Powering PV(solar) panel with LED lights

bal_jop

I have seen some questions and responses in this regards in this forum. I am not very convinced, so I shoot the question again.

Is it possible to feed PV cells with LED lighting?

With LED's coloured lighting is easier, and getting a frequency in the range rated for a panel is easier. But would we be producing more energy than what we supply for the LED panel?

I am considering a model as follows;

1 PV panel open to sun. This would power my LED panels and temperature regulators(Not an AC but some fan setup).

1 or 2 PV panel working with the LED panels, inside a temperature regulated closed environment.

Would this be an efficient design?

pantaz

Think this through --
1) Your first PV panel (solar) generates electricity (what's the efficiency?)
2) That power goes to the driver circuits for the LEDs (Don't forget to calculate losses here.)
3) Then, using the light output of the LEDs (efficiency?) to illuminate another PV panel (efficiency?) to generate electricity for some other end-use.

bal_jop

well... I was thinking, Having a 1KVA solar inverter setup, covering around 60 sqft(PV panel); I can illuminate atleast twice the area continuously(120 sqft). Considering the secondary panel in the illuminated area, with even a 80% efficiency of the primary ones, I would have higher energy than what I stared with and with temperature controlled, the efficiency and panel life may even increase.

Any thoughts on this?

bal_jop

hello...

Is anybody around....?

Is the idea worth pursuing? Are my assumptions right or wrong?

DaleSwanson

Are you proposing powering LEDs with a PV cell which then illuminates a PV cell generating some power which is fed to LEDs and etc?

If so, no that absolutely will not work. It would be a perpetual motion machine. If such a simple setup could produce infinite power then it would be in widespread use.

jack action

I'm not sure I understand the purpose of your apparatus, but here are some questions and answers:

Is it possible to feed PV cells with LED lighting? Yes.

Is it an efficient way of transporting electricity? No, a simple wire would be more efficient.

Why would you use such apparatus? You can use this apparatus to act as an on/off switch. When an object passes between the PV cells and the light source, it cuts off the power transmission.

Is it the most efficient and affordable switch? Probably not, but purpose will dictate your design choice.

Not an expert on the subject, just my 2¢.

jim hardy

Are you suggesting that:
by collecting 80% of the energy that falls on 60 square feet,
then spreading that energy over 120 square feet at unspecified efficiency,
then recollecting it again at unspecified efficiency,

you'll net a gain in energy?


Sounds to me like perpetual motion. Or current economic policy .

bal_jop

okay! probably I should have framed the question differently...

1. What is the luminous intensity/flux/flux density required for a PV panel to operate at its rated efficiency?

2. Is it possible to illuminate twice(or area more than the size of the PV panel) the area from the power obtained from one PV panel with the luminous intensity/flux/flux density stated for question 1?

DaleSwanson

Still not sure I understand what you want, but it sounds like a perpetual motion machine.

Anyway, from wikipedia:
http://en.wikipedia.org/wiki/Energy_...ion_efficiency

Energy conversion efficiency:
Solar cell 6–40% (technology dependent, 15% most often, 85–90% theoretical limit)
Light-emitting diode (LED) 4.2–14.9%, up to 35%

Let's say 100 watts hit a square meter of PV cells. If the PV has 25% efficiency it will output 25 watts power. Now if the LEDs also have 25% efficiency they will output 6.25 watts light. So you end up with 6.25% efficiency, and that is with 25% for both the PV and LEDs, which is probably high.

mfb

I doubt that you understand what efficiency means here. It is the rate of outgoing power divided by the incoming power. While this efficiency can depend on the flux for real cells, this is not relevant here. The issue is that the efficiency is always less than 1.

If you use the electric energy gained in an area A to illuminate an area of 2A, the maximal light intensity is 50% of that which reaches the initial area.

Another simple, realistic system:

Take 1m^2, exposed to sunlight (~1kW/m^2). It receives ~1kW. With an efficiency of 40% (let's be optimistic here, and hey this was reached in labs), you get 400W of electric power.

