Electronics > Beginners
Hooking up 4 x 3v LED string lights to a 6v PSU
gf:
--- Quote from: bob21 on February 04, 2020, 11:38:06 pm ---How would be best to test this? Probe at start and end of string?
--- End quote ---
Measuring both ends is a good ieda, since there might be a voltage drop along the string
(at only 40mA I would not expect too much, though).
--- Quote ---Each string only has 2 wires. If you removed 1 LED, the entire string would fail.
--- End quote ---
Only if you destroy the cable during removal. The LEDs of a string are not connected in series, but in parallel, therefore the removal of a LED is not supposed to break the circuit. Only the current consumption would decrease.
Note, if the LEDS of a string were connected in series, then you would need a voltage of about 3V*10=30V in order to drive 10 LEDs - too much for 2x AA cells (unless a switching regulartor would be used to boost the voltage).
--- Quote ---I’m not sure what you mean by daisy chained.
--- End quote ---
With daisy chained I mean concatenated behind each other, i.e. the second string connected to the end of the first one (+ to +, and - to -), which is electrically a parallel connection.
--- Quote ---If you mean line 1 pos to psu pos, then line 1 neg to line 2 pos, then line 2 neg to psu neg, then yes.
This is how they are connected. Same for 3 and 4.
--- End quote ---
OK, you really mean an electrical series connection.
Again, this would only be possible for strings 1+4, which draw the same amount of current, and for strings 2+3 which also draw the same amount of current. If you connect two strings in series then the same amount of current flows through both ones.
If you would connect strings 1+2 in series, then you'll either need to drive string 1 with a lower current than 41mA or strings 2 with a higher current than 36mA, since there is "only one current" flowing through both strings then. The whole amount current which leaves string 1 enters string 2, since there exist no path where the current could escape. At each connection point in the circuit, the sum of incoming and outgoing currents is the same (Kirchhoff's law).
gf:
--- Quote from: gf on February 02, 2020, 08:14:25 pm ---[...]
Is it a SMPS or a traditional PSU?
I.e. does it provide clean DC (filtered, or even stabilized), or just pulsating DC directly from the rectifier?
[...]
--- End quote ---
If the PSU provides just pulsating DC, then one consequence to be expexted will be flickering of the LEDs.
bob21:
--- Quote from: gf on February 05, 2020, 12:46:35 am ---If the PSU provides just pulsating DC, then one consequence to be expexted will be flickering of the LEDs.
--- End quote ---
Well in that case I would have to guess that the DC does not pulsate. The LEDs definitely do not flash when hooked up to this. They are solid on.
--- Quote from: gf on February 05, 2020, 12:25:37 am ---Measuring both ends is a good ieda, since there might be a voltage drop along the string
(at only 40mA I would not expect too much, though).
--- End quote ---
This might be tricky as I’ve have to start stripping other ends of cables which I ideally wanted to avoid, but what I can do this morning is take a voltage reading before and after each string is turned on (when connected to its cell pack) and provide the voltage drop figures.
--- Quote from: gf on February 05, 2020, 12:25:37 am ---Only if you destroy the cable during removal. The LEDs of a string are not connected in series, but in parallel, therefore the removal of a LED is not supposed to break the circuit. Only the current consumption would decrease.
Note, if the LEDS of a string were connected in series, then you would need a voltage of about 3V*10=30V in order to drive 10 LEDs - too much for 2x AA cells (unless a switching regulartor would be used to boost the voltage).
--- End quote ---
Ah, I understand now. Thanks. :-+
--- Quote from: gf on February 05, 2020, 12:25:37 am ---With daisy chained I mean concatenated behind each other, i.e. the second string connected to the end of the first one (+ to +, and - to -), which is electrically a parallel connection.
OK, you really mean an electrical series connection.
--- End quote ---
I have both, 1 and 2 are electrically in series, as are 3 and 4. Then these are in parallel to the PSU to share the 6V.
--- Quote from: gf on February 05, 2020, 12:25:37 am ---Again, this would only be possible for strings 1+4, which draw the same amount of current, and for strings 2+3 which also draw the same amount of current. If you connect two strings in series then the same amount of current flows through both ones.
If you would connect strings 1+2 in series, then you'll either need to drive string 1 with a lower current than 41mA or strings 2 with a higher current than 36mA, since there is "only one current" flowing through both strings then. The whole amount current which leaves string 1 enters string 2, since there exist no path where the current could escape. At each connection point in the circuit, the sum of incoming and outgoing currents is the same (Kirchhoff's law).
--- End quote ---
Got it. So I need to do some reorganising: 1+4 in series, 2+3 in series, both connected in parallel to PSU.
Just out at the moment, will provide voltage drop readings for each unit in a bit.
Edit: Changed title to reflect current aims.
bob21:
More info:
With a 2 x AA cell pack start reading of 2.93V, I hooked up each string of lights and checked the voltage again from the point where the pack was reconnected to the wires.
Line 1: 2.86V
Any visible resistors in LED heat shrink? Difficult to tell, as opaque, but my guess is no, it's too thin. Ie. it only looks as wide and long as the LED legs
Line 2: 2.85V
Any visible resistors in LED heat shrink? Difficult to tell, as opaque, but my guess is no, it's too thin. Ie. it only looks as wide and long as the LED legs
Line 3: 2.87V
Any visible resistors in LED heat shrink? Definitely not, the heat shrink is transparent, and it's just the LED legs and wire.
Line 4: 2.86V
Any visible resistors in LED heat shrink? Definitely not, the heat shrink is transparent, and it's just the LED legs and wire.
The size of each LED heat shrink part is the same size in 1 and 2 as in 3 and 4. I know resistors can be tiny, but I honestly don't believe there's any in there. The heat shrink would be bigger or show the outline of an additional component, it's tight against the LED legs and wire.
bob21:
Having worked out some more numbers (again, please do not shoot me, these might all be junk!), we are now using:
Line 1 + 4 = Lets call this Circuit C
Line 2 + 3 = Lets call this Circuit D
Updated Diagram:
Pink and Purple dots are resistor locations? Is this correct?
C's total draw is 82mA
6V - 3V = 0.082 x R ==> R = 3V / 0.082 = 0.246 Ohm
We only have 22 or 27 Ohm resistors, so can we use a 22 for this?
D's total draw is 73mA
6V - 3V = 0.073 X R ==> 3V / 0.073 = 41.09 Ohm
We only have R39 and R47
What is the formula to adjust a resistor value and work out what the resultant current will be?
Is it:
Circuit C: R = 3V x C ==> 3V / R22 = C ==> 0.136mA so if using a R22 resistor on C, it will get 136mA not 82mA.. problem.
Circuit D: R = 3V x C ==> 3V / R39 = C ==> 0.076mA so if using a R39 resistor on D, it will get 76mA not 73mA... no big deal?
So D is pretty ok at this point - how to go about C? I thought about using 2 x R47's for 1/2 R so 23.5R:
R = 3V x C ==> 3V / R23.5 = C ==> 0.127mA so if using a 2 x R47 resistor on C, it will get 127mA not 82mA.. still a problem.
Am I over thinking this, could use some wisdom :)
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