# Powering a laser partially with Photovoltaic Cell

1. Jul 21, 2013

### Ezekial

Hello, I am working on a system where a laser does something and then there is a lot of wasted laser radiation. I want to capture this waste radiation in a small array of solar cells then feed the resulting current back into the laser to make it more efficient. The output from the solar cells will be put through a switching DC-DC converter to match the lasers input voltage requirements. Now I will have two power supplies: the output from the switching converter that is the right voltage but with insufficient current to power the laser, and another to supply the rest of the current.

My Question: I know I can't just hook these two power supplies up in parallel, so how do I connect them so that all of the Solar power goes into the laser and the rest comes from the main power supply? Is there some kind of charge controller IC or something I can buy? (I do not want to use a battery).

Thanks so much for your help!

2. Jul 21, 2013

### Staff: Mentor

If they have very similar voltages, a small resistor could work.

I doubt that you can gain a lot with that setup.
If the laser power is a few watts, where is the point? If you have a kW-laser, you can burn a hole in your solar cells unless you build an expensive setup to spread it out.

Edit: Oh right, diodes are better.

Last edited: Jul 21, 2013
3. Jul 21, 2013

### Baluncore

Lasers can be very inefficient, maybe 1%. Most energy, (99%), is released by the power supply or laser drive as heat. You would be better trapping and converting that waste heat than the remnant scattered light.

But to recover scattered light with a panel you would use two power diodes. One would feed current from the panel to the laser. The other would feed current from the mains supply to the laser. You would arrange the voltage of the mains supply to be slightly less than the voltage from the solar panels. That way the panels would provide what current they could, while the mains supply would make up the difference.

4. Jul 21, 2013

### Ezekial

Thanks for the reply. How would a small resistor help? In my system, an IR laser is absorbed by a crystal, then re-emitted at a higher frequency, cooling the crystal (anti-stokes fluorescence). It is this diffuse re-emitted light I will be taking power from. I just need the simplest way to dump the PV cell generated power back into the laser so that all of the PV power is used and the rest is supplied by the main supply.

5. Jul 21, 2013

### Integral

Staff Emeritus
You need to look at the efficiency of your solar cells at the lasers wavelength.

6. Jul 21, 2013

### Ezekial

Thanks Baluncore. Will the output voltage from switching converter (which I would use to make the voltage slightly higher than the mains supply) drop under load automatically or are there more necessary components? Is this how the main supply would 'know' to make up the difference?

7. Jul 21, 2013

### Ezekial

Integral, why is this important? Since my light is nearly monochromatic, I should be able to achieve nearly 80 % with gallium arsenide cells.

8. Jul 21, 2013

### Baluncore

A solar panel should be treated as a current source with a series resistance. So find out how much current the panel will be generating by doing a short circuit test with an ammeter. Then measure the voltage remaining after the internal resistance drop at that current. Set your mains supply just below that voltage, combine them with two diodes and use that combined current into the switch-mode laser supply.

Driving a switch-mode supply directly from a solar panel is not a good solution because the switch-mode supply expects a low input impedance. Here, the low impedance will be provided by the mains supply. The panel will provide a high-impedance current, it's voltage being pulled down by the laser drive to the voltage of the mains supply that can then make up the difference.

I think you are planning to do something that can be done, but that is going to be incredibility inefficient and therefore uneconomic.

Last edited: Jul 21, 2013
9. Jul 21, 2013

### Ezekial

Baluncore, Thank you so much for your replies! You have been very helpful. I am almost ready, but just have 2 more questions.

1) My main power supply is a fixed voltage. If I leave the switching DC converter connected to the PV array, but put a capacitor to ground on the input to the converter (connected to the pv array), current would still flow into the capacitor when the transistor in the switching converter is open, thus lowering input impedance. Does this sound feasible?

2) Could I instead put the switching converter between the main supply and junction with diodes to achieve the voltage level you previously described?

10. Jul 21, 2013

### davenn

because your solar cells may not be very efficient at the wavelength of the laser, that's why

you still haven't told us all what the power of the laser is.
and as others have suggested, by the time that tiny bit of stray light get captured by the panels and converted to power the output of the panels is likely to be extremely low, probably milliWatts
seriously its hardly worth the effort !

Dave

11. Jul 21, 2013

### Baluncore

There are too many unknowns to be specific.
What is your AC supply voltage? 115V, 230V ?
What type of laser ? make & model number? Power input? power output?
What voltage does the laser need? DC? maybe 3V, 300V or 3kV?
What current will it then draw? Or is it a constant current device?
That information would make it possible to complete the design.

12. Jul 21, 2013

### Ezekial

DC Supply is 12v. Not sure exactly what the pv output will be. I don't have a part number for the laser but it is a 10w IR laser diode requiring 6vdc. The laser driver module would limit current. The main power supply is 12vdc. The combined supply needs to supply 3 amps.

13. Jul 22, 2013

### Staff: Mentor

It would help to avoid reverse currents, but diodes are certainly better.

A 10W laser... assuming this is the laser power and neglecting all other losses, the solar cells might be able to get ~5W back. To save 1kWh, you need 200 hours of operation. How much does 1kWh cost in your country?

Solar cells cost about $1/Wpeak (give or take a factor of 2), but this assumes bright sunlight. If you use scattered light, you probably need something like 100Wpeak to cover the whole light, which costs ~$100. This does not even include the DCDC converter and other electronic components.

I wonder how many years you have to operate this to get the investment back. I guess putting the solar cells into sunlight is more efficient.

14. Jul 22, 2013

### sophiecentaur

If you want to avoid losing energy when the laser is not required, then why not just turn it off? What sort of mechanism are you going to need, to divert the laser beam onto the PV array? How reliable, efficient and costly is that going to be, compared with a simple switch? I would go for an electronic solution to the problem rather than a semi-mechanical solution, any day.

15. Jul 22, 2013

### Baluncore

If the laser you use is a 10W pulsed laser diode, such as the LD-1550-10WP, then it will need 50A for 200nsec, (with a 5% duty cycle). Average current will therefore be 2.5 amps, average radiated IR power is 0.5watt.

“LD-1550-XXWP series of pulsed laser diodes are MQW structure devices with InGaAsP active layers fabricated using advanced MOCVD epitaxial growth techniques. Devices are designed with pulse width FWHM of 50~100ns and repetition frequency up to 15 KHz. Standard TO(9mm) package with wavelengths of 1550 nm, different emitter sizes and stack layers are available for various applications such as range finder and optical measurement.”

The solar panel will therefore only be in use while the laser is pulsed on, for 5% of the time,.
Even if 100% efficient the panel will generate less than 0.5watt. Average DC input power to the laser is more than 25watt, so the solar panel cannot recover more than 2% of the energy.

Is your laser diode pulsed like the LD-1550-10WP ?

16. Jul 22, 2013

### Ezekial

Yes the laser is pulsed, but the fluorescence from the crystal (which I would be collecting) is pretty much continuous. That said, you all have made some very good points about economic feasibility. I think I may focus my efforts toward efficiency elsewhere. A sincere thank you to everyone who replied.

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