Electronics > Beginners
wiring new PSU to a flickering LED lamp
mariush:
What's the point of your circuit?
It's a badly designed one ...
You basically have 2 strings of leds and in each string, the leds are connected in parallel. this means each string will only require the power supply to provide at least the forward voltage of an individual led. And, because the leds are connected in parallel, the current adds up, so each string because like a single led that needs 8 leds x 20mA = 160mA
So, both strings in total would require 320mA.
Now, if the leds are arranged this way, you would need to limit the current going into each string with a resistor.
You have the formula V = I x R so
Vin - (forward voltage led x number of leds ) = Current x Resistance where number of leds = 1 because you connected all 8 leds in parallel and current is 8 x 20mA again because parallel
so 12v - 10v = 0.160 x R => R = 2v / 0.160 = 12.5 ohm
And the power dissipated in the resistor would be P = IxIxR = 0.16 x 0.16x12.5 = 0.32w so you'd need a 0.5w rated resistor at minimum, ideally a 1w resistor for EACH string of leds.
In your picture both strings receive a limited amount of current ... 12v - 10v = I x 100 ... so I = 2 / 100 = 0.02A or 20mA , so each led would only consume 20 mA / 8 = 2.5mA
You can see here the proper way you should do it : http://tinyurl.com/ybjmzqut - it's an online circuit simulator, you can right click on components to edit the values, and you can hover mouse over components to see voltage, current, power dissipated etc
If you don't want to click on shortened links or in case link dies, go to http://www.falstad.com/circuit/ and use Import from text option in menu and paste this:
The exact current values and power in components in simulated circuit is just a tiny bit different than the math above, it's just the simulated leds consuming slightly different current (~22mA instead of 20mA each)
--- Code: ---$ 1 0.000005 10.20027730826997 50 5 43
162 256 96 256 160 1 10 1 1 1 0.16
162 288 96 288 160 1 10 1 1 1 0.16
162 320 96 320 160 1 10 1 1 1 0.16
162 352 96 352 160 1 10 1 1 1 0.16
162 384 96 384 160 1 10 1 1 1 0.16
162 448 192 448 256 1 10 1 1 1 0.16
162 416 96 416 160 1 10 1 1 1 0.16
162 448 96 448 160 1 10 1 1 1 0.16
162 480 96 480 160 1 10 1 1 1 0.16
162 480 192 480 256 1 10 1 1 1 0.16
162 256 192 256 256 1 10 1 1 1 0.16
162 384 192 384 256 1 10 1 1 1 0.16
162 288 192 288 256 1 10 1 1 1 0.16
162 320 192 320 256 1 10 1 1 1 0.16
162 352 192 352 256 1 10 1 1 1 0.16
162 416 192 416 256 1 10 1 1 1 0.16
w 256 96 288 96 0
w 288 96 320 96 0
w 320 96 352 96 0
w 352 96 384 96 0
w 384 96 416 96 0
w 416 96 448 96 0
w 448 96 480 96 0
w 480 160 448 160 0
w 448 160 416 160 0
w 416 160 384 160 0
w 384 160 352 160 0
w 352 160 320 160 0
w 320 160 288 160 0
w 288 160 256 160 0
w 480 192 448 192 0
w 448 192 416 192 0
w 416 192 384 192 0
w 384 192 352 192 0
w 352 192 320 192 0
w 320 192 288 192 0
w 288 192 256 192 0
w 480 256 448 256 0
w 448 256 416 256 0
w 416 256 384 256 0
w 384 256 352 256 0
w 352 256 320 256 0
w 320 256 288 256 0
w 288 256 256 256 0
r 208 96 144 96 0 12.5
r 208 192 144 192 0 12.5
w 256 96 208 96 0
w 256 192 240 192 0
w 240 192 208 192 0
w 256 256 80 256 0
w 256 160 80 160 0
w 80 160 80 256 0
w 144 96 144 192 0
w 144 192 144 288 0
w 80 256 80 288 0
v 80 352 144 352 0 0 40 12 0 0 0.5
w 80 288 80 352 0
w 144 288 144 352 0
--- End code ---
But anyway the problem with this kind of layout is that if one led dies shorted, each string is STILL limited to 160 mA so now instead of having 160 mA divided between 8 leds, so 20mA per led, you now have 160mA divided among 7 leds, which means each led in that string with a dead led will receive around 23mA .... If one of those 7 remaining leds is particularly sensitive to over current, it could die and then you have 6 leds which receive 160mA so each led now gets 160mA / 6 = 26.6mA ... you get the idea? You risk a cascade failure, where a tiny number of dead leds in string of parallel leds can kill the whole string of leds.
