Electronics > Beginners
Hooking up 4 x 3v LED string lights to a 6v PSU
bob21:
--- Quote from: Brumby on February 02, 2020, 02:45:34 pm ---
--- Quote from: bob21 on February 02, 2020, 01:52:22 pm ---If I understand correctly, a 300mA 6V PSU should be dishing 150mA to A and B.
--- End quote ---
No.
This is a very common misconception. The current rating of a supply is the maximum current it is rated to provide. The actual current drawn will be a function of the load. If the load only takes 100mA (for example), then that is all the supply will dish out. This is the same for any and all power sources that you would generally come across.** Ohms Law describes this for the simple case of a resistive load.
** This is the "constant voltage" type. There are supplies that operate on a "constant" current or current limiting basis - but we're not talking about these just here.
--- Quote ---Would I just need one resistor where the purple dot is? Or would I need 2, where the green dots are?
--- End quote ---
One where the purple dot is would be enough to provide basic protection - but one for each of the green dots would be considered a better option as it would allow independent "finding the operating voltage" for each arm.
--- Quote ---I am thinking the answer to this is that both A + B need one?
--- End quote ---
Need? - Not really - but that would be the "better" way, in my book.
--- End quote ---
Ok, understood. So I need to work on the assumption that each series could receive the full 300mA load. Like if there is a loose connection or all LEDs in a series fail.
In which case, that would be a max of 15mA per LED
6V - 3V = 0.015 (15mA) x R ===> R = 3V / 0.015 = 200ohm
So I need to place a 200ohm resistor on each series for the best possible safety?
--- Quote from: gf on February 02, 2020, 02:49:36 pm ---Line 3 (6 LEDs) and line 4 (15 LEDs) in series is not a good idea, since the two lines likely require different drive currents (assuming the LEDs are all the same type).
--- End quote ---
How not good are we talking? Broken LEDs or Fire?
These things are near furniture which Is why I am trying to make this as safe as possible. If the LEDs just fail, they were cheap. If they're going to burn the house down, that's different.
mariush:
First be sure that your 6v 300 mA is actually 6v. And is it DC or AC? How are you measuring the voltage?
If it's AC voltage, you'll have to convert that to DC using a bridge rectifier and a capacitor.
You'll get a DC voltage with a peak of around 1.415 x Vac - ~ 1.5v ,,, so if you have 6v AC power adapter, you'll get around 7v DC.
The current will be lower, around 0.62 x Iac ... so the 6v AC 0.3A becomes ~7v DC 0.2A
If it's DC voltage... some cheap power adapters will have a slightly higher output voltage when there's no load (or very low load). For example, that 6v adapter may be 6v when you measure using a multimeter because the multimeter is a very low load, it consumes less than 1mA measuring the voltage. However, if you connect something that consumes a bit more power (like an incandescent 6v bulb for example), the voltage may go down a bit, let's say to 5.5v
If your chain of leds has no resistors or anything to control the current going through them, you WILL damage the leds.
Such a chain will such as much current as the power supply can provide so you'll want to limit the current in some way.
If the chain is basically multiple leds in parallel like I assumed in my previous post, a single resistor at the start of the chain could be enough.
You can use that formula: Voltage input - Forward voltage of leds = current x resistor.
With 20 leds in parallel, each consuming 10mA, you have 200mA or 0.2A
So with 6v DC and 3v for the led, you have
6v - 3v = 0.2 x R => R = 3 / 0.2 = 15 ohm
What I forgot to mention is that you also have to be aware how much power is lost in the resistor, so that the resistor won't burn.
The formula for that is Power = Current x Current x R
In the above example, the power is 0.2 A x 0.2A x 15 = 0.6 watts, so you'd want a resistor rated for at least 1w otherwise the resistor will be super hot and may even burn up.
If you have 2 chains each with 20 leds in parallel... you could connect them together to form a single chain of 40 leds in parallel.
BUT, you'll have to keep in mind how will the extra long wire will affect things and you'll also have to adjust the resistor to account for that.
So for 40 leds, you'd have R = (6v-3v) / 0.4 A = 7.5 ohm (you'd probably want to use standard 6.8 or 8.2) , and the power on resistor P = 0.4 x 0.4 x 7.5 = 1.2 watts, so a 1w resistor is no longer suitable, you'd have to use a 3w rated resistor
The extra long wire could cause some of the leds towards the farthest end to be less bright than the leds closer to your power supply.
Brumby:
--- Quote from: bob21 on February 02, 2020, 02:57:37 pm ---So I need to work on the assumption that each series could receive the full 300mA load.
--- End quote ---
I'm going to stop you right there.
You seem fixated on the 300mA rating of the supply. Forget about that while you play with all the other numbers.
The 300mA ONLY comes into consideration when you work out everything else and come up with the total current that your circuits will want to draw. Then all you need to do is compare that load current with the 300mA.
