LED Calculations, resistors, forward voltage and current etc. No Data Sheets - Page 1
Forum section: electronics / beginners

author: PetrikNZ
author_rank: Newbie
timestamp: December 15, 2021, 10:40:56 pm
content: I have been through a few screen fulls of Topic Search to see if this has been covered (I went back several pages worth of results, but not all the way) So it may have been covered, but I have not found it.I am slowly learning about Ohms law, calculating resistors needed for LEDs etc. However, all the info I find on the "how to" articles I have come across, refer to getting some figures from data sheets.So, if you have a random LED with no info (data sheets) whatsoever, are you stuck? Surely there's a way to work it out.I hesitate to give numbers as I want to work out the answers, but say we have a Power Supply of X Volts. (Pick a number, it's the one thing we DO know) How would one proceed to then work out the required resistor for this unknown Unicorn coloured LED. (I say Unicorn coloured as I know there are accepted ball park numbers for standard LED colours, eg. Red, Blue, Yellow, etc.) Let's assume we're colour blind. So we can't tell what colour the LED is, therefore not able to rely on generally accepted ranges, when Data Sheets are not available.So with a multi-meter, and Power Supply of X Volt available (5, 9, 6, 12, etc. take your pick for the example). How can I get all the figures I need and as a result, be able to select the correct resistor for that LED to work with said Power Supply.Additionally I am also interested in figuring out how many LEDs (assuming they all are that same spec) I could run on X Volt 1Amp or X Volt 2Amp (or any other Amp power supply of said X Volt) I suspect from answering the first part that this answer will fall into place from there. Making the assumption that by the time I am done with the first part I will have got the current as part of the answer, so then it's a bit more math to work out, all things being equal, how many of that combination of LEDS I can get on a Y Amps at X Volts Power Supply.I hope I have not confused everyone too much here.

author: retiredfeline
author_rank: Frequent Contributor
timestamp: December 15, 2021, 10:51:09 pm
content: First of all you need to know the LED forward voltage. For red, green, yellow LEDs they are in the vicinity of 2V. For the superbright green, blue and white, it's closer to 3V. Those are ballpark figures. You can measure it in your test circuit.Next the current. Obviously this depends on the type and application, but for THT type LEDs, in the single digits of mA. Modern LEDs are efficient, so the typical maximum 20 mA stated in datasheets will be really glaring. For board mounted SMT types, could even be under 1 mA. You need to experiment to judge if the brightness is suitable for your application for the ones (even the batch) you have. Start with 1 mA and go from there. The higher voltage colours are more efficient, I often have to keep them below 1 mA.With a power supply, a voltmeter, some resistors of known values, and Ohm's law you will get there.

author: Benta
author_rank: Super Contributor
timestamp: December 15, 2021, 11:42:01 pm
content: First, WelcomeNo, you're not confusing anyone here, but perhaps yourself a bit.Everyone understands voltage sources: 1.5, 9, 12 V batteries, 5 V USB, and 120, 230 VAC.The limitations are: how much current can be supplied?Current sources are the complement of voltage sources.Current sources deliver a certain current, and will do this "whatever it takes". This means that the output voltage of a current source is theoretically unlimited.Except that, just like a voltage source (current limit), the current source has a voltage limit.Now, if you supply a constant current to your LEDs, you'll be able to measure the voltage drop across them.Most lab power supplies will work in costant current mode.

author: PetrikNZ
author_rank: Newbie
timestamp: December 16, 2021, 12:06:59 am
content: @BentaRight, so saying I have an X Volt Power supply is really not meaning anything. It's the Y Amps on the Power Supply that's what matters.So that means I then need to decide how many of those Amps I wish to give to each LED, or have I got that wrong and am I now confusing myself even more?For argument sake, if we say the Power Supply I have is 9V 1Amp, for my scenario I gave. How do I proceed. (I have the multi-meter, just not sure how to use it. In other words, how and what to measure for completing my puzzle.@retiredfeline has answered as well, but assumes I have numbers (ball park or otherwise) that are not available in the scenario I laid out. That was the whole point of my scenario. Having only the details of the power supply, eg. 9V 1AmpI have a Lab Power Supply on order. My scenario references one of those plug in the wall jobs X Volts Y Amps, types.

