With a 4.5V battery the LEDs must be in parallel although it's possible each LED has a resistor in series with it to balance the currents between them.
(1) measure the voltage across the battery
(2) measure the voltage across the resistor - the voltage across the (parallel) LEDs is the difference.

For more information, put a 10 ohm resistor in parallel with the 15 Ohm resistor and measure again and measure the voltage across the battery with no load (if there is no switch, insert a slip of insulating plastic between the cell and the contact with the battery holder) to get an idea of the battery internal resistance. Probably more information than is really needed. The wires going in and out of the LED globes must be insulated.

 

The battery voltage will drop when there is a heavy load current.

 

It sounds like they are in parallel, but could you please confirm that the wire you are holding in your hand on the left is actually two wires?
[/CITAZIONE]

yes, those are two wires I am holding in my hand

 

Don’t assume the battery voltage is 4.5V. Measure it with the lights on.

And sorry if I missed it, but what color are the LEDs?

From the attached photo they look yellow to me (I could be absolutely wrong), but on the packaging it says they are all white (with the coloured glass around them)

 

With a 4.5V battery the LEDs must be in parallel although it's possible each LED has a resistor in series with it to balance the currents between them.
(1) measure the voltage across the battery
(2) measure the voltage across the resistor - the voltage across the (parallel) LEDs is the difference.

For more information, put a 10 ohm resistor in parallel with the 15 Ohm resistor and measure again and measure the voltage across the battery with no load (if there is no switch, insert a slip of insulating plastic between the cell and the contact with the battery holder) to get an idea of the battery internal resistance. Probably more information than is really needed. The wires going in and out of the LED globes must be insulated.

battery voltage = 3.7V (it consists of 3 AA alkaline batteries placed in series)
voltage across the resistor = 1V

So if the LEDs are in parallel (assuming there is not a resistor in series with each of them) there should be 2.7V across each LED.

 

I have 3 AA 1.5V batteries powering a strip of 40 LEDs that I bought (4.5V total)
I would like to calculate the current and/or voltage on each LED to find out approximately what values we are talking about.

I think the strip of LEDs is in parallel (I attach the photo) but I would like your confirmation.
View attachment 283358

If they are in parallel, I would like to find the current on the first LED.
Drawing the schematic with Vdc, 15 ohm resistor (it's in the box with the batteries) and all the LEDs in parallel I apply ohm's law to the first link (the rectangle you see in this figure)
View attachment 283359
I = (4.5 - Vdiode_forward) / 15

How can I calculate the value of the current on D1 if I do not know the value of its forward voltage ?
Similarly ... how can I calculate the D1 forward voltage if I don't know the value of the current?

I could assume that, normally, a yellow LED supplied with 5V has a Vf=2V ... so from the calculation I would have 166mA of current ... seems a bit too much!

This is the setup, I cannot measure voltage between these two points.
Maybe because they are isolated?


@WBahn
@Jerry-Hat-Trick

 

battery voltage = 3.7V (it consists of 3 AA alkaline batteries placed in series)
voltage across the resistor = 1V

This is the setup, I cannot measure voltage between these two points.
Maybe because they are isolated?

So if the battery voltage is 3.7V when turned on and there is 1.0V across the resistor your voltage across the LEDs must be 2.7V - where else can it go? With 1.0V across the 15 ohm resistor the total current is 66.7mA which suggests current through each of the 40 LEDs averages out at 1.67mA. Probably not very bright but visible

 

Three alkaline AA cells should produce more than 3.7V if they are fairly new and are not overloaded. Maybe each LED draws 20mA then the total current at 4.5V is 45 LEDs x 20mA= 900mA which is too much for the battery and heats the 15 ohms 1W resistor with 0.9A squared x 15 ohms= 12.15W!

If each LED draws only 5mA then the total for 45 of then is 225mA.
The datasheet for an Energizer AA alkaline cell shows that for it to have a voltage of only (3.7V/3=) 1.23V then it has been drained by 225mA for 5 hours.
Measure each battery cell when they are powering the 45 LEDs to see if one cell is almost dead. The red middle cell?

EDIT: 66.7mA from an AA alkaline cell shows an almost dead cell.

 

From the attached photo they look yellow to me (I could be absolutely wrong), but on the packaging it says they are all white (with the coloured glass around them)

The color you see in a picture (with the LED off) doesn't tell you anything about the color of the light the LED produces. If the package says they are white LEDs, then they almost certainly are. The voltage drop should be somewhere in the 3 V to 4 V range, which is pretty consistent with your reading of the voltage across the resistor.

