How do you use ohms law if you dont know current draw of LED (pwr ON ind light) - Page 1
Forum section: electronics / beginners

author: raspberrypi
author_rank: Frequent Contributor
timestamp: February 18, 2017, 10:43:35 am
content: I have 15VAC and want to make a power indicator light.I noticed that different color LEDs seem to pull more power then others. I can hook up two green off the same resistor but when I try a red and green just the red comes on and the green goes out. I also have some weird one I want to try like pink and white. The pink seems to be a UV led lighting up a phosphorous? One cool thing about my eye condition is I see the spectrum of things. With most LEDs its monochromatic, but on this pink one I see the pink color and a deep blue band thats at much less intensity. Also white LEDs don't look like RGB, but don't look like when I look at a incandescent light bulb either (I will see a solid rainbow smear with bright yellow and very little after green). Some white LED's look like the spectrum of HID's where I will see sharp bands of color, but not what you would expect if you held a prism up to it.Anyways. I know the E = 15V- 0.6V for the 1/2 wave rectifying diode but this s not enough to figure out ohms law. Do you just put in some arbitrary resistor then figure out the mA for each LED? MY meter doesn't do current even with a new fuse.

author: Brumby
author_rank: Unknown
timestamp: February 18, 2017, 11:18:47 am
content: If you want to connect more than one diode to the same supply, here are the proper ways to do it....LEDs in series:LEDs in parallel:If you put two LEDs in parallel with each other and then run those through a single resistor, then you are going to have uneven illumination as the Vfof LEDs is never exactly the same. You might have two of the same colour from the same batch that are really close and you may not see any difference, but the chances that they are exactly the same is pretty low.When you start looking at different colours, then the Vffigures are going to be vastly different. If you put a red LED (Vfaround 1.8V) and a green LED (Vfaround 2.4V) directly paralleled this way, the red LED will pull the voltage across both down to 1.8V - which is too low for the green LED to light.To calculate the required resistor for a power indicator, the simplest way is to use a single LED and a series resistor. To do the calculation, you need a supply voltage figure, the Vfof the LED and the current you want it to run at.* For 15VAC, the voltage I would use is the peak - which is 15 * 1.414 = 21.2V (peak)* Find the Vfof the LED ... let's try a Vfof 2.4V for a green LED* With 2.4V across the LED, there will be 21.2 - 2.4 = 18.8V across the resistor.* If we want 20mA through the LED (and the resistor), using Ohms Law:18.8 = 0.020 * Ror R = 18.8 / 0.020= 940 ohmsI would use a 1K resistor. This would drop the current to 18.8 mA, which isn't going to make the brightness any noticeably different.The only problem with this is that you will get half cycle conduction, which might give you an annoying flicker. You can reduce this by putting a capacitor across the LED.

author: Ian.M
author_rank: Super Contributor
timestamp: February 18, 2017, 11:28:14 am
content: From your comment "I can hook up two green off the same resistor but when I try a red and green just the red comes on and the green goes out." it sounds like you are attempting to parallel dissimilar LEDs. That doesn't work. If you parallel LEDs, the one with the lowest Vf hogs the current. If they are exactly the same type and batch, it can more or less work*if you are running them far enough below their max If rating, but if they are different colours, the Vf difference is so great that the highest Vf LED gets almost no current.Prolonged or repetitive reverse voltage in excess of the datasheet rating (typically 5V) can degrade luminous output, even though the reverse breakdown voltage is often an order of magnitude greater than the max reverse voltage in the datasheet. You also have to consider the photocurrent - a bright light will increase the reverse leakage and a camera flash could even trigger device destruction.If you are running LEDs from an AC supply,you either need a bridge rectifier to produce DC across the LEDs or you need an antiparallel diode to limit the reverse voltage or a low leakage series blocking diode. The second and third options feed the LEDs with half wave rectified pulsing DC at your local mains frequency so tend to flicker objectionably. The bridge rectifier outputs full wave rectified DC which doubles the frequency and makes the flicker much much less obvious.Connect your LEDs in series, and connect them to the + and - terminals of a small bridge rectifier (or a bridge made of individual diodes), anode positive. Put your current limiting resistor in series with one of the bridge's AC terminals and connect the resistor and the other AC terminal across the supply. Use the supply's PEAK voltage minus the rectifier drop minus the sum of all the Vfs, and Ohm's law to calculate the series resistor.Its also possible to put the current limiting resistor on the DC side of the bridge rectifier, but then the bridge must be rated for the full supply voltage. If you put the bridge after the resistor it only needs to be rated for the total LED Vf, which can save both on parts and required creepage distances if e.g. you want a panel indicator to run off 240V AC.* but is usually a reliable indicator of a poor design

