Can nomeone explain this to me: current(AC vs DC)

[John Paul]

I've got some of these small Halogen "Under-the-cabnet" lights rated at 20 watts and 12 Volts each.<p>If I'm understanding correctly, 20 watts divided by 12 volts is about 1.7 amps. However on AC 12V I'm reading about 1 amp, and on DC 12V I'm reading almost 2 amps.<p>What am I missing?<p>
Thank you,
John

[Trejep]

John Paul,
Here is a website showing you the waveforms of the different current flow (dc vs. ac) http://physics.mtsu.edu/~phys2020/Lectu ... s__ac.html<p>I would imagine your multimeter is measuring the RMS (root mean square) of the current flow which would explain the lower value for your AC current flow.

[John Paul]

So you're saying that it actually it drawing about two amps AC?<p>Thanks,
John

[jimandy]

<blockquote><font size="1" face="Verdana, Helvetica, sans-serif">quote:</font><hr> So you're saying that it actually it drawing about two amps AC <hr></blockquote><p>Who know exactly what the lamp draws, The lamp mfg. has to give a very liberal spec in case somebody plugs in a bunch of lamps and it starts blowing breakers. And "about two amps AC" is very close to your measured 1.7 amps.

[Trejep]

Actually no...I guess at times during the cycle it would be drawing around 2 amps and other times there is no current flow. The measurement will "approximately" be the effective current draw over a period of time.<p>[ March 12, 2005: Message edited by: Trejep ]</p>

[Chris Smith]

It’s a sinusoidal percentage or average that we use for the equation. <p>We add up the missing or Zero points with the sine wave portions of "on", then average it out over one cycle, or time. <p>From memory its around 66% of the total?

[Ed B.]

The average of a sine wave - over one cycle - is "Zero."

[Chris Smith]

Not true......<p>The average under the curve, is consumption. <p>The only ZERO function, is in pure theory, and only if,... no secondary windings consume such amperages.

[sofaspud]

Using your DC figures, the percent of error is about 12% or slightly more. Not outrageous I wouldn't think. Or perhaps the halogen bulb is actually 22 or 23 watts at 12VDC. In that case the amperage doesn't seem out of the ordinary at all. I admit, though, to knowing next to nothing about how tightly bulbs are spec'd.

[josmith]

Where are you finding dc in the circuit? Most low voltage lighting uses a transformer but no dc rectifier.

[terri]

Most of my technical life has been spent hunting down and mercilessly killing extraneous uncontrolled variables.<p>Some questions to ask:<p>(0) What frequency of AC is being applied? (Lamp filament L-C ratio? [Yes, Virginia, lamps sometimes exhibit self-resonace.])<p>(1) What kinds of meters are you using? Sensitivity (loading effects)? RMS? P-P? "Average" under the curve? Accuracy? (Many inexpensive meters will drop off in sensitivity above audio f.)<p>(2) Does the lamp light as brilliantly on both?<p>(3) Are you using a true sinusoidal AC, or is it from some kind of inverter? (Again, many inexpensive meters will drop off in sensitivity above audio f.)<p>(4) Are there other appliances which could be trashing your measurements? (E.G, Lamp dimmers, etc?)<p>(5) Can the sampling rate of your meter be heterodyning ("beating") with the frequency of the measured parameter? (Are your readings steady or do they dance around?) <p>(6) Is (are) the rectifying circuit(s) (diodes, etc) in your meter(s) OK?<p>(7) Is the AC superimposed on a DC level?<p>(8) Is the DC relatively ripple-free?<p>Just a sample, mind you, based on memory. I have run afoul of every one of those uncontrolled variables, and I have often recited "The Scientist's Prayer" on account of them: "Give me the strength to probe this question until my observed results equal my calculated results and in pious glee I discover and assault my error."<p>Actually, given an affirmative answer to Item 2 above, these readings are extreme, but barely within the limits of variability of accuracy, non-linearity in the meter(s), lamp spec variables (as jimandy has suggested), etc.<p>[ March 13, 2005: Message edited by: terri ]</p>

