High Voltage Swiitch Question

Discussion in 'The Projects Forum' started by moocrow, Nov 1, 2011.

  1. moocrow

    Thread Starter New Member

    Apr 3, 2011
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    Hi,

    A few months ago some guys on the forum helped me with a circuit for a xenon flash, thanks! In the end I opted for cannibalising a disposable camera flash and now have a working circuit, triggered by a button press.

    The main aim of my project is to get to a flash duration of <10μs for the purpose of high-speed photography. I plan to do this with a variation of the following circuits:

    [​IMG]

    On the right hand side the optocoupler which is common to both circuits closes the trigger circuit which powers the tigger transformer for the xenon flash tubes.

    In the top version the main circuit, connecting the main 330v capacitor to the flash tubes, is also closed with an optocoupler. In the bottom circuit it is closed with a MOSFET.

    The MOSFET circuit (bottom) has the advantage of better responsiveness but I can't work out how to do it without sharing the earths of the battery-powered circuit and the circuit connected to the flash's main capacitor at 330v. This seems dangerous?

    The optocoupler only circuit (top) doesn't share the earths but is no where near as responsive, with an internal LED rise-time of 5μs.

    Is there a better way of doing it altogether, with or without using a 555 IC which has a lower limit for the monostable pulse generation of 6μs +/- 20%?

    Thanks for any help.
     
    Last edited: Nov 1, 2011
  2. KMoffett

    AAC Fanatic!

    Dec 19, 2007
    2,574
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    No images or links! ???

    ken
     
  3. SgtWookie

    Expert

    Jul 17, 2007
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    The image didn't "take", but I dug the URL out of the OP's post:

    [​IMG]
     
  4. SgtWookie

    Expert

    Jul 17, 2007
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    Consider using digital isolators instead of optocouplers.

    Digital isolators provide a similar level of isolation, but are a great deal faster than optocouplers. You could drive the input of a digital isolator using something like a CMOS 555 timer, and use the isolated output to control a MOSFET gate driver to give you very fast turn-on/turn-off times.

    Here's a digital isolator design guide from Texas Instruments:
    http://www.ti.com/lit/an/slla284/slla284.pdf

    Searching on Google for "digital isolator" will result in many hits.
    Digital isolators are relatively new.
     
  5. moocrow

    Thread Starter New Member

    Apr 3, 2011
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    Thank you for digging the picture out and the advice.

    Would this digital isolator be suitable: http://www.ti.com/lit/ds/symlink/iso721.pdf?

    I'm afraid I don't understand the pinout, even having read a couple of tutorials. Does the digital isolator fit in the circuit like the MOC3020 would? I can't work out what the equivalent pins are?

    Going back to my original question, is it right that sharing the earths is either dangerous or would prevent the circuit from working? I.e. is there a way of having the 555 output switch the MOSFET without an isolator?

    Thanks.
     
  6. moocrow

    Thread Starter New Member

    Apr 3, 2011
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    Hi, I've managed to find a few variations on switch circuits, with or without isolators, including two here: http://www.excelitas.com/downloads/APP_PFL.pdf which use either two SCRs or one IGBT.

    I have a capacitor charged to 300V and four flash tubes. I can't find a way of calculating the max current each flash tube will take as I can't calculate their minimum resistances. I have found two formulae for calculating the minimum resistance, one in the document above and one here: http://www-math.mit.edu/phase2/UJM/vol1/KAMALICF.PDF. Both require the maximum current however, which is exactly what I'm trying to find.

    Without knowing these, how do I select an IGBT or SCR which can handle the high current as the flash tubes discharge?

    Thanks,

    Moocrow.
     
  7. SgtWookie

    Expert

    Jul 17, 2007
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    I can't answer your reply #5, as I don't know how your flash units are powered. If they are all isolated supplies (like batteries) then you might be able to do that.

    However, you're going to have a problem getting a lot of power down a cable and back to the flash unit.

    You basically need something like twisted pair cable to the flash unit, with line drivers on the camera end and line receivers on the flash ends to minimize ringing on the cables; then drivers and IGBT, SCR or MOSFETs to complete the flash tube's current path.

    The amount of current will be related to how much capacity in uF the cap(s) store at what voltage, and the discharge time (which is probably around a millisecond).

    I really don't know offhand what the current is.
     
  8. moocrow

    Thread Starter New Member

    Apr 3, 2011
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    Apologies for not being clear in my previous post - I appreciate it's not possible to work out the current from the information I provided but wondered how I go about calculating it?

    I'm using a circuit from a disposable camera flash. I have connected four xenon flash tubes to it and am using an 80μF capacitor. Other than the wires, there's nothing else to the main discharge circuit so the flash tube really is the only source of resistance.

    The resistance of the flash tube falls as the current increases but I can't work out where it bottoms out and therefore what the maximum current is?

    Cheers,

    Moocrow.
     
  9. SgtWookie

    Expert

    Jul 17, 2007
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    If you had a DSO (digital storage oscilloscope) you could capture the waveform of a voltage drop across a 1 Ohm resistor in series with the cap return when the strobe fires.

    If you don't, you might consider making a peak detector from an opamp and a few assorted components recording the peak voltage across the above mentioned resistor, firing the flash, then measuring the voltage that the peak detector recorded.

    As far as calculating it - I don't know what the discharge characteristics are for a Xenon flash tube, nor your tubes in particular. However, one normally doesn't flash the tubes themselves; they provide current to the primary of a HV transformer.

    The rapid change in the voltage across the primary causes a much higher voltage to be developed on the secondary (which has many more turns than the primary).

    However, it's going to be difficult to calculate what that primary current is at a given point in time without knowing what the inductance of the primary is. The smaller the inductance of the primary, the faster the rate of change of current in the primary.
     
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