I keep referencing a series connection because it looked like the TS was seeking a solution to a problem. Initially the problem looked like a current overload. A series arrangement only needs one current control and thus there is less power wasted in whatever is producing the voltage drop that controls the current. So usually a series connection of LEDs is more efficient. Aside from that, often it uses fewer components and so is less expensive.
So these are the main reasons for my suggesting a series LED arrangement.

 

I got the impression you were thinking the TS had a series LED issue.

The video I made showed that you can light several LED's all through a single current limiting resistor. You can't tell me it won't work - the video is self evident that it does. Some people like AG believe I would have a cascading failure. But the video proved that with the right resistor you can't have a cascade. Albeit - AG is correct, powering multiple LED's through a single resistor is poor design; and I don't advocate for doing so. My video was just an experiment to see if it could be done. When I learned it could be done I made the video. Yes, that's MY video. Not something I found on the web. And there are no tricks, no hidden gimmicks, nothing to fool the viewer. It's legitimate. What you see is what you get.

The illustration in post #16 is the impression I immediately took from what the TS was saying. I also offered another solution drawing more current, but never drawing more current than the power bank was capable of. And quite possibly enough current being drawn to keep the bank powered. The video does not cause the source to deliver more current because it's all limited by the single resistor. And though I demonstrate it can be done - the illustration clearly shows individual resistors for each LED. Let's call it an LEC (Light Emitting Circuit) where you have a single LED and a single resistor in series to limit the current. Each LEC draws 23mA. The more LEC's you put in parallel the more current you draw. Ten LEC's will draw 230mA. That might be more current than the PB (power bank) can deliver. So somewhere between 2 LEC's and 10 LEC's would be the place where the PB switches on and stays active until it's reserve power is drained away.

 

This video is great thank you! Also, I will soon be able to connect a few more LEDs to demonstrate whether the PB stays on

 

Thanks for your reply. If the rated output of the powerbank is 1000ma should I put enough leds on that draw 1000?

No. That would drain and possibly damage the battery in an hour.

You have three design constraints. The voltage supplied from the power bank. The current required by one LED. The number of LEDs.

Match the voltage required to the voltage supplied.

Ensure the current available (in each series string) matches the current required for each LED.

Check that the current required in each LED string times the number of strings is small enough to run the LEDs long enough given the mAH rating of the power bank.

 

A powerbank should be smart enough to turn itself off when the battery voltage is low.
1000mA into LEDs will produce a very bright light. Why?

 

Ok, so I tried 13 LEDs in parallel and cheated with one resistor at the beginning at attached to the power bank. (I checked the forward voltage of each, they all have the same which was 2.7. I tried different resistors to understand what happens, a 120, a 330 and a 1k. The power bank still switches off after about 5 seconds.

(I did it the hotch potch way as it's super lockdown here and everything I need is stuck in the post - New Zealand if you're interested)
I have 6 330 ohm resistors which I'll try again with.

 

It may also be that the power bank device is a one function only item, not useful for anything else. It might be looking for a response from a phone.

 

MissOrange, you did not do the math.
(5.3V - 2.7V)/1K= 2.6mA (0.0026A) which is almost nothing.
(5.3V - 2.7V)/120 ohms= 22mA (0.022A) which is a low current the same as one fairly bright LED.

If you used 13 LEDs, each with a 120 ohms resistor then the total current would be 282mA (0.28A) and the powerbank might stay turned on.

 

Ok, so I tried 13 LEDs in parallel and cheated with one resistor at the beginning at attached to the power bank.

