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I have a general question - I have two sets of ccfl lights 12v dc 5.0 ma.

I found an old transformer that's 12v 500 ma

When i hook up one light it looks great I'm very happy. When I hook up the second set they both dim.

Why?  Isnt there enough power for both sets?  I really don't want to have two transformers hooked up for this.

I am assuming this is an old 'wall wart' type of transformer. The actual voltage out of these is only gong to be approximate. A 12v dc unregulated supply will probably measure around 15-18V with no load. Connecting one lamp may not be loading it much and the lamp may actually be getting a higher voltage than it should (thus is it bright). Connecting 2 lamps may be enough of a load to bring the transformer closer to it's rated voltage and thus the dimming. The only way to tell for sure it to put a meter on it and measure the voltage with no lamps, with one lamp and with 2 lamps.

Also I am assuming you are connecting the lamps in parallel and not in series.

I'll give that a try.    I have a multimeter. 

His ballast inverter board has two 2-pin headers which allow
for two tubes to be hooked up to it - there are in parallel.

I think they need 700mA for 2 tubes.

Specification

    * Input Voltage: DC 12 Volt.
    * Input Current: 700mA
    * Input Connector: Molex 4pin connector with pass-through.
    * Output Connector: 2 Output socket for dual 4", 12” or 15” cold cathode lights. Inverter: Sound control activation, sensitivity adjustable, with on/off switch to override sound control feature.

Here's how I have them hooked up:

Plug ===== light set on +-
                    |= light set on +-

The wire is cut halfway up and I have one set spliced in. Then on the End of the wire I have the second set.

If stayout is right then the 500 mA is not enough power for the two sets

On a savings note so far I have saved 89w of power by unplugging those lights. The heat saved is huge too....!!!

I can save 60 more with 3 more sets of ccfls

Yes, IF you are both talking about the same lamps. In this thread
 http://newlifegames.net/nlg/index.php?topic=10246.15
there was talk of replacing the fluorescent lamps with after-market CC tubes. The specs in the picture say those tubes only draw 5 ma. I assumed that was what we were talking about since 5ma was mentioned here. Some clarification is needed as I'm not sure where the 700ma number is coming from.

I'm not sure either -- the package does say 5.0 mA


Yes they are the same lamps as the other thread.

If you have a meter, you can take those voltage reading as mentioned above. You should also be able to measure the actual current with it.

I copied & pasted the specs from the Logisys website.
In fact, they sell a multi-output 12v/24v converter or inverter that's rated at up to 900mA per output.
I have fairly dim lights too when using 2 tubes on one inverter as I posted that problem in
one of the threads about these cold cathode light conversions last year.

The solution was to get a stronger rated power transformer "wallywart"... :96-
The 2 tubes draw more amperage than listed on the box.
Be careful when around these little lights - the inverter shoots out about 680volts! >>>


http://www.logisyscomputer.com/viewsku.asp?SKUID=PD10&D (http://www.logisyscomputer.com/viewsku.asp?SKUID=PD10&D)

There may be some 'fudging' of the numbers than. If the tubes assemblies are rated at 5 ma, then 2 tubes should only be 10 ma. That is far below the 500 ma rating of the transformer. HOWEVER, if the 5 ma is the rating of the tube it's self, then what might be happening is that they are rating the 5 ma at the tube voltage level. So if the inverter is raising the voltage to let's say 1000 volts to light the tube, and the tube is drawing 5 ma (at 1000 v), that mean that at 12V input to the inverter, it would be drawing around 500 ma (+ ~10% for inverter inefficiency). Someone really needs to take a meter and get an accurate current reading of these things (i.e. the current at the power supply voltage). It would makes a big difference in that other thread talking about power savings if they are actually drawing 500 ma @ 12v vs 5 ma @ 12v. (6 watts vs .06 watts). Actually the 6 watt figure sounds more realistic and would still be about 1/3 of a 15 watt bulb. But the payback period goes up about 100X making any actual $ savings nonexistent.

That's what might be happening Poppo...interesting numbers...good stuff!
It may be why my wallywart transformer is rated at 12v and 1A
to power up the 2 tubes nice and bright!

Question - if I use a 12v 2a wallwart does that waist electricity if only 1a is needed?   

It'll only draw what it needs.

OK - I'm not quite sure what I'm doing so ---

When I put the tester (10A DC Setting)  - on the transformer without a load I get 2.18  :103-

Now when I hook up one light - (it lights obviously)  - as soon as the tester goes across the wires it goes out and I get a reading of 2.10.

One thing -  I had both sets of lights on all night long and ------ the transformer was hot hot hot this morning  :81-

The transformer being hot hot hot is because it was being overworked
by the 2 tubes being on and drawing more than the transformer can handle.
Poppo may be dead on about the 5.0mA numbers being "fudged".
I'll bet the number of 5.0mA is for only one tube.
I'd install the stronger transformer.

Start your multi-meter on the 1000 DCV setting then work your way down
to no lower than the 20 DCV setting watching the display on the multi-meter.
I don't know why you're getting 2.10 with it on the 10A setting?
What are you touching the multimeter probes with?


