Measuring power off a ATX power supply 5

[vicw]

Hi! Any suggestions as to how to measure the power given off by a ATX power supply going to a PC Motherboard?

 

[Rich2001]

This site should help

http://www.compute-aid.com/atxspec.html

 

[vicw]

I actually looked at that site, but it doesn't help. I'm actually trying to find how much power my system uses up... I have built a jig using 1ohm resistors to measure the voltage across them, but the voltage drop is too high. I have limited resources. Any suggestions?

 

[rhatcher]

I have never tried to do what you are trying to do before, but based on common sense I assume the one ohm resistors were to be used as a DC current shunt to measure current flow by voltage drop. Resistors are cheap so you may select a smaller value, say 0.1 or 0.01 ohms. I would pick the smallest value that will yield a measureable result with the type of voltmeter you have and the expected current flow. If the voltage is (guessing...) less than 5VDC, causing a voltage drop of greater than 0.25-0.5V will not only cause the measured values to be off but may also cause damage to the PC and/or the power supply.

As another suggestion, if the amperage is expected to be less than 10 amps, you can use the ammeter function of a common multimeter. The multimeter ammeter function will have very low resistance and should not cause a voltage drop.

I hope this helps.

 

[vicw]

According to ATX specs. The max amp for the +5V is 30A and +3.3V is 14A. Do you know of any meters or measuring techniques that can handle over 10A? I agree with your concept.
I, too, agree that a smaller resistor would work, but I would need resistors that can handle that amount of power. My budget is limited... Is there a way to work around this... using the 1ohm resistors(i have 14 1 ohm resistors 50W)?
I currently am using 11 resistors for my current setup.

 

[jbartos]

Suggestions:
1. A shunt that would produce volt drop for voltmeter with current scale
2. A regular voltmeter
3. Watts= volt x amps
4. The voltage for the motherboard is supposed to be within a certain tolerance and relatively stable. Then, the voltmeter from 1. above could have the scale in Watts.

 

[vicw]

Hi jbartos!
Thank you for your suggestion, but how would I go about using a shunt? Would I place the shunt in series with the multimeter? My multimeter can only handle up to 10A. The max. for the +5V source is 30A. or Do I place the shunt in parallel with the multimeter and measure the voltage across the shunt?

 

[vicw]

Is it possible to setup my jig such that it is in parallel with the board? This is a suggestion from a friend of mine. I do not believe it's possible with what I have(14 x 1 ohm resistors). Any suggestions?

 

[electricpete]

I'm not sure I followed the whole thread... but I can advise on the normal use of a "shunt".

"Would I place the shunt in series with the multimeter?"
No.

"Do I place the shunt in parallel with the multimeter and measure the voltage across the shunt?"
Yes. "Shunt" is actually another way for parallel.... intended to be connected in parallel with your multimeter (voltage scale) in this case.

Also, normal application of a shunt resistor for current measurement demands that the resistance be much lower than system impedance so that it doesn't affect the circuit current we're trying to measure. Therefore decreasing the resistance should decrease power requirements of the resistor (assumes constant I) and give you a cheaper alternative. But I'm not 100% sure this application is a current-measuring shunt resistor.

 

[nbucska]

I assume you have an Ammeter up to max. 10 A.

Connect your Ammeter in series with one 1 Ohm 50W resistor.
Connect these to the 5V, with the PC disconnected.
It will show about 5 A.

Get some iron wire about the thickness of a coat hanger.
Find the lenght which -- when connected parallel to the Ammeter --
reduces the reading exactly to the one forth.
This is your shunt. Connect it parallel to the Ammeter.

Now your max current is 40A-- measure your PC. Your actual current is
4 times of the indicated amount.

<nbucska@pcperipherals.com>

 

[electricpete]

As previously stated, I'm a little lost as to exactly what we're trying to accomplish.

nbucska suggested a novel alternate use of a shunt - to divert a fixed portion of the current around the ammeter. That might be a good approach. I would point out that you would be relying on current division among two low-resistance parallel elements (the ammeter and the shunt) and a small change in resistance (as due to temperature variation) might wildly affect the current division.

 

[nbucska]

Yes, Electricpete, you are right -- this is why I recommended thick iron wire. I think this is the cheapest
and still usable solution. <nbucska@pcperipherals.com>

 

[vicw]

Hi nbuscka! I very much appreciate your suggestion. Just a quick question, do I still need the 1 Ohm resistor? I can detach this resistor from my setup. Originally, I believe that there was a voltage drop across my 1 Ohm resistor that resulted in a failure to my test.
Do you think a 12AWG wire will do or should it be thicker?

 

[vicw]

Just one more question... How can I read the current if the PC is disconnected? I would still need some power, so do I connect the power supply to my jig, but without turning the power supply on?

 

[vicw]

Is there an equation I can use to measure the length of wire needed?

 

[nbucska]

Hi Vicw : You use the 1 Ohm to calibrate the shunt i.e. you
replace the PC with the 1 Ohm & ammeter. This will give you
about 5 A i.e. less than 10 A which you can read. After you
calibrated it, connect ir parallel to the ammeter and
connect them in series with the PC input, of course you don't need the 1 Ohm any more.

If you knew the resistance of the ammeter and the actual specific resistance of the wire you could calculate the appr.wire lengh.

AWG 12 should be OK. for the accuracy you need. <nbucska@pcperipherals.com>

 

[jbartos]

Suggestion: Power rheostats and potentiometers are convenient to have for any testing. To use shunts, one may also need low resistance rheostats of resistors. However, the clue is to use a proper methods that have already been addressed above.