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..
Be sure to click 'more' and select 'suggest as answer'!
If you're the thread creator, be sure to click 'more' then 'Verify as Answer'!
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..
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.
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.
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.
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.
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.