# Need help in figuring out why voltage drops under load

#### Aleksey Shurtygin

Joined Dec 21, 2018
62
Hi everyone! I am trying to find out more about why a voltage drops when you put a load onto the source. Also when using Ohms Law formula what voltage should I use. The V with or without load.

I have a basic battery powered LED strip that uses 3 AA batteries. So, when using voltmeter it shows 4.5 volts without load. However voltage drops to 3.5 when LED is connected. So assuming I know the current, do I use 4.5 or 3.5 to find the resistance, given I know the current.

I've been trying to research it but nothing useful is coming up. All text books and sites simply seem to refer to rated voltage of the battery and do not mention anything about this "drop". I did find more about voltage drop or voltage sag. These appear to be slightly different..

Could someone please either explain me why this happens or let me know what this behavior is called so that I can maybe try to find something relevant.

Thanks!!
I appreciate it!!

#### WBahn

Joined Mar 31, 2012
26,037
Welcome to AAC and welcome to the world of real electronics.

There are two issues at play (although I'm guessing at one of them).

You may or may not be able to answer this question, but does that LED strip you are using have any kind of built in current limiting, such as a resistor? I'm going to assume it doesn't -- but I could be wrong.

Let's first set the stage with some question to make you consider what is reasonable.

Your car likely has a 12 V battery in it. When you turn the key that 12 V battery is essentially connected directly to a starter motor to turn the engine until it starts up. You can take eight AA batteries and put them in series to get a 12 V power source. What do you think would happen if you tried to start your car using your AA-based 12 V power source? Do you think it would work? If not, would you still expect the voltage across this source to be 12 V while connected to the starter motor?

Now let's consider something else. Imagine taking your 4.5 V power source made up of your three AA batteries and treating it as a "load" to be powered by your car's 12 V battery by connecting positive to positive and negative to negative. What voltage would you expect to see between the terminals? 12 V? 4.5 V? Something in between?

These are thought experiments -- don't do them for real, especially the second one, as they can be quite dangerous.

A given power source has a limited ability to supply current. If asked to supply more current than that, the terminal voltage will fall. For some power sources this is a very abrupt change in that they will maintain very close to their rated voltage output up until you reach that point and then the voltage will drop rapidly as you try to exceed it. This is particularly true to active supplies. Supplies like batteries exhibit similar behavior but tend to be much smoother in that their terminal voltage starts to drop as soon as you start drawing any current and the more current you draw the more the voltage drops. It is as if there were a resistor in series with an ideal battery and you can only connect to the series combination of the two. In fact, we call this phenomenon the "source resistance" of the battery.

The source resistance of a battery is not a physical resistor that is intentionally built into the battery -- the phenomenon is due to the physics and the chemistry involved and we simply have to live with it by understanding how it behaves, at least well enough to allow us to design circuits that do what we want in spite of it.

You can get a feel for the value of the source resistance for your three AA batteries by plotting the voltage vs current for several different loads. Start of with the open-circuit voltage and plot this value (on the vertical axis) at a current of zero. Now put a load resistor that is fairly high-resistance, perhaps 10 kΩ. Measure the voltage and then calculate the current by using that voltage and dividing by the load resistance. Now use a smaller resistor, perhaps 5 kΩ (by putting a second 10 kΩ resistor in parallel with the first) and repeat. Keep doing this. Try to make the measurement as quickly after connecting the resistor as you can so that you don't give it or the battery time to heat up and let the battery rest for a bit between measurements. Work your way down to the 10 Ω range or even a bit lower.

If the line that you plot is reasonably straight (or at least the first part of it is), then you can model the source resistance as a fixed resistor whose value can be estimated by dividing the drop in terminal voltage by the current for the smallest resistor along the linear portion.

One thing you need to keep in mind is that the source resistance of a battery is not some static, constant value. It changes with temperature (which is why many batteries appear dead when they are cold but then are fresh when they warm up) and it generally increases as the battery discharges.

With this notion of how a battery works, we can now talk about how an LED works. As you apply a voltage to it, initially very little current flows until you start getting close to the "knee" voltage. At that point, the current increases significantly as you increase the voltage. For rectifier diodes, this is a very sharp, exponential increase. For light-emitting diodes it is much softer and quickly looks like there's a resistor involved. But in either case the voltage vs. current curve is highly nonlinear and so you can't just take the voltage value and figure out how much current is flowing.

