How do I affordably measure LOW wattages ?

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Old 05-05-07, 01:19 AM
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How do I affordably measure LOW wattages ?

OK I am trying to figure out to go about (relatively affordably) measuring the WATT consumption of a light.

I have a few Killawatts but they do not appear to be accurate with extremely low wattage items (10watts and under)

How does one measure watts when it consumes so little?

Suggestions?

Chris Taylor
 
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Old 05-05-07, 03:25 AM
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Depends on just how accurate you need to be. I would use an amp meter, a volt meter and a pocket calculator.

A clamp on ampmeter will give you 1/10 of an amp accuracy +/- 1 %
It will also not tell you much at all below 1.0 amps.

A bench meter can be used in series with the load, but be very carefull that you are sure not to try to measure too high an amperage this way. The meter will tell you it's limit. Also be carefull to not be holding the leads while you test anything higher than 24 volts. You could use wirenuts to hold the wires to your test leads, and turn power on just long enough to read the meter.
 
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Old 05-05-07, 05:31 AM
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Can you plug in several identical bulbs? This will increase the total load so the Killawatt can get a more accurate measurement. Divide that number by the number of bulbs.
 
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Old 05-05-07, 06:11 AM
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Another way also with a clamp on ammeter should be accurate enough for your purposes.

You can make a simple transformer out of a test cord.

You would make a very short grounded extension cord out of stranded conductors.
You make the hot wire longer so that you can wrap this conductor around the jaws of the meter.
If you put ten wraps around the jaws the reading will be ten times the current draw.
You then divide your reading by ten to get the actual amperage draw.

This is a trick used in the refrigeration trade to check fan motors that draw less than an amp when applying fan blades.
It's the most accurate way to be sure you don't over-blade a motor.

Amprobe company at one time made a plug in version of this principle but it wasn't practical for hard wired motors.

Oh, I stand corrected..............they still do make it:

http://www.contractor-books.com/AM/AM-Clamp-on_Acc.htm

One for + 10x readings:

http://www.contractor-books.com/AM/AM-A50_CT-50.htm

Still simple to make your own.
Just make sure your wire wraps are neat and bundled.
 
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Old 05-05-07, 11:17 AM
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If you don't have a clamp on ammeter and cannot connect additional lights, another option is to set your Kill-A-Watt to measure cumulative usage and run the light for many many hours. Divide the usage by the time and you have a more accurare reading.
 
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Old 05-05-07, 11:28 AM
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Very cool guys thanks. I bet I can get an amp meter. to do the volts amp thing at this low a wattage I think I need better than .1 resolution since they only draw about .1amps :-) (I am estimating about 4 watts per tube)

Multiple tubes is an issue because I do not know at WHAT point these things (kill-a-watts) become accurate. 20 40 100 watts?

but the difference between say 3.9 and 4.1 amps I thing is important. I need to figure out the proper resistor to use for these lamps (the guy I bought them from did not make them right WAY overpowering the LED's burning them out in weeks!) but there VERY cheap so I want to make them work. I got a Variac and had to reduce the voltage to something like 80 volts to stop them from generating too much heat) now I need to figure out "how many watts that is" so I can figure out what the proper resistor is. I have no idea how to do that but I do know I need to measure the current to start :-)

I am going to try the wrap the cord and amp meter trick. That sounds easy and accurate. I was hoping there would be something like a killawatt thats more accurate and capable of lower measures.

Thanks SO much for the responses VERY helpful !! I appreciate it!

If you think of anything else please let me know!
 
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Old 05-05-07, 11:42 AM
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So, now we know "the rest of the story"!
You are dealing in near milliwatt range and a digital multitester with an amps function would likely be better.

Also, since you are referring to LED's is the power you are measuring perhaps DC.
If so the clamp on ammeter will not work unless it specifically can measure DC current which most common ones can't.
 
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Old 05-05-07, 12:02 PM
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no no these are "fixtures" they are AC and have 198led in them.

Whoever designed them used the WRONG resistor way over driving them. So based on replies in this forum it was suggested that I could lower the voltage to lower the current.

80volts was just right :-) so now I need to figure out what resistor to replace the one in the units with so I can run them at 124v at the proper current. I need to figure out what that proper current is.

Its not as simple as 198xrated ma because there is entirely too little cooling in place (80% of the watts goes to heat) So it must be even lower than rated current to compensate for lack of proper cooling. 80volts and the current resistor used gives me this proper current.

I need to figure out what the current is so I can go back to using 120v and still end up with the same current I now get at 80v

IE I do not want to have to buy Variac's for for all the rooms as they cost a lot of money (though they are cool :-)
 
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Old 05-05-07, 01:02 PM
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I thought we covered this, but I will post it again.

