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Thread: ma/volt confusion

  1. #1

    Question ma/volt confusion

    Ok if it's just me, sorry, i'm stupid...


    first of all, I've been "modding" my leadlight 110 apc board like mad, i've ripped off the APC modual, i've shorted out the pot, i've done a resistor jump to bypass the APC completely..

    i've removed a 220 resistor from the back of the driver board, shorted it out, and of all these mods, i've gained very little brightness... it's probably up around 20 - 25mw of power now, which is great and all but........ . . .. . ..... .

    lol


    I measured the 220 resistor, it's actually around 32 ohms of resistance, that's very little...

    now the confusing part

    when doing all these mods, people seem to worry about the amperage not the voltage.. what's a resistor do? a Resistor drops the voltage, not the amperage right?

    a light bulb at 3.5 volts, let's say it draws 500ma, in other words, a 3.5volt battery (1000ma) can supply a 3.5volt bulb X 500ma for approx 2hours before the battery goes dead 3.5v x 250ma would last for 4 hours on a 100ma 3.5v battery.. and so on..

    a laser modual, it draws however many MA it needs, where on the driver board does it control the ma? i mean if you stick a resistor on a light bulb at 3.5volts x 300ma, you could lower the voltage to say 3.0volts, but it would still be pulling 300ma right?

    So by dropping the resistance by bypassing something, decreasing the resistance, you get a brighter laser, what determines how many ma's it pulls, all i see is Voltage mods, not amperage modifications.

  2. #2

    Default Re: ma/volt confusion

    Quote Originally Posted by tobjectpascal
    when doing all these mods, people seem to worry about the amperage not the voltage.. what's a resistor do? a Resistor drops the voltage, not the amperage right?
    A resistor simply resists the flow of currrent. The voltage across the resistor is a function of the famous ohm's law equation V=I*R.


    Quote Originally Posted by tobjectpascal
    a light bulb at 3.5 volts, let's say it draws 500ma, in other words, a 3.5volt battery (1000ma) can supply a 3.5volt bulb X 500ma for approx 2hours before the battery goes dead
    In theory anyway. But in reality it's usually less. Some of the battery's energy will be wasted as I˛R losses due to the battery's own internal resistance. Basically at higher current draw, this wastage will be more and it's capacity will "appear" to be less than ideal and might not last 2 hours at all.

    Quote Originally Posted by tobjectpascal
    a laser modual, it draws however many MA it needs, where on the driver board does it control the ma? i mean if you stick a resistor on a light bulb at 3.5volts x 300ma, you could lower the voltage to say 3.0volts, but it would still be pulling 300ma right?
    Wrong. Inserting a resistance in series with a load (light bulb in your example) will further limit the current in the circuit and it will not pull 300ma. I'm not familiar with drive electronics for lasers but resistors are used to limit the current through an LED to a safe level.

    Quote Originally Posted by tobjectpascal
    So by dropping the resistance by bypassing something, decreasing the resistance, you get a brighter laser, what determines how many ma's it pulls, all i see is Voltage mods, not amperage modifications.
    Again I'm not familiar with laser diodes per se but if they work like LEDs then the using a lower limiting resistance will result in more current and maybe a brighter laser. I say maybe because most LEDs have a saturation point at which the gains in brightness per some unit of current are negligable and only serve to cause heat and decrease it's life.

    Regards,

    Dave

  3. #3
    *Flashaholic* James S's Avatar
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    Default Re: ma/volt confusion

    Good info from NicklePlate there.

    But none of that is the same when modding a current limited power supply. That driver board is a lot more complicated than just a resister. The resister and pot that you have modified are just used by the circuit to sense the amount of current actually going through the laser module, they provide a feedback not actual current limiting themselves.

    Most power supplies work by using a transistor switch to cut up the power at a high frequency and dump it through a coil or a capacitor so that pulses can be added together to provide a higher voltage. Those bits have a maximum amount of power they can store, so even if you could fake it into thinking that it still needed to send more power by the various mods, there is an upper limit that the board simply can't go beyond.
    -James

    E=sqrt((mc^2)^2+(pc)^2)

  4. #4
    Enlightened
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    Default Re: ma/volt confusion

    This used to trouble me as well. What you have to realize is that a resistor is not a pure voltage or current device, it is a POWER device, so both voltage and current are affected.

  5. #5
    Flashaholic* WildRice's Avatar
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    Default Re: ma/volt confusion

    Just a quick SMD note here. The 220 resistor is 22 ohm. so reading 32 ohm on the meter is possible, given 10% tolorance in the resistor, 5-10% in the meter, and prob loads more in the leads, esp reading LOW values. The resistor numbers are #,#,zeros. meaning 220= (2), (2), and zero zeros.
    AND if there is an "R" in there, ie 1R0, it means (1), (.), (0), or 1.0ohm.

    Jeff
    "We cannot learn from one another, until we stop shouting at one another, until we speak quietly enough, so that our words can be heard, as well as our voices." R.M.Nixon 1969

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