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Problem understanding led's behavior

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sakishrist

New Member
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
 

Tipsy

New Member
Google "Ohms law", it will give a hint that Voltage isn't the only element that makes up electricity.
 

sakishrist

New Member
After I googled "ohm's law" I got to this:

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.

So since I have just that led and I apply more "pressure" shouldn't it be brighter?
 

audioguru

Well-Known Member
Most Helpful Member
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.
 

sakishrist

New Member
Thanks for your reply.

The thing is I measured the batteries without any load as well and that leaves me a bit confused again.
 

audioguru

Well-Known Member
Most Helpful Member
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.
 

sakishrist

New Member
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.
 

crutschow

Well-Known Member
Most Helpful Member
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.
E stands for electromotive force and is the voltage (V).

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.
 

audioguru

Well-Known Member
Most Helpful Member
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.
 

sakishrist

New Member
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?
 

QuietMan

Member
V = i1* r1 .........
 
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Sceadwian

Banned
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.
 

QuietMan

Member
I treat an LED like I would a zener diode, subtract its Vf and then use the remaining voltage with Ohm's Law.
 

Sceadwian

Banned
Make sure you use the VF at the current you expect to be going through it.
 

audioguru

Well-Known Member
Most Helpful Member
Fully charged Ni-MH cells have an extremerly low internal resistance since they can supply 5A or 10A.

The brightness of an LED depends on how much current you allow through it limited by the value of the current-limiting resistor and the voltage of the battery and of the LED. Each LED has a maximum allowed current.
 

Tipsy

New Member
Sorry to butt in - what would be the result of four x 3v LED's in series across a 12V 60Ah car battery? Do the LED's in series provide the correct resistance or will they too burn out?
 

audioguru

Well-Known Member
Most Helpful Member
Leds do not have a resistance like a light bulb. They are diodes and if you apply a voltage that is slightly too high then the current will be extremely high causing the LED to burn out. If the voltage is slightly too low then the LED will be dim or will not light. You must add a series current-limiting circuit or resistance.

Nobody makes an LED that needs exactly 3.0V. They make LEDs that have a range of voltage from about 2.7V to about 3.3V for one colour. Some of them will be 3.0V.

If you connect four 2.7V LEDs in series then they need only 10.7V and a 12V battery that is actually 12.6V to 13.8V will blow them up instantly.
If the LEDs are actually 3.3V then four in series need 13.2V and would be dim, normal or will burn out depending on the actual battery voltage.

You must have a current-limiting circuit or resistor in series with LEDs and enough extra voltage to create the proper amount of current in the current-limiting circuit.
 
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