 

A white LED normally looks yellow when not powered because the phosphor on top of its blue LED appears yellow.

 

Three alkaline AA cells should produce more than 3.7V if they are fairly new and are not overloaded. Maybe each LED draws 20mA then the total current at 4.5V is 45 LEDs x 20mA= 900mA which is too much for the battery and heats the 15 ohms 1W resistor with 0.9A squared x 15 ohms= 12.15W!

If each LED draws only 5mA then the total for 45 of then is 225mA.
The datasheet for an Energizer AA alkaline cell shows that for it to have a voltage of only (3.7V/3=) 1.23V then it has been drained by 225mA for 5 hours.
Measure each battery cell when they are powering the 45 LEDs to see if one cell is almost dead. The red middle cell?

EDIT: 66.7mA from an AA alkaline cell shows an almost dead cell.

Yes the batteries are not new, they are a bit used.
I measured ... each battery is about 1.3V, but they still total 3.7V, which does not explain why I cannot display the voltage at the ends of a generic LED

 

Three alkaline AA cells should produce more than 3.7V if they are fairly new and are not overloaded. Maybe each LED draws 20mA then the total current at 4.5V is 45 LEDs x 20mA= 900mA which is too much for the battery and heats the 15 ohms 1W resistor with 0.9A squared x 15 ohms= 12.15W!

If each LED draws only 5mA then the total for 45 of then is 225mA.
The datasheet for an Energizer AA alkaline cell shows that for it to have a voltage of only (3.7V/3=) 1.23V then it has been drained by 225mA for 5 hours.
Measure each battery cell when they are powering the 45 LEDs to see if one cell is almost dead. The red middle cell?

EDIT: 66.7mA from an AA alkaline cell shows an almost dead cell.

Now my question is:
is it possible that on each LED there is a very low potential difference (in the mV scale) and therefore difficult to read by my multimeter? As far as I know, it is the current that determines the light intensity.
Even if this were the case, this does not explain where the 2.7 circuit voltage ended up (Vbattery - Vresistor = 2.7V)

 

I have 3 AA 1.5V batteries powering a strip of 40 LEDs that I bought (4.5V total)
I would like to calculate the current and/or voltage on each LED to find out approximately what values we are talking about.

I think the strip of LEDs is in parallel (I attach the photo) but I would like your confirmation.
View attachment 283358

If they are in parallel, I would like to find the current on the first LED.
Drawing the schematic with Vdc, 15 ohm resistor (it's in the box with the batteries) and all the LEDs in parallel I apply ohm's law to the first link (the rectangle you see in this figure)
View attachment 283359
I = (4.5 - Vdiode_forward) / 15

How can I calculate the value of the current on D1 if I do not know the value of its forward voltage ?
Similarly ... how can I calculate the D1 forward voltage if I don't know the value of the current?

I could assume that, normally, a yellow LED supplied with 5V has a Vf=2V ... so from the calculation I would have 166mA of current ... seems a bit too much!

I think the wire is probably enameled .. so copper, coated with insulating enamel.
This should explain why I cannot detect the voltage at the ends of the LEDs

Am I wrong?

@WBahn
@Audioguru again

 

Yes the batteries are not new, they are a bit used.
I measured ... each battery is about 1.3V, but they still total 3.7V, which does not explain why I cannot display the voltage at the ends of a generic LED

What, exactly, do you mean by the voltage at the ends of a generic LED?



If you want to measure the voltage across the LED, measure the voltage between points B and C.

Do NOT put both probes somewhere on B, or both somewhere on C. In either case, you will be measuring the tiny voltage drop across a low-resistance wire -- a reading that will have NOTHING to do with the voltage across an LED.

 

I think the wire is probably enameled .. so copper, coated with insulating enamel.
This should explain why I cannot detect the voltage at the ends of the LEDs

Am I wrong?

@WBahn
@Audioguru again

If the wire is insulated (which is a pretty sure bet), then you have to get through the insulation to make the measurement. One way to do that is to just use a sharp knife (or similar) to gently scrape away the insulation. This is going to expose the wire so that you can get your probe on it, but it is also going to expose the wire so that other bad things can happen (and risks the wire breaking, either now or down the road due to the damage). For the voltages involved here, you can use something like clear fingernail polish to cover the exposed wire (though any damage to the wire will still be there.