Glad you got something out of the video. However, with a single resistor limiting the current, the more matched LED's you put on circuit the more current they share from what is available through the resistor. Using a single resistor - while that will work - will not draw more current the more LED's you put on. Suppose with one LED and one resistor you have 25mA. Then suppose you add a second LED parallel to the first. You haven't added a resistor, just put two matched LED's parallel to each other in series with a single resistor. The circuit still sees the same 25mA but each LED shares that milli-amperage, meaning each will use 12.5mA (assuming they are perfectly matched - which doesn't happen). One LED WILL draw a little more than the other. If it draws enough the second LED will go out. As you saw in the video, the green LED worked fine by itself and the one resistor. But when I put the red LED in parallel with the green LED - the green LED went out. That's because the red LED hogged all the current. When I was putting the blue LED's in parallel each had a 2.82Vf average. The minimum Vf out of 100 LED's was 2.80Vf and the max was 2.83Vf. Very close to each other. Close enough that they all would light up. If you know anything about statistics, the 100 LED's had a standard deviation of 0.01Vf. 10mVf is close enough that they will all light up.

When you added the additional LED's to your circuit - you didn't increase the current draw. All you did - as I did in the video - is get LED's to share current relatively equally. The current didn't change. As in my example above, 25mA divided between 2 LED's was 12.5Vf. When I put 7 LED's parallel through the single resistor each LED saw only 3.86mA. If you paid close enough attention you could see that seven LED's the light was greatly reduced. Still, "Super-Bright" LED's (such as the ones I used) will still produce significant light at 3.9mA of current. Even then, when I plugged the red LED back into the circuit the blue LED's all went out. Again, it was the lower Vf that drew all the current.

AudioGuru expected a cascading failure - and that's exactly what can happen. Suppose I planned on 6 LED's all through a single resistor. Whatever current was available would drive all six equally divided amongst them. Suppose I wanted all six to see 20mA. I would have had to choose a resistor that would provide 120mA of current. Then, as AG expects, if one LED goes out, the rest would be dividing that 120mA (amongst 5 remaining LED's). Each would then see 24mA. While the expected max milli-amperage might be 30mA, 24mA will not likely blow any of them out. But suppose one more went out. Now there's only 4 LED's in circuit. 120mA ÷ 4 = 30mA (the max) and each would be getting hot. Like AG said, they will begin to thermally run-away; overheating and drawing more current. One will fail before the rest, leaving 120mA to be shared by the now 3 remaining LED's at 40mA. This is where the cascade begins - at 40mA it's expected that none of the LED's will survive. 3 LED's quickly turn into 2 working LED's, turn into 1 working LED, into none. They will fail one after the other faster than you could react to shut off the circuit. That's the whole reason why you put a single resistor with a single LED. There IS an exception to that - when you have a high enough voltage you can string LED's in series with a single resistor. Suppose you have a 12 volt source with three LED's whose Vf = 3Vf. 3 LED's subtract from the voltage source leaving 3 volts to be dropped by the resistor. A 120Ω resistor will limit the current to just 25mA. That's a lot of light, but that's also the correct way to build that circuit. If you don't fully understand - just ask for a drawing. will take me less than 10 minutes to bang out a drawing explaining what I just said. But overall, you don't generally put a bunch of LED's on a single resistor. And remember, more parallel LED's doesn't mean drawing more current. Each LED must have its own resistor to create an LEC (Light Emitting circuit). The more LEC's you add (in parallel) the more current you will draw, just as I showed in the diagram. Let me know if you have more questions.

TR

 

Ok, so I tried 13 LEDs in parallel and cheated with one resistor at the beginning at attached to the power bank. (I checked the forward voltage of each, they all have the same which was 2.7.

How did you measure the forward voltage of the LEDs? If you measured when they were connected in parallel, of course they would all have the same voltage.

The forward voltage of an LED is not a constant, it depends on the current through it. To measure the forward voltage correctly, you must specify a current, and you need two meters, one to measure current and one to measure voltage at the same time. To be pedantic, the forward voltage also depends on temperature, but for low power LEDs at room temp you can ignore that effect.

Bob

 

Oh thanks for persevering, it's been so helpful. Fundamentally this is all left brain stuff and I'm a right brain person.