author: mariush
author_rank: Super Contributor
timestamp: December 16, 2021, 12:19:39 am
content: The most basic way to limit the current going through a LED is by adding a resistor in series with the led.For this, the only formula you need is Ohm's law : V = I x R voltage = current x resistanceFrom this formula, you can get to this : Input voltage - ( number of leds in series x forward voltage of 1 led ) = Current x ResistanceSo let's say you have a 5v source (from let's say a USB port) and you have 2 red leds with a forward voltage of 1.9v and let's say you want 5mA (0.005A) of current going through the leds:5v - (2 x 1.7v) = 0.005A x Resistance => Resistance = 1.6 / 0.005 = 320 OhmThis is not a common value in the E series - seehttps://en.wikipedia.org/wiki/E_series_of_preferred_numbers#Listsif you don't know what I mean - so you'll probably choose to use 330 ohm resistor, which means the current will be slightly less (1.6v/330 ohm = 0.0048A or 4.8mA)You would also need to know how much power is lost in this resistor, and you can do that using formula Power = I2xRIn the example above, Power = 0.0048 x 0.0048 x 330 = 0.0076 watts which tells you that you can use a very small and simple resistor rated for maximum 0.1 watts without any worries.Things change with bigger values.For example, let's say you have 5v input , 3.2v forward voltage of a white led, 100mA current...You apply the formula: 5v - 1 led x 3.2v = 0.1A x R => R = 1.8 / 0.1 = 18 Ohmand the power dissipated in resistor would be P = 0.1 x 0.1 x 18 = 0.18w so you could barely do with a 0.25w rated resistor, but a 0.5w rated resistor would be better.

author: retiredfeline
author_rank: Frequent Contributor
timestamp: December 16, 2021, 12:22:16 am
content: That amp rating on the power supply is a maximum that you can safely draw. For your purposes you can assume it will be adequate for lighting LEDs.I see people ask about say laptop PSU bricks and think that a 5A replacement is unsuitable because the original stated 4A.It's the same with household AC circuits. Your house sockets are rated at say, 13A. That doesn't mean you actually draw that much current. The appliance takes what it needs at the rated voltage (230V in AU/NZ) and as long as it's under the max, you won't blow the fuse or trip the breaker.

author: PetrikNZ
author_rank: Newbie
timestamp: December 16, 2021, 12:42:37 am
content: I'm really new to all this so E Series, is something I never heard about, let alone what it means. New thing to study.However, the explaining, again assumes one has some basic details. NOT following my scenario where there are ALMOST no details, other than the Power Supply.My scenario assumed that all you knew was that you had a X Volt Y Amp power supply. NO details on the LED, including knowing the colour of it.So I can only guess what I am asking is just not possible.

author: retiredfeline
author_rank: Frequent Contributor
timestamp: December 16, 2021, 12:51:08 am
content: Because it depends on the type of LED. Obviously ones for lighting rooms are going to take more current than an indicator in a piece of equipment. However within a class of LED you can assume rough figures such as those I gave, or use datasheets for similar types.Where are these LEDs that you know nothing about, not even the colour? Do you actually have a use case you can write about?Look, forget about the max amp rating for the power supply, work on the basis of the output voltage, work out how much current you need to supply to light the LEDs you have, and then if it exceeds the current rating of the PSU, then you need a bigger one.

author: PetrikNZ
author_rank: Newbie
timestamp: December 16, 2021, 01:31:16 am
content: The project would be a diorama. The circuit would be a simple parallel circuit with single (but multiple) THT LEDs or even SMT LEDS. As such, if my understanding is correct, I would need to calculate the resistor for one LED (of each variant used) and then each LED variant in the circuit would get that same specification of resistor.The LED is plucked out of a mixed bag of LEDs, so in the scenario I put forward there is no data available on the LED, other than it's a THT or a SMT.What I am trying to ascertain is if this can be worked out (with the aid of a multi-meter) with such limited info.I can rewrite the scenario to try and make it clearer as to what I am asking if it helps.I guess really what I need to know is the HOW to your earlier comment. When we know nothing regarding the LED other than the type."With a power supply, a voltmeter, some resistors of known values, and Ohm's law you will get there."

author: retiredfeline
author_rank: Frequent Contributor
timestamp: December 16, 2021, 01:39:39 am
content: Look, if the current is decided on then you can calculate the limiting resistor with the information available, agreed?So the problem is how to decide on the current, right? You'll find that even with a datasheet you will get typical and maximum safe values for the current with no easy correlation to the brightness. In theory you could use the mcd output stated but even then it's very subjective, due to ambient light and so forth. That's why I say use an empirical method to decide how much current to put through the LED, staying within maximum rating.If you are wiring up lots of LEDs you may want to put groups in series as the typical power supply voltage is several times the LED voltage. Then you waste less power in the limiting resistor.

