Laser Diode - Does it work this way?

Discussion in 'General Electronics Chat' started by hoppyumr, Apr 8, 2011.

  1. hoppyumr

    Thread Starter New Member

    Apr 6, 2011
    2
    0
    Hi all,

    First post on here, so go easy. I've tried to find what I'm looking for, but am not 100% satisfied just yet.

    Generally, what I'm looking to do is have several (50+) low power laser diodes (labeled as 5mW, with output more like ~3-4mW as I understand it for diodes of this type) powered from the same voltage source. I know this is typically an issue as they will not have the same characterisitics, and without limiting the current bad things happen to your diodes and they die off one by one - This, I'd like to avoid. A continuous current source driver appears to be the most common method of protecting the diodes, but I've only really seen them for one diode at a time. I simply do not want to buy/make that many true "drivers," and wasn't really interested in purchasing the lasers as completed modules (i.e. complete with a brass case/heatsink and quasi current limiting circuit) as the cost goes up considereably for the overall project, and most of the modules out there are just huge. The diode I'm looking at is an NEC NDL3210, which on the datasheet is rated at 6mW, but all of the values correspond to 5mW output.

    The TYPICAL diode operating characteristics at 5mW, electrically speaking, are as follows:
    Operating voltage: 2.2V
    Max Operating VOltage: 2.5V
    Threshold Current: 40mA
    Max Threshold Current: 60mA
    Operating Current: 50mA
    Max Operating Current: 70mA
    Monitor Current: .5mA (not even sure what this is, or if it is something I can even control as the diode is essentially a 2 terminal device excluding case)

    Then, there are some ratings that don't make sense to me such as:
    Reverse Voltage: 2.0V
    Reverse Voltage: 30V
    Forward Current: 20mA
    I understand what reverse voltage is, but why do they have it listed twice and why are they soooo different in magnitude?


    Now that you have the background I'll tell you where I'm stumped. I was using the Circuit Simulator Java Applet found on Falstad dot com, and noted that they don't have a "laser diode" choice available - not that I really expected it anyway. As such, I started off by giving a go at using just their regular diode, but didn't really believe what was going on. Then I figured I'd try to simulate it with a zener diode, and am more satisified with the results; however, I'm still not certain of the operation.

    The circuit I simulated was fed by a 6V DC source, to a resistor in series with (the diode, a capacitor, and another resistor). In practice the 6V source would be a voltage regulator fed from 9V DC, lets say. The values I've come up with are 28ohms on the two resistors, and the value of the capacitor is mostly irrelevant is it would serve to hold my voltage more steady during quick voltage fluctuations such as powering it on (lets say it was a 470uF cap) - so without doing the math this would level out my voltage in roughly 10ms - kind of a soft start if you will. Again, the capacitor could be omitted for the basis of my question here, as it doesn't affect anything steady state. I'm not saying the circuit I came up with is remotely correct, but I started off thinking about KVL, thus the two resistors. Knowing that a diode isn't linear, I assumed the diode would take over and force the voltage for that half of the circuit once it gets to its operating voltage, which is something I can't control on the simulator (or I just don't know how), and then first resistor would serve to limit the current to the diode from there on.

    So for the diodes:
    Method with Zener Diode - The applet requires some information on the zener diode, which I know fundamentally operates completely different from a regular diode when reverse biased, but I am using both in the forward direction. So looking at the datasheet and trying to determine what values to use I've set the zener diode to a forward voltage of 3V at 1A (I don't get a choice to reduce the current to a realistic value for the diode that wont burn it up, so I've had to extrapolate the value from looking at the graphs - not sure if this is the correct approach or not, though it seems logical to me), and because I'm not using the zener in reverse mode it does not matter what I set the zener voltage to. This methode give me 2.27V, and 51.93mA across the diode, which is perfectly in range of the laser diodes operating characteristics. BUT, I don't believe the voltage to current relationship can be established by the single 3V to 1A approach taken by the simulator. I like the results since they align with my LD operating characteristics, but I don't know if they are remotely correct at all.

    Method with Regular Diode - Using the extrapolated 3V value at 1A that the applet requires, which was used in the zener diode method gives me, I get the following: 2.66V, and 24.62mA. This whole thing worries me because the method using a regular diode in the simulator gets me close to the max voltage of the diode, and is right at half of the operating current which is less than the turn on (threshold current) so the device wouldn't work anyway. I've played with the resistor settings and nothing seems to change things in my favor. In fact, using only the first resistor in the circuit still never gets me the values I need for the LD to work properly no matter how much I play with it. I'm not sure I can draw much of a conclusion here.

    Does anyone have any advice on how to better model this?
    Does anyone have any advice on a better circuit, that I can easily duplicate 50+ times?
    Per my train of though and previous schooling, the LD should actually get to its operating voltage and then more or less follow its curve depending on the current available, correct?
    If you said I was correct on the question directly above, then would you tend to reasonably believe the results I've obtained by simulating it with the applet's zener diode because those two points are generally on the diodes curve, or maybe they still aren't close enough?

    Hopefully, there are a few smart people out there that know what really happens. Many thanks for spending time reading this.

