Question about Powering a lighting unit: 14 VDC @ 3.2A / 28VDC@ 1.6A

In summary, the light unit might try to draw too much power from the battery if it is connected to a power supply that delivers a different voltage than what the unit can work with. This is why the battery head bubbled up in this particular case.
  • #1
exp6
4
0
TL;DR Summary
Help reading to decide how to power units without burning them up
New to electricity so take it easy on me...

I have a light unit: 14 VDC @ 3.2A / 28VDC@ 1.6A

2 questions:

I was curious to see if a 9V Duracell battery will power it up: and it turned it on but I noticed my battery head bubbled up. From my understanding correct me if I am wrong on a 9V Duracell - 550mAH = 900 Milliamps in an Hr. that converts to .55A? Comparing 14 VDC @ 3.2A and my 9V @ .55A the light unit should have not turned on but it did ...why? And why did it cause the battery to bubble up?

Last,

How would I read and know what power cord to use to turn it into a lampstand?

Thank you for your time!
 
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  • #2
If your light can work both with 14 V and 28 V then it has some electronics inside that converts the input to a different voltage. The given currents are what the lamp will use at these voltages. We don't know what it will do at 9 V because it's not designed to be connected to 9 V, but it's likely that it will try to draw at least 3.2 A, maybe even more.

The mAh value for a battery tells you the total charge it can deliver which is linked to its lifetime. It doesn't tell you anything about the current that will flow. The battery will provide 9 V (at least for reasonable currents) and the current flow is determined by whatever you connect to it.

If you connect a 9V battery to a circuit that leads to a current flow of 3.2 A then you damage your battery. You shouldn't do this. The battery is not designed for these currents.
exp6 said:
How would I read and know what power cord to use to turn it into a lampstand?
Get a power supply that delivers either 14 V or 28 V and can provide at least 3.2 A or 1.6 A, respectively.
 
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  • #3
exp6 said:
Summary:: Help reading to decide how to power units without burning them up
I have a light unit: 14 VDC @ 3.2A / 28VDC@ 1.6A

Could you tell me if you are talking about this light?
https://www.knots2u.net/aeroled-par36-landing-light-4-950-lumens/

If it is this LED light, it has a very complicated electronic circuit control inside it. Otherwise it cannot work in the wide voltage range of 9-40 VDC.

Assuming that the light output can be controlled and kept constant, this implies that it should draw approximately constant 45W from the power supply, which means it may try to draw 45W / 9V = 5A from the 9V battery, which is a very frightening situation. Because it is about ten times the specifications of the 9V Duracell battery, it is no wonder the battery head bubbling in this case.
 
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  • #4
mfb said:
If your light can work both with 14 V and 28 V then it has some electronics inside that converts the input to a different voltage. The given currents are what the lamp will use at these voltages. We don't know what it will do at 9 V because it's not designed to be connected to 9 V, but it's likely that it will try to draw at least 3.2 A, maybe even more.

The mAh value for a battery tells you the total charge it can deliver which is linked to its lifetime. It doesn't tell you anything about the current that will flow. The battery will provide 9 V (at least for reasonable currents) and the current flow is determined by whatever you connect to it.

If you connect a 9V battery to a circuit that leads to a current flow of 3.2 A then you damage your battery. You shouldn't do this. The battery is not designed for these currents.Get a power supply that delivers either 14 V or 28 V and can provide at least 3.2 A or 1.6 A, respectively.

Thank you
 
  • #5
alan123hk said:
Could you tell me if you are talking about this light?
https://www.knots2u.net/aeroled-par36-landing-light-4-950-lumens/

If it is this LED light, it has a very complicated electronic circuit control inside it. Otherwise it cannot work in the wide voltage range of 9-40 VDC.

Assuming that the light output can be controlled and kept constant, this implies that it should draw approximately constant 45W from the power supply, which means it may try to draw 45W / 9V = 5A from the 9V battery, which is a very frightening situation. Because it is about ten times the specifications of the 9V Duracell battery, it is no wonder the battery head bubbling in this case.
Not that specific light but you're close.

Just what you are trying to say is the unit itself is like A MAGNET and it will suck whatever power it needs to make it work?
 
  • #6
exp6 said:
Just what you are trying to say is the unit itself is like A MAGNET and it will suck whatever power it needs to make it work?

What I mean is that there is a kind of high efficient LED driver/ LED lamp with a wide power supply voltage range and approximately constant output current/ brightness. In this case, if the power supply voltage is low, it will automatically reduce the input resistance to draw more current from the power supply, that is, trying to maintain an approximately constant input power.
 
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Likes Keith_McClary
  • #7
exp6 said:
battery to bubble up
12-year-old female patient referred to our emergency room as a result of battery explosion caused by short circuit seen in homework designed by her with 3 batteries (each 1.5 V, AA).
They have a picture "after 2 years" (unsightly scars on thighs, otherwise OK) but I won't post it because it is half of the same JPG with the post-operation image.

You can get a 4-pack of 100w equivalent LED bulbs with 1600 lumens each for $20. It is not worth messing with the light you have unless you happen to also own the appropriate power supply and know what you're doing.
 
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  • #8
Keith_McClary said:
You can get a 4-pack of 100w equivalent LED bulbs with 1600 lumens each for $20. It is not worth messing with the light you have unless you happen to also own the appropriate power supply and know what you're doing.
Funny coincidence. Today I took a neighbor shopping. The first stop was a large-chain, national hardware store. One item searched for was a 100Watt equivalent light bulb. The lowest price was $8.45 for one piece of LED bulb. She decided it wasn'r needed THAT badly. Location: Southern California.
 
  • #9
Tom.G said:
price
Light bulb prices could be a thread. Walmart has 8-pack of 60 watt eq for $16 Canadian ~US$12. Six of those bulbs (US$9) would equal @exp6 's light.
 

FAQ: Question about Powering a lighting unit: 14 VDC @ 3.2A / 28VDC@ 1.6A

What does 14 VDC @ 3.2A and 28VDC@ 1.6A mean?

These numbers refer to the voltage and current requirements for the lighting unit. 14 VDC @ 3.2A means that the unit requires a direct current (DC) voltage of 14 volts and a current of 3.2 amps to function properly. Similarly, 28VDC@ 1.6A means that the unit can also be powered by a DC voltage of 28 volts and a current of 1.6 amps.

Can I use a power source with different voltage and current ratings?

No, it is important to use a power source that matches the voltage and current requirements of the lighting unit. Using a power source with different ratings can damage the unit and potentially cause safety hazards.

Can I use a battery to power the lighting unit?

Yes, as long as the battery's voltage and current ratings match the requirements of the lighting unit. It is important to also consider the type of battery and its capacity to ensure it can provide enough power for the unit.

What is the difference between VDC and VAC?

VDC stands for direct current, which means the voltage and current flow in one direction. VAC stands for alternating current, which means the voltage and current periodically change direction. In this case, the lighting unit requires a direct current (DC) power source.

Can I use a power adapter with a higher voltage or current rating?

No, it is not recommended to use a power adapter with higher ratings than the requirements of the lighting unit. This can cause the unit to overheat and potentially damage it. It is always best to use a power source with matching or slightly higher ratings.

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