Electronics > Beginners

Hooking up 4 x 3v LED string lights to a 6v PSU

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gf:

--- Quote from: bob21 on February 05, 2020, 11:45:03 am ---C's total draw is 82mA
6V - 3V = 0.082 x R ==> R = 3V / 0.082 = 0.246 Ohm

--- End quote ---

I think you still didn't get it. If you put two components in series, then the currents don't add up.

41mA flow from the PSU into the "+" connector of line 1, 41 mA come out of line 1's "-" connector and flow into line 2's "+" connector, and again 41 mA come out from line 2's "-" connector and flow back to the PSU.

So C's "total draw" is still 41mA.

But the voltage drops of the components add up. If one string has a voltage drop of 2.86V, then two strings in seris will have a voltage drop of 2 * 2.86V = 5.72V.

This leads to the following calculation for your sub-circuit C:

Vpsu = 6 V (measured)
Vled = 2.86 V (measured)
Iled = 0.041 A (measured)
Vresistor = Vpsu - Vled - Vled = 0.28V
R = Vresistor / Iled = 0.28 / 0.041 = 6.8293 Ohm ==> 6.8 Ohm
Presistor = Vresistor * Iled = 0.011480 W

The problematic thing here is that after subtracting 2 * Vled from Vpsu there are only 0.28V are left for the resistor (which leads to the low resistor value of only 6.8 Ohm).
Under these conditions the resistor is not supposed to ensure a very stable working point for the circuit.

Note, however, the above calculation applies only if the PSU output is 6V clean DC!
For "pulsating DC" (i.e. full-wave rectified sine-wave) we will get different numbers!
In this case 6.8 Ohm will be ways too low and would lead to a significantly higher (average) current than 41mA!
I don't know details about your PSU, but I have the feeling that it does not deliver clean DC.

bob21:

--- Quote from: gf on February 05, 2020, 06:09:17 pm ---I think you still didn't get it. If you put two components in series, then the currents don't add up.
--- End quote ---

You are correct, I didn't get it, but I do now --> Current does not add up when in series. Thanks for being patient with me.  :-+


--- Quote from: gf on February 05, 2020, 06:09:17 pm ---But the voltage drops of the components add up. If one string has a voltage drop of 2.86V, then two strings in seris will have a voltage drop of 2 * 2.86V = 5.72V.
--- End quote ---

Just to clarify, this is a drop to 2.86V from 2.93V so a loss/drop of 0.07V. Or have I somehow measured this incorrectly? If I have measured this incorrectly, and I need a PSU with more voltage and current, I have a whole box of them here. I am looking at one here now that is 12V 1.5. Would that be more suitable?


--- Quote from: gf on February 05, 2020, 06:09:17 pm ---Note, however, the above calculation applies only if the PSU output is 6V clean DC!
For "pulsating DC" (i.e. full-wave rectified sine-wave) we will get different numbers!
In this case 6.8 Ohm will be ways too low and would lead to a significantly higher (average) current than 41mA!
I don't know details about your PSU, but I have the feeling that it does not deliver clean DC.
--- End quote ---

How do I test if it provides clean DC? You said the LEDs would flicker if unclean, but they do not......

gf:

--- Quote from: gf on February 05, 2020, 06:09:17 pm ---Note, however, the above calculation applies only if the PSU output is 6V clean DC!
For "pulsating DC" (i.e. full-wave rectified sine-wave) we will get different numbers!
In this case 6.8 Ohm will be ways too low and would lead to a significantly higher (average) current than 41mA!
I don't know details about your PSU, but I have the feeling that it does not deliver clean DC.

--- End quote ---

Now the same calculation for unfiltered, pulsating DC with an average voltage of 6V:
(The calculation below is Matlab/Octave code. I'm doing the integration in a very simple way numerically.)


