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Start a led on a sertent voltage

magnus2112
magnus2112 over 11 years ago

Hi!

Do enyone know how i can make a led turn on that uses 3 volts, when a voltage has reached 330volts?

I am trying to make a indicator turn on when a capasitor is full..

Thanks for eny help! image

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  • jw0752
    jw0752 over 11 years ago +1
    Hi Magnus, Just some additional information that may be of help. In the old days I used to work on battery powered camera flashes. They would run off 3V or 6V battery power. An oscillator would charge…
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  • magnus2112
    0 magnus2112 over 11 years ago

    Thanks Don! But your explanation was way to advanced for me..image

    Eny way you can explane it in a "noob friendly" way?

    Does it exist a mosfet with a gate that has to to be 330v or higer?

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  • D_Hersey
    0 D_Hersey over 11 years ago in reply to magnus2112

    No, but you don't need one, I got a little business to attend to, then I can drop by my office where I can draw pictures for you, that will probably make things clear.  Wait a couple of hours.

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  • D_Hersey
    0 D_Hersey over 11 years ago in reply to D_Hersey

    Hi, I'm back.  Let me say that you are brave for starting off in HV, something I've avoided for years.  An LED is not a good electrical match for 330 V.  You should either establish an LV, or use an incandescent or neon bulb to establish a visual indicator.  Let's suppose we are going to use a resistor to ballast an imaginary indicator LED.  Let's say that our current through the LED should be 30mA and its drop is 1.2V, consistent with a red LED.  The amount of power our LED is using is the product, 36mW.*  The drop across the ballast is 330 - 1.2V, 328.8V.  328.8 / 1.2 = 274.  That's how much more power the ballast is using as regards the LED.  The ballast uses almost 10W!  There are other reasons to go to an auxiliary LV as well.  I will presume a 5V LV, as a design point of departure.

    image

    Can you make sense of this circuit as I have drawn it so far?  Assume 0V and gnd are connected.

    ----------

    *Per definition, all two-terminal elements strung together in series have the same current running through them.  P(W) = Potential(V) * Current(I).  Hence, the proportionality is the ratio of the drops.

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  • D_Hersey
    0 D_Hersey over 11 years ago in reply to D_Hersey

    Hi, I'm back.  Let me say that you are brave for starting off in HV, something I've avoided for years.  An LED is not a good electrical match for 330 V.  You should either establish an LV, or use an incandescent or neon bulb to establish a visual indicator.  Let's suppose we are going to use a resistor to ballast an imaginary indicator LED.  Let's say that our current through the LED should be 30mA and its drop is 1.2V, consistent with a red LED.  The amount of power our LED is using is the product, 36mW.*  The drop across the ballast is 330 - 1.2V, 328.8V.  328.8 / 1.2 = 274.  That's how much more power the ballast is using as regards the LED.  The ballast uses almost 10W!  There are other reasons to go to an auxiliary LV as well.  I will presume a 5V LV, as a design point of departure.

    image

    Can you make sense of this circuit as I have drawn it so far?  Assume 0V and gnd are connected.

    ----------

    *Per definition, all two-terminal elements strung together in series have the same current running through them.  P(W) = Potential(V) * Current(I).  Hence, the proportionality is the ratio of the drops.

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