Maximizing output voltage of 9v battery with simple components

Discussion in 'General Electronics Chat' started by sixstringartist, Apr 8, 2008.

  1. sixstringartist

    Thread Starter Member

    Apr 8, 2008
    18
    0
    Hello all,

    Could someone help me understand the behavior of one of the capacitor/diode elements shown in the circuit below, and how these behave when put in parallel? Thanks.

    [​IMG]
     
  2. sixstringartist

    Thread Starter Member

    Apr 8, 2008
    18
    0
    Also the input is a 9V dc source that gets changed into a square wave.
     
  3. scubasteve_911

    Senior Member

    Dec 27, 2007
    1,202
    1
    I may be completely wrong, since I have never seen this before.

    By looking at it, you will immediately notice the push-pull stage via the two complementary? transistors. Upon the 9V that is allowed through the top transistor, your horizontal capacitors are charged, since the right terminal allows a path to ground. Upon the bottom transistor conducting and the top transistor is off, then this charged capacitor is allowed to transfer it's charge to the vertical capacitors. These charged vertical capacitors are in series, thus additive at the output. You can expect there to be 17X((9V-1.4-Ids*RDSon) at the output. Obviously, low current draw must be expected, since it relies on capacitance!

    Steve
     
  4. sixstringartist

    Thread Starter Member

    Apr 8, 2008
    18
    0
    Thanks, your reply helps.

    I see that when the 9v gets through it charges the horizontal caps, and I also see that the vertical caps, once charged, will act as batteries in series, but I cant seem to visualize how charge gets from the horizontal caps to the vertical caps.

    Also, what is the purpose of IC2 here?
     
  5. scubasteve_911

    Senior Member

    Dec 27, 2007
    1,202
    1
    You're welcome, hopefully I am correct..

    If you draw it out in steps, you will understand it, as long as you know the way current can flow through diodes.

    The LM311 is a voltage comparator. It seems to be designed to trigger when the output has hit a certain voltage. You can see this by the pot being used along with the voltage being divided down to the other input. Check out the LM311 datasheet.

    Steve
     
  6. SgtWookie

    Expert

    Jul 17, 2007
    22,182
    1,728
    Yes, the comparator triggers when the threshold voltage has been reached, inhibiting the astable mode of the 4047, thus stopping it from charging up any more. When the output voltage drops, the comparator output toggles, thus starting the charging cycle back up.

    That's a heck of a lot of wiring! It would've been easier (and I'll bet more efficient) to do with a flyback transformer. Even with low-ESR low-value caps to minimize losses, that's a lot of pumping, and the Vf across 34 1N4148 diodes ... (let's just say 22v) ... is a lot of overhead. Also, ALL of the horizontal caps (those tied to the MOSFET outputs) would have to have a voltage rating exceeding the max desired output voltage - and good luck troubleshooting it if one decides to short out, likely taking the MOSFETS with it.

    In contrast, have a look at this page:
    http://www.dos4ever.com/flyback/flyback.html
    The circuit ideas could be adapted to a 9v supply. Ronald's done a pretty good job of a tutorial, and thrown in a few extra handy circuits to boot.
     
  7. thingmaker3

    Retired Moderator

    May 16, 2005
    5,072
    6
    For an understanding of how the diodes and capacitors work together, try an internet search on the phrase "Cockroft Walton"
     
  8. mrmeval

    Distinguished Member

    Jun 30, 2006
    833
    2
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