High Speed LED Camera Flash - How To Get Short Duration

Discussion in 'The Projects Forum' started by moocrow, Apr 11, 2011.

  1. moocrow

    Thread Starter New Member

    Apr 3, 2011

    I posted this thread, http://forum.allaboutcircuits.com/showthread.php?t=52468, to get some help with using a transformer to produce a high enough voltage to fire a flash tube. The aim is to get a low flash duration of <20μs for some high speed photography.

    PackratKing suggested trying to achieve the same effect using LEDs and a 555 timer. I have put the following monostable 555 circuit together.

    The combination of C1, D1 and R2, included between pin 2 and the trigger switch make the 555 edge triggered. Without these the LED remains lit as long as the trigger button is pressed down, even if this is for longer than the duration specified by the combination the timing resistor (R3) and timing capacitor (C3). This is inevitably the case when the desired duration is a small fraction of a second.

    So far, I have successfully lit the LED array using an 11μs (and higher) combination of R3 & C3 (1kΩ & 0.01μF) but, not surprisingly, it is very dim, as it is at 110μs.


    Will raising the voltage across the LEDs cause them to light brighter?

    Assuming the LEDs are rated for 3.5-4V forward voltage what is roughly the highest they could take for 11μs or 110μs?

    Should I consider raising the current?

    Will firing the LEDs from a capacitor make any difference?


    S1 - Push to make, non-latched, used to trigger the pulse from the 555.
    S2 - Flip switch, switches the 555 circuit on and off.

    All 1/4W:
    R1 = 10KΩ
    R2 = 4k7Ω
    R3 = Timing resistor, values 1kΩ to 1MΩ tried
    R4 = 150Ω (so the 555 can be used to power a single LED directly, for testing)

    C1 & C2 = 0.01μF (ceramic)
    C3 = Timing capacitor, values 0.01μF (ceramic) and 10μF (electrolytic) tried
    C4 = 220μF (electrolytic)

    D1 = 1N4007

    MOSFET = 75337P

    LEDs - An array of 72 'superbright' white LEDs, i.e an 'Inspection Lamp' from Maplin (http://www.maplin.co.uk/72-led-inspection-lamp-345045)

    4.5V Source - Three AA batteries. 4.5V chosen to leave enough voltage coming out of the 555 to power a single LED, which I successfully tested before connecting to MOSFET and LED array.

    6V Source - The four AA batteries included in the inspection lamp. One terminal from the battery compartment and one from the LED array are connected to the MOSFET.

  2. Wendy


    Mar 24, 2008
  3. someonesdad

    Senior Member

    Jul 7, 2009
    If you can, get your hands on a scope and a function generator. Most function generators can output roughly 10 volts into 50 ohms, meaning a current on the order of (10000 mV)/(50 Ω) = 200 mA. I'll assume that's enough to drive your LED. If not, use it to switch your MOSFET.

    Then you can hook the function generator across the LED, set the function generator's output appropriately, and monitor the current through and voltage across the LED. You'll be able to measure how much current at a given pulse width and rate you can put into the LED. It's mostly controlled by how fast the heat dissipates from the die. You monitor the output of the LED with your eye to decide when it's bright enough. BTW, this makes a handy strobe -- I once used it to find the fan that was intermittently making noise inside my computer.

    You'll have to use engineering judgment about what's too much current. Sometimes the LEDs data sheet might give you an idea of the maximum current you can use for a short (microseconds) pulse, but often not. Thus, if your application isn't a one-off thingy where it's easy to replace the LEDs, you'll probably want to do an experiment to see that you get adequate life running a sample of LEDs at the pulsed current levels you choose.
  4. moocrow

    Thread Starter New Member

    Apr 3, 2011
    Sorry, forgot to include the resistor (actually two) as it's part of the LED inspection lamp. Corrected schematic below.


    Thanks someonesdad: are you saying that increasing the voltage and/or current WILL increase the brightness? Is it one or both and which has the bigger effect?
  5. retched

    AAC Fanatic!

    Dec 5, 2009
    LEDs are current controlled.

    With LEDs, once you satisfy the Vf, you control brightness with PWM or current control.

    Have you read through the bills' article he linked to in post #2?
  6. someonesdad

    Senior Member

    Jul 7, 2009
    Your question will be answered by reading about the voltage-current relationship (part c of the picture) of a diode. It's a fundamental curve that everyone should know about. The actual numbers for a particular diode depend on lots of things (especially the materials of construction), but the basic curve shape is similar for most diodes (there are some special diodes that don't have curves that qualitatively look like this).

    That curve to the right grows essentially exponentially with voltage. That steep rise is current is the fundamental reason why a resistor is always needed to be in series with the diode unless some other mechanism is used to limit the current. Otherwise, the diode burns up when the voltage increases too much.

    Here's a paper that shows you how the light output varies as a function of DC voltage, current, and power.
  7. Kermit2

    AAC Fanatic!

    Feb 5, 2010
    I think you are creating a problem here, when there isn't one.

    The 'speed' of high speed photography has little to do with the duration of the flash. The flash itself(xenon tube)is going to take a finite time to complete. Ionization is based on actual physical interaction in the atoms themselves. This cannot be 'speeded up', but takes time to complete due to distances between gas atoms and the happenstance/random nature of the electron interactions between atoms of gas.

    For extreme shutter speeds, all that is needed is sufficient photonic flux. The shutter will open and close several times(in high speed photography) before a normal flash event has completed.

    you don't need a shorter duration flash for high speed photography, just a higher level of illumination during the actual 'shoot'. All based on shutter speed and aperture size.
  8. Adjuster

    Well-Known Member

    Dec 26, 2010
    Depending what type of camera you are using, and on the exposure method, a short flash may be useful (or not).

    In otherwise dark conditions, a flash that is actually shorter than the shutter time can be used to obtain very short exposures.

    On the other hand, some types of camera shutters, such as the focal plane types used in some SLRs, take more than their rated exposure time to expose all of the image. Using a short flash in this case can result in part of the image being unexposed.

    Another limitation is the response time of the light source. The last poster has explained the situation with gas tubes. Monochrome LEDs could be pretty quick by photographic standards, but the white variety generally utilise phosphors to obtain the longer wave part of their spectra. The phosphor afterglow can be expected to slow down the response.

    I don't know the order of magnitude of the decay time, but I would guess milliseconds rather than microseconds. There would be no point in making the electrical drive pulse shorter than the LED response time.
    Last edited: Apr 12, 2011
  9. John P

    AAC Fanatic!

    Oct 14, 2008
    I wonder if in this exact case, you could get away without using a resistor in series with the LEDs. You would store a certain amount of energy in a capacitor, then have a transistor that dumps it through the LEDs. What a resistor does is limit the power (rate of delivery of energy) that gets dissipated in the LEDs, but if the power is limited by the total energy available, your LEDs would inherently be protected. But you would need to know the acceptable overload power through your LEDs; if you were looking them up in a catalog, this might be listed. What they would tell you would be highest d.c. current rating, and the highest pulse current for a particular duration and repitition rate. That's the key figure.