Its simple. Just check the voltage drop of the LED and calculate the resistor value for 10mA current. For indicators even 2 mA current is enough. Even you dont have to be exact, just find nearest value of resistor you calculated.

LOL!

This is like our first physics lesson in 7th grade. My grandpa taught me this when I was about 7 though beacause he was a physics teacher.

Sounds like you are missing the most basic of electronics skills. You should have bought and read a basic electronics book for $10. Then you would have known in an hour or two and not years.

Glad I can help. best wishes

Thank you very much for your kind comment.

Thank you very much Tom. If I had more room I would love to build a model railway but there is not enough room to swing a cat where I live. Best wishes

@@markusfuller “Not enough room to swing a cat!” I love it! Ha!

Exactly the same reason as I am glad I learned this. Indicator lights for a display panel for my HO railway.

Ft Gordon?

Good for you, hmmm, um put your finger over the value you want, so if its V then IxR, if it is if you want Current (I) then it's V over R (V/r or V divided by R) because V is over R. Sorry didn't watch the vid maybe he/she says this. Then you have Watts so P Power (at the peak of the triangle) = I x V. Type in Ohm's Law to your browser then go to images and you should find a wheel image that uses more conventional maths. Stops my brain falling apart like a chocolate orange. You'll get the hang of it. :) Many electronics shops have them as a bookmark. Such as I=√P/R..... Hi from Oz.

Thank you

Also he should have stated that the current rating of the LED is the maximum current spec before the LED fails. I usually go for half the rated current value. It will be bright enough.

Or a topology compromise somewhere between the two cited extremes

Hello. firstly thank you for your kind comment, I must rewatch the video and have a look at what I said at 19min41s . I made this video a while ago and cannot remember what I said but I have made mistakes before in other videos so I may be able to add some text on the video to correct any mistake or if not place it in the about section under the video. thanks for pointing this out to me. best wishes.

There is a typo on the screen. The LEDs are in parallel and each want to take 20mA , whereas the previous series example they were sharing the same 20mA. LED (indeed all diodes) should not be connected in series like this a fraction of a mV difference is all that is needed for ALL the current to go through ONE LED and cause it to very quickly fail (sometimes dramatically). The better solution is to add a limiting resistor in series with each diode and break the relationship. Use the series configuration if you have a lot of voltage and the parallel if you are running from a low voltage. Efficiency of design is also understood by looking at the power wasted by the LED, you obviously want that to be as small as possible

@@deansmith4752 You mean "all diodes should not be connected in _parallel_ like this ..." and that's because whichever diode is taking more current, it will heat up more and its negative temperature coefficient will cause it to draw more of the current, and so on. The power consumed by the LED is fixed for a given brightness, so the efficiency is governed by the power wasted in the dropping resistor (not the LED). To maximise efficiency, you want the supply voltage not too far above the on-voltage of the LED, but that makes the dropping resistor small and can lead to greater variations in current consumed for changes in temperature or variations in samples of the LED.

