How figure out current draw (of LEDS) using a DMM and resistors: - Page 1
Forum section: electronics / beginners

author: raspberrypi
author_rank: Frequent Contributor
timestamp: February 19, 2017, 10:09:49 pm
content: I have a bunch of LEDs and I want to figure out their current draw. IF I have a bunch of resistors a power supply (in this case 3.3 5 ad 12 volts) ohms law and a meter how do we figure out current draw just by measuring voltages?I know we need to measure the drop in voltage when we hook them up, but can we just use the resistance of the resistor and assume the LED is just acting as a wire? I tried to measure the resistance of the LED using the meter but you either get infinity or some really small number (0.01).When measuring the voltages we do it with one end of the meter hooked up to the negative at the power source then probe around the circuit? We wouldn't connect the leads to the ends of individual parts because hat will just give us the value of the rest of the circuit cutting out hat we want to measure, I think.

author: elecman14
author_rank: Unknown
timestamp: February 19, 2017, 10:43:35 pm
content: If you have a known voltage drop across a resistor and known resistance you can use ohms law to determine the current. Current equals voltage divided by resistance. If you have a series circuit that goes positive voltage then a resistor then a led and finally to ground the current will be the same throughout the circuit. You may want to read about Kirchhoff's circuit laws.

author: AG6QR
author_rank: Frequent Contributor
timestamp: February 19, 2017, 11:25:38 pm
content: An LED doesn't have a fixed resistance. When driven in the forward direction, the voltage is close to (but not exactly) constant, at the Vf value. The LED will take as little or as much current as you want to give it, until it exceeds its current handling capacity and stops acting as an LED. An ohmmeter may give any sort of nonsense value when measuring a nonlinear load like an LED. The ohmmeter normally works by either passing a fixed current through and measuring the voltage, or by putting a fixed voltage across the terminals and measuring the current. Once it has a voltage and current, it divides voltage by current and displays the result. That strategy works consistently well for resistors, but gives wildly varying results for an LED, depending on the particular value for voltage or current used.Most small LEDs used for indicators can take up to 20mA or so safely. Larger LEDs intended for bright illumination might handle hundreds of mA. To get the accurate answer for how much current your LED can handle, you'll need to look at the data sheet, but if you don't have access to the data sheet, you can probably try driving it with 15-20mA or so. Or you can increase the current until it quits working, throw that LED away, and then don't drive the next LED so hard.You can start by using Ohm's law to calculate a resistance that will produce approximately a 15mA current from your supply. A 1K resistor across a 12V supply should draw 12mA, which is close enough to start with. Put the 12V power supply, 1K resistor, and LED in series, and the LED should light up. You can measure the voltage across the LED to measure its Vf. You can measure the voltage across the resistor to determine the current through the resistor -- it should be a bit less than 12mA, because it won't have the entire 12V across it. If your meter has a suitable mA scale, you can measure the current through the circuit directly. Once you've done this for the 1K resistor, you can repeat the experiment slowly reducing the resistance, which will increase the current. You can keep reducing the resistance until you either reach the desired brightness or burn up your LED. Or you can increase the resistance and watch the LED get dim as the current is reduced.

author: alsetalokin4017
author_rank: Super Contributor
timestamp: February 19, 2017, 11:26:33 pm
content: Quote from: raspberrypi on February 19, 2017, 10:09:49 pmBasically, yes. You measure the voltage drop across the resistor (voltmeter connected across resistor) and use Ohm's Law I =Vdrop/Rresistor. This will give you the current through the resistor, and since your LED is connected in series, the same current flows in the LED as well.QuoteThis is because the LED is a _non-linear_ circuit element. It is generally useless to try to measure the "resistance" of the LED. Instead, use the "diode check" function of the DMM to determine the "forward voltage" of the LED (or look it up in the Data Sheet), and then use this value to calculate the series resistor value for your _desired_ current through the LED (and the resistor) using Ohm's Law like this:R=V/I so (Vsupply- VfwdLED)/Idesired= RseriesresistorThe Vfwdparameter of an LED or other diode is that voltage at which a typically small current begins to pass through the LED.For example, say you want 20 mA current through your LED (a typical _maximum_ practical value for jellybean 5mm LEDs, very bright) and you are using a regulated supply of 5VDC. You look at the Data Sheet for the Vfwdof the LED, and read 2.1 Volts. So you calculate 5-2.1=2.9, and 2.9/0.020 = 145 Ohms. So now you pick the nearest standard R value as 150 Ohms. Connect your components then measure the Vdropacross the 150 Ohm resistor and again use Ohm's Law I=V/R to find the _actual_ current, which should be close to your desired 20 mA (0.020A). Adjust power supply or resistor value as necessary to get your measured current to equal your desired LED current. Remember, the LED current is something _you choose_ and set by the voltage of the supply and the series ("current-limiting") resistor.QuoteYou can measure whatever you like, but does it make sense? In a simple series circuit the current everywhere will be the same. The voltages you measure by probing around the circuit will depend on the voltage drops of the individual components, according to Kirchhoff's circuit laws and Ohm's law.