E = I x R.

this is called ohms law. this is not the kind of law like speeding that you can break. This is a law of nature.

No matter how hard you try, you cannot break ohms law.

So you cannot change the watts without changing the other stuff. It is impossible. You are following the wrong train of thought.
 
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Old 05-05-07, 01:11 PM
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a 240 to 120 v transformer would have the right ratio for your drop from 120 to 80 volts.

this is not as simple as plug this to this. you will have to wire stuff up.
 
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Old 05-05-07, 01:38 PM
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while I'm not sure about jwhites calc (a 240 to 120 volt x-former is a 2 to 1 winding (and would give you 60 volts froma a 120 input) and for a 120 to 80 volt transformation you need a 3 to 2 ratio (don't know offhand what would give you that but it may be around)) But, using a transformer to lower the voltage would be the best route.

A resistor uses power therefore, it consumes watts. A transformer simply changes the voltage to what you design. A x-former does consume some power but it would be less than a resistor used to "eat" voltage, which is what you are designing.

nearly all the wattage consumed would be from actual led use rather than led + resistor.
 
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Old 05-05-07, 01:43 PM
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if you want to remove the origianl resistor as well, youneed to measure the voltage from the load side of the resistor to the very end of the circuit. That is the voltage being used by the leds themselves. If you measure the line side of the resistor to the load side of the resistor, that will tell you the voltage being used by the resistor.

Both those numbers added together shuold give you the output voltage of the power source you are using.
 
  #13  
Old 05-05-07, 03:46 PM
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well what I was thinking of doing was putting in a Variable resistor.

Adjust till I get the same watt consumption on the meter and then measure the resistance of the pot.

Replace with a resistor of that value. Or just leave the danged variable thing in there. They can't be that much more than a resistor can they relatively speaking anyway. I am talking maybe 10 of them.

Yes I remember ohms law and I can break the rule if I want :-) no seriously if it proves easier I plan to REMOVE the existing resistor and put in the PROPER resistor.

I know enough about this to know you can not tell me thats not possible. these are burning out BECAUSE he is using the WRONG resistor. he is permitting too much current and frying the LED's If you were right it would not be possible to use LED's at all.

You said I could not alter the current BEFORE the lamp but I can alter the current INSIDE the lamp if I am willing to disassemble them and replace components. THAT is what I am talking about.

for arrays where I use 4-6 in a room I will just use a variac but for single installations like in the bathroom doing the work to replace the resistor makes more sense than installing a $60 variac for a single 4watt fixture.

I am going to order a nice meter off ebay. Found one for $70 that measures amps volts and temperature and it can even jack into USB so I can plot all this info :-) cool !
 
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Old 05-05-07, 04:17 PM
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you don;t need to measure wattage. You don;t actually have to measure current either.

since you already know what voltage you want to apply, you can place the variable resistor in circuit and adjust it until you get the correct voltage (as measured from the line side of the fixture(not including the variable resistor) to the opposite connection of the led fixture.(apparently 80 volts is the target).

then measure the resistance of the variable resistor and viola`, you have your reisistor rating.

you can do that with a $10 meter.

I am curious as to what type of meter you are getting for $70 that has a PC interface built into it. I have a couple meters that were several hundred dollars and all the do is give me numbers on the face of them.
 
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Old 05-05-07, 06:04 PM
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Forget about replacing the existing resistor of your lamp assemblies, you can do the job easier by installing an additional resistor in the power supply leads. Yes, you may use an adjustable resistor if you wish.

You want to "drop" about 40 volts across the resistor. To calculate the necessary resistance you need the amperage draw of the lamp assembly.

Take your digital multimeter and set it for AC Volts. Connect up your lamp assembly and the Variac. Adjust the Variac output to 80 volts applied to the lamp assembly. Turn off the power to the Variac without changing the voltage adjustment.

Set the multimeter to AC Amps and connect it in series with one of the output leads from the Variac to the lamp assembly. Reapply power to the Variac and read the amperage from the multimeter. It will be a small fraction of an ampere. You may need to set the multimeter to a lower range to get a meaningful reading.


You now know two of the three terms to apply to Ohm's law. You want to find resistance so the formula is R=E/I or Resistance equals Voltage divided by Current. Take the 40 volts you want to drop from the line supply and divide it by the current you found in the test. Remember that the meter may have (probably did) give a readout in milliamperes so be sure to convert that to amperes before the division. The result will be the necessary resistance needed to drop the voltage. You also need a figure for watts to properly size the resistor. For this multiply the current (in amperes) by the voltage being dropped by the resistor (40) and the result will be the MINIMUM watt size of the needed resistor. I personally would at least double this watt size and remember that the resistor will get hot dduring operation and must be suitably mounted and perhaps even cooled.

