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Reducing voltage in a circut for the layman. Answered

I understand something, but for some reason I am unable to use that knowledge in application to find another answer.

Let's say I am building something: (This setup is completely imagingary, so I will use strange numbers.) I have a battery box that has 4 AA batteries rated at 1.5V and 2.7amps (according to wikipedia).

Now lets say I want to hook that up to a LED that has a max 4 volt 20mA forward. I know that I need a resister. I know that V = IR. What I can't wrap my head around is this: How do I change that equation so that is tells me how many Ohms resistor do I need? Because according to that equation if I increase the resistance, I increase the voltage. Becuase were talking about the same source. Do I need to use two V=IR equations and solve for one then place that value in the original.

I just can't get it....maybe I have had too many beers since graduating, or using algeba? I need a layman's way of understanding this...(exasperated sigh...)?

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OK ..I did all the math I'm using 3 leds...in series. They are 2.4 Volts (max)..20 ma.
20ma Divided by 1000 = .02...
3 led's 2.4 x 3= 7.2 volts
12v (car supply) subtract 7.2 total volts of led's
12 - 7.2 = 4.8 volts
4.8v / .02 = 240 ohms
So I need a 240 ohm resister.....do i need a certain type. do I need a certain watts

In laymans terms, as you wanted.  Lets pretend you are using a 9 volt battery and an LED with a forward voltage of 3 volts and 20 mA  (mA is divided by 1000, so 20mA would be .02).

Take your battery voltage (9volts) and subract from it your LED voltage (3 volts), that leaves you with 6 volts.   Divide your remaining voltage by the LED amperage (.02).   So you would have 9 - 3 = 6 divided by .02.  That gives you 300.  You would need a 300 Ohm resistor.

"Because according to that equation if I increase the resistance, I increase the voltage."  The important thing to remember is that the voltage drop across the whole circuit is equal to the voltage of your source.  Lemonie's math below is correct; what you use V=IR for is to calculate the resistance required to drop the voltage the extra 2V that the LED doesn't need, at 20 mA.

R=V/I
I=V/R

If you've got 4x1.5V in series, you've got 6V, you want 20mA.

Take out 4V forward across the LED, you'll have 2V across the resistor.

V=2, I=0.02
R=2/0.02 = 100 Ohm

L

Darn you, Quickdraw!

I've got to sleep sometime, and I'm working tomorrow...

L

Holy crap! I knew I was missing something! Take out the voltage that you want to let through and THEN figure out the resistance....DUH. I actually laughed when I read it it was so simple. Thanks for helping me out with my brain fart gentlemen.