author: Brumby
author_rank: Unknown
timestamp: February 19, 2017, 11:38:11 pm
content: I presume your circuit wil look something like this:Firstly let me tell you, working out the current draw is easy.1. Find out the value of the resistor2. Hook up the circuit and measure the voltage across the resistor3. Use Ohms Law to calculate the current through the resistorSome basics about measuring:A. While it is true that in circuit diagrams, voltages around the circuit are given with respect to the negative terminal - the zero volt reference - that is not the only way you can measure voltages. Voltages are measured between ANY two points of a circuit. Which two points you choose depend on what you are trying to measure.B. The current through the resistor is the same as the current through the LED. Kirchoff's Current Law formalizes this in a way that can be applied to much more complex circuits than this.Now comes the bigWARNING....LEDs - and any diode for that matter - are NOT resistors. The current through them is NOT easily calculated and Ohms Law does not apply. If you have a look at a datasheet for an LED, you will see the relationship between voltage and current is NOT a straight line (as you would get for a resistor.) Here is such a chart which shows typical curves for a few different LEDs:The important thing to notice here is that, as the voltage across the LED increases, there is a point at which the current starts increasing DRAMATICALLY. Once you are in this region, you are taking the LED into dangerous territory, where the slightest increase in voltage can result in a massive jump in current ... and the possible destruction of the LED.Using a resistor lessens the chance of sudden and massive current increases,but if the value of the resistor istoo low, it could still allow enough current to pass that the LED is pushed past its limits and its life will be shortened - perhaps to less than one second.This approach comes to mind...Quote from: AG6QR on February 19, 2017, 11:25:38 pmI am still a little unclear as to what your objective is with this experiment - as an LED does not have a "typical" current draw. What it does have, is a "knee" in the voltage vs current characteristic (as you can see in the previous graphic). It is the value of the current at this point which IS of significance - as it is close to the point of the maximum current of the LED. However, the only way to find this "knee" is to make a number of measurements** with different resistors and/or power supply voltages ... and NOT use a combination that will fry the LED.** For the creation of a voltage vs current chart using the techniques discussed in this thread, the voltage measurement is the voltage measured directly across the LED. The current measurement is calculated using Ohms Law, using the value of the series resistor and the voltage across that resistor.

author: hamster_nz
author_rank: Super Contributor
timestamp: February 19, 2017, 11:43:33 pm
content: Quote from: raspberrypi on February 19, 2017, 10:09:49 pmYou can work out the current through an LED by measuring the voltage over current limiting resistor (I = V/R), and you can find out the forward voltage of the LED, but I don't think you can work out the "safe current" for an unknown LED.

author: IanB
author_rank: Super Contributor
timestamp: February 19, 2017, 11:43:51 pm
content: Quote from: raspberrypi on February 19, 2017, 10:09:49 pmI think you have a misunderstanding here.You cannot measure the maximum permitted operating current of an unknown LED. You need to obtain this information from the datasheet.If you don't have the datasheet and you don't know anything about the LED then the maximum safe operating current will have to be estimated from the physical appearance and comparison of the LED type with other known LEDs. For example, most miniature LEDs have a maximum operating current of about 20 mA. If you know nothing else, you should assume this.

author: rstofer
author_rank: Super Contributor
timestamp: February 20, 2017, 12:45:09 am
content: Quote from: IanB on February 19, 2017, 11:43:51 pmWithout a datasheet, the process is very complicated and the resistor will still be a function of the specific LED.Here is a datasheet for some unknown LED. The important bit is on page 4 where it defines Vf in terms of a prototypical (ie recommended) If. In this case, the forward voltage drop is somewhere between 3.0 and 3.6V at 20 mA. So, just for giggles, take Vf=3.3V (right down the middle) at 20 mA.http://www1.futureelectronics.com/doc/EVERLIGHT%C2%A0/334-15__T1C1-4WYA.pdfSo, if you have a 5V supply the calculation if (5.0 - 3.3) / 0.020 or 85 Ohms for the ballast resistor. I would use 100 Ohms because I can get along with a little less current. We have a 5V source dropped by a resistor to give 3.3V across the LED. The resistor, therefore, drops 1.7V at 0.02A or 85 Ohms. Add it up! 5v - 3.3v - 1.7v = 0 - we have met the requirements for Kirchoff's Voltage Law - the sum of the gains and losses of voltage around a circuit = 0. The battery is a gain, the resistor and LED are losses.Notice on page 3 of the datasheet, there is a maximum continuous current of 30 mA.Once you have the power supply, resistor and LED assembled, measure the voltage drop across the resistor and calculate the current. The same current is flowing through the LED.If you calculate both given voltage values (3.0..3.6) you will find that the resistor values vary between 70 and 100 Ohms.You need to find a representative datasheet. The one above is for a white LED and they have higher Vf.Here is one for a more typical red LED. Note that Vf is much lower (1.65-2V but the forward current is still 20 mA.http://www.nteinc.com/specs/3000to3099/pdf/nte3019.pdfI might use 160 or 180 Ohm resistor if I had them but since I don't, I'd just grab a 220 Ohm. I''m not concerned whether I have 20 mA or 10 mA as long as it's less than max.As an experiment, grab some resistors and vary the current between 10 mA and 20 mA to see what you like. You may be surprised that the 10 mA value works quite well - and consumes less power.