... and here is the ONLY time you use the 300mA rating of the supply:
If the load current <= 300mA then all is well.
If the load current > 300mA then you will be overloading the supply.
The load circuit determines the current drawn. For a particular circuit, the current it will draw from a supply of a given voltage will be the same, whether that supply is rated at 300mA or 300,000A. If that circuit requires 1mA, then that's all that a power supply will put out.
bob21:
--- Quote from: Brumby on February 02, 2020, 03:55:48 pm ---
--- Quote from: bob21 on February 02, 2020, 02:57:37 pm ---So I need to work on the assumption that each series could receive the full 300mA load.
--- End quote ---
I'm going to stop you right there.
You seem fixated on the 300mA rating of the supply. Forget about that while you play with all the other numbers.
The 300mA ONLY comes into consideration when you work out everything else and come up with the total current that your circuits will want to draw. Then all you need to do is compare that load current with the 300mA.
... and here is the ONLY time you use the 300mA rating of the supply:
If the load current <= 300mA then all is well.
If the load current > 300mA then you will be overloading the supply.
The load circuit determines the current drawn. For a particular circuit, the current it will draw from a supply of a given voltage will be the same, whether that supply is rated at 300mA or 300,000A. If that circuit requires 1mA, then that's all that a power supply will put out.
--- End quote ---
You'll have to forgive me for being as thick as two short planks when it comes to this. I am trying to digest as quickly as I can though, so forget the 300mA for now. Got it. The only reason I was going back to the 300mA is because I logiced that if 300mA / 41 LEDs = 7.31mA, then I could not use the below calculation with a single LED current of 10mA, as the final currrent draw would be 410mA, which is overloaded.
--- Quote from: mariush on February 02, 2020, 03:34:36 pm ---First be sure that your 6v 300 mA is actually 6v. And is it DC or AC? How are you measuring the voltage?
If it's DC voltage... some cheap power adapters will have a slightly higher output voltage when there's no load (or very low load). For example, that 6v adapter may be 6v when you measure using a multimeter because the multimeter is a very low load, it consumes less than 1mA measuring the voltage. However, if you connect something that consumes a bit more power (like an incandescent 6v bulb for example), the voltage may go down a bit, let's say to 5.5v
If your chain of leds has no resistors or anything to control the current going through them, you WILL damage the leds.
Such a chain will such as much current as the power supply can provide so you'll want to limit the current in some way.
If the chain is basically multiple leds in parallel like I assumed in my previous post, a single resistor at the start of the chain could be enough.
You can use that formula: Voltage input - Forward voltage of leds = current x resistor.
With 20 leds in parallel, each consuming 10mA, you have 200mA or 0.2A
So with 6v DC and 3v for the led, you have
6v - 3v = 0.2 x R => R = 3 / 0.2 = 15 ohm
What I forgot to mention is that you also have to be aware how much power is lost in the resistor, so that the resistor won't burn.
The formula for that is Power = Current x Current x R
In the above example, the power is 0.2 A x 0.2A x 15 = 0.6 watts, so you'd want a resistor rated for at least 1w otherwise the resistor will be super hot and may even burn up.
If you have 2 chains each with 20 leds in parallel... you could connect them together to form a single chain of 40 leds in parallel.
BUT, you'll have to keep in mind how will the extra long wire will affect things and you'll also have to adjust the resistor to account for that.
So for 40 leds, you'd have R = (6v-3v) / 0.4 A = 7.5 ohm (you'd probably want to use standard 6.8 or 8.2) , and the power on resistor P = 0.4 x 0.4 x 7.5 = 1.2 watts, so a 1w resistor is no longer suitable, you'd have to use a 3w rated resistor
The extra long wire could cause some of the leds towards the farthest end to be less bright than the leds closer to your power supply.
--- End quote ---
The PSU is a 6V DC and I checked it by switching my Multimeter to '20' on the DC side and the reading was 6.01V.
Because I think the LEDs at the end of a super long chain of 40 would suffer (as you've suggested), I think it may be better to place 2 resistors at the green dots and have the 2 x chains connected individually as per the last picture I posted (unless you mean this isn't possible?). If it is possible, I need to find 2 resistors - one for each green dot, right?
I am just repeating this so you can see my thinking, I want to make sure I understand.
So with each LED consuming 10mA:
Circuit A - 20 LEDs: 6v - 3v = 0.2 x R => R = 3 / 0.2 = 15 ohm
Circuit B - 21 LEDs: 6v - 3v = 0.21 x R => R = 3 / 0.21 = 14.28 (so safe to just use 15?)
For the resistors losing power and getting hot, I would like them to be as cool as possible as these are near a sofa. So:
Circuit A - 20 LEDs: 0.2 x 0.2 x 15 = 0.6 watts (as you've said)
Circuit B - 21 LEDs: 0.21 x 0.21 x 15 = 0.66 watts (still ok to use 1W and remain cool?)