I put together 7 LEDs with 330 ohm resistors on each, so 5.3 - 2.7/330 x 7 = 55ohms (is that correct?) And therefore is still the possible reason why the power bank switches off.

My plan is; wait for resistors to arrive in the post, and second option - think of a way of installing an extra connection direct to the battery. The end goal is a bunch of arty outdoor party lights - (with 5 or so less each) each light would connect to a power bank for ease of recharging, so this first step has been learning to understand a little of batteries and electricity.

 

(5.3V - 2.7V)/330 ohms= 7.88mA (0.00788A). Then 7.88mA x 7= 55.2mA (0.0552A). It is not much current.
Voltage divided by resistance equals current, not resistance again.

 

5.3 - 2.7/330 x 7 = 55ohms

First - not "Ohms", "AMPS". And no - it's not 55 amps, it's 55mA (milli-amps).

Why did you go with 330Ω? If I wanted to light an LED with 5 volts (5.3 volts in your case) I'd settle first on what amperage I wanted it to operate at. Then when knowing the accurate Vf (forward voltage) I would calculate for the resistor needed to achieve the amperage. So (5.3V - 2.7Vf) ÷ 25mA = 104Ω. [(5.3V - 2.7Vf) ÷ 0.025A = 104Ω]. 25mA X 7 parallel sets = 175mA total. Since there are no such animals as a 104Ω resistor - unless you search through a bunch of them to find which one is on the plus side of their tolerance, I'd just use a 100Ω resistor. The resulting current would be (5.3V - 2.7Vf) ÷ 100Ω = 26mA. [ 0.026A ]

 

Thank you. I chose 330ohms because I have very limited choice and am waiting for resistors in the post (it's lockdown here and people just went back to work yesterday)

 

An LED will light with less than the rated voltage, and it WILL draw lewss than the rated current at that lower voltage. The voltage/current/ light output relationship is very non-linear, but they will light up some. So a string of two rated at 2.3 volts each, in series on five volts, will light up and draw just a bit more than the rated current. They will not fry instantly, but probably would have a slightly shorter life. So you can add a bit of series resistance abd make the current just what the spec says it should be.

 

Thank you. I chose 330ohms because I have very limited choice and am waiting for resistors in the post (it's lockdown here and people just went back to work yesterday)

You DO realize that 330Ω dropping 2.6 volts (5.3V - 2.7Vf = 2.6V) results in 7.9mA (0.007878••• Amps)

Yeah, Bill is right about running an LED under its rated voltage, but that's not the way to do it. Just because you can doesn't mean you should. Recall my video - you CAN run a bunch of parallel LED's through a single series resistor. But it's just not the right way to do it.

 

You DO realize that 330Ω dropping 2.6 volts (5.3V - 2.7Vf = 2.6V) results in 7.9mA (0.007878••• Amps)

Yeah, Bill is right about running an LED under its rated voltage, but that's not the way to do it. Just because you can doesn't mean you should. Recall my video - you CAN run a bunch of parallel LED's through a single series resistor. But it's just not the right way to do it.

If you do not need allof the rated light output then you have no need to drive it at max. Just like driving a car, you do not need to run at max HP all the time, unless you are racing on a track.

 

Well, I haven't tried it, but others have said if your Vf is greater than your source then the LED won't light up. I don't know how true that may be, but I'll accept that others know more about it than me.

 

Well, I haven't tried it, but others have said if your Vf is greater than your source then the LED won't light up. I don't know how true that may be, but I'll accept that others know more about it than me.

I believe this is true. That’s why with a red and blue LED in parallel with a common resistor, only the red LED illuminates.

 

Vf varies with current. A typical LED will have a Vf rated at some specific current, for common 3 or 5 mm through hold LEDs this is normally 20mA. It is not true to say that if you power it with a constant voltage less than that it will not light up. It will light up visibly for voltages of several hundred mV lower.

Bob