What went out?
The lights? or the multimeter?
If it was the multimeter, I'd suspect your finger
is hitting the "on/off" switch on the multimeter?>>>

I just did a little test on mine.
The red and black wires of the inverter give me a reading of 8.1Vdc
when my multi-meter is set on the 200 DCV setting with two tubes lit...

Across? You do mean in series, right?

one probe on one wire and the other probe on the other wire -  when I tested the voltage I got a constant 11.8 with or without the lamp connected.

Hard to tell from the picture, but if you are measuring the voltage, the dial needs to be pointed 180 degrees at the 20V DC range and you measure across the + and -.  If measuring current, you are on the right setting (10A), but you connect the probes in series with one of the lines like this
(http://scienceshareware.com/images/electricity/how-to5.jpg)
http://scienceshareware.com/how-to-measure-DC-current-with-a-dmm.htm (http://scienceshareware.com/how-to-measure-DC-current-with-a-dmm.htm)

What about with both lamps connected? In any case, it is sounding more like those tubes draw more than the 5ma they claim. As suggested, best to just get a bigger power supply.

What are the manufacturer's actual specs. for operating the tubes?  Are these wired like the other ones in the thread that require 680 volts to operate?

according to your specs. 5.0ma is actually .005 amps. "millie" is -3, your transformer is rated at 500ma or .5 amp  should be more than enough to operate the tubes.

you CANNOT measure current  in a circuit as you do voltage.  you have to measure current in series with the load.  

a standard transformer will only provide an AC voltage out. as its name implies it transformers a AC voltage. i.e. 115 vac  input to a transformer, the output would be 12 vac  @ a current rating.  to get a operating DC voltage you have to rectify it and filter it .

Jim  

That is the big question. In that other thread, this is what the specs show. But it may be misleading if the 5 ma is for the tube - at the tube voltage, and not the current of the supply voltage. Current is inversely proportional to voltage. So 5 ma at several hundred tube volts is going to be much higher current at the supply voltage of 12v. At 12v, it would be approx 280 ma (slightly higher due to inverter inefficiency). So two tubes would be at least 560 ma which is over the specs of his wall wart.

okay,
I decided to plug it into my benchtop DC power supply which has an ampmeter on it.
With one bulb at 12Vdc ,the inverter draws just over 400mA.
With two bulbs on the inverter, it draws over 600Ma.
Logisys specs on the packaging is way off.
Look at these pictures from my power supply needles>>>

 :25- Well somehow I fried one of my inverters  :(

Stayout - I now have two bulbs hooked up to a 12V 1000mA transformer.  That should keep it happy.   

Looks like the electric savings is going to be minimal.

You must of put the negative wire onto the 12V live wire to fry it...
You can't be doing that with dc voltage!

I was modifying my post while you guys posted.
Take a look at my power supply amperage readings.

-  Thanks for the time to check this  !!

That came pretty close to my guess.  :89-

How can they get away with such inaccurate information on the package?

It's not wrong if it is the tube current. But it is misleading. 

I wish my BK Precision Bench Top PS station had digital readings rather than the old needles... :8-

I'm not even sure it's that misleading, as the current spec is listed directly under the output voltage of the inverter, which would lead me to believe the current measurement was made on the output side as well.

If the draw was listed directly under the input voltage, I would assume the draw was measured on the input side as well.

Why? My PS does have digital readout, but only .1 resolution.

You should post those findings in the other thread, or link to this one since it does change a lot of assumptions in the other one.

I'm wondering about something...
If I add voltage, for example 13Vdc, would the amperage draw be lower?

I just found out something...I put 10 volts on the PS and it only draws just over 500Ma.
Yet, the tubes appear to be just as bright.

When I put 15 volts into it, it draws about an extra 100mA.

These were done with 2 tubes attached.

I agree, but how many people really know which is which? If only one current value is given, the average Joe would think it was talking about the whole unit. In the other thread, even though I thought it was low, I assumed it was the total current, and I'm a tech guy. But without one in my hands, I had no way to actually check it.

Short answer is no. However, depending on what the load is, it may. Current=voltage/resistance. So if the voltage goes up, the current goes down. But most circuits are not simply resistive, so depending on what is being powered, as voltage goes up, so may the level of smoke.  :98-

The inverter is probably regulated and anything over it's 'nominal' voltage will just be dissipated as heat and hence the extra 100ma with no increase in brightness. And once you reach a certain brightness level, the human eye does not notice subtle changes.

I agree..I couldn't really tell the "brightness" without a photographer's light-meter in my hand.
Let's just say it would be "good enough"?    lol
I posted a couple of more pics and readings from 10 and 15 volts above ^^^

That is good info though as there is no sense in over-driving them for no perceivable increase in brightness and only end up using more power. However, 12V supplies are going to be a lot more common than a 10V one.

Yeah...when I adjusted the benchtop power supply to the other popular 9Vdc, they were pretty dim.