How this calculation works
Subtract the assumed total forward drop for one series string. If positive headroom exists, compare the ideal target resistance with the separately entered candidate and calculate current and dissipations without choosing ratings.
Positive headroom = Vs − count × Vf; Rtarget = headroom / target amperes; PR = I²RA worked example
5 V minus a 2 V assumed LED drop leaves 3 V. A 20 mA target gives an ideal 150 Ω and 0.06 W resistor dissipation. The separately entered 180 Ω candidate gives about 16.666667 mA and 0.05 W under the same fixed-drop assumption.
Good to know
One ideal fixed-drop series string. No color lookup, nonlinear I–V/temperature model, standard-value selection, parallel sharing, current-limit approval or power-rating/derating choice. Absent positive headroom does not determine the real diode current.
Reference: Reference: model definitions and scope (opens in a new tab) ↗
The resistor receives the supply minus the entire string drop
With a 5 V supply and one assumed 2 V forward drop, the resistor has 3 V across it. The entered 20 mA target is 0.02 A, so the ideal resistance is 3/0.02 = 150 Ω. The resistor’s modeled dissipation is 0.02² × 150 = 0.06 W.
For two identical assumed 2 V LEDs in one series string, the total modeled drop is 4 V, leaving 1 V. The same target current would give 50 Ω. This page handles one series string and does not use a single resistor model to infer sharing among parallel LED branches.
Ohm’s law applies to the resistor after the assumed LED-string voltage has been subtracted. The complete supply voltage is not automatically the resistor voltage.
The entered candidate is a separate operating scenario
The default candidate is 180 Ω, separate from the exact 150 Ω target result. At a fixed 3 V resistor drop, it gives about 16.666667 mA and 0.05 W resistor dissipation. Those calculated powers are not selected component ratings or approvals of a current limit