Rookie Ohm's question...

Rookie Ohm\'s question...

Ok, I'm still a rookie. So I apologize if this has been asked a gazillion times. [That's more than a gabillion, right?] Did a search but didn't really find the answer.
Well, I have done a few newbie mods. One of the 9V ones here. And my first PR Bulb Mod here.

The question I have is, in the few mods I have done, I supplied the power supply and resistor connecting them to a single LED. Very basic and not using any regulation, yet. What I want to know is... what determines at what current/voltage is the LED being driven?

Say I use 6V. [4AA] and a 100 Ohm resistor. At what current/voltage is the LED being driven? 4V at 20mAhs or 3.4V at 30mAh, since both value combinations fit the formula. Does the LED consume as much current as it can? I used LED_Pro to get and verify the values.


I guess I can open the circuit and hook up my multimeter as an ampmeter. Just thought you guys could shed some light [pun intended] the subject. /ubbthreads/images/graemlins/help.gif
Danno

Re: Rookie Ohm\'s question...

The voltage across the LED changes very little over a wide range of current values. For all practical purposes, the LED is being driven at its Vf.

Paul

Re: Rookie Ohm\'s question...

Danno,

The current passing through the LED will depend on its Vf at a given temperature. The Vf of the LED will be slightly different lot to lot, and there is a small temperature gradient (can someone post what this is? I can't remember...).

The simple way to determine current is to divide the voltage drop across the resistor by the value of the resistor. So, in your case, if The Vf of your LED is 3.4V and your supply 6V, that means 2.6V/100ohm=.026A, or 26mA.

Again, the current in this case will depend purely on the Vf of the LED. The voltage drop of the LED will increase slightly as the current increases, but nowhere near the linear fashion that a resistive load exhibits, so these measurements are probably close enough.

Re: Rookie Ohm\'s question...

Since you have a multimeter, you can measure the Vf across the LED. This would not change the current running through the LED by any noticeable amount. Might as well measure the V across the resistor while you're at it. V = I*R. You now know the current flowing through the resistor. Since there is only 1 pathway for the current to flow, that must be the current flowing through the LED too.

There. That must be easier than all that tiresome if and maybe. If you do open up the circuit and put the ammeter in series, you introduce an extra element which, theoretically should consume no power, but really would consume just the slightest bit and skew your readings slightly. With only mA to play with, you don't really want that as the percentage error would be quite large. Would be alright in high current flows since the percentage error would be smaller.

Re: Rookie Ohm\'s question...

I think Steelwolf just blew past a really important concept. Those of us who understand and use it forget how key this basic is. If there is a voltage across a resistor, there is a current through it causing that voltage. The voltage (in Volts) and the current (in Amps) are related to each other by the resistance (in Ohms). In general, more current (or voltage) means more voltage (or current) for the same resistance. If we have values for two of the three, Ohm's Law will calculate the third (this is the basis for the various 'programs').

If you change one of these three values, hold another constant, the third has to change so Ohm's Law stays satisfied.

Since you know the value of the resistor and can measure the voltage across it when the LED is running, dividing that voltage (in Volts) by that resistance (in Ohms) will give you the LED current in Amps. Multiplying that current (in Amps) by 1000 will give you more convenient miliamps. This is the real current through the LED at the time (since it's a simple series circuit, the current is 'everywhere the same').

This is a far more accurate way to set the current than the 'programs' and 'calculators' as it measures the real current (indirectly via voltage drop); the 'computer values' cannot take advantage or real world values for Vbat and Vf.

Doug Owen

Re: Rookie Ohm\'s question...

Duh! Man do I feel dumb. /ubbthreads/images/graemlins/banghead.gif
I guess I am a rookie. I shoulda thought about just
measuring the voltage drop across the LED. Then just
calcute w/the known resistance. [since it is a very simple circuit]

Hey, thanks you guys for the great info. You guys are
right, it is basic. Thanks again fellas! You guys rock! /ubbthreads/images/graemlins/buttrock.gif

Danno

Re: Rookie Ohm\'s question...

Danno: Measure the V across the resistor, divide by the value of the resistor. This gives you current flowing through the resistor which because there is only 1 path, is also the current flowing through the LED.

The way you wrote that last post, I'm afraid that you will measure the voltage across the LED and divide that by the value of the resistor. That would give you the wrong values.

Re: Rookie Ohm\'s question...

Oops, typo. Meant to type "I shoulda thought about just
measuring the voltage drop across the resistor. Then just
calcute w/the known resistance. [since it is a very simple circuit]"

Doh! Thanks Steelwolf!

Danno