Continue to Site

Welcome to our site!

Electro Tech is an online community (with over 170,000 members) who enjoy talking about and building electronic circuits, projects and gadgets. To participate you need to register. Registration is free. Click here to register now.

  • Welcome to our site! Electro Tech is an online community (with over 170,000 members) who enjoy talking about and building electronic circuits, projects and gadgets. To participate you need to register. Registration is free. Click here to register now.

cold cathode lighting project

Status
Not open for further replies.

airbrush

New Member
Hi there,

I am wanting to make a lamp using the cold cathode lighting kits you use for computer case modding. I may want to have more than just the two that comes with the kit. They operate off of 12v...so say for example i want to run two pairs of these off of a 12v wallwart...what capacity(Ma) would i need the wallwart to be? How would i calculate this for two or more? Also is it possible to put these kind of lights on a dimmer and function correctly(ie. dim from off to full brightness)...if so how? and what would i require to do so?

Suggestions and help much appreciated.

the specs for the cathodes are as follows:
# Tube diameter: 3.0mm
# Tube length: 300mm/100mm
# Input voltage of inverter: 12v
# Output voltage of inverter: 680v
# Current draw: 5.0mAv
# Brightness: 28,000~30,000 cd/m?
# Lifetime: 30,000 hours
 
airbrush said:
what capacity(Ma) would i need the wallwart to be?

You mean mA. You need an inverter, I assume it comes with the kit, you should measure the current with the tube connected using a multimeter.

But I'll give an estimate if you like.

The inverter output is 680V (I assume this is with the tube connected).
5mA per tube, and two tubes, so that's 10mA.

Inverter output power = 0.01 * 680 = 6.8W
Inverter Input power (assuming an effiency of 80%) = 6.8/0.8 = 8.5W.
Inverter input current = 8.5/12 = 0.70833A

So for one inverter yourt adaptor needs to have a rating of at least 708.3mA, you need two inverters so that's 1417mA. I'd go for a 12V 1.5A mains adaptor, but I suggest you measure the current consumption first, you could also check the datasheet.
 
Okay thanks...
Yes they come with invertors already, i would assume the current draw would be for one tube only? but would have to check it.
Also each set would be on a switch if that makes any difference? So i can turn each set on or off.

Does anyone have any knowledge of these and whether the can be dimmed? I'm assuming not, since they are like a flourescent or neon tube...
 
airbrush said:
Okay thanks...
Yes they come with invertors already, i would assume the current draw would be for one tube only?
The 5mA current it output current, not the input current which I have estimated for you.

airbrush said:
Does anyone have any knowledge of these and whether the can be dimmed? I'm assuming not, since they are like a flourescent or neon tube...
That depends on the inverter and tubes, some will regulate the current through the tube if you reduce the voltage keeping birghtness constant, others will simply let it fall and the tube will dim. My guess is yours will dim to some extent if you reduce the voltage and might even work with a PWM dimmer, however dimming will probaly only work reliable down to about 4.5V 6V.
 
Your LM317 circuit has a 2.2k resistor that is supposed to be 120 ohms, and a 500k pot that is supposed to be a 1k pot.
The LM317 needs an input voltage that is at least 2.5V higher than its output voltage so you need a 14.5VDC or higher power supply.
There will be a fairly high current flowing through the LM317 that will cause it to be too hot unless it has a pretty big heatsink.
 
hmm...okay so technically this setup would work for dimming the cathodes, but since its outputing 12v instead of the 6v(or so) as it was in the LED setup its going to give off some heat. I did have some heatsinks(not very large...maybe 3/4 inch length) on the led project as well but never noticed any heat build up in the enclosure its in. What are you considering "large" heatsinks? Since i would want this one in a much smaller enclosure (inside of 2" aluminum tube) i may have to give up the dimmer idea then..although it would be a nice touch instead of just a simple on/off switch.
Basically I am wanting to make a standing room lamp, aluminum and frosted glass, contemporary design. I was actually going to do it with LEDs like the other lamp, but after seeing the brightness and size of the cathodes..they would make an interesting light source.
 
The heat from a voltage regulator is not caused by how much is its output voltage. The heat is caused by the voltage across the regulator (input minus output voltages) times the current through it.

It will have an input voltage of about 15V and an output voltage of about 8V with about 350mA which is a power dissipation of about 2.45W. Try your little heatsink.
 
hey...maybe the whole lamp could be the heatsink :p....i think i'll have to stick with the on/off only on this one then
 
Okay, so i'm about ready to start putting this little project together...i decided to go with three sets of the cold cathodes, each with an on/off switch. What rating of power would i need for the power adapter...also does this cause a problem if say only one set of the lights are on...while the others off...would it be receiving too much power or would it just draw what it needs??


Suggestions and help much appreciated.

the specs for the cathodes are as follows:
# Tube diameter: 3.0mm
# Tube length: 300mm/100mm
# Input voltage of inverter: 12v
# Output voltage of inverter: 680v
# Current draw: 5.0mAv
# Brightness: 28,000~30,000 cd/m?
# Lifetime: 30,000 hours
 
The current specified is the output current.

But working back from the figures you've given me.

Assuming no losses.
[latex]P = 680 \times 0.005 = 3.4W[/latex]
[latex]I = \frac{3.4}{12} = 0.283A[/latex]

Total current is, [latex]0.283 \times 3 = 0.85A[/latex], but there will be losses so I'd go for a 1A minumum but I would measure the current for one tube then multiply it by three to be sure.
 
Okay, i was going to go for a 1.5A or a 1.2A.

So in regards to the second part of my question:
...also does this cause a problem if say only one set of the lights are on...while the others off...would it be receiving too much power or would it just draw what it needs as i was going to power all 3 sets off of it??
 
thanks for the links..will read up on it.

My knowledge of electronics and electricity is limited...all the help is much appreciated.
 
Status
Not open for further replies.

Latest threads

New Articles From Microcontroller Tips

Back
Top