[Noisebridge-discuss] A simpler circuit for ... [driving multiple LEDs with minimal batteries]
cm.hardware.software.elsewhere at gmail.com
Sun Jan 16 20:15:10 PST 2011
On Sun, 2011-01-16 at 18:05 -0800, Jonathan Foote wrote:
> Ooh! Another teaching moment. How Switching Power Supplies Work.
> Switching supplies are efficient because the transistor "switch" is
> either fully OFF, (in which case negligible current) or in saturation
> (fully ON) in which case there's negligible voltage across it. In
> either state the power consumption is tine (recall P=V*I), compared to
> linear mode which has both significant current and voltage drop. The
> JFET in that current limiting circuit is precisely in linear mode, and
> it will dissipate precisely the same power as a resistor (minus a
> little for the feedback resistor). In fact a popular use of JFETs are
> as voltage-variable resistors.
> > It's simpler and easier to use than a resistor (you don't even have to
> > calculate a value- you just get one that's got a lower millamp rating than
> > the target LED and make sure the battery voltage exceeds the sum of the
> > voltage in the LED string)
> It's not that much simpler seeing as how it's not a discrete component
> that I'm aware of. It's a handy circuit if your supply voltage or load
> is variable (though I would use a LM317 for any appreciable current,
> as Igs for JFETs is rarely better than 50mA). Otherwise, simpler and
> easier to pick a resistor using Ohm's Law.
For all switching regulators, you need some element that can store
energy (a capacitor or, in these cases, an inductor), not just dissipate
it (like a resistor or a switch).
So Ohm's Law becomes a little more complex.
If you don't want to use a boost converter,
maybe you get more bang for the buck if you use something like
[ from http://www.diodes.com/datasheets/ZXLD1352.pdf from
but you need a few extra components ... in addition to the inductor, you
need a capacitor, a resistor and a Schottky diode.
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