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Comparing power ratings between CW and pulsed laser

  1. Dec 4, 2011 #1
    I'm trying to compare some figures of power for continuous wave vs. pulsed lasers.

    I have a known application which uses pulsed nitrogen lasers (λ = 337.1nm, such as this one) to produce fluences of ~20-200Jm^-2 and pulse widths on the order of ~2ns to produce irradiance figures of roughly 10^11 Wm^-2.

    I would like to see how close I can get to these figures of power using an InGaN laser diode (even though I understand these types of laser are generally much less powerful). If I have a multiquantum well InGaN laser (λ = 405nm, a blu-ray player laser), with an output power of 1-2.5W and a pulse width of (as low as) 60 ps, spot sizes on the order of 0.3-0.5μm, and repetition rates of 20-80 MHz, what is the maximum power I could produce from this setup?

    When I do the math as I understand it, I get
    power = 1 J/s (W)
    spot size = (0.363E-6m)2
    Irradiance = 7.754820E6 J/m2·s (Wm^-2)

    I get a little lost when incorporating the pulse width. How do I calculate fluence (Jm^-2) from these figures? Even if the diode laser is a CW laser can I get more power from it by pulsing it at 60ps pulse widths? Does pulsing a CW laser yield any possibility of a power boost? Would there be any other conceivable means of increasing the power of the InGaN laser?

    Any help on this calculation and any help on understanding the benefits/downsides of pulsing a CW laser is much appreciated!
     
  2. jcsd
  3. Dec 5, 2011 #2
    Nitrogen based lasers being gas mode are known to have extremely high current in the ionized condition; They tend to self destruct faster than say, slow flow CO2 lasers, but the color UV/blue is good for cutting copper and some metals which CO2 will reflect off of. I'm not familiar with that particular spectra physics laser, but many dye lasers also have an interference shutters so that they may build up power in their internal cavity over a period of time before releasing it all in one burst -- which makes the wattage delivered very high. Joules/millisecond --> 10000XJoules/nanosecond.

    Total power is obviously limited to the input power. If you are asking about the peak power capability if the laser diode is overdriven, I'd caution you that Laser diodes have a damage threshold where the surface polish can be easily destroyed which will result in the diode becoming non-functional. Every diode is unique, and each manufactured batch has variances so that if your manufacturer doesn't specify it -- you will need to have several to destroy.

    Overdriving a diode is not something I would recommend. I have done IR based laser diodes with fiber coupling in the past, and the secret to success is keep it *VERY* cold, and minimize the reflection experienced at the outlet of the fiber -- because *any* light reflected increases the stress on the laser diode and reduces total power by a little over double the reflected light (eg: 1-5% -> 2 to 10% loss... and worse). I assume a blue ray would have the same issue, though it isn't a fiber optic -- it does have an output lens which may not be optimized with an anti reflection coating as those tend to be expensive.... So your maximum power is likely limited by your optics more than your electronics.

    Usually when people specify spot size, its the diameter of the spot -- not the area.

    Again, the pulse width should not affect the flux (watts/m^2), unless you are overdriving it which is risky.
    I am not aware of electronics that can drive at the 60ps time interval except for avalanche transistors; and I think you would loose more power overall from the few setups I have seen.

    Most industrial laser processes are ablative vaporization of materials at high energy densities; there are two primary factors in whether a laser will replace another -- the material's sensitivity to wavelength, and the heat dissipation capability of the material. power vs. color tradeoffs;

    Copper, for example, can be hit pretty hard with CO2/micron wavelengths and even NdYAG survive with 50W +CW beating down on it; and although the NdYAG can destroy the surface of the copper using a pulse shutter to boost the flux by accumulating the light energy and releasing it all in a burst -- it was seldom very clean, and was always slow. (unless one adds oxygen to cause copper to burn -- and the Guy I knew who tried this made the mistake of doing it under plexi-glass..... WOW! weeks of work up in *SMOKE*) But the same laser system with a tripler crystal which wastes a *LOT* of power to convert from IR to near-UV is able to cut copper like a dream and !! fast !!.

    The first problem you need to determine, experimentally, is whether the color change in the laser diode improves or reduces the power required to make your project go. This effect is often more important than people give it credit for...!

    If your irradiance figures are right, then the total joules is simply the effective lasing time of the pulse (the whole pulse does not produce laser light, as it takes time for the diode charge to form properly) multiplied by the irradiance; the irradiance already gives you the power density -- so the peak energy density is going to be on the order of irradiance x time = Joules/area.
    (longer is better....)

    If repeatability of power is more important than peak power, (and often it IS!), then you will be forced to pre-charge the diode to a low power idle state (It emits light but not coherent....) and provide bursts of power to cause it to lase -- this keeps the diode at a fairly warm temperature, unfortunately, but it makes the power output repeatable.

    The longer the diode is off, the more power the actual diode lasing can do per pulse -- with a catch, that it still has a damage threshold which I wouldn't know. But, if IR diodes are at all similar -- a 10W diode won't give more than 30W peak; and it will be short lived. So, I wouldn't expect more than a peak wattage of 7W.... and I would expect you to loose diodes at that power level....

    Since the laser is 2.5W, and assuming very optimistically that *all* of that went into the spot size of 0.5 micron **diameter**, gives (0.5e-6**2)/4.0*pi --> 196fm**2 around -- 2.5W / 196f m^2 ~= 13o.E12W / m^2 irradiance. Now that might be off for various reasons, such as 75.% of the power doesn't make it out of the diode, and of that power 10% is focused improperly by the lenses so that more of the energy is reflected into the packaging around the lens (heating it...) But, I think irradiance in the "spot" there is potential to be near the same order of magnitude if the blue ray diode is a "perfect" one, but likely you will be 10-1000 times less powerful once the optics are all considered.

    Best wishes, but unless you know that a longer wavelength is acceptable -- chances are that you will meet unexpected problems.

    :redface:
     
    Last edited: Dec 5, 2011
  4. Dec 6, 2011 #3
    Wow andrewr,
    Thanks for the response! This will take me some time to digest. I very much appreciate it!
     
  5. Dec 6, 2011 #4
    Luckily, I have no immediate plans of performing these experiments. I am trying to formulate an idea for a new use for this 405nm laser diode and was curious what the hypothetical limit of power was for this type of laser to see whether it is even worth pursuing but got held up with some of the theory and calculations. I think the power (13E12 W / m^2 irradiance) is high enough for what I was looking to do. Thanks again for the help!

    Got this info from here:
    BDL-405-SMC
     
  6. Dec 10, 2011 #5
    I'm not totally sure what the 60ps is being used for;

    405nm ~= 1.35 femto seconds/wave.
    60ps / 1.35 fs ~= 44,000 wavelengths ~ 17.9mm

    It seems too long to be doing much of photon / QM sqeeze effect tests; That short of a pulse, I would guess, is aimed at fluorescence type experiments more than vaporization of material.

    It's kind of interesting that they have the thread with inch pitch rather than metric, being from Germany. Rothen Baasel, and others I have played with usually are fully metric.

    Good luck... :) and you're welcome.
    --Andrew
     
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