author: MrAl
author_rank: Super Contributor
timestamp: February 18, 2017, 12:36:33 pm
content: Quote from: raspberrypi on February 18, 2017, 10:43:35 amHi,We can still approximate using Ohm's Law but we have to subtract the LED voltage first. That means we have to know or can measure what is called the "characteristic" voltage of the LED.For example, a small 5mm white LED might have a characteristic voltage of 3.5 volts. If we want to know the current draw with a 5v power source and a 100 ohm resistor we first subtract the 3.5 from the 5 and then use Ohm's Law.5-3.5 equals 1.5, then 1.5/100=0.015 which is 15ma.For a red LED it might have a characteristic voltage of 2v. Now we do the same thing but use 2v instead:5-2 equals 3, and then 3/100=0.030 which is 30ma which may be too high for this LED so we might increase the resistor to 200 ohms and now we have:5-2 equals 3, and then 3/200=0.015 which is 15ma and now we are more in the right range of current.In a formula this would look like this:i=(Vcc-Vd)/Rwherei is the current through the LED,Vd is the characteristic voltage of the LED,R is the resistor value we chose.Now many times we want to know the required resistor (R) value, so we turn that around and get:R=(Vcc-Vd)/iand this directly calculates the resistor value.Note that this is always an approximation because the LED voltage is not exactly what we use in the formula, but it's close. To know for sure it is a good idea to measure the current.Lastly, when using higher power LED's it is required to calculate the power dissipation in the resistor. The power dissipation is:Power=R*i*iwhich is simply the resistor value times the current times the current a second time, or just R*i^2.The power rating of the resistor should be as a min about two times that value, so the resistor power rating should be at least:ResistorPowerRating=2*R*i*iand that is a min. If the resistor will be confined in a closed container with no air flow the power rating must be increased more. The final project must be tested to make sure the resistor does not overheat.

author: tpowell1830
author_rank: Frequent Contributor
timestamp: February 18, 2017, 12:53:07 pm
content: I am going out on a limb here, but I assume that you have a transformer of some type that is rated at 15VAC RMS. If this is true, then your equation is incorrect. You would multiply an AC RMS voltage by 1.414 in order to get the peak voltage of the transformer.The best solution is to convert the AC voltage to DC by adding a full wave bridge rectifier and a capacitor for smoothing the ripple. You would then have the effective DC voltage for the system.Next, in order to estimate the current for a given LED, if you have a 5 K Ohm potentiometer, connect in series with the LED and starting from full resistance, slowly turn the pot until the LED appears at its brightest. Stop, do not move any more because you may burn out the LED. Check the voltage across the pot, and then remove power and measure the resistance of the pot. Plug the numbers into your equation E=IR, or I=E/R. Voila, you have the ideal current for the specific LED. Write the numbers into a spreadsheet using the part number for the LED, the color, the current, and even measure the voltage drop across the LED. You have a library. Add any pertinent data that you wish.

author: Brumby
author_rank: Unknown
timestamp: February 18, 2017, 01:16:55 pm
content: Quote from: raspberrypi on February 18, 2017, 10:43:35 amI just realised I missed that 0.6V drop from my calculations.My calculations should have been:* For 15VAC, the voltage I would use is the peak - which is 15 * 1.414 = 21.2V (peak)* Then deduct the rectification diode voltage drop: 21.2V - 0.6V = 20.6V* Find the Vf of the LED ... let's try a Vf of 2.4V for a green LED* With 2.4V across the LED, there will be 20.6 - 2.4 = 18.2V across the resistor.* If we want 20mA through the LED (and the resistor), using Ohms Law:18.2 = 0.020 * Ror R = 18.2 / 0.020= 910 ohmsI would use a 1K resistor. This would drop the current to 18.2 mA, which isn't going to make the brightness any noticeably different.

