IanB
author_rank: Super Contributor
timestamp: January 08, 2012, 08:13:07 am
content: Here's a conversion I did:https://www.eevblog.com/forum/index.php?topic=4541.msg60384#msg60384The 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.

author: danz409
author_rank: Contributor
timestamp: January 08, 2012, 08:37:28 am
content: 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,

author: onewatt
author_rank: Contributor
timestamp: January 08, 2012, 03:42:54 pm
content: 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.QuoteI 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 DMMor a thermocouple temperature probe

author: IanB
author_rank: Super Contributor
timestamp: January 08, 2012, 05:06:32 pm
content: Quote from: onewatt on January 08, 2012, 03:42:54 pmWell, 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 CQuite toastyIt'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.

author: Unknown
author_rank: Unknown
timestamp: January 08, 2012, 08:53:23 pm
content: Quote from: onewatt on January 08, 2012, 03:42:54 pmSo put them in a box with plenty of ventilation to avoid touching them. 1/4 W resistors can also get quite toasty near their rated power by the way, it's just that we tend to run them at << 1/4 W. Even though they may get 100 °C instead of 200 °C for a power resistor, 100 °C is still plenty to burn your skin.Quote from: onewatt on January 08, 2012, 03:42:54 pmAs IanB I will assume you mean temperature and not heat, since heat equals power in this case.The data sheet will sometimes contain a graph of temperature rise vs dissipated power. Add the temperature rise to the ambient temperature and you've got an estimate for the temperature it will reach. This is usually defined for the resistor sitting in free air, so it will be higher for an enclosed space. The maximum power is usually derated with ambient temperature, that 5 W resistor won't be able to handle 5 W at 150 °C ambient. If the data sheet does not contain a load vs temperature curve, check other data sheets of resistors with similar materials, construction and dimensions.Another way would be to estimate the thermal resistance from the temperature derating curve. Temperature derating often just shows the maximum power that will keep the temperature below the max. operating temperature, so if a 20 W resistor can handle 100% of the power at 70 °C ambient, linearly decreasing to 0% at 270 °C, you might estimate the thermal resistance to be 200 K / 20 W = 10 K/W. This assumes that the thermal resistance is constant, which may not be accurate, so it may be too conservative (high).

author: danz409
author_rank: Contributor
timestamp: January 08, 2012, 09:39:53 pm
content: my luck sucks... was starting day 2 of my power supply build. bought some new test leads that have a clip on attachement from radioshack.. (yea i know.. horrable place..)but honestly the price wasn't too bad for once $15 AND they are fairly decent qualityhowever i went to go do some testing with my $4 Cen-Tech and apparently now there is a constant short between the leads somehow. i opened it up and i see nothing on the board that would cause the issue... that's 2 of these chepos down and 1 left.. problem is... i spent the last 2 hours looking for it and i can't find it.thankfully i got another coming in tomorrow with a heap of components from Jameco. which i have work so.yay for not soo productive weekend

author: onewatt
author_rank: Contributor
timestamp: January 08, 2012, 10:46:56 pm
content: QuoteI remember from college physics, for example, that a swimming pool with 100 °C water temperature has a lot more heat than a cup of water at 100 °C - even though both have the same temperature.Thanks for explaining that Watts relates to heat. I think I have a better understanding of what's going on.

author: IanB
author_rank: Super Contributor
timestamp: January 08, 2012, 11:01:58 pm
content: Quote from: onewatt on January 08, 2012, 10:46:56 pmThere's a difference between stored heat and power. Power is the flow of energy, and heat flow is the flow of thermal energy. Thermodynamics says that all forms of energy are equivalent and energy is conserved when it changes from one form to another.So in the case of a resistor, the electrical power consumed by the resistor (volts x amps) is equal to the thermal power (heat flow) being generated by the resistor. The heat being generated inside the resistor will cause it to heat up until the heat leaving the resistor balances the rate of heat generation. If you wrapped the resistor inside a thick layer of insulation so the heat could not escape, the resistor would get very hot indeed and might even melt.

author: Unknown
author_rank: Unknown
timestamp: January 09, 2012, 12:37:05 am
content: There's no difference between a heat sink and a power resistor as far as the thermal behavior is concerned. The only difference is that the thermal resistance of a power resistor is often not specified, you have to estimate it from other data like temperature rise or dimensions. Dave did some videos about thermal calculations for heat sinks, and I'm sure there are plenty of other resources.

author: IanB
author_rank: Super Contributor
timestamp: January 09, 2012, 02:09:07 am
content: The important thing to know with any design is that the "thermal resistance" between the device (heat sink or whatever) and the surroundings is not a fact. It is an estimate based on certain assumptions or presumptions about the environment and air flow around the device and how it is installed. For example there may be thermal conduction through the leads or soldered joints of the device and for SMT devices this could be critical to good performance.Huge differences in heat transfer can exist when considering enclosed and open spaces, free convection or forced convection, nearby hot components, and so on. Just because the manufacturer quotes a number or provides graphs in a data sheet, it doesn't mean you can rely on those numbers without doing your own due diligence.

author: Unknown
author_rank: Unknown
timestamp: January 09, 2012, 03:52:46 am
content: Absolutely, but this due diligence likely consists of doing actual measurements, especially with the limited thermal specifications that you usually get with power resistors.

author: vk6zgo
author_rank: Super Contributor
timestamp: January 09, 2012, 04:51:00 am
content: This seems to be a practice confined to computer power supplies.Most switchmode supplies used in other service don't care if they have a load or not!Early Philips TV supplies needed a load,or they would go into the dreaded "hiccup" mode.Philips,being a power unto themselves,never stopped gazing at their own navel long enough to realise nobody else had this problem,& Techs could reduce the load with abandon,in the search for a fault in other brands.Computer PSU's may have more legitimate reasons for this practice--for instance,more critical voltage regulation,but they are definitely out of step with the rest of the Electronics industry.VK6ZGO

author: Unknown
author_rank: Unknown
timestamp: January 09, 2012, 06:56:12 am
content: Most current Mean Well SMPSs with multiple outputs specify a minimum current for each output. Some switchers in Tek scopes also needed a minimum load to start reliably. I don't think it's limited to computer power supplies.

author: amyk
author_rank: Super Contributor
timestamp: January 09, 2012, 09:47:23 am
content: In many cases the minimum load is already provided by resistors internal to the PSU, next to the output.(They're also known to cook the capacitors next to them.)

author: danz409
author_rank: Contributor
timestamp: January 09, 2012, 10:23:37 pm
content: mine may need a load D:http://youtu.be/yujYeMnXlNQ

author: danz409
author_rank: Contributor
timestamp: January 10, 2012, 12:07:11 am
content: i did some experiments WITH a load. and it still does the same thing... it may be some sort of safety feature