If you desire some adjustment (probably a good idea) then you will want to use a variable resistor with the same (or greater) watt rating as calculated above and a resistance rating of at least 25% to a maximum of 100% greater than calculated. You would then insert this resistor in series with one of the power supply leads.
 
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Old 05-05-07, 06:08 PM
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Nap, I picked up a digital multimeter from MCM Electronics a month or two ago that has an optical output and cable to connect to a serial port. I paid about $50. for it and it included the software to be used for graphing purposes.

I haven't done anything with it yet other than load the software into my laptop.

MCM has several of multimeters with serial ouput capabilities. I don't recall any specifications but I doubt that Fluke has any worries.
 
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Old 05-05-07, 08:27 PM
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Ok I am slightly confused. I put a resistor on the 120line coming in ?? I did not know I could use a resistor on AC lines ???

Or desolder one end of the existing resistor and solder in the variable? To eliminate cooling issues (and since they draw SO little power I figure I will just use a multiple amp resistor I am only drawing .1 amps) The reason I want to know the current is to make it easy to set up. Once I know what current its drawing now I can put in the variable plug into 120 instead of 80v and just turn down the resistor till I get the same value as I got on the 80v line. Or am I missing something. Still unsure of exactly where to put the variable?

For the meters I am looking at this one with the USB 190108354116 (ebay) and this is the one I really want but it does not have USB I am hoping they have this second one with USB as it does LUX and SOUND as well !! 190109206156
 
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Old 05-05-07, 09:41 PM
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wow, things sure have changed. Kind of funny side note. Under the DMM section is lists "analog multimeter".


Digital Multimeters

Accessory Analog Multimeter Benchtop Clamp Handheld LCR Meters
================

the resistor will be "in series" in the power line.

in my method, you do not need to know the current since it is going to be the same when you feed the circuit with 80 volts or use a resistor to reduce the voltage used by the light to 80 volts.

a volt meter and an ohm meter are the only 2 things needed to test with and there isn't even any calculation, simple taking readings.

your setup would be like this

L1-------resistor------lamp-------L2

L1 and L2 are the hot and neutral from the receptacle. if you measure the voltage from a point in between the resistor and the lamp for one point and on the lamp to L2 connection, then adjust the variable resistor so you get an 80 volt reading. Then dissassemble the circuit and read the resistance the variable resistor is set at. This is the resistance rating of the new resistor you will need to change the voltage available to your lamps to 80 volts.

If you study series electrical circuits, you will understand why my readings will give you what you want to know. Simply put, in a series circuit, the voltage is divided proportionately according to the relationship of the resistances of the componants of the circuit. You are currently applying 80 volts to the circuit. You need to determine what resistance when put in line will eat up the 40 volts and leave 80 volts for the light. My method does just that.

and as always, current flow is a function of the applied voltage and the resistance of the circuit.

I'm not trying to discount or infer Furd's method is incorrect. It's not. You will end up with the same result either way after all is said and done.
 
  #19  
Old 05-06-07, 02:49 AM
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This whole thing sounds a little hokey. Series resistors is not the best way to control current to a lamp. The resistor will dissipate heat, thus wasting energy. Seems like a simple regulated power supply would make more sense. This is what they do for LED emergency exit signs, for example.
What is this light used for?
 
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Old 05-06-07, 05:21 AM
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Lighting up my Rooms in my house - I do not mind wasting a little energy to heat. We are talking about a mere 4 watts :-)

I will try one and see if it appreciably raises the watt consumption. So what range resistor should I use? It consumes around 11 watts at 120v and roughly 4 watts at 80v
 
  #21  
Old 05-06-07, 06:26 AM
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Using your target of 4 watts at 80 V and ohms law, I get 50 ma

Using ohms law again I get 1600 ohms across the lights.

Next drawing a complete ciruit with a 120 v source and two resistors in series I incert the values above for one of the resistances.

Since I want the lights to drop 80 volts from the 120, I want the other resistor to drop the remaining 40 volts.

Using ohms law agian, I get 800 ohms for the needed resistor. But I get 2 watts. That resistor will not be as easy to find as a 1/4 watt resistor.

I looked at radioshack.com and did not find one.

They do have it listed in the digikey cataloge, but I do not think they sell retail.

I suggest a guitar store. Those people can sometimes get this type of stuff.
 
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