--- Code: ---t = [0:99] * 0.01 * pi;       % 100 steps 0 <= x < pi for numerical integration
Vpsuavg =  6                  % measured average voltage
Vpsupeak = Vpsuavg / 2 * pi   % peak voltage of the rectified sine wave
Vpsu = sin(t) * Vpsupeak;     % PSU voltage as function of time
Vled = 2.86
Iledavg = 0.041
Vresistor = Vpsu - Vled - Vled;
Vresistor(Vresistor<0) = 0;                     % set to 0 in the off-time where Vpsu < 2 * Vled
Vresistoravg = mean(Vresistor)                  % avarage voltage across resistor
Vresistorrms = sqrt(mean(Vresistor.*Vresistor)) % RMS voltage across resistor
R = Vresistoravg / Iledavg
Presistor = Vresistorrms * Vresistorrms / R

--- End code ---

This gives the following results:

--- Code: ---Vresistoravg =  1.4235 V
Vresistorrms =  2.0477 V
R =  34.720 Ohm
Presistor =  0.12077 Watt

--- End code ---

So we would now require 33 Ohm (while we had only 6.8 with clean DC).

bob21:
Ok, but see my post above. Did you think the voltage drop was by 2.86V or to 2.86V? Because it was a drop to the values listed, so:

Line 1: loss = 0.07
Line 2: loss = 0.08
Line 3: loss = 0.06
Line 4: loss = 0.07

I appreciate your time explaining the differences in ohms when the DC supply is dirty, but all of this is going way over my head now. If the PSU is not suitable, I would rather:

1. Test my existing PSU to see if it really is / is not clean - you specifically mentioned a few posts back that unfiltered DC would cause flickering - and I replied that my PSU does not cause even slight flickering
2. Try another PSU (I have tons here), most are from gadgets like phones, power banks, speakers etc.. surely they can't all be dirty?
3. Buy a PSU that is clean - is this an example? https://tinyurl.com/uutxd9v

I just want the simplest easiest way to power these 4 strings off a wall outlet that won't catch fire to my house.

If I need to buy something else I will. I've just bought a new DMM and a ton of resistors. What else do I need?

I am an electronics n00b. I knew nothing about any of this 4 days ago and my brain hurts. I am trying my best to take all this in as quickly as possible. I've even tried calculating values myself and not just expecting other people to do it all for me. Yes, some have been wrong, but I am posting in the beginners section, after all.

gf:

--- Quote ---I appreciate your time explaining the differences in ohms when the DC supply is dirty, but all of this is going way over my head now.
[...]
I am an electronics n00b. I knew nothing about any of this 4 days ago and my brain hurts. I am trying my best to take all this in as quickly as possible. I've even tried calculating values myself and not just expecting other people to do it all for me. Yes, some have been wrong, but I am posting in the beginners section, after all.

--- End quote ---

OK, I see, you primary aim are obviously quick results.
Here's one of several possible approaches:

* Don't connect strings in series, but 4 all strings in parallel to the PSU.
* Use a separate current limiting resistor in series with each string.Determine resistor values experimentally, for one string at a time:

* For string #1 start with 82 Ohm (>= 1/4 W), connect to the PSU and measure the current
* If it is lower than 41 mA reduce to 75 Ohm and measure the current again
* Use the same resistor value for string #4
* For string #2 start with 100 Ohm (>= 1/4 W) and measure the current
* If it is lower than 36 mA reduce to 91 Ohm and measure again
* If it is still lower than 36 mA reduce to 82 Ohm and measure again
* Use the same resistor value for string #3
EDIT:
Note, if you renounce the desire to connect two stings each in series, there is significantly more voltage headroom for the resistor to establish a stable working point. Additionally, clean vs. "dirty" DC does no longer make such a huge difference, so that resistor values in the range 82...75 Ohm should fit with the 41mA strings, and 100...82 Ohm should fit with the 36mA strings. When you determine the value experimentally, then start with larger resistor values, measure the current, and decrease R until you hit the string's current target (+/- 5% or even +/- 10% is perfectly OK).

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