author: PetrikNZ
author_rank: Newbie
timestamp: December 16, 2021, 03:09:43 am
content: I'll have to do some more research in the series setup, as I understand the resistor works differently in such configurations.So with the parallel circuit, I can then decide what current to feed each LED and get the resistor info from that. I guess then if there is an issue with the LED not being bright enough I can recalculate with a higher current amount or reducing the resistance.Or have I got it wrong again?

author: retiredfeline
author_rank: Frequent Contributor
timestamp: December 16, 2021, 03:13:42 am
content: Look, here's a hypothetical worked example for you:Say by experimentation with a 9V power supply and various limiting resistors you determine that 10 mA through one red LED gives you a satisfactory brightness. But the blue LED is more efficient and you get roughly the same subjective brightness with 3 mA.Now at this point you have the resistor values for one LED, so that would be too easy. So say you want to put 3 red LEDs per group in series. So they will take 6V and the resistor has to drop 3V. By Ohm's law the value you need is 3/0.010 = 300. Closest E12 values are 270 or 330.For the blue LEDs you wire up 2 per group in series. So they will take 6V and the resistor again has to drop 3V. By Ohm's law you need 3/0.003 = 1k. 1k is a standard value.Now the actual LED drops will vary a little with current but not that much. So you could measure the actual voltages and tune the resistors, bearing in mind that the E12 series has 10% tolerance. You could go to the E24 series (5%). Or accept that a 10% variation in brightness won't be noticed. The LEDs will also exhibit variation within the batch, and across batches. Did I mention that design is iterative?Then sum up all the currents of the LED (chains) and if it exceeds the max rating of the power supply, get a higher rated one.PS: In a series setup the current is the same though all the LEDs in the chain. That reduces the number of resistors needed and also the power wasted.

author: PetrikNZ
author_rank: Newbie
timestamp: December 16, 2021, 03:34:17 am
content: I'll need to take some time to get my head around that.The other thing with series, as I understand it, is (probably not a worry for so few lights) that as you get further down the chain the brightness drops off. Where in parallel each one (of the same type/resistor combo) is of identical brightness and does not drop off.

author: retiredfeline
author_rank: Frequent Contributor
timestamp: December 16, 2021, 03:41:46 am
content: No, the brightness does not "drop off" further down the chain. As mentioned the current is the same through the chain, so all other things being equal (the LEDs being the same type and the batch being sufficiently uniform), the brightness will be equal.Series LEDs are used all the time. Some of those large 7-segment displays have more than one LED per segment.

author: james_s
author_rank: Super Contributor
timestamp: December 16, 2021, 03:47:31 am
content: No, that isn't right at all. In series Kerchoff's law guarantees that each node (LED in this case) has exactly the same current through it, and the sum of the voltage drop is always exactly equal to that of the source. In fact it was once common (in the USA at least) from around the 1920s through the 1950s for street lights to be wired in series using special lamps and fed from a high voltage constant current regulator. This was done partly so that a large number of lights could be controlled from a single location back before individual photocontrols were reasonably priced, and partly because even if you have a loop that is miles long with dozens or even hundreds of lamps, the brightness will be constant.In a parallel circuit loads further from the source will see a lower voltage due to drop in the wires, but for a handful of LEDs this is not going to come into play unless you are using ridiculously hair thin wire.Now regarding calculating current a good rule of thumb is that your average 5mm through-hole LED can safely take up to 20mA without overheating and the voltage drop will be anywhere from 1.5-4.5V. You are generally safe to start with a 1k resistor on a supply of around 12V or less and measure the voltage across the LED, and then use that with Ohms law to calculate the resistor you need for the desired current.*edit: My response is to the OP, not to the reply above that came in while I was typing it.

