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Lowering Voltage

Discussion in 'Electronic Basics' started by Alan Smithee, Nov 17, 2003.

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  1. Alan Smithee

    Alan Smithee Guest

    Hi,

    I'm trying to use this circuit diagram
    (http://ourworld.compuserve.com/homepages/Bill_Bowden/page4.htm#whistle.gif)
    to control a laser. The circuit runs using +12v, but I'd like to use a
    simple +9v square battery. What do I have to consider to drop the voltage?

    Can I just drop the resistance? 9 volts is 3/4 of 12. Therefore reduce
    resistance by 3/4 - is this right?

    My main concern is that the filters will be thrown out and the signal needed
    to trigger the flip flop will be too great - I can only whistle so loud.

    Also, do I need a relay? The laser I plan to use operates at 3 volts -
    could I just add resistance to drop the voltage? or does it not work like
    this.

    Many thanks
     
  2. I don't see anything that depends on the voltage, really. The filters depend
    on the frequency, but not the voltage.

    Thus, I'd say you could probably just run it off of 9V.

    Regards,
    Bob Monsen
     
  3. Alan Smithee

    Alan Smithee Guest

    Thanks Bob,

    Apparently at 9 volts I'll get about 7.3 vots out of the emitter at 50-100
    mA - the laser I plan to use runs at 3volts and 85mA - what do I need to
    know to make the change?

    Thanks
     
  4. John Fields

    John Fields Guest

    ---
    Instead of an emitter follower driving a solid-state relay as the
    output, I'd blow off the relay and use the output transistor as a
    saturated switch. Like this:

    9V
    |
    [7R5]
    |
    |A
    [LASER]
    |
    C
    Q>--[1K]---B 2N4401
    E
    |
    GND

    Q is the output from the 4013, and for a 9V supply what you need to
    know to limit the current through the LASER diode to 85 mA is that you
    subtract the LASER diode voltage and the transistor saturation voltage
    from the supply voltage and then divide that difference by the LASER
    diode current. That number will be the reqired resistance.

    So, R = (Vcc - (Vlaser+Vcesat))/85mA = (9V-3.3V)/0.085A ~ 67.06 ohms.

    The closest standard 5% value available which will keep the current from
    going over 85mA is 68 ohms, which will limit the current to
    I = E/R = 5.7V/68R = 84mA. You will be dropping 5.7V across the
    resistance and allowing 84mA to flow through it, so the power it will
    dissipate will be P = IE = 84mA*5.7V ~ 479mW, so I'd use a 1 Watt
    resistor.

    The 9V battery will be supplying 9V*84mA = 756mW, so looking at the
    curves for Service Life VS Power Drain for a Duracell alkaline 9V
    battery at

    http://www.duracell.com/oem/Pdf/MX1604.pdf

    reveals that you can expect about 4 hours of continuous operation until
    the battery voltage decays to 6V.
     
  5. Alan Smithee

    Alan Smithee Guest

    Thanks for that John - I'm almost there !!!! Just a couple more questions if
    you don't mind
    Curiously Maplin doesn't stock a 1w resistor - they do a 2w and a 0.6 -
    what's best?
     
  6. John Fields

    John Fields Guest

    9V So, I take a feed directly from the 9Volt
    |
    [7R5] Is this a resistor? Sorry for being dumb.
    |
    |A
    [LASER] The laser of course
    |
    C
    Q>--[1K]---B 2N4401 Q and Data to base?
    E
    |
    GND

    ---

    1. Yes, take the feed directly from the battery

    2. Yes, it's a resistor, but it should read 68R, (_not_ 7R5)
    since it's the 68 ohm resistor referred to in the text.
    Sorry about that...

    3. Yes, the LASER. The 'A' refers to the anode connection.

    4. Yes, through a 1000 ohm 1/4 watt resistor.

    5. Any NPN capable of dissipating 50mW while carrying a collector
    current of 84MA and sporting a beta of 100 or better should work.
    A 2N2222A (PN2222A) should be fine.

    6. you can make the equivalent of a 68 ohm 1 watt resistor by connecting
    four 68 ohm 1/4 watt resistors in series-parallel:

    +-----+---->
    | |
    [68R] [68R]
    | |
    + +
    | |
    [68R] [68R]
    | |
    +-----+---->
     
  7. John Fields

    John Fields Guest

     
  8. Alan Smithee

    Alan Smithee Guest

    Your an absolute star John - I can now begin to design the PCB layout - gulp
     
  9. Alan Smithee

    Alan Smithee Guest

    +-----+---->
    I take it that the 1 watt resistor would be better for battery consumption?

    This looks like 2 lots of 2 resistors in series - is this right? Would just
    like to be sure.

    Thanks again
     
  10. FYI, the max power value of a resistor doesn't affect anything except the
    heat it can expect to dissipate safely. So, the power consumption will be
    identical for the 2W or 4 1/2 watt resistors. Thus, you can use the 1/2W
    resistors to build a 1W resistor wihtout affecting the power consumption.

    Regards,
    Bob Monsen
     
  11. The LM1458, 741, etc. opamps do poorly at low voltages, they just
    can't put out much. You could replace the 1458 with a LM358, which
    has identical pinout. The rest of the circuit looks like it should
    work okay off a lower voltage.

    --
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    My email address is whitelisted. *All* email sent to it
    goes directly to the trash unless you add NOSPAM in the
    Subject: line with other stuff. alondra101 <at> hotmail.com
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  12. [snip]
    Q and Data to base? Is there an alternative to 4401?
    Maplin don't seem to stock it

    Use a BC337-25.


    --
    @@[email protected]@[email protected]@@[email protected]@[email protected]@[email protected]@@[email protected]@[email protected]@[email protected]@,@@[email protected]@[email protected],@@[email protected]@[email protected]@[email protected]@
    ###Got a Question about ELECTRONICS? Check HERE First:###
    http://users.pandora.be/educypedia/electronics/databank.htm
    My email address is whitelisted. *All* email sent to it
    goes directly to the trash unless you add NOSPAM in the
    Subject: line with other stuff. alondra101 <at> hotmail.com
    Don't be ripped off by the big book dealers. Go to the URL
    that will give you a choice and save you money(up to half).
    http://www.everybookstore.com You'll be glad you did!
    Just when you thought you had all this figured out, the gov't
    changed it: http://physics.nist.gov/cuu/Units/binary.html
    @@[email protected]@[email protected]@[email protected]@[email protected]@[email protected]@[email protected]@@[email protected]@[email protected]@@[email protected]@[email protected]@[email protected]@[email protected]@[email protected]@[email protected]@@
     
  13. Alan Smithee

    Alan Smithee Guest

    Thanks to everyone!

    I now have my circuit completely planned - save the layout :) or rather
    :-( since I've not done this before !

    Thanks anyhow
     
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