If the knee voltage for your LEDs is somewhere around 3.5 V (not uncommon for white LEDs), then what is happening is that in order to supply the current needed to get much more than 3.5 V would require much more current than an AA battery can supply. So the battery terminal drops until a balance is reached where the battery is at a voltage where the current it can supply matches the current the LED requires in order to operate at that voltage.

If you want to know how much current is flowing, you need to put a current sensing resistor in series between the batteries and the LEDs. Pick a value such that the voltage drop across it is small compared to the other voltages, say no more than 100 mV. Since you don't initially know how much current is flowing, this will take some trial and error (although you could go back to the curve you plotted and look at the current that results in that terminal voltage to get a good starting point).

#### KeepItSimpleStupid

Joined Mar 4, 2014
4,147
Very nice explanation.

Very short version. Look at the datasheet for the batteries your using. You will likely find that it can;t supply the current your after. A simple MODEL of a battery consists of an ideal voltage source and a resistor.

With that model, the voltage will drop as you draw more current.

As Wbahn said, it's not quite adequate, The manufacturer of the battery publishes some curves that will help you decide if the battery is appropriate.

Car batteries are rated in Cold Cranking Amps which is a way of expressing the effect of internal resistance with bigger numbers.

The website www.batteryuniversity.com has a wealth of information.

#### ScottWang

Joined Aug 23, 2012
6,926
The simple way is to use the OHM range multimeter to measures the input resistor of LED strip without power, and to calculates the I = V/R.

#### WBahn

Joined Mar 31, 2012
26,037
The simple way is to use the OHM range multimeter to measures the input resistor of LED strip without power, and to calculates the I = V/R.
There may not be an input resistor and even if their is it might not be accessible in isolation.

Plus, I don't think this will give the TS what they think they are looking for. I think they are looking for a value of resistance that they can take the battery voltage and divide by that resistance to get the current.

@Aleksey Shurtygin : When you say "assuming I know the current", DO you know (or believe you know) the current? If so, how are you arriving at that value?

#### LesJones

Joined Jan 8, 2017
2,687
You say to find the resistance but you don't say which resistance you are talking about. It could be the apparent resistance of the LED strip. It could be the resistance value of the current limiting resistor in the LED light strip. (On the assumption that it is not some active form of current control.) It could be the internal resistance of the battery. WBahn's post #2 has given you a detailed explanation so you should now be able to say which resistance value you want to calculate.

Les.

#### Danko

Joined Nov 22, 2017
963
Could someone please either explain me why this happens or let me know what this behavior is called so that I can maybe try to find something relevant.
1V is dropped on internal resistance of battery.
Connect ammeter in series with LED light strip and measure current I_strip.
Then battery internal resistance R_battery is (4.5V - 3.5V) / I_strip.
Assume I_strip = 0.5A, then R_battery = 1V / 0.5A = 2 Ohm.
EDIT:
When you know R_ battery (2 Ohm in this case), you easily can calculate battery voltage drop (V_drop).
For example:
Battery with V_bat = 4.5V and R_battery = 2 Ohm feeds red LED (V_LED = 2.2V),
connected in series with R = 44 Ohm,
then
V_drop = ((V_bat - V_LED) / (R + R_battery)) * R_battery =
= ((4.5 - 2.2) / (44+2)) * 2 = (2.3 / 46) * 2 =0.05 * 2 = 0.1V.
So, under load, voltage of battery V_load = V_bat - V_drop = 4.5 - 0.1 = 4.4V.

Last edited:

#### vengateshd

Joined Dec 21, 2018
1
Hi Aleksey Shurtygin I too have this doubt for long days but after some experiment with different things i got cleared.
First thing the batteries what you referred here the AA batteries ! yes you heard correct!! the AA battery and other lithium ion battery are based on the
internal chemical reactions. The voltage that appear at the output terminal of the battery is the net potential generated between the anode and cathode inside
the cell. So when small load or any load is connected the electrons transfer from anode to cathode to supply the required current via load that you have connected. As a consequence of this there occurs sudden depletion of electrons in anode . Due to this the output voltage drops to 3.5volts from 4.5Volts till the depletion on both the side balances. After that the voltage remains constant. If you left the battery for long days without charging the voltage further reduces and not able supply the required current. Thus to calculate the current you must take the voltage that appears when the load is connected. Thus the current will
be Current=(3.5Volts) / ( Resistance of Led + Series resistance(if any)). Not only this case , but also in our home sockets where the AC voltage
appears to be 230/110VAC
when we connect load the "Voltage sag appears " but this sag in voltage are corrected by separate transformers installed in the nearby substation thus compensating the Drop or sag in voltage.