So I need to place 2 x 15 Ohm Resitors rated at least 1W at each green dot point in my diagram?
Just a side question, would it be possible to use a 15ohm @ 3W on these smaller chains to keep the heat down even more? Or will this not work? I am thinking about minimum heat due to proximity to fabrics.
Now going back to the PSU mA rating - can this load cope with it? 10mA x 41 = 410mA, so no...
So is it not at all possible to just adjust the calculation so it fits within the confines of the PSU (Yes I know I am meant to be forgetting this... but this was my thought process)
If I have a 300mA PSU and 41 LEDs, I do not want each LED to draw more than 7.31mA. So recalculating the above using that value:
Circuit A: 7.31mA x 20 = 146.2mA or 0.146A
Circuit B: 7.31 x 21 = 153.51mA or 0.15351A
Circuit A - 20 LEDs: 6v - 3v = 0.146 x R => R = 3 / 0.146 = 20.54 ohm (so safe to just use 20?)
Circuit B - 21 LEDs: 6v - 3v = 0.15351 x R => R = 3 / 0.15351 = 19.54ohm (again, just use 20?)
For the resistors losing power and getting hot:
Circuit A - 20 LEDs: 0.146 x 0.146 x 20 = 0.426 watts - so 1W well and truly safe?
Circuit B - 21 LEDs: 0.15351 x 0.15351 x 20 = 0.494 watts - again, 1W well and truly safe?
So this is why I kept going back to the 300mA.. 41 LEDs x 7.31mA = 299.71mA
Question is, is it wise to run it at it's limit? Would it not be better to take it down a notch and say, for example:
Each LED 7mA
Circuit A: 7mA x 20 = 140mA or 0.14A
Circuit B: 7mA x 21 = 147mA or 0.147A
Circuit A - 20 LEDs: 6v - 3v = 0.140 x R => R = 3 / 0.140 = 21.4285 ohm (so 22ohm?)
Circuit B - 21 LEDs: 6v - 3v = 0.147 x R => R = 3 / 0.147 = 20.4081 (so 21ohm?)
For the resistors losing power and getting hot:
Circuit A - 20 LEDs: 0.140 x 0.140 x 20 = 0.392 watts - so 1W surely not getting hot at all now?
Circuit B - 21 LEDs: 0.147 x 0.147 x 21 = 0.453 watts - again, 1W well and truly safe?
Now we come back to the 41 total LEDs each drawing a maximum of 7mA = 287mA so we aren't running the PSU to the max.
So I need to put this on the 21 LED string: https://www.reichelt.com/gb/en/metal-oxide-resistor-1-w-5-22-ohm-1w-22-p1799.html?PROVID=2788&wt_guka=92495611608_406948187429&PROVID=2788&gclid=EAIaIQobChMI15_ok7Oz5wIVSLTtCh3rvgI9EAQYASABEgJIjfD_BwE&&r=1
Can't seem to find 21 ohm 1W resistors easily, so would a 22ohm just work on both? Would be way easiser to grab one pack.
If I haven't understood it now, I don't think I ever will.
mariush:
You can slightly adjust the resistor values to get something easy to buy. You'll just get a bit less or a bit more current through the leds.
Reuse the formula just put R instead of current, for example
Circuit B - 21 LEDs: 6v - 3v = 0.21 x R => R = 3 / 0.21 = 14.28 (so safe to just use 15?)
Rewrite it like this
Circuit B - 21 LEDs: 6v - 3v = Current x 15 => Current = 3 / 15 = 0.2A (200mA / 21 leds = 9.52mA per led)
The PSU is a 6V DC and I checked it by switching my Multimeter to '20' on the DC side and the reading was 6.01V.
Yes, well that's no load.
Repeat the measurement with 20 leds connected to it and see if you still get 6v. You may find out that the closer you are to its maximum advertised output of 300mA, the voltage may go down a bit. So you may see 5.5v at 250mA or something like that.
The power supply may be conservatively specified, you may find out that it can actually output 500mA while the voltage goes down to around 5.5v ... not saying it would a good idea to use such a power supply for long time above its advertised capabilities... it's up to you.
If you go above 300mA, it may or may not output more, but there's no guarantees.
The power supply may lower the voltage a bit (so the total power will be the same, for example 6v x 300mA = 1800mW or 5.5v x 325 m = ~1800mW, same amount of power)... The power supply may have some protection mechanism built in, where it would constantly reset itself for brief moments of time.
Yes, you can use resistors rated for higher wattage than needed. You can also parallel resistors to spread the heat across a larger surface. When you parallel two resistors, you get half the resistance but 2x the area to dissipate heat.
So for example, instead of using a 10 ohm resistor, you could have two 22 ohm resistors in parallel ending up with 11 ohm
You can use 3w resistors but they're big and ugly. I'd look into paralleling 2 1w resistors first.
Navigation
[0] Message Index
[#] Next page
[*] Previous page
Go to full version