hey /diy/, i got an electronics 101 question
connected a 3V battery, a 100 Ohm resistor, and a green LED in series. expecting I = 3 / 100 = 30 mA
connect an old "Fluke 29" multimeter, and it's current measurement reads "06.75 DC" (assuming it means 6.75mA)
am i doing something wrong, is the multimeter bad?
>>919638
Coin batteries have very high internal resistance, it's like you've put the100 ohm resistor in series with it. To calculate it short the battery with the ampmeter, then divide the voltage by the current measured.
You're doing the math wrong. Using I=V/R like that gives you the current that would be going through the resistor if it were connected directly across the battery with nothing else in the circuit.
The purpose of an LED series resistor is to produce a voltage drop so the LED gets the recommended voltage. When the LED gets the recommended voltage it will draw the recommended current.
If the LED is 3v 20mA (0.02A) and the supply is 5v then you need the resistor to drop 2v. Vdrop = R / I = 2v/ 0.02A = 100Ω. Putting a 100Ω resistor in series with the LED will make the LED get 3v from the 5v supply. The current through both the resistor and LED will be 20mA.
I'm assuming your green LED is 3v, so it's going to get the correct voltage from the 3v battery and draw the correct current without any need to produce a voltage drop with a resistor.
>>919660
sorry, i should've been more specific. my 3V battery is actually two 1.5V AA batteries in series.
when i short them with the ampmeter it reads 0 on its "10A" setting and OVERLOAD on its 40mA setting.
>>919664
thanks for the explanation
Electrical engineer here, LEDs always have a roughly 1.2-1.4 volt drop across them, the rest of the voltage is dropped on the other components I series with the LEDs. That would leave you with 1.6 volts, 1.6V/100ohm=16mA. Also when you read current with multimeter, you have to put the meter in series (parallel for voltage).
>>919664
>>919674
No, the purpose of the resistor is not to drop the voltage. The purpose of the resistor is to reduce the current across the diode. When LEDs operate, their internal resistance is very low, and will draw far too much current and burn out in a flash. Rule of thumb is to use 100-300 ohm resistor with a LED. If you want to get the max brightness, you need to look up the max current allowable through the LED, and then choose the correct resistor for your power source.
>>919868
You're not an EE.
>>919638
OP, you need to know what voltage and current the LED is rated for. You can find that on it's datasheet. You subtract the voltage across the LED from your voltage source, and that gives you the voltage drop across the series resistor. Then you divide the voltage drop of the series resistor by the current the LED is rated for, and you get the series resistor value. Multiply the voltage drop across the series resistor by the current through it, and you get the power the resistor needs to be rated for. Pick a resistor of a standard value closest to the resistance value calculated, and at least of the power rating calculated.
>>919666
You NEVER, EVER put a current meter directly acrosss a voltage source! You can damage your meter that way (or at least blow the fuse).
>>919868
>LEDs always have a roughly 1.2-1.4 volt drop across them
No. An LED with that voltage drop is very rare. The overwhelmingly common parameters of LEDs like OP pictured is: IR, Red,Yellow: 2v, 30mA. Green, Blue, White, UV, everything else: 3v,20mA
>>919875
> When LEDs operate, their internal resistance is very low, and will draw far too much current and burn out in a flash.
> internal resistance
You should not be thinking of LEDs like resistors. I've attached an LED current vs voltage curve for reference. The "resistance" of an LED is effected by the voltage across it. It's only "very low" (producing too much current draw) when the voltage exceeds the recommended value. When voltage drop across an LED is at the recommended value its "resistance" is just right to produce the recommended current draw. You add a series resistor to produce a voltage drop so the voltage across the LED is correct which causes its current draw to be correct.
>>919949
thank you, that clarified a few things
pic related, LEDs
>>919638
Internal resistance of the cell * V= I (R +r) *
