<div><br></div><div><br></div>Wrong. You are limiting the power rather than burning a portion of it off as heat. Yes, there *is* an inegrated resistor, but it's very low resistance- only enough so the JFET can sense the current (V=IR) and turn off when there's too much current and turn back on when there's not enough. <div>
<br></div><div>It's a switching power supply with just two leads.<br><div><br></div><div>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)</div>
<div><br></div><div>And it's way more efficient anytime the supply voltage is higher than the the LED voltage. Like if you want to drive LEDs off 110V AC and a bridge rectifier. </div><div><br></div><div>Sure, if you happen to match the supply voltage closely to the LEDs, then a resistor is fine. Like in the LED throwies- a resistor can even be zero.<br>
<br></div><div>But if you want a circuit that works with white and red LEDs, that works when you add more in series, etc., you want current limiting not voltage limiting.</div><div><div><br></div><div>T</div><div><div><div>
<div><br><div class="gmail_quote">On Sun, Jan 16, 2011 at 11:27, Jonathan Foote <span dir="ltr"><<a href="mailto:jtfoote@ieee.org">jtfoote@ieee.org</a>></span> wrote:<br><blockquote class="gmail_quote" style="margin:0 0 0 .8ex;border-left:1px #ccc solid;padding-left:1ex;">
As in all engineering solutions, optimizing one variable (efficiency,<br>
say) comes at the cost of another (simplicity).<br>
<br>
A little teaching moment here: there will be a voltage drop V across,<br>
and a current I through, the CLD.<br>
Power = V x I. How much power is this? Where does it go?<br>
<br>
And how much power would an equivalent resistor use?<br>
<br>
Seeing as how neither the battery voltage nor the load is changing<br>
appreciably, what's the advantage to using a CLD over a resistor?<br>
<div><div></div><div class="h5"><br>
<br>
<br>
On Sun, Jan 16, 2011 at 10:08 AM, T <<a href="mailto:t@of.net">t@of.net</a>> wrote:<br>
><br>
><br>
> Here's another idea in the thought of not getting overwhelmed by building<br>
> complex circuits.<br>
><br>
> You can get a device called a "Constant current diode (also called CLD,<br>
> current limiting diode, constant-current diode, diode-connected transistor<br>
> or CRD,current-regulating diode)"<br>
><br>
> <a href="https://secure.wikimedia.org/wikipedia/en/wiki/Constant_current_diode" target="_blank">https://secure.wikimedia.org/wikipedia/en/wiki/Constant_current_diode</a><br>
><br>
> As long as your battery has a higher voltage than your string of diodes (add<br>
> up the voltage drops if you run them in series as others have advised) and<br>
> is capable of producing the current (milliamps), picking a constant current<br>
> diode that has a current rating at or below the rating of your LEDs should<br>
> do the job in a very simple circuit:<br>
><br>
> --- - battery + ---- CLD |> ----- LED ---- LED ---- LED .... ---<br>
> | |<br>
> ----------------------------------------------------------------<br>
> So something like a 12V camera battery should be able to drive up to 3 white<br>
> 3V LEDs or a few more of the lower-voltage colored variety, a 9V "transistor<br>
> battery" should be able to drive 2.<br>
> And it lends itself to experiment too... you can hood up the battery and the<br>
> CLD and one LED, and it should work fine (since it's current-limited it will<br>
> limit voltage too), and you can hook up two, and you can hook up three, and<br>
> if you hook up too many they just won't light up, no harm done.<br>
><br>
> Best Regards.<br>
</div></div></blockquote></div><br></div></div></div></div></div></div>