[John Paul]

[QUOTE=josmith]Where are you finding dc in the circuit? Most low voltage lighting uses a transformer but no dc rectifier.[/QUOTE]<p>That is when I test the light with an AC>DC transformer(not the one supplied)<p>
[QUOTE=terri]Some questions to ask:[/QUOTE]<p>(0) I would assume around 60 Hertz.<p>(1) a cheap Digital MultiMeter, testing (AC amps) and (DC amps) up to 20A.<p>(2) It seems to.<p>(3) Right from the plug and into their 12V AC>AC transformer.<p>(4) Umm... not realy.<p>(5) The readings are steady.<p>(6) I have no reason to think otherwise.<p>(7) No clue <p>(8) It's comming from a 12V "Wall-wart".<p>
All I'm trying to do is have a setup where I can have several of these lights hooked up either to a car battery, or to 120VAC. So I'm wondering how big(amps) a transformer I realy need to go from 120VAC to 12V(?AC?DC?).<p>[ March 13, 2005: Message edited by: John Paul ]</p>

[sofaspud]

If you have a 12V wall wart that was supplied with the bulb, it shouldn't be too difficult to determine whether it outputs AC or DC.
I think your replacement power supply should have an amp rating of 2 times the number of bulbs to be used.

[terri]

OK. Two amps is a lot from a Wall Wart, and this one is probably not well-filtered or very stiffly regulated, especially when it's delivering two amps. I would suspect that it is delivering essentially half-wave rectified AC to your lamp.<p>That is, almost 100% ripple.<p>The DC AVERAGE voltage for a half-wave rectifier is 45% of the Root-Mean-Square value of the AC out of the wall wart's nominal 12V transformer, or about 5.4 volts.<p>Essentially the same can be said for the measurement of current in an inexpensive meter.<p>So you are measuring Average Values when you measure the DC current, but very close to RMS values when you are measuring the AC current, which should be close to a pure sine wave.<p>This, along with the inherent non-linearities in the rectifying circuits in the meter, would easily explain the difference in readings.<p>Bear in mind that with most meters (even expensive ones), the meter actually measures average AC voltage of a pure sine wave, and the factory calibration is set so that this "average" corresponds to the pure sine wave's RMS value.<p>Therefore, all bets are off if the measured signal is NOT a sine wave. (Nevertheless, this average value is still useful in many respects.)<p>On the other hand, "true RMS" meters actually sample the instantaneous variations throughout the wave's cycle, and actually calculate an RMS value for the meter's readout, even if the applied signal is NOT a sine wave.<p>RMS readings, by the way, are the only ones which will reflect the true "heating power" of a power source, regardless of the wave shape. This is why some power meters actually measure the temperature change of a heating element within the meter to arrive at this true heating power. These are sometimes called thermocouple meters, and their readings are independent of the wave's shape.<p>I do not advise on anything involving safety issues, especially when it comes to home-rigged appliances, so I will not advise on the current rating of the single transformer needed to light however many lamps you will use. However, I will say that you should use wiring of sufficient carrying capacity and FUSE THE HECK out of your application. I formally recommend using only properly installed CE and UL approved appliances in all cases.<p>[ March 13, 2005: Message edited by: terri ]</p>

[John Paul]

<blockquote><font size="1" face="Verdana, Helvetica, sans-serif">quote:</font><hr>Originally posted by sofaspud:
If you have a 12V wall wart that was supplied with the bulb, it shouldn't be too difficult to determine whether it outputs AC or DC.
I think your replacement power supply should have an amp rating of 2 times the number of bulbs to be used.<hr></blockquote><p>
The transformer supplied was AC>AC 12V@60W. The Wall wart is AC>DC [email protected] that I used for testing. the bulb was reading 2A on 12V AC>AC, and a little more than 1A on 12V AC>DC. That's what originaly confused me because I thought an amp was an amp AC or DC.<p>John<p>[ March 14, 2005: Message edited by: John Paul ]</p>