Zero999:
--- Quote from: smile on October 18, 2018, 09:55:15 am ---Thank you for your circuit :)
I simulated my 300mA 12V minimum PSU and LED lamp online and got this result:
--- End quote ---
Why did you simulate that circuit? It's about the most pointless thing ever. It won't tell you how well or badly it will work.
The LEDs will all have slightly different forward voltages, so the current sharing will be imperfect. Some LEDs will draw more current than others. The simulator will just use exactly the same model for each LED, which will show perfect current sharing, which is not a true representation of reality.
If you plan to power this from the 230V mains, then use a capacitive dropper (Google it) and a rectifier for the LEDs. Of course the LEDs will all be live at mains voltage, so put them in a well insulated enclosure.
If the LEDs need to be powered from 120V mains, then use a voltage doubler circuit to power the LEDs. Again, the LEDs will need an insulated enclosure.
If you must run the LEDs at a 'safe' voltage, then splitting them into separate strings is the correct thing to do, but put a low value resistor in series with each string, to ensure good current sharing. It will only need to drop half a volt or so to give decent results.
You could use your constant current driver board to drive eight strings of two LEDs in series, each with a 22R series resistor, to aid current sharing.
smile:
Thank everyone for replies.
--- Quote ---Why did you simulate that circuit? It's about the most pointless thing ever. It won't tell you how well or badly it will work.
--- End quote ---
Well my PSU can't output 175V, so I thought that I can connect parallel and drop the 2V, from minimum 12V to 10V and be happy. My simulation was for perfect world I know it. I simulated 20mA current when the actual is 26mA so even if 5 LEDS will fail the circuit should use 26mA only then and should not immediately fail.
--- Quote ---If you plan to power this from the 230V mains, then use a capacitive dropper (Google it) and a rectifier for the LEDs. Of course the LEDs will all be live at mains voltage, so put them in a well insulated enclosure.
--- End quote ---
But that will make leds flicker as far as i know. The stock led driver is just that capacitor based.
--- Quote ---You could use your constant current driver board to drive eight strings of two LEDs in series, each with a 22R series resistor, to aid current sharing.
--- End quote ---
I rewired the 16LEDS parallel in one string.
I tested this connected to my lab psu. I set voltage to 10.4V and was increasing the current. The CC switches to CV at 35mA I thought parallel would make it want to use 300mA? I tried to give more current like 100mA but it was limited to CV and obviously current did not increase and the LED brightness was low. :-//
mariush:
Your voltage of 10.4v may actually be too small and your leds may not be fully turned on.
Why don't you actually determine the forward voltage of your leds first?
Your power supply may not have enough precision at such low currents, often they're something like +/- 5mA precision or something like that.
Set the maximum current output of your psu to something small like 50mA and then you can fine tune the current limit by simply connecting the probes of the multimeter (set on mA current range) to your power supply - the current will go through the current shunt inside your multimeter so the circuit is complete and the multimeter will show the actual current. Now if you have the meter set on a small range like 100mA or something like that, you can easily fine adjust the current limit of your power supply to 15mA or whatever desired value you want.
Note that there's going to be a tiny voltage drop on the current shunt resistor inside the meter but it's relatively small (on sub 1A range on a meter, typically you have a 1 ohm current shunt, so at something like 20mA you have V = I x R = 0.020A x 1 ohm = ~ 0.02v drop inside the meter... practically nothing.
The probes of your multimeter will also have some resistance but that's usually small, maybe 0.2 ohm or thereabouts ... so in general it's small enough to ignore it.
OK, once you set your current limit to known threshold on your power supply, like 15mA or 20mA for example, you can lower the output voltage to a value you know it's way below the forward voltage of a single led, and then you can connect the led in series with the multimeter.
Once that's done, you can slowly raise the voltage and watch your multimeter... the led will start to be somewhat bright and use a bit of current (between fully off and fully on) and at some point when the voltage is high enough, you'll see the led hitting that current limit you set.
smile:
--- Quote ---Your voltage of 10.4v may actually be too small and your leds may not be fully turned on.
--- End quote ---
Yes that seems to be the case.
--- Quote ---Why don't you actually determine the forward voltage of your leds first?
--- End quote ---
The DP832 is more precise then my meter I think.
I increased voltage to 10.98V and current is now 300mA, quite an increase from 10.4v at 35mA
However the PSU in my first post seems to be bad quality in that it's not silent it produces typical whine noise of low quality inductors make. Hmm, now if only it was silent then that would be OK. I measured 8% flicker compared to stock PSU at 123%. The capacitor LED PSU is about 20-40% flicker.
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