    -mark-
     
    Last edited: Apr 8, 2011
  2. hoppyumr

    Thread Starter New Member

    Apr 6, 2011
    2
    0
    So, I've played with it a little bit more and found that I can post the code for creating the circuit on the simulator. I've played with some of the values, but based on what I told you the applet easily allows changes to be made. Just copy the entire text below and select File/Import and paste it in the popup window then select the Import button. Voila.

    $ 1 5.0E-6 10.20027730826997 50 5.0 43
    v 112 368 112 96 0 0 40.0 6.0 0.0 0.0 0.5
    w 112 96 208 96 0
    r 208 96 208 224 0 56.0
    w 208 224 272 224 0
    w 272 224 336 224 0
    c 272 224 272 368 0 4.7E-4 5.946400515888106
    w 272 368 336 368 0
    w 272 368 208 368 0
    w 208 368 112 368 0
    r 208 272 208 368 0 28.0
    s 432 272 432 224 0 1 false
    s 336 272 336 224 0 1 false
    d 432 272 432 368 1 2.2
    w 336 224 432 224 0
    w 336 368 432 368 0
    z 336 272 336 368 1 3.0 2.2
    s 208 272 208 224 0 1 false
    o 9 64 0 35 7.62939453125E-5 9.765625E-5 0 -1
     
  3. SgtWookie

    Expert

    Jul 17, 2007
    22,182
    1,728
    Use one LM317L and one 27 Ohm resistor per laser.

    See the attached.
     
  4. Adjuster

    Well-Known Member

    Dec 26, 2010
    2,147
    300
    The drive current level for a laser diode is more critical than it is for an LED. Until the drive current exceeds a minimum level referred to as the threshold current,t he laser diode produces very little optical output (and what it does produce is not coherent). The threshold current is the level at which laser operation commences, and further increases of drive produce rapidly rising output.

    Unfortunately, the threshold current and the drive level required for a given power output are temperature dependent, and vary somewhat between devices even though they may be of the same type. This requires caution, as what may be a satisfactory drive level for one device may produce little output from a second, but may be high enough to burn out a third.

    For this reason, it may be preferable to operate the lasers under some form of closed-loop control. This is what the monitor current is about. Frequently, laser manufacturers package their devices with an integral monitor device, normally a photodiode. The monitor diode is arranged to pick up some of the laser output, often from the opposite facet of the laser to that used for the main output. It may be that the extra forward current and reverse voltage ratings refer to this monitor device.

    To stand any chance of success with this, you must obtain a datasheet for your lasers showing correct connection details and operating levels.

    See link: http://www.fiber-optics.info/articles/laser_diodes

    Edit: Your application may not have anything to do with optical fibres, but the principles of operation are the same.
     
  5. retched

    AAC Fanatic!

    Dec 5, 2009
    5,201
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    Outside of the whole simulating problem, laser diodes are most efficently driven with a CONSTANT CURRENT DRIVER.

    As things heat up, and Vf changes with temperature, it would throw off the math and change the current limited by the resistor.

    Using a constant current driver, like Sgt.Wookie recommended, will adjust to keep the diode on setpoint throughout changes.

    And, with 5 components total, 2 decoupling caps, LM317L, resistor to set LM317L current output, and your Laser Diode

    LM317L is around 38 cents each for 50 quantity, 2 to 3 cents for the resistor, and a 2 to 5 cents for the decoupling caps.

    http://search.digikey.com/scripts/DkSearch/dksus.dll?Detail&name=296-17221-1-ND
     
  6. SgtWookie

    Expert

    Jul 17, 2007
    22,182
    1,728
    When you use the LM317L as a constant current source, you don't use a bypass cap on the output. Voltage sources are low impedance (ideally, zero impedance) and current sources are high impedance (ideally, infinite impedance). Adding a bypass cap on the output would lower the output impedance.
     
  7. retched

    AAC Fanatic!

    Dec 5, 2009
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    Listen to the man, he knows what he is talking about. ;)
     
  8. mjhilger

    Member

    Feb 28, 2011
    119
    16
    Also, aside from the electrical considerations, those brass housings provide the thermal conduction to heat sink the energy not converted into light. If you run several diodes together in their laser output region, they will get hot. You will need some kind of thermal conduction to carry the heat away or the VI characteristic changes even with feedback will either destroy or shut down the device.
     
  9. Wendy

    Moderator

    Mar 24, 2008
    20,766
    2,536
    I doubt for a 50ma LD heating will be a problem. Figure it is just over 100mw disapation.
     
  10. Adjuster

    Well-Known Member

    Dec 26, 2010
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    Unfortunately, laser diodes can be very sensitive to temperature, much more so than, say, a silicon transistor. They might be better regarded rather more like the old germanium transistors.

    Even a rise of a few tens of degreess can notably reduce the efficiency, and the maximum temperature may be only 80°-90°C or so.

    Example: http://www2.renesas.com/opto/en/pdf/PL10643EJ04V0DS.pdf

    For this reason, pretty extensive heat-sinking is common and the use of thermoelectric temperature control is not unknown, particularly if very good wavelength stability is required - some lasers can be tuned over a limited range by temperature control.

    If a laser has to be subject to some temperature variation, closed- loop control of the optical output is a possibility. In this case a very definite maximum drive current limit is required, perhaps combined with a thermal shutdown - otherwise the laser may be destroyed by a kind of thermal runaway.
     
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