2 pulses sent to control the beam shutter

Discussion in 'The Projects Forum' started by caterinec, Sep 15, 2009.

  1. caterinec

    Thread Starter New Member

    Sep 15, 2009
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    Hi all,

    I'm currently trying to a pulse which first send out 24 V for 5 ms pulse width and stay 10 V for 10 ms pulse width after that in order to control the laser through beam shutter . Please refer to the attached drawing. The 24 V is to give the beam shutter a big push to open up quickly in order to let the laser go through, but, continuously passing 24 V into the beam shutter will fry it. So, 10 V is required to hold the beam shutter open for 10 ms.

    I created 2 pulses with 2 different set of circuit and tried to combine them together. However, I'm not sure how to combine both pulses together as one and sent into a load, the beam shutter. I tried to connect two collectors of each TIP102 and send in to a beam shutter as a load at one end, on the other end, I connect one to 24 V source and another one to 10 V source. It turned out incorrect.

    Can someone please tell me what's the best way to combine 2 pulses together before entering the load or correct the circuit in order to perform as required. Thank you!

    [​IMG]
     
    Last edited: Sep 15, 2009
  2. rjenkins

    AAC Fanatic!

    Nov 6, 2005
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    A common, simple method is to use a resistor to set the lower 'hold' current and put a capacitor across that to give the initial higher current pulse.

    You are then only switching a single circuit.

    The actual component values depend on the resistance of the load (solenoid?) and the supply voltage in use; presumably just 24V for this method.
     
  3. caterinec

    Thread Starter New Member

    Sep 15, 2009
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    rjenkins,

    Thanks for your sharing and reply. I wonder if this is applicable for "lower" hold voltage? I forgot to mention that all outputs are only referring to the set voltage but any current as long as it is lower than 200mA.
     
  4. rjenkins

    AAC Fanatic!

    Nov 6, 2005
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    As long as the load has a fixed resistance (eg. a solenoid coil), you can treat that plus a series resistor as a potential divider.

    eg.
    To give the 10V on it from a 24V supply, the series resistor would be 1.4 x the load resistance.

    To give the 10V on it from a 48V supply, the series resistor would be 3.8 x the load resistance.

    Knowing either the load resistance or load current at a specific voltage (which allows you to calculate the resistance) is important to be able to design an effective drive circuit.

    If you must do it with two separate supply voltages, you need to add two PNP transistors to your circuit, one with the emitter fed from +24V and one with the emitter fed from +10V.

    Each needs a base-emitter resistor to ensure it turns off, then a resistor from base to the appropriate output of your existing NPN switching circuit.

    The collector of the transistor supplied from 24V goes direct to the positive of the load, with the load negative to 0V.

    The collector of the transistor supplied from 10V goes to the anode of a diode (1N400 series or similar) with the diode cathode to the load positive. The diode prevents the higher voltage feeding back through the 10V circuit.

    You only need one trigger pulse circuit for both 555s as they are running on the same supply, just connect the pin 2 inputs together. If you build it as in your diagram, don't forget the 10n trigger cap in the right half, it's omitted from the drawing.

    Due to the diode forward drop, the '10V' output will be reduced by about 0.7V, if that is not acceptable you will have to adjust the supply voltage to get a true 10V output.
     
  5. gerty

    AAC Fanatic!

    Aug 30, 2007
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    Will the shutter have time to move with only a 5 or 10 ms pulse?
    We used to control our laser by leaving the shutter open and firing the laser.
     
  6. caterinec

    Thread Starter New Member

    Sep 15, 2009
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    Robert, thank you so much for explaining into details. This definitely helps a lot.

    For some reason, we want to use two separate voltage sources in the circuit. So, I've made some modification referring to your suggestion (hopefully it looks like what you've described :p). However, I still have a question about the ciruit. How to determine the value of the resistors that marked Ra & Rb around the new added PNP transistor? Is there any specific equation to calculate them?

    p/s: Oh yea, thanks for catching the missing 10n capacitor!! Good eyes ;)

    [​IMG]
     
  7. caterinec

    Thread Starter New Member

    Sep 15, 2009
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    Gerty:
    Yes. In fact, the 24V of the 5ms pulse is the initial force to open the shutter quicker.
     
  8. rjenkins

    AAC Fanatic!

    Nov 6, 2005
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    Hi,
    I was thinking two high-side switches but your method should work just as well - in fact better as the 24V cannot stay active when the 10V is off; and it's simpler!

    I'd try RA and RB at 1.5K also, however move RA directly across the base & emitter of the transistor switching the +24V, leaving RB as drawn.

    Edit: Just a thought, as it is (presumably) an inductive load.
    Add two diodes, one with anode from the bottom of the load and cathode to +24V, and another with cathode to the top of the load and anode to 0V.

    When you switch off an inductive load, the current attempts to keep flowing and the polarity reverses, to extreme voltages if precautions are not taken.

    The conventional way it to fit a 'flywheel diode' directly across the load, but this allows the current to keep circulating for some time and delays the turn-off. This is OK with a typical relay, but not good with something timed to the millisecond.

    Putting the diodes back across the 24V supply will limit the voltage spike to around 25V, whilst giving a fairly fast turnoff.

    Another alternative is to put a VDR ('Zenamic') rated at about 30V across the load, which will give an even quicker turnoff.
     
    Last edited: Sep 16, 2009
  9. caterinec

    Thread Starter New Member

    Sep 15, 2009
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    Thank you again, Robert!

    Is the layout looks like what you were mentioning before? 2 diode added in red line across the load which is a 28 ohm solenoid.

    [​IMG]
     
  10. rjenkins

    AAC Fanatic!

    Nov 6, 2005
    1,015
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    Hi,
    almost what I was thinking, but with the top of the diode on the left connected to +24V rather than back to the top of the solenoid.

    That should give a quicker 'off' response.
     
  11. caterinec

    Thread Starter New Member

    Sep 15, 2009
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    It's WORKING!! Yay~!! Thank you sooooo much, Robert!! You're very helpful, patient and details-oriented. Once again, I appreciate your quick response & effort. ^^
     
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