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[Reply #60 on:]

Basically correct but again your words show the wrong thinking:-

im assuming each LED will suck whatever voltage they need from the 12volts.

They don't suck voltage out of anything. They develop voltage across them. So with 12 volts supply you have 12V to spread over everything in series. If the LEDs drop 3.6V each then 3 of them will drop 3.6 * 3 = 10.8V, your resistor will take a voltage = I * R and the rest will drop across the regulator. You just have to make sure that this is enough so that the regulator still works (it's in the data sheet)

[Reply #61 on:]

Should i be thinking of voltage like potential energy ...

Actually the term 'voltage' is the unit and it shouldn't really be used the way we all use it.  After all we don't talk about the miles-per-hourage of our automobiles or the degreeage of our ovens.  The correct term is Electromotive Force and the correct symbol is E.

If I plug another LED or whatever else with say, 2v vF that would be 9.2v on the 338 vOut.

Is this correct?

More or less (actually less due to the 1.24 v across the resistor), if the forward voltage of 2v is what is actually there with 2 amps of forward current.  You can't pursue this much further without running into problems with the regulator dropout voltage.

Don

[Edit] I didn't see Mike's post.  I see he has already mentioned the dropout problem.

[Reply #62 on:]

I guess plugging-in 2 more LED's is out of the question, at least feeding them all with 3.6v.
The LM338 needs at least 2v  more on the vIn compared to the vOut.

[Reply #63 on:]

...at least feeding them all with 3.6v.  

AARRGGHH

You 'feed' LEDs with current, the voltage that appears across them is dependent upon the current that you 'fed' them.

Don

[Reply #64 on:]

Argh x_x Im thinking the right way but typing wrong  

My best chance would probably be making 2 of these circuits and control 4 LED with 2 different circuits.
The circuits would be parallel to each other connected to the 12V Lead-Acid Battery.

[Reply #65 on:]

Made the circuit, packed everything nice and tight and here we go:




Thanks everyone

[Reply #66 on:]

...at least feeding them all with 3.6v.  

AARRGGHH

You 'feed' LEDs with current, the voltage that appears across them is dependent upon the current that you 'fed' them.

Don

That's what happens with a Norton current source.  With a Thevenin voltage source you feed a voltage and the circuit responds with a current.  With a real circuit its neither a perfect voltage nor a perfect current source, it has an impedance and voltage offset and the circuit responds accordingly - effectively the circuit solves two simultaneous equations.