OT Question for any EE's out there: R in Parallel circs

eric · Since 13 Jan 2006 1872 Posts XTreme Poster

Let's say I want to build a simple flashlight out of three 1.5V batts is series (4.5v), and a 3.7v 20mA LED. Using Ohm's law it looks like I need about a 40 ohm resistor--easy enough.

What if I want to use the same 4.5V source and connect 3 of the above LEDS in parallel. Can I still put one 40 ohm resistor in before the negative terminal of the first LED, or do I put a 40ohm on EACH LED, or a 13.33 on EACH, or....

I have this fuzzy memory of 1/R from long ago.

Help
Thanks
Eric

Nak · Wed Sep 17, 08 7:52 pm

Your math assumes 0 ohms internal resistance in the batteries. You need to find the internal resistance and add it to the circuit if you need to be precise.

eric · Since 13 Jan 2006 1872 Posts XTreme Poster

I don't need precise, but I do want a better understanding of what my moving from one, to three LED's in a parallel circuit does to my ohm requirements---how many ohms, and where should they go, is what I am looking for in a general sense (forget source resistance, and so forth).

E

PDXF · Since 10 Sep 2008 116 Posts Stoked

To make it simple you add up the voltage drop of all of the LEDs subtract that from your source (4.5v) divide your forward current (125mA to better than 1.5A, check data sheet) into that to give you the required resistance. If you have super bright LEDs check your wattage.

PDXF · Since 10 Sep 2008 116 Posts Stoked

Um. maybe I should read the post better.
a single 40 ohm 1/2w resistor would do. Nak is right about the internal resistance of the batteries to get the most out of the LEDs but you won't let the smoke out ignoring it.

PDXF · Since 10 Sep 2008 116 Posts Stoked

um, maybe I shouldn't reply... make that an 1/8w. Let me go hide under a rock or something.

eric · Since 13 Jan 2006 1872 Posts XTreme Poster

so, if I read your reply correctly, putting a resistor of x ohms before the lead LED for three identical LED's in parallel is the same as putting three x/3 resistors on each anode of the three LED's in parallel?

genek · Since 21 Jul 2006 2165 Posts East Po KGB

I think you got it backwards. When splitting resistors into parallel paths you want to increase resistance to maintain the same equivalent resistance.

http://physics.bu.edu/py106/notes/Circuits.html

So having resistor X in series with batteries before the split to go to LEDs should be like having three 3X resistors if each is connected in series with each of the LEDs.

Sounds like having 3 LEDs in parallel will require 3 times as much current for each to be fully turned on. Voltage drop will still be the same so you'll want 1/3 the resistance to generate 3x the current. You could either put 1/3*40 =13.3 ohm resistor in series with batteries before split. Or just use three 40 ohm resistors as long as you put one of each AFTER the split hooked up directly to each LED.

Hope this clarifies it. Do a bit of research on the internet about Ohms law and parallel circuits if you'd like to learn more. Tons of websites like the one I linked.

pkh · Since 27 Feb 2005 6549 Posts Couve / Hood Honored Founder

holy nerdfest batman! Laughing

eric · Since 13 Jan 2006 1872 Posts XTreme Poster

YES! This makes total sense. You provided the missing piece of logic for me--3 pathways demands 3 times the current, thus requires r/3 in series before split, or r at each leg.

I see the light--thanks!

Eric

Hein · Since 08 Mar 2005 1314 Posts Possessed

A schematic for each of your proposed circuits can help you visualize the current flow which in turn will help you set up and solve the appropriate equations.

Teach your students the fundamental step of drawing schematics for their circuits. It makes the math easier.

Wind Slither · Since 04 Mar 2005 2626 Posts The 503 METAL

12 Volt Bible!

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Nak · Thu Sep 18, 08 6:42 am