Thank You Aleksey Shurtygin

#### Aleksey Shurtygin

Joined Dec 21, 2018
62
These are all great and helpful responses. Thank you!

I'll try to respond to all comments and answer all questions. Currently my aim is to try to attempt to figure out current in the LED strip circuit because my multimeter has a limit of 250 mA. I'll provide some more context and what I am trying to achieve in order to provide a whole picture.

I am currently trying to build a circuit for a vanity mirror lighting so that it turns on only when person stands in front of it and when there is ambient light present (do not want it to turn on suddenly in our faces if its dark in the bathroom). For those two sensors I am planning to use some kind of IR sensor and some light sensor. I also want to add brightness control and replace batteries with an AC adapter.

I have already picked power supply that should be able to provide enough current to the LED strip. It provides 5V and 2.5 Amps. However this is where I am starting to loose it, quite possibly to misleading or incorrect information provided by the LED strip seller. The description shows 4.5V as the power source (I am assuming because of the 3 AA batteries) and rated at 2.5Watts/M. However the markings on the strip itself show "DC 6V". Assuming I can use Ohms law here, I can estimate that the current would be around 5W/6V=0.8A or 5W/4.5V=1.1A depending which voltage I should be using in the formula (I will be using whole 2M strip so it will be 5W). With my multimeter maxing out at 250 mA I can't really confirm either number. So, with these numbers I figured 5V/2.5 A power supply should be sufficient with some extra room to spare. Let me know if my calculations and estimations are not correct so far. Also, here is the LED product link if it helps.

https://www.amazon.com/gp/product/B01D303Z6E

So, as a first step I am planning to setup the circuit that would allow me simply to be able to control brightness. Initially I was thinking about using simple potentiometer but got a bit uneasy when I attempted to find range for the resistor. With a 10 Ohm resistor the brightness was about how I would like it to be on the low end. That would mean that my pot's range should be in 0-10 Ohm range. But 10 Ohm resistor becomes a bit hot to the touch. The current with that resistor and LED strip was about 150 mA which I was able to measure. However it would put it a bit over resistor power rating. When I tried to find a trimming pot that would be able to take 1A, I could not find any. So, I decided to look into PWM circuitry as an alternative.

So, at this point the uncertainty about current going thru the LED strip caused me to try to figure out all of the formulas and calculations and try to get to the bottom of it. Which brought me here with my initial question.

Now with all of this out of the way I will try to respond to all of you.

#### Aleksey Shurtygin

Joined Dec 21, 2018
62
Welcome to AAC and welcome to the world of real electronics.

There are two issues at play (although I'm guessing at one of them).

You may or may not be able to answer this question, but does that LED strip you are using have any kind of built in current limiting, such as a resistor? I'm going to assume it doesn't -- but I could be wrong.

Let's first set the stage with some question to make you consider what is reasonable.

Your car likely has a 12 V battery in it. When you turn the key that 12 V battery is essentially connected directly to a starter motor to turn the engine until it starts up. You can take eight AA batteries and put them in series to get a 12 V power source. What do you think would happen if you tried to start your car using your AA-based 12 V power source? Do you think it would work? If not, would you still expect the voltage across this source to be 12 V while connected to the starter motor?

Now let's consider something else. Imagine taking your 4.5 V power source made up of your three AA batteries and treating it as a "load" to be powered by your car's 12 V battery by connecting positive to positive and negative to negative. What voltage would you expect to see between the terminals? 12 V? 4.5 V? Something in between?

These are thought experiments -- don't do them for real, especially the second one, as they can be quite dangerous.

A given power source has a limited ability to supply current. If asked to supply more current than that, the terminal voltage will fall. For some power sources this is a very abrupt change in that they will maintain very close to their rated voltage output up until you reach that point and then the voltage will drop rapidly as you try to exceed it. This is particularly true to active supplies. Supplies like batteries exhibit similar behavior but tend to be much smoother in that their terminal voltage starts to drop as soon as you start drawing any current and the more current you draw the more the voltage drops. It is as if there were a resistor in series with an ideal battery and you can only connect to the series combination of the two. In fact, we call this phenomenon the "source resistance" of the battery.

The source resistance of a battery is not a physical resistor that is intentionally built into the battery -- the phenomenon is due to the physics and the chemistry involved and we simply have to live with it by understanding how it behaves, at least well enough to allow us to design circuits that do what we want in spite of it.

You can get a feel for the value of the source resistance for your three AA batteries by plotting the voltage vs current for several different loads. Start of with the open-circuit voltage and plot this value (on the vertical axis) at a current of zero. Now put a load resistor that is fairly high-resistance, perhaps 10 kΩ. Measure the voltage and then calculate the current by using that voltage and dividing by the load resistance. Now use a smaller resistor, perhaps 5 kΩ (by putting a second 10 kΩ resistor in parallel with the first) and repeat. Keep doing this. Try to make the measurement as quickly after connecting the resistor as you can so that you don't give it or the battery time to heat up and let the battery rest for a bit between measurements. Work your way down to the 10 Ω range or even a bit lower.

If the line that you plot is reasonably straight (or at least the first part of it is), then you can model the source resistance as a fixed resistor whose value can be estimated by dividing the drop in terminal voltage by the current for the smallest resistor along the linear portion.

One thing you need to keep in mind is that the source resistance of a battery is not some static, constant value. It changes with temperature (which is why many batteries appear dead when they are cold but then are fresh when they warm up) and it generally increases as the battery discharges.

With this notion of how a battery works, we can now talk about how an LED works. As you apply a voltage to it, initially very little current flows until you start getting close to the "knee" voltage. At that point, the current increases significantly as you increase the voltage. For rectifier diodes, this is a very sharp, exponential increase. For light-emitting diodes it is much softer and quickly looks like there's a resistor involved. But in either case the voltage vs. current curve is highly nonlinear and so you can't just take the voltage value and figure out how much current is flowing.

If the knee voltage for your LEDs is somewhere around 3.5 V (not uncommon for white LEDs), then what is happening is that in order to supply the current needed to get much more than 3.5 V would require much more current than an AA battery can supply. So the battery terminal drops until a balance is reached where the battery is at a voltage where the current it can supply matches the current the LED requires in order to operate at that voltage.

If you want to know how much current is flowing, you need to put a current sensing resistor in series between the batteries and the LEDs. Pick a value such that the voltage drop across it is small compared to the other voltages, say no more than 100 mV. Since you don't initially know how much current is flowing, this will take some trial and error (although you could go back to the curve you plotted and look at the current that results in that terminal voltage to get a good starting point).
Awesome response with some in-depth details. Thank you!
I will have to digest all of it but overall I think I got the idea. I suspected that this has something to do with power source not being able to provide enough current.

By the way similar behavior happens when I use power supply. Voltage drops from 5V to around 4V. With that explanation and AC adapter ratings wouldn't you expect voltage to stay at least close to 5V or AC adapter ratings indicate only the absolute max above which AC adapter simply stops working (in the best case)?

Also, I see some elements with a 820 marking present on the strip itself that look like flat resistors, but I may be wrong.

#### Aleksey Shurtygin

Joined Dec 21, 2018
62
There may not be an input resistor and even if their is it might not be accessible in isolation.

Plus, I don't think this will give the TS what they think they are looking for. I think they are looking for a value of resistance that they can take the battery voltage and divide by that resistance to get the current.

@Aleksey Shurtygin : When you say "assuming I know the current", DO you know (or believe you know) the current? If so, how are you arriving at that value?
When I said "assuming I know the current" I was just referring to the general use of the Ohms law formula. But to answer your question directly I would say "I believe I know, but not 100% sure". See my full story above for the details but in short I don't know and have no good way to measure neither resistance nor current. And not knowing which voltage to use in formulas I am at a loss...

#### Aleksey Shurtygin

Joined Dec 21, 2018
62
The simple way is to use the OHM range multimeter to measures the input resistor of LED strip without power, and to calculates the I = V/R.
I am suspecting that resistance of the LED strip is not in linear proportion to the voltage. When I attempted to measure resistance with a multimeter it would show around 500 Ohm. But got suspicious when Ohms law calculations did not add up. So, at this point I realized that Ohms law may not be applicable here or at least the way I was attempting to use it. Which I think I am kind of answering my own question now. Maybe I gotta use the actual voltage reading. This way the DC adapter voltage shows 4V and with 5W I think I can find the current I = P/V = 5/4 = 1.25A. And to find resistance I can use R = V/I = 4/1.25 = 3.2 Ohm..

Hmm, that does start to come together. I think. Also I am assuming I can trust sellers specifications.

#### KeepItSimpleStupid

Joined Mar 4, 2014
4,147
Currently my aim is to try to attempt to figure out current in the LED strip circuit because my multimeter has a limit of 250 mA.
Most multimeters have a series resistor that usually drops <0.6V at maximum current. You can't use the multimeter to measure voltage and then configure it to measure current. The multimeter changes the circuit.

Markings of 820 on resistors will usually mean 82 ohms. 82 with zero 0's.

#### Aleksey Shurtygin

Joined Dec 21, 2018
62
1V is dropped on internal resistance of battery.
Connect ammeter in series with LED light strip and measure current I_strip.
Then battery internal resistance R_battery is (4.5V - 3.5V) / I_strip.
Assume I_strip = 0.5A, then R_battery = 1V / 0.5A = 2 Ohm.
EDIT:
When you know R_ battery (2 Ohm in this case), you easily can calculate battery voltage drop (V_drop).
For example:
Battery with V_bat = 4.5V and R_battery = 2 Ohm feeds red LED (V_LED = 2.2V),
connected in series with R = 44 Ohm,
then
V_drop = ((V_bat - V_LED) / (R + R_battery)) * R_battery =
= ((4.5 - 2.2) / (44+2)) * 2 = (2.3 / 46) * 2 =0.05 * 2 = 0.1V.
So, under load, voltage of battery V_load = V_bat - V_drop = 4.5 - 0.1 = 4.4V.
Thanks! Unfortunately my multimeter takes up to 250 mA, so I can't really measure it. I will try to use my other calculations (see my other responses) and will see whether they will add up. However this may not be what I am looking for. I do not really care about the internal voltage drop of the battery itself. I do not care if the actual cells provide 4.5V and internal resistance drops it to 4.4. I care about 4.4V dropping to 3.4 V when I connect it to the LED strip.