You can use this power to operate LEDs. They will give you something like ~200W of light (again quite optimistic). As this light has a nice wavelength distribution, you might be able to reach a higher efficiency here, let's say ~60%. Therefore, your "secondary solar cells" now give you 120W.
You wasted 400W of electric power to get 120W of electric power. Do you see the problem?

Even if LEDs and solar cells would have an efficiency close to 100%, you would still waste power (and heat the setup).

D H

No, (No) x 1000.

Don't go there. Can you lift yourself by your own bootstraps? (No.) There is no way for a device to produce more energy than comes in. Learn some physics. Learn to realize that thinking this way is just wrong.


Always. bal_jop, this is why a device such as the one you envisioned cannot be created. It's the laws of thermodynamics. Think of it as playing a game.

Law #0: You have to obey the laws of the game.
Law #1: You can't win. The best you can do is break even.
Law #2. You can't break even, except on a very, very cold day.
Law #3. It never gets that cold.

bal_jop

Sorry, if you feel I am still persisting with this...

I was doing some calculation on LED's at http://led.linear1.org/led.wiz

for 10,000 LED's

each 1 ohm resistor dissipates 0.225 mW
the wizard thinks 1/4W resistors are fine for your application
together, all resistors dissipate 281.25 mW
together, the diodes dissipate 450000 mW
total power dissipated by the array is 450281.25 mW
the array draws current of 18750 mA from the source.

A typical 1KVA panel measures close to 1sq.m; in one example it measures 120x80 sq.cm.

Any further thoughts on this?

bal_jop

Sorry, if you feel I am still persisting with this...

I was doing some calculation on LED's at http://led.linear1.org/led.wiz

for 10,000 LED's

each 1 ohm resistor dissipates 0.225 mW
the wizard thinks 1/4W resistors are fine for your application
together, all resistors dissipate 281.25 mW
together, the diodes dissipate 450000 mW
total power dissipated by the array is 450281.25 mW
the array draws current of 18750 mA from the source.

A typical 1KW panel measures close to 1sq.m; in one example it measures 120x80 sq.cm.

Any further thoughts on this?

mfb

Sorry, this does not lead to anything. You use ~450W to power 10000 LEDs to get something like 200W of light (again, optimistic performance). So what?

A typical panel of what?

russ_watters

A handful of things:

1. Do you recognize that solar panels are rated based on exposure to average sunlight?
2. You're misreading the spec sheet if you think a 1 sq.m solar panel can produce 1 kW, since the average solar irradiance at the top of the atmosphere is 1.37 kW. Assuming 80% of that reached the surface (1.37*.8=1.1 kW), and the panel was 40% efficient, you'd get 1.37*.8*.3= 0.33 kW.
3. As said above, the LED isn't 100% efficient either - more like 20%. So that .45 kW LED array would only put out .09 kW of light. So even if you could focus it all on the solar panel, .09 kW is a lot less than 1.1 kW.

bal_jop

I see in most of the spec that the panel rating assumes temperature not more than 25 degree and light at 1000 lux. Producing a 1000 lux of light in a photographer's studio with LED lighting is a child's play.

Refer;
http://answers.yahoo.com/question/in...6164802AA3VNeZ

This means I can produce 2.5 times the lux at a fraction of my previous calculation.

The power produced by a solar panel depends on the lux and the lux is inversely propotional to distance from the light source.

bal_jop

okay, I got it here;
http://answers.yahoo.com/question/in...4212857AAiPmUA

The rating is not with lux assumption but with KW of light/m^2; which is like

http://answers.yahoo.com/question/in...4212857AAiPmUA
668,449 lux = 1000 watts/square meter

But looking at this it seems like earth can never receive such light from sun. The illumination table in
http://en.wikipedia.org/wiki/Lux

states the max lux we can receive is; 32,000–130,000 lux Direct sunlight.

That seems very very small compared to 668,449 lux. :(