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Can someone help me work out total amp draw for LEDs with an in-series resistor?

I am getting conflicting answers depending on where I look.

  1. One way tells me the amp draw is the total amp draw of all LEDs combined, same as supply voltage.
  2. Another way seems to indicate that the total amp draw is singular, as in 3 LEDs with an in-series resistor (each LED being 20mA) will still only be 20mA.

EG:

12v + ---->|---->|---->|---///---- -

12v DC
LED = Vf3.8/If20mA
R = 33ohms

Would the amp draw be 20mA or 60mA?

Further on from that, for multiple chains like that running from the same source, EG, 20 chains in parallel of x3 LEDs (each with in-series resistor), 400mA or 1.2A?

Hopefully I have explained my question right, but probably got some vernacular wrong, so I apologise in advance.

Thanks

PS: Apologies for the Resistor not showing correctly, having trouble getting the ASCII to display correctly

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    \$\begingroup\$ Please use the circuit editor to provide a schematic instead of ASCII-art. \$\endgroup\$ – Arsenal Nov 23 '17 at 12:04
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    \$\begingroup\$ If LEDs are connected in series, the same current flows through all LEDs. Therefore 3 LEDs will need the same 20 mA. But the voltage drop is 3.8 V per LED and 11.4 V for all three. The voltage drop at the resistor will be 12 - 11.4 = 0.6 V, with 20 mA the resistor should be 0.6/0.02 = 30 Ohm. Taking the next larger one is OK. If 20 of those chains ( 3 LEDs in series with 33 Ohm) are connected in parallel, they need 20 * 20 mA = 400 mA \$\endgroup\$ – Uwe Nov 23 '17 at 14:56
  • \$\begingroup\$ Thank you. I was being advised from another source that it would be the total LEDs instead, which was giving me something like 4 or 5amps. 1.7A seems more reasonable, thank you. \$\endgroup\$ – Dominic James Sibthorp Nov 23 '17 at 15:12
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Lets look at your setup with voltage vectors..

schematic

simulate this circuit – Schematic created using CircuitLab

Since, NOMINALLY, you have three 3.8V forward voltage drops in your chain, that leaves only 0.6V remaining from the original 12V across your resistor. The current in the chain is therefore 18.182mA. That current is common to the entire chain.

If you have N chains the same in parallel the total current will be N times 18.182mA.

HOWEVER:

This simple math does not take into account any variance in the forward voltage of the LEDs or any variance in the supply voltage. Further, the forward voltages will change with temperature.

The forward voltage of LEDS can vary significantly. If the forward voltage is one single point less at 3.7V the current will be 27mA, If it drops to 3.5V the current will be 45mA... and on and on.

That variance can mean your actual LEDs may burn out, or have a short life with this setup.

LED strings are best driven with a constant current source, not a voltage source. If that is beyond your capabilities I suggest you reduce your strings to two LEDs per string and use a larger resistor, 220R, to soak up the variance in the LEDs.

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  • \$\begingroup\$ OK, so if I have understood you correctly, the current is the voltage left over / the value of the resistor. So that being the case the typ/max values of my LED (20mA Typ/30mA Max) are just the nominal/max values and not, as I thought, what the LED actually draws ? RE using a constant current source, I've thought about that, but wouldn't know where to start. All I know is I will be using a 12v DC (Amps at whatever the load ends up requiring). I assume there is a way, once I know exactly what amps I require, of delivering exactly that amount all the time. Thanks. \$\endgroup\$ – Dominic James Sibthorp Nov 23 '17 at 14:02
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    \$\begingroup\$ @DominicJamesSibthorp with LEDs, since they are current driven beasts, we PICK, or use a specified current, and select the resistor that will setup that current at the nominal forward voltage. We size everything so with tolerances and temperature effects the LED stays within it's recommended operating conditions. Having the spec sheet of the LEDs helps A LOT. Using a constant current driver is of course better because of all that. \$\endgroup\$ – Trevor_G Nov 23 '17 at 14:07
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    \$\begingroup\$ @DominicJamesSibthorp then go with two in a string and 220R and you should be fine. \$\endgroup\$ – Trevor_G Nov 23 '17 at 14:22
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    \$\begingroup\$ @DominicJamesSibthorp a cheap old 12V 2A wall-wart would be simpler and safer. \$\endgroup\$ – Trevor_G Nov 23 '17 at 14:40
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    \$\begingroup\$ Which is what I have already got, thats great thank you. As long as my working out for load is correct I can breath easier now. Thanks again. \$\endgroup\$ – Dominic James Sibthorp Nov 23 '17 at 14:42
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The answer SHOULD be obvious to you.
You need to "worry it to death" until it becomes obvious.

Multiple LEDs plus a resistor, all in series, will have non linear behaviour as applied voltage changes BUT as there is CLEARLY only one current path the current in each element MUST be the same.

A series string of a resistor plus multiple LEDS is effectively a resistance at a given applied voltage.
Just as the current drawn by N parallel resistors all connected to a common source voltage is the sum of the individual resistors, so too, connecting multiple LED strings across a common fixed voltage results in the current drawn being the sum of the currents in the individual strings.

Draw simple versions of the above and convince yourself that the answer is as stated above.

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