Another thing you could do is expose the wires between the last and next-to-last LED in the chain and make your measurements there. Then you can just cut the wire close to the the next-to-last LED and, if you want, dab some polish over the end just to be safe. Now you will have a string with 39 instead of 40 LEDs in it.

But the measurement you are looking for can be had by just measuring the voltage between the negative terminal of the battery pack and the end of the resistor that is not connected to the positive end of the battery pack.

 

Cosa intendi esattamente per tensione ai capi di un LED generico?

View attachment 283477

Se vuoi misurare la tensione attraverso il LED, misura la tensione tra i punti B e C.

NON mettere entrambe le sonde da qualche parte su B, o entrambe da qualche parte su C. In entrambi i casi, misurerai la minuscola caduta di tensione attraverso un filo a bassa resistenza - una lettura che non avrà NIENTE a che fare con la tensione attraverso un LED.
[/CITAZIONE]

Sì, intendevo alle estremità dell'a led.

But the measurement you are looking for can be had by just measuring the voltage between the negative terminal of the battery pack and the end of the resistor that is not connected to the positive end of the battery pack.

You are right! I didn't think about.
I had focused on measuring the enameled cables outside the box when I could have measured as you have shown in the picture.

 

You are right! I didn't think about.
I had focused on measuring the enameled cables outside the box when I could have measured as you have shown in the picture.

It WAS something that was recommended back in Post #8.

What do you mean that you measured the voltage at the ends of any LED?

You want to measure the voltage ACROSS the LED. Put one lead (the red lead, most likely) at the junction between the resistor and the LEDs and the other lead (the black one, most likely) at the other end of the LED (if they truly are in parallel, that will be the negative terminal of the battery).

It sounds like you are measuring the voltage across a piece of wire going from the cathode of the LED to the battery.

 

Three alkaline AA cells should produce more than 3.7V if they are fairly new and are not overloaded. Maybe each LED draws 20mA then the total current at 4.5V is 45 LEDs x 20mA= 900mA which is too much for the battery and heats the 15 ohms 1W resistor with 0.9A squared x 15 ohms= 12.15W!

If each LED draws only 5mA then the total for 45 of then is 225mA.
The datasheet for an Energizer AA alkaline cell shows that for it to have a voltage of only (3.7V/3=) 1.23V then it has been drained by 225mA for 5 hours.
Measure each battery cell when they are powering the 45 LEDs to see if one cell is almost dead. The red middle cell?

EDIT: 66.7mA from an AA alkaline cell shows an almost dead cell.

Can you explain this reasoning to me? I understand the calculations above.
Thanks!

 

1V/15 ohms= 66.7mA. You have 45 LEDs in parallel then each one draws a current of only 66.7mA/45= 1.48mA.

If the battery was new at 4.5V then the LEDs would have a voltage of about 2.9V and the resistor would have 4.5V - 2.9= V= 1.6V.
Then the total current is 1.6V/15 ohms= 106.7mA and each LED would have 106.7mA/45= 2.37mA.

 

Measure the total battery voltage two ways:

First, measure it with the LEDs turned off (this is known as the open-circuit, or no-load voltage).

Then measure it with the LEDs turned on (this is known as the closed-circuit, or loaded voltage).

Also with the LEDs turned on, measure the voltage across the resistor in order to determine the closed-circuit current.

With these numbers, you can calculate an estimate of the battery's internal resistance.

R_internal = (V_open - V_closed) / I_closed

Note that this is the combined internal resistance of all three cells in series. If we assume that they are all the same (shaky assumption, but we'll go with it), then divide that result by three.

Here is the current datasheet for the Energizer AA primary alkaline cell:

https://data.energizer.com/pdfs/e91.pdf

Sadly, it has been stripped of nearly all of the useful information that they used to provide.

Also note that this is for the MAX battery, which is the only primary alkaline they list on their site now.

The Duracell site still has decent technical documentation (though it, too, has degraded over the years).

https://www.duracell.com/wp-content/uploads/2016/03/MN15USCT0122-A.pdf

Those performance curves are probably similar to the Energizer's, but there will be differences.

If I am reading your picture correctly (and I may well not be), it looks like those Energizers have an expiration date of Feb 2004. How old are these batteries? Why not just get three fresh batteries?

If your internal resistance, per cell, is over about 300 mΩ to 400 mΩ, then your batteries are probably well on the way to being dead.