@@RexxSchneider I'd say you're both right. To summarize it; a series connection *guarantees* that the same amount of current (A) runs through each LED. A parallel connection of LED's *guarantees* that the same voltage potential is applied across them. LED's are inherently a bit "tricky" in ohm's law, as they don't follow a linear relationship between voltage and current since it's an "active component", being a semiconductor. Since ohm's law doesn't really work on active circuits, we have to use static/idealized values for semiconductors. LEDs, and regular diodes for that matter, are simple enough since when surpassing their forward voltage, (or reverse breakdown voltage), they'll be idealized as essentially having resistance that IS the proportional of voltage and current, and likewise, when below the threshold, we could in a way say the resistance is infinity. But naturally a LED that's ON will heat up, and that's the culprit that makes LEDs so "hard to calculate" with Ohm's law, cause that changes it's forward voltage. The hotter it gets, the lower the forward voltage gets. So dean is correct in that when LEDs are in series, one single LED with different forward voltage would drop less voltage, allowing more current to pass in total, and also leaving more potential for the next LED. It should be noted that the next LED would then sink more total power, decreasing it's forward voltage faster than the first, and kinda "balancing" it out. The solution proposed isn't in effect changing this fact, but does make it look a lot easier to calculate individual LED in a diagram. I think what Dean meant by "current to go through ONE LED" was that one led would drop all the voltage, in effect "consuming all the current", since the current would be the same across all LED's in series what. (disregarding inductance etc.) Rexx is absolutely correct that LEDs shouldn't be connected in parallel, but the "like this" doesn't show in the video, since they all have a limiting resistor. But if the LEDs were indeed in parallel, this is when Dean's comment would happen. One LED would be a fraction out (inevitable) of the forward voltage of the others, thereby making it drop less voltage and in turn let more current flow. This is thermal-runaway waiting to happen, as that LED would consume more power, getting hotter, dropping even less voltage but passing even more current. I wouldn't say that "the power consumed is fixed for a given brightness", since brightness tends to get weaker when over-driven (more power used for heat than for producing photons). But I get the idea. The reason a parallel circuit could be stable a while with very similar LEDs, is that power consumption is equal to voltage times current. A LED is dropping less and less voltage, but taking more and more of the current, making it effectively use close to the same amount of power. But when it drops enough voltage to drag another LED into it's forward voltage threshold, that other LED is drastically affecting the total current in the circuit, since it's basically "turning off" letting all it's current flow through the other LEDs. This is when they'll get over driven and blow. (sometimes dramatically) ^^ So to summarize: use a LED driver to maximize efficiency. It changes voltage in accordance with current drawn, thereby limiting the total current and not exceeding the rating of any LEDs, and use a resistor as a "current limiter" that would account for a certain drop in forward voltage. It's easier, but simplicity has it's tradeoffs. Wow, sorry for the wall of text... I got on a roll there..

@@arsenic1987 You have a decent grasp of what is happening when diodes are connected in parallel, but underestimate the effect of the exponential relationship between voltage and current. When you say, for example, _"A LED is dropping less and less voltage, but taking more and more of the current, making it effectively use close to the same amount of power"_ you miss the point that a forward voltage increase only needs be around 20mV to produce a doubling of the current through it. That's equivalent to as little as 10°C rise in temperature. That means that the power consumed by the diode (or LED) does increase quite dramatically as its temperature increases; the increase in current is is far greater than the minute change in forward voltage. It definitely does not _"effectively use close to the same amount of power"._ The whole point of the dropper resistor is to stabilise the current against changes in temperature since an increase in current as the diode heats up will cause more voltage to be dropped across the resistor, thus reducing the voltage across the diode and significantly reducing the increase in current because of the exponential relationship.

Oh, and since using power as well, could have drawn the corresponding "power triangle". P = V * I (Or P=U*I). Where P is power, measured in Watt (w).

Thankyou Alan

Yes. That's the purpose of the resistor. Just perform the calculation for 12v

that works for some DVMMs

say you have a 5V power supply, connect a LED and a 200 Ohm resistor as done here, Use a multi-meter (20 VDC setting) and measure the voltage across the LED - that gives the Vf

Another way of thinking is that a diode will always stay at almost the same voltage, while the current will adapt according to the resistor. The more difficult question is, how much power can a scavenged diode take? Use 200ohm like suggested and see how long it takes for it to die .. it's a scavenged diode after all ;) Most 'normal' diodes will live "forever" with a 1k resistor, most even with a 100ohm resistor .. the high power diodes are the ones that are more sensitive.

And yet another way of looking at the problem is to consider that almost every LED ever made can take a current of 20mA and will have a forward voltage somewhere between 1V and 3.5V. So get a 12V supply and a 470 ohm 1/2 Watt resistor, hook them up with the LED and measure the voltage across the LED with a voltmeter. You can then calculate the current. If you only have a 5V supply, then use a 180 ohm 1/4 Watt resistor, although that will give more variability in the current depending on the type of LED. If the LED can't take 20mA, it's not going to be much use to you anyway.

Do you set up a circuit and measure the voltage across the LED - yes

Thankyou very much.

I just use a variable power supply or variable resistor when theres lots of unknown leds to work out.