author: Ian.M
author_rank: Super Contributor
timestamp: February 18, 2017, 01:38:14 pm
content: If you don't have a datasheet for the LED, the suggested procedure with the pot can be used, except the brightness doesn't peak - its rate of increase just flattens out, and it gives you no idea of the LED current till you do the maths afterwards.Ideally, stick a multimeter on a 50mA or greater current range in series with the LED, but if you only have a voltmeter, I would strongly recommend adding a 100R resistor in series with the LED and pot. You can then monitor the voltage across the 100R resistor which will be proportional to the current through the LED at 100mV/mA.In the absence of any data, don't exceed 30mA through any 3mm, or 5mm or other small LED. Obviously you need a supply voltage at least 3V higher than the LED Vf @If=30mA while testing. If that's a problem use a 10R resistor for a 10mV/mA readout. Measure the voltage across the LED, (Vf) and you then have the expected voltage drop at the current you want, and can calculate the series resistor for any supply voltage >Vf.Use a well smoothed PSU or a battery while testing because a pulsing supply (e.g. unsmoothed half or full wave rectified) will give you readings that depend on how your meter averages such waveforms, and the peak current through the LED may be excessive.* <pedantic> its actually wired as a variable resistor - a potentiometer is when you use it as a three terminal voltage divider. </pedantic>

author: SeanB
author_rank: Super Contributor
timestamp: February 18, 2017, 01:44:02 pm
content: If you are using modern LED's I would suggest running them at 10mA, as they are going to be blindingly bright at 20mA. thus use a 2k2 resistor there, giving around 9mA of current. Will be plenty bright, not perhaps bright enough to compete with direct sunlight, but will definitely show up as lit in any room, and might be very bright at night in a dark room. I have some with blue led's runnig off 5V, and they are still very bright with a 10k series resistor, i replaced the original 220R one with a 10K one there, just so I could see the casing in daytime without having eyestrain from the blue burning point.

author: tpowell1830
author_rank: Frequent Contributor
timestamp: February 20, 2017, 02:48:32 pm
content: Quote from: Ian.M on February 18, 2017, 01:38:14 pmYes, if the OP is unfamiliar with this, simply tie 2 of the legs of potentiometer together in order to create a variable resistor.Quote from: SeanB on February 18, 2017, 01:44:02 pm* <pedantic>Yes, the current of the LED in this case will not be 9mA as the effective DC voltage, after rectified and filtered will be around 20VDC (*approximation), so 20VDC=(voltage drop R1)+(voltage drop LED1), current = voltage drop R1/resistance R1. Since the voltage drop of R1 is not known until the circuit is tested, we do not know the current.</pedantic>*Just measure the voltage drop across R1 under load of LED. That is the only voltage we are interested in.Also the OP states that his/her meter does not measure current, which is the original problem.

author: rstofer
author_rank: Super Contributor
timestamp: February 20, 2017, 05:05:00 pm
content: Quote from: tpowell1830 on February 20, 2017, 02:48:32 pmWell, if we're being pedantic, one of the legs needs to be the center terminal.QuoteNot being able to measure current may be a blessing. As stated, just measure the voltage drop across the resistor and do the math. I = E / R.

author: CraigHB
author_rank: Regular Contributor
timestamp: February 20, 2017, 05:45:17 pm
content: Quote from: SeanB on February 18, 2017, 01:44:02 pmYes those blue LEDs are bizarrely bright. I have a device that came with blue LED power indicators and they were blinding at 5mA bias current. I upped the resistors on that unit to drop the bias to 1mA and they're still really bright.