author: fourfathom
author_rank: Super Contributor
timestamp: December 16, 2021, 04:19:01 am
content: Somebody draw a schematic! OK, here's one:

author: PetrikNZ
author_rank: Newbie
timestamp: December 16, 2021, 08:21:11 am
content: Being a visual learner, I found a great video that has really helped with this.https://youtu.be/2vRn7MBz1g0

author: rstofer
author_rank: Super Contributor
timestamp: December 16, 2021, 04:20:54 pm
content: Let's bring things up a notch and talk about the idea of a 'curve tracer'. I'm not suggesting we actually have one but somebody does and they create Ifversus Vfgraphs so we can get some idea of what the LED looks like electrically. The point is, there isn't just one operating point (20 mA, 2V for a mythical Red LED), we can operate the LED anywhere we want subject to some limiting values which we don't know. We will know we went to far when the LED pops.So, we just eyeball the LED as we try to build our own graph with just a few data points. For the Red LED and 5V source, we know we need a resistor value of 150 Ohms for that mythical 2V 20mA LED so we just pick values nearby and see how bright the LED becomes. 150, 220, 270, 330 and so on. There is absolutely no reason to run as much as 20mA through a red LED. With a few experiments we can end up with a chart similar to thishttp://lednique.com/current-voltage-relationships/iv-curves/Check out the linked videos...As to resistor values, there are several standards. All we do is pick the value digits (say 2.2) and apply a multiplier like, say, 100 to get a 2200 Ohm value in that standard. I tend to stock E-24 values.https://www.electronics-notes.com/articles/electronic_components/resistors/standard-resistor-values-e-series-e3-e6-e12-e24-e48-e96.phpThe idea of using a multifunction component tester (really cheap from China) seems interesting. I have one and hardly ever use it. Some DMMs will also include a diode test function that does the same thing. But it's just a single point, not a continuum of values.

author: Martian Tech
author_rank: Regular Contributor
timestamp: December 16, 2021, 05:00:36 pm
content: Put a 10K potentiometer in series with the LED ans set it for maximum resistance. Then turn on the power supply and adjust the pot until the LED gives you the brightness you want. Shut off the power, measure the resistance and find a fixed resistor with approximately the same value that you measured. For the most part, you don't need to worry about what the forward voltage of the LED is or how much current you're giving it...

author: rstofer
author_rank: Super Contributor
timestamp: December 16, 2021, 05:02:48 pm
content: Quote from: PetrikNZ on December 16, 2021, 08:21:11 amThat's a great video and covers all of the points made in the thread. But...It only counts when you do the experiment yourself. Just an ordinary LED and a few values of resistors. You're not just lighting the LED, you are showing Ohm's Law at work and gaining familiarity with the tools of the trade.If you have two DMMs, you can measure Vfand If.You will be able to slowly increase the power supply voltage to the resistor and gain a sense of the IV curve. Put the values of Vfand Ifin a spreadsheet and plot a graph!At that point, you will really understand what is going on. You can try parallel and series configurations for extra credit.

author: rstofer
author_rank: Super Contributor
timestamp: December 16, 2021, 05:07:34 pm
content: Quote from: Martian Tech on December 16, 2021, 05:00:36 pmI would want a fixed resistor in series with the pot to limit the maximum current should I set the pot to zero by mistake and drive the entire capability of the power supply through a zero ohm resistor and LED. I'm not sure which would pop first. Maybe the fixed resistor would limit the current to something like 30 mA. The pot would adjust the current down from there.Yes, I learned this the hard way...

author: Martian Tech
author_rank: Regular Contributor
timestamp: December 17, 2021, 04:11:32 am
content: Quote from: rstofer on December 16, 2021, 05:07:34 pmWell OK, but if you don't blow things up from time to time you're not trying hard enough

author: Ian.M
author_rank: Super Contributor
timestamp: December 17, 2021, 04:57:03 am
content: Its safer and easier to simply use a variable PSU and a 1K series resistor. Set the output voltage to 5V before connecting the LED (safe reverse voltage for any normal LED), then once you have confirmed the correct polarity of connection, adjust the output voltage to get the desired brightness, and measure If (from the voltage drop across the 1K resistor) and Vf for use in your actual circuit design. You can also match LEDs for intensity if you put two (or more) in series.

author: rstofer
author_rank: Super Contributor
timestamp: December 18, 2021, 05:09:30 pm
content: OK, I'm an Analog Discovery fan - big time.I hadn't thought about it in a while but I have the Transistor Tester adapter board and it does a nice job of plotting Ic/Ib curves and a host of others.One thing it will do is plot the characteristic curve for an average red LED - see attached.Note that when the Ifis 20 mA, the Vfis about 2V - more or less average but reasonable.https://digilent.com/reference/test-and-measurement/transistor-tester-adapter/reference-manualI have another curve tracer attachment from Knack Supply. It requires an external app to run on the AD but it does a pretty good job.https://www.tindie.com/products/smartislav/ad2tm-curve-tracer-add-on-board/