Problem understanding led's behavior
Hello
I have a problem understanding a led's behavior and generally some things about electricity.
I have a led, I connect it to two AA batteries (2,4V) and it lights up. When I connect the same led to a COM port (6,4V - that is what I measured) it lights up again but not as brightly as before. Why is that?
Thanks
Google "Ohms law", it will give a hint that Voltage isn't the only element that makes up electricity.
I did so but didn't understand what each one actually is ... that's why I came here.
Thanks
After I googled "ohm's law" I got to this:
So since I have just that led and I apply more "pressure" shouldn't it be brighter?Current is what flows on a wire or conductor like water flowing down a river. Current flows from negative to positive on the surface of a conductor. Current is measured in (A) amperes or amps.
Voltage is the difference in electrical potential between two points in a circuit. It's the push or pressure behind current flow through a circuit, and is measured in (V) volts.
The com port supplies less current than the battery. You measured the voltage of the com port when it was not loaded.
You should never connect an LED directly to a battery without a series current-limiting resistor. If your AA cells were alkaline and were brand new then their voltage would be 3.2V and their current into the LED would be enough to burn out the LED.
Uncle $crooge
Thanks for your reply.
The thing is I measured the batteries without any load as well and that leaves me a bit confused again.
Two AA alkaline batteries that have a total unloaded voltage of only 2.4V are dead.
Since you didn't use a current-limiting resistor then you are lucky that the batteries had a high internal resistance.
If the batteries were new then your LED would have burned out.
Uncle $crooge
Mm ... I see. About the internal resistance, I found this:
I = E / (r + R)
So what exactly is E and how can I calculate it and the internal resistance?
Thanks.
By the way, the batteries are rechargeable (Ni-MH).
E stands for electromotive force and is the voltage (V).Originally Posted by sakishrist
To measure the internal resistance of a battery, connect two different resistors, say 2Ω and 4Ω (R1 and R2), to the batteries and measure the voltage across the load for each load (V1 and V2). The internal resistance is then (V2-V1) ÷ (V1/R1 - V2/R2). The internal resistance equals the change in voltage divided by the change in current.
Carl
Curmudgeon Elektroniker
Ni-MH cells (I wish you told us earlier) have a very low internal resistance. They can supply 5A or 10A of current. They are about 1.25V each at low current so two make 2.5V.
You are lucky that your LED works on 2.5V because many red LEDs are 1.8V and the current in them at 2.5V is enough to melt them.
Uncle $crooge
So, is it because of the internal resistance that the led does not light up that brightly?
And you said: "the current at 2.5V"
So what is practically the connection between Current and Voltage?
V = i1* r1 .........
Last edited by QuietMan; 28th November 2009 at 10:47 PM.
"Good enough is enemy of the best." An old engineering saying, Author unknown.
sakishrist with semi conductors there is no practical connection it's VERY complex. V = I * R can be used to calculate the instantanious values in an active circuit but circuits do no respond like static loads to.
"Because I be what I be. I would tell you what you want to know if I
could, mum, but I be a cat, and no cat anywhere ever gave anyone a
straight answer, har har."
I treat an LED like I would a zener diode, subtract its Vf and then use the remaining voltage with Ohm's Law.
"Good enough is enemy of the best." An old engineering saying, Author unknown.
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