author: Cubdriver
author_rank: Unknown
timestamp: February 20, 2017, 07:46:16 pm
content: Quote from: rstofer on February 20, 2017, 05:05:00 pmWell, technically you'd get a variable resistor if you wire the outer terminals together, too. It'll go from zero resistance up to about half of the pot's full resistance value then back down to zero as you turn it from one extreme to the other. (Assuming it's a linear taper pot, of course.) I'm not sure what you'd use such a configuration for, but itwouldbe a variable resistor...-Pat

author: Nerull
author_rank: Frequent Contributor
timestamp: February 21, 2017, 02:25:49 am
content: Blue leds are uniquely irritating for a handful of reasons - they actually are brighter, producing more light output per unit power. Our eyes also can't focus properly on blue light, and it gets scattered more by fluid in the eye, which causes those halos around blue signs at night. Plus our rod cells, which give us monochrome night vision, are much more sensitive to blue light, so they get perceptually even brighter in the dark. (Conversely, which is why red light preserves night vision - our rod cells don't really pick it up, so they don't become desensitized.)

author: james_s
author_rank: Super Contributor
timestamp: February 21, 2017, 03:03:03 am
content: Glad I'm not the only one. When blue LEDs first came out I thought they were so cool, for years it was a color that was completely absent. Then they dropped in price and showed up on *everything*. Just because blindingly bright blue LEDs are available doesn't mean they have to be used in every possible application.

author: hfleming
author_rank: Regular Contributor
timestamp: February 21, 2017, 07:16:19 pm
content: Quote from: Nerull on February 21, 2017, 02:25:49 amThanks for that info. i always thought It was just me, but my eyes just cannot focus on a blue LED. I do have a brain tumor that makes things very difficult for me sometimes, so I just thought that not seeing a blue LED as a single spot was just one of those things due to the tumor.

author: Audioguru
author_rank: Super Contributor
timestamp: February 22, 2017, 02:08:01 am
content: Have you seen a white LED when it is turned off? It looks yellow inside. You know why?It is a blue LED with a yellowy phosphor on top. You can make a much better white color with RGB.

author: james_s
author_rank: Super Contributor
timestamp: February 22, 2017, 05:51:45 am
content: It's hard to properly mix the light from the RGB chips though, and the red in particular is less efficient. That's why all of the high efficiency LED lightbulbs use phosphor type white LEDs, some also add a few red emitters to boost the CRI into the 90+ range. For a while the better LED bulbs used a remote phosphor but more conventional monolithic emitters seem to have caught up. The most recent LED bulbs I've got are around 95 lm/W, pretty spectacular when you consider compact fluorescent bulbs are barely more than half that and old fashioned incandescent lamps are not even 20 lm/W.

author: CraigHB
author_rank: Regular Contributor
timestamp: February 22, 2017, 06:06:31 pm
content: And that's one thing about LED lighting that gives me trouble, they tend to have a very blue color temperature. You have to seek out the ones that generate a warmer color temperature, but when they can be found they are excellent lights. I found some Cree LED floods like that to use in my desk lamps and they're the best lights I've used for that. Otherwise I avoid LED light bulbs and stick to the flouro bulbs for home lighting. Even those can be hard to find in a comfortably warm color temperature.

author: james_s
author_rank: Super Contributor
timestamp: February 22, 2017, 07:45:56 pm
content: By far the most widely available LED bulbs are 2700-3000K color temperature, most with a CRI in the low 80s. 2700K is what a standard incandescent lamp is, halogen tend to be 2850-3000. You can get colder 5000-6500K "Daylight" bulbs but most are meant to mimic incandescent. A low Color Rendering Index is responsible for the "icky" light from many fluorescent and lower cost LED lamps. Incandescent and sunlight have a CRI of 100. The old Cool White fluorescent is about 60 IIRC.I was an early adopter of LED bulbs and my house has been 95% LED since late 2011 and my favorite so far have been Philips. I have some of the Philips remote phosphor bulbs that have been in dusk till dawn service in my porch lights since 2011. A few months ago I pulled one and ran it side by side with a new unused identical bulb and they were indistinguishable.

author: Cubdriver
author_rank: Unknown
timestamp: February 22, 2017, 09:17:54 pm
content: Quote from: CraigHB on February 22, 2017, 06:06:31 pmAre you talking about replacement A19 style lamps, or raw LEDs? I've had no issue with finding what I think are good quality output warm white LED A19 replacement lamps - Home Depot stocks plenty of them. I have the Crees and some made by Philips. For small halogen style reflectors the options seem to be fewer, but at the same time I haven't looked at those for a few years now so that situation may well have improved too.Most of the lights at my house have now been changed over to LEDs, and I'm quite happy with them.-Pat