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schematic

simulate this circuit – Schematic created using CircuitLab

I have a series circuit with LED1 (2.0 Vf), LED2 (2.0 Vf), a 50 Ω resistor (1 V) and a power supply of 5 V. How much of an overhead voltage should I give for this scenario, meaning should we leave some voltage of the total voltage untouched when try to see how many LEDs the power supply can manage? I also wanted to verify that when the forwarded voltages are added up they must also include the resistor voltage.

Some work:

  • Vs-VLED1-VLED2-VR = 0
  • -VR = -1 V
  • VR = 1 V
  • V = IR
  • V/I = R
  • 1 V/0.02 A
  • R = 50 Ω
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    \$\begingroup\$ It depends how much tolerance there is on the LEDs at the current you want to drive them. Also the resistor and power supply have some tolerance. So you must know all those tolerances and then you need to decide if you want to have more or less margin for number of LEDs in series. Do you have any numbers for the ranges you want to be in? \$\endgroup\$
    – Justme
    Commented Mar 11, 2023 at 20:58
  • \$\begingroup\$ Thanks for answering. I'll have to research the ranges. You've given me a good start. \$\endgroup\$
    – user121903
    Commented Mar 11, 2023 at 21:04
  • \$\begingroup\$ LEDs are not fussy about the current - you may find that they will produce plenty of light at much less than 20 mA. I once had to set the current to under 1 mA to get a LED sufficiently dim for my needs. The "advertised" 20 mA may be the Absolute Maximum rating for the LED - if so, you should use a lower current. \$\endgroup\$
    – Peter Bennett
    Commented Mar 11, 2023 at 21:22
  • \$\begingroup\$ For lighting applications it is pretty common to drop about 20% of voltage across the resistor. There is a tradeoff, with smaller voltage across the resistor wasting less power, but also making the circuit more sensitive to temperature and fluctuations in supply voltage. For an indicator light where you don't care about these things, it usually doesn't matter. \$\endgroup\$
    – user1850479
    Commented Mar 11, 2023 at 21:24
  • \$\begingroup\$ So I am overthinking this the indicator light: there is no established guideline about how much of the power supply you can and cant use, except for using too much going out of range, of course. \$\endgroup\$
    – user121903
    Commented Mar 11, 2023 at 21:37

1 Answer 1

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The important part in choosing the voltage for a LED or diode circuit is to make sure that:

  1. Voltage across the diodes are satisfied
  2. The recommended current is satisfied

For (1), the formula is simply

$$ V_{source} = V_{R1} + V_{LED1} + V_{LED2} $$ $$ V_{source} = V_{R1} + 2 + 2 $$

So based on this, the \$ V_{source} \$ should always be greater than 4V. The \$ V_{R1} \$ voltage is simply an offset, \$ (V_{source} - 4) \$, and this voltage doesn't matter as long as \$ V_{R1} > 0 \$ or \$ V_{source} > 4V\$ .

However, the LED current (2) is the second thing that matters, and this is where the \$ V_{source} \$ and \$ R_1 \$ will be restricted.

$$ I_{R1} = \frac{V_{R1}}{ R_1} = \frac{V_{source} - 4}{R_1 }<= 0.02 A $$

And based on this, if you apply a random Voltage above 4V, you must apply the corresponding R1 resistance so that the current will be 20mA. A higher voltage overhead means you also need a higher resistance to satisfy the 20mA current requirement and vice versa.

Take note though that applying an exact 4V as Voltage source means that the resistor voltage would be 0V and thus we cannot use any resistor to limit the LED current, so 4V is not included to the solution.

For a general solution, for any arbitrary number of LEDs, you just need to make sure that:

  1. \$ V_{source} \$ is larger than the sum of \$ V_{LED} \$ voltages. $$ V_{source} > \sum{V_{LED}} $$

  2. Calculate the \$ R_1 \$ resistance so that \$ I_{LED} \$ is satisfied. $$ R_1 = \frac{V_{source} - \sum{V_{LED}}}{I_{LED}} $$

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