#### Danko

Joined Nov 22, 2017
963
Every LED in your strip connected to power supply in series with resistor 82 Ohm,
so at 5V every LED takes current I_LED = (5V - V_LED) / 82.
For white LED V_LED = 3.3V, so for one LED current is: I_LED = (5V - 3.3V) / 82 = 0.021A.
For 10 LEDs strip current equals 0.21A, for 20 LEDs - 0.42A, and so.
EDIT:
You do not inform us how many LEDs contains whole strip you test.
Assume it is 80 LEDs, then current at 5V should be 1.68A,
so you should use 5VDC, at least 2A (10W) adapter.
EDIT:
Cut shortest piece (one LED + one resistor) from strip, connect it to 5V adapter and measure current for one LED. Hope it will lower than 250mA which is maximum for your multimeter. Then multiply it by number of LEDs in your strip.
EDIT:
For PWM, as WBahn recommended in post #17, you can use $1.66 PWM controller. Last edited: #### WBahn Joined Mar 31, 2012 26,037 Also, here is the LED product link if it helps. https://www.amazon.com/gp/product/B01D303Z6E Looking at the information on Amazon I think this is the situation. All of the LEDs are singletons with their own current-limiting resistor. In another post you said that the resistors were marked 820, which would make them 82 Ω resistors. White LEDs come in a variety of forward voltage drops, with something between 3.3 V and 3.8 V seeming to be the most common. The product description says that there are 60 LEDs/meter (they should use 'm' and not 'M' in their specs). Since they spec a voltage of 4.5 V, their power spec is probably at that voltage. With 2.5 W/m, that would be about 42 mW/LED which, at 4.5 V, would be 9.3 mA/LED. With an 82 Ω current limiting resistor, the resistor would be dropping 765 mV, placing 3.73 V across the LED, which is quite reasonable. Let's call them 3.7 V LEDs. Normally that would mean that if you put 3.5 V across it, that there wouldn't be enough voltage to get any appreciable current through the LED. But remember that LEDs have a soft knee, so they very likely are producing light at something well under 3.5 V. Without having the V-I characteristic for those LEDs, we can't really even guess too well. #### WBahn Joined Mar 31, 2012 26,037 So, as a first step I am planning to setup the circuit that would allow me simply to be able to control brightness. Initially I was thinking about using simple potentiometer but got a bit uneasy when I attempted to find range for the resistor. With a 10 Ohm resistor the brightness was about how I would like it to be on the low end. That would mean that my pot's range should be in 0-10 Ohm range. But 10 Ohm resistor becomes a bit hot to the touch. The current with that resistor and LED strip was about 150 mA which I was able to measure. However it would put it a bit over resistor power rating. When I tried to find a trimming pot that would be able to take 1A, I could not find any. So, I decided to look into PWM circuitry as an alternative. I would definitely recommend using a PWM approach. Note that 150 mA with 120 LEDs would be 1.25 mA/LED, which is probably reasonable for a low-light operating condition. Was power source were you using when you had the 10 Ω resistor in the circuit? Did you measure the voltage across the strip under those conditions? That would give us a useful data point regarding the forward voltage of the LEDs at low light levels. Also, you can make a little circuit to measure larger currents than the range of your multimeter by using two parallel paths in which both paths go through the device but only one goes through the meter. This is known as a shunt. It does require that you know a bit of information about your meter. Since you are working with such small overhead voltages, you would need to limit the current through the meter to about 25 mA to keep its impact on the circuit minimal. We can talk about that more if you would like. Thread Starter #### Aleksey Shurtygin Joined Dec 21, 2018 62 You do not inform us how many LEDs contains whole strip you test. Assume it is 80 LEDs, then current at 5V should be 1.68A, so you should use 5VDC, at least 2A (10W) adapter. For the number of LEDs its mentioned on the product page but by now you probably already seen posts that its 120 LEDs. This would bring current to about 2.5A using these calculations. However this does not add up to their specs of 2.5W/M. And would 3 AA batteries be able to provide enough juice at that current for a reasonable time? Btw I already use the same full 2M strip for the pantry closet with the batteries as intended. Cut shortest piece (one LED + one resistor) from strip, connect it to 5V adapter and measure current for one LED. Hope it will lower than 250mA which is maximum for your multimeter. Then multiply it by number of LEDs in your strip. This is a nice idea actually. But if only I had a spare cut. Unfortunately I will have to use full strip. Its precisely enough to cover the mirror frame. I may be actually short an inch or so. For PWM, as WBahn recommended in post #17, you can use$1.66 PWM controller.
Its not fun to use existing product. Its always more fun to create it yourself. But that would be an option if all other fail. Was power source were you using when you had the 10 Ω resistor in the circuit? Did you measure the voltage across the strip under those conditions? That would give us a useful data point regarding the forward voltage of the LEDs at low light levels.
I used that same AC adapter when I had 10 Ω resistor in place. I believe I did and which surprised me at that time was he the voltages were roughly about the same (a bit over 2 volts across resistor and LED strip). Can't remember for sure now, but I will be able to try it again sometime later or over the weekends.

On a side note, how do you insert an Ohm symbol? For some reason it feels weird to type out "Ohm" every time and copy/pasting it from somewhere feels a bit cumbersome.

Also, you can make a little circuit to measure larger currents than the range of your multimeter by using two parallel paths in which both paths go through the device but only one goes through the meter. This is known as a shunt. It does require that you know a bit of information about your meter. Since you are working with such small overhead voltages, you would need to limit the current through the meter to about 25 mA to keep its impact on the circuit minimal. We can talk about that more if you would like.
This sounds interesting and probably useful for future reference. What information would be needed about the meter?

• Danko

#### Danko

Joined Nov 22, 2017
963
1. Measure voltage on resistor 82Ω and divide it in 82, then you will have current of one LED.
2. Click on "S" here: and choose symbol: EDIT:
At 1 A the service life to 1 V per cell for fresh Duracell Quantum batteries is only about an hour.
Current of one led at V_battery = 3.5V:
I_LED = (3.5V - V_LED) / 82 = (3.5V - 3.3V) / 82 = 0.2 / 82 = 0.0024A.
Current of 120 LEDs is equal 0.0024 * 120 = 0.288A.
Seems it can lights relatively long time from battery...

Last edited:

#### WBahn

Joined Mar 31, 2012
26,037
For the number of LEDs its mentioned on the product page but by now you probably already seen posts that its 120 LEDs. This would bring current to about 2.5A using these calculations. However this does not add up to their specs of 2.5W/M. And would 3 AA batteries be able to provide enough juice at that current for a reasonable time? Btw I already use the same full 2M strip for the pantry closet with the batteries as intended.
They won't last very long. If you are actually delivering 5 W then your total current is about 1.67 A.