Hello,

I am just in the starting stage, please go easy on me.

I have bought a multimeter like the one attached. I have connected 5 to 6 LED lights in series to a 12V lead acid battery. Please help me to measure the watts.

when I check with the A at the right hand side, with the meter in 200m, the meter read 192. when I used an online calculator to find watts, it said it was 2.9 watts. will LEDs draw only that much power? I took the SMD leds with the board from a phillips LED bulb.

 

1. How did you connect the meter to measure current?
2. What formula are you using to calculate power?

 

Measure the current drain from the battery, measure it's voltage under load, multiply them together Bingo! Wattage.

 

Is there anything to limit the current through the LEDs besides the internal resistance of the battery?


If these are common LEDs such as those shown in the picture above, they should be connected with a resistor in series with them, otherwise you could have excessive current through the LEDs (Here, "excessive current" means enough current to significantly reduce the life of the LED, maybe down to zero.). With that, power measurement is easy -measure the voltage drop across the resistor and divide by the value of the resistor to get amps, then measure the voltage across the diodes to get the watts into the LEDs. Then just multiply them per Dodgydave.

 

Hello salai ezhil Mathi, welcome to the forums.

Power used in some thing can be measured many ways. The basic way is to measure the voltage across the thing and the current thru the thing. Multiplying the voltage by the current gives the power:

Power [Watts] = Voltage [Volts] x Current [Amps]

Note the units of Volt, Amp and Watt. One Watt is one Volt times one Amp.

The "200m" current scale means the maximum reading is 200 milliAmps. When you see 192 it means 192 mA, or .192 amps.

You should also measure the voltage your 12 V battery is actually supplying. Let us assume the 12 V is accurate. In that case the power is:

12 Volts x .192 Amps = 2.304 Watts

I do not know why this does not agree with your online tool. I tend not to use these, just work with the basic equations.

The next equation you will need is Ohm's Law:

Current [Amps] = Voltage [Volts] / Resistance [Ohms]

(Edited to correct typo as Dick suggested)

 

Slight typo perhaps.
12 Volts x .192 Amps = 2.304 Watts

 

You stated the following:

Power [Watts] = Voltage [Volts] x Current [Amps]

I totally agree as this is a basic rule. Then you additionally state the following:

Note the units of Volt, Amp and Watt. One Watt is one Volt times one Amp.

(Emphasis added)

I may be obsessing over this? But this is not totally true. A better statement for me would be one Volt times one Amp is one Watt. This is an unambiguous statement, unlike the original.

The first statement is not uniquely true. There are multiple solutions to the proposal, such as one Watt is two Volts times 500 milli-amps.

 

Is there anything to limit the current through the LEDs besides the internal resistance of the battery?

View attachment 125169
If these are common LEDs such as those shown in the picture above, they should be connected with a resistor in series with them, otherwise you could have excessive current through the LEDs (Here, "excessive current" means enough current to significantly reduce the life of the LED, maybe down to zero.). With that, power measurement is easy -measure the voltage drop across the resistor and divide by the value of the resistor to get amps, then measure the voltage across the diodes to get the watts into the LEDs. Then just multiply them per Dodgydave.

They are SMD LEDs. I cut them with the circuit board from a 4 feet led light battern.

 

That probably explains why they are drawing such a high current. Things should be ok if you put a resistor somewhere bear 470 ohms in the series string.

 

That probably explains why they are drawing such a high current. Things should be ok if you put a resistor somewhere bear 470 ohms in the series string.

I would request your help to design a led circuit please.

I have a lot of and led connected in series. Please let me know if I should connect 3 or 4 leds to a 12 v lead acid battery and what resistor I should use. I want to design them for 3 and 6 Watts. I can parallel them in sets with 3 or 4 leds in each set as per your suggestion

I got the LEDs from a 4 feet 18 Watts light battern. The LEDs were not very hot when the voltage across them was 3.02 volts.

 

Are these LEDs all the same color and therefore the same voltage?
What are the power ratings of each of the LEDs?
In their original circuit were the LEDs connected in series or parallel?

I think that is all that is needed to design your circuit.

 

Are these LEDs all the same color and therefore the same voltage?
What are the power ratings of each of the LEDs?
In their original circuit were the LEDs connected in series or parallel?

I think that is all that is needed to design your circuit.

Thank you for replying.
Yes, all are the same color and size. The battern I took these leds was cool day light battern. In the original circuit, the voltage across each LED was 3.02 volts DC.
This is the led, please ignore the resistor. I did that.

I am not sure what is power rating. I can find it if you could guide me how to

The original circuit was 4 feet long. They were not totally in parallel or series. There were around 87 leds I think. For eg, 3 sets of 32 serial leds each were in parallel to each set. I hope I am not confusing.

However I have now broken the 4 feet down. The ones I have now are in series. So I can cut the track and place a resistor or anything as per your guidance.

I appreciate your time and patience. Thank you.

 

From the photo they look like they can handle 30 ma safely if you leave them on the board because the board is a heatsink.

One LED would draw 30 ma x 3.02 volts = 90.6 milliwatts -call it 91mw. A string of three will take 272 milliwatts, so you would want 11 series stings in parallel for 3 watts and 22 strings in parallel for six watts.

The sting will consist of 1 resistor in series with three LEDs. Let's find the value of the resistor:
The voltage across the string will be about 3 x 3.02V = 9.06V. The remaining voltage across the 12 volt battery is 2.94 volts. The value of the resistor needs to be 2.94 volts/.03 amps = 98 ohms. Call it 100. The dissipation of the resistor will be .03 amps x 3.02 volts = 90 milliwatts. You can safely use a 1/4 watt resistor, and even a 1/8 watt resistor if you don't mind having a smaller safety margin -and that should probably ok for a hobby project.

To summarize one string would be thee LEDs in series with a 100 ohm resistor. Each string will draw about 1/10 watt.

 

From the photo they look like they can handle 30 ma safely if you leave them on the board because the board is a heatsink.

One LED would draw 30 ma x 3.02 volts = 90.6 milliwatts -call it 91mw. A string of three will take 272 milliwatts, so you would want 11 series stings in parallel for 3 watts and 22 strings in parallel for six watts.

The sting will consist of 1 resistor in series with three LEDs. Let's find the value of the resistor:
The voltage across the string will be about 3 x 3.02V = 9.06V. The remaining voltage across the 12 volt battery is 2.94 volts. The value of the resistor needs to be 2.94 volts/.03 amps = 98 ohms. Call it 100. The dissipation of the resistor will be .03 amps x 3.02 volts = 90 milliwatts. You can safely use a 1/4 watt resistor, and even a 1/8 watt resistor if you don't mind having a smaller safety margin -and that should probably ok for a hobby project.

To summarize one string would be thee LEDs in series with a 100 ohm resistor. Each string will draw about 1/10 watt.

Thank you. The board is not made of metal. It is some kind of semi transparent thing. Is it still ok?

 

Probably. We need to assume that the board was designed so the LEDs could run near their rating while on it. Otherwise they would have been wasting LEDs or selling an assembly with a shorter lifetime than would be necessary.

 

Probably. We need to assume that the board was designed so the LEDs could run near their rating while on it. Otherwise they would have been wasting LEDs or selling an assembly with a shorter lifetime than would be necessary.

In the original circuit, there were 87 leds and the battern was rated as 18 Watts. I will first try the above and last you know how it worked. Thank you.

 

Probably. We need to assume that the board was designed so the LEDs could run near their rating while on it. Otherwise they would have been wasting LEDs or selling an assembly with a shorter lifetime than would be necessary.

Hello,

Is there a way I can limit the voltage that is supplied to the LEDs. We have connected a solar panel to the lead acid battery and the lights sometime are very bright and very hot. I think when the sun shines, there is a lot of voltage.

The panel is connected to the Battery and we have connected the lights to the same terminal.

Thank you.

 

Yes, you can put a voltage regulator between the solar panel and the LED strip. Do you have resistors in series with the LEDs?

LEDs are primarily constant voltage devices. The current can change over a wide range while the voltage varies only a little. It is common practice to regulate the current to LEDs rather than the voltage. In many cases merely putting a resistor between the voltage source and the LED string is sufficient.

 

Hello,

Is there a way I can limit the voltage that is supplied to the LEDs. We have connected a solar panel to the lead acid battery and the lights sometime are very bright and very hot. I think when the sun shines, there is a lot of voltage.

The panel is connected to the Battery and we have connected the lights to the same terminal.

Thank you.

The "very hot" and the brightness indicate that led are receiving a lot of current. It seems to me that you are concerning yourself with the wrong thing.

 

The "very hot" and the brightness indicate that led are receiving a lot of current. It seems to me that you are concerning yourself with the wrong thing.

Ok. Please guide me what I should do.