Switchable voltage booster?

Discussion in 'The Projects Forum' started by Razor Concepts, Oct 7, 2008.

  1. Razor Concepts

    Thread Starter Active Member

    Oct 7, 2008
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    I'm looking into making a voltage booster than can boost the voltage going to a motor by two or three volts, triggered by a switch.

    Two power leads (ground and positive) will be inputs, and will have a good amount of amps flowing through them (30-50 amps, 7 volts). It should output those directly, except when the completion of a circuit (switch) triggers a large capacitor to boost the voltage to maybe around 9 or so. The switch will be a Bit-Switch, enabling me to use a servo pulse to trigger it. The Bit-Switch can only handle about a fourth of an amp, so all the current cannot flow through that...

    Right now all I want to know is if it's possible to do. Thanks :D
     
  2. kubeek

    AAC Fanatic!

    Sep 20, 2005
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    So you need a circuit to boost the voltage, or you want some switch that will temporarily add the capacitor in series?
     
  3. DickCappels

    Moderator

    Aug 21, 2008
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    I think the way he wants to do boost the voltage is by switching a capacitor in series with the supply. At 30- 50 amps, that's got to be a carefully selected capacitor.

    Yes, it is possible. Not necessarily achievable unless you have a large R&D budget or a lot of time and patience (From experience, you will probably still need a large budget to cover the expense of all the burnt parts).
     
  4. Razor Concepts

    Thread Starter Active Member

    Oct 7, 2008
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    Yes, while the switch is on the capacitor will drain. I was thinking of putting the capacitor in parallel, just in case the user has the switch on after the capacitor has drained (so then no power will be going through).

    Hmm, I never realized this would be a cost issue. What will be the part burning out? I was assuming if the capacitor is temporarily added in parallel for the boost, it wouldnt be damaged.
     
  5. scubasteve_911

    Senior Member

    Dec 27, 2007
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    Okay, let me get this straight.. What kind of motor are you using? Secondly, how much inertia do you have connected to the motor? Why are you doing this?

    I might be failing to see the point of adding so much instantious current to the motor. It will not result in instant torque development, nor speed. Your inductance will limit the rise of your current, as will, in turn, your torque. Then, the speed increase is governed by the inertia attached and the torque being applied.

    If you create a dynamic model of your system, you may find the 'instantanious' current pointless. The electrical time constant is always magnitudes smaller than the mechanical.

    Steve
     
  6. Razor Concepts

    Thread Starter Active Member

    Oct 7, 2008
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    This would be for use in radio controlled cars/trucks, as sort of an electric nitrous system, where it delivers a short burst of speed.
     
  7. beenthere

    Retired Moderator

    Apr 20, 2004
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    What kind of volume and weight constraints are there? If the boost puts a 75% weight penalty in the car, it's not much of an advantage.
     
  8. Razor Concepts

    Thread Starter Active Member

    Oct 7, 2008
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    I guess maximum could be about 1/2 to 3/4 pound, and volume maybe the size of a candy bar (or more).

    EDIT: Did some more reasearch and found that a capacitor would need to be HUGE for my application. Oh well...
     
    Last edited: Oct 8, 2008
  9. beenthere

    Retired Moderator

    Apr 20, 2004
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    You could maybe put some NiMH cells in parallel and use a servo to close a switch.
     
  10. Razor Concepts

    Thread Starter Active Member

    Oct 7, 2008
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    The rc/car truck would be using nimh/lipoly to power it already, so might as well just replace the main power system with something bigger. I was hoping a capacitor maybe the size of a candy bar would provide some extra power, but with all the farads I need it wont be practical.
     
  11. scubasteve_911

    Senior Member

    Dec 27, 2007
    1,202
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    Yeah, this is an energy density problem, where the capacitor clearly loses in comparison to any battery.

    Steve
     
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