Inductor as a current drop???
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EvilLithiumMan
Enlightened
Inductors typically have an extremely low DC resistance; in the milli-Ohm range. But worse than that, think about what will happen when you turn off your light: an inductor will oppose a change in current - when you kill the current, the inductor will generate a voltage spike in an attempt to maintain current flow. Given a large enough inductor, you could destroy the emitter with the resulting voltage spike.
I am working on a light that uses an extraneous diode-- the non-light-emitting kind-- to drop the voltage enough to direct drive. This should be more power-efficient, I think, than a resistor providing an equal voltage drop.
For an series of ordinary diodes operating at ambient temperature, the voltage drop is about 0.7V per diode. I think this is a better alternative than using an inductor for the job.
Chalo Colina
For an series of ordinary diodes operating at ambient temperature, the voltage drop is about 0.7V per diode. I think this is a better alternative than using an inductor for the job.
Chalo Colina
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chalo said:
I am working on a light that uses an extraneous diode-- the non-light-emitting kind-- to drop the voltage enough to direct drive. This should be more power-efficient, I think, than a resistor providing an equal voltage drop.
For an series of ordinary diodes operating at ambient temperature, the voltage drop is about 0.7V per diode. I think this is a better alternative than using an inductor for the job.
Chalo Colina
[/ QUOTE ]
It won't be more effecient at all. Dropping 0.7V at X current means the same power dissipated, regardless if it's a diode or resistor. The advantage of the resistor is that it will allow you to use your batteries to a lower voltage.
The only way to use an inductor to drop the voltage is in buck circuit. The input voltage is switched on and off rapidly, with the duty cycle dependent on how much the step down is required. The inductor helps to maintain a constant current into the LED, smoothing out the input ripple.
chalo said:
I am working on a light that uses an extraneous diode-- the non-light-emitting kind-- to drop the voltage enough to direct drive. This should be more power-efficient, I think, than a resistor providing an equal voltage drop.
For an series of ordinary diodes operating at ambient temperature, the voltage drop is about 0.7V per diode. I think this is a better alternative than using an inductor for the job.
Chalo Colina
[/ QUOTE ]
It won't be more effecient at all. Dropping 0.7V at X current means the same power dissipated, regardless if it's a diode or resistor. The advantage of the resistor is that it will allow you to use your batteries to a lower voltage.
The only way to use an inductor to drop the voltage is in buck circuit. The input voltage is switched on and off rapidly, with the duty cycle dependent on how much the step down is required. The inductor helps to maintain a constant current into the LED, smoothing out the input ripple.
[ QUOTE ]
EvilLithiumMan said:
Inductors typically have an extremely low DC resistance; in the milli-Ohm range. But worse than that, think about what will happen when you turn off your light: an inductor will oppose a change in current - when you kill the current, the inductor will generate a voltage spike in an attempt to maintain current flow. Given a large enough inductor, you could destroy the emitter with the resulting voltage spike.
[/ QUOTE ]
The inductor will only generate a voltage spike if you switch out its load, not its source. And since you've switched out its load, what's it going to spike into?
If switching off the input to an inductor as you state were so dangerous, than buck DC-DC converters wouldn't work, and would be blowing up LEDs left and right, since buck DC-DC converters work by repeatedly sending a pulse of current through the inductor, then cutting off the input source.
EvilLithiumMan said:
Inductors typically have an extremely low DC resistance; in the milli-Ohm range. But worse than that, think about what will happen when you turn off your light: an inductor will oppose a change in current - when you kill the current, the inductor will generate a voltage spike in an attempt to maintain current flow. Given a large enough inductor, you could destroy the emitter with the resulting voltage spike.
[/ QUOTE ]
The inductor will only generate a voltage spike if you switch out its load, not its source. And since you've switched out its load, what's it going to spike into?
If switching off the input to an inductor as you state were so dangerous, than buck DC-DC converters wouldn't work, and would be blowing up LEDs left and right, since buck DC-DC converters work by repeatedly sending a pulse of current through the inductor, then cutting off the input source.
EvilLithiumMan
Enlightened
[ QUOTE ]
evan9162 said:
The inductor will only generate a voltage spike if you switch out its load, not its source. And since you've switched out its load, what's it going to spike into?
If switching off the input to an inductor as you state were so dangerous, than buck DC-DC converters wouldn't work, and would be blowing up LEDs left and right, since buck DC-DC converters work by repeatedly sending a pulse of current through the inductor, then cutting off the input source.
[/ QUOTE ]
evan9162,
Thank you for your reply. You correct on your first point. But you are assuming where the inductor would be connected in relation to the switch. Some folks place resistors in the head, others in the tailcap. Without being specified, it's anyone's guess how it would be connected.
Second point, I should have been a little more clear about my concerns. I did not mean to imply an inductor in and of itself was 'dangerous'.
Third point, a great majority of DC-DC converters, whether buck or boost, rely on the charactoristics of an inductor to function. But a DC-DC converter is a lot more circuitry than just a switch and an inductor.
Best regards, ELM
evan9162 said:
The inductor will only generate a voltage spike if you switch out its load, not its source. And since you've switched out its load, what's it going to spike into?
If switching off the input to an inductor as you state were so dangerous, than buck DC-DC converters wouldn't work, and would be blowing up LEDs left and right, since buck DC-DC converters work by repeatedly sending a pulse of current through the inductor, then cutting off the input source.
[/ QUOTE ]
evan9162,
Thank you for your reply. You correct on your first point. But you are assuming where the inductor would be connected in relation to the switch. Some folks place resistors in the head, others in the tailcap. Without being specified, it's anyone's guess how it would be connected.
Second point, I should have been a little more clear about my concerns. I did not mean to imply an inductor in and of itself was 'dangerous'.
Third point, a great majority of DC-DC converters, whether buck or boost, rely on the charactoristics of an inductor to function. But a DC-DC converter is a lot more circuitry than just a switch and an inductor.
Best regards, ELM
Thanks for straightening me out on that. /ubbthreads/images/graemlins/ohgeez.gif
I was under the mistaken impression that the 0.7V drop across a diode was a threshold voltage unaccompanied by a corresponding amount of resistance. Now that I think about it, that doesn't really make sense, does it?
Chalo Colina
I was under the mistaken impression that the 0.7V drop across a diode was a threshold voltage unaccompanied by a corresponding amount of resistance. Now that I think about it, that doesn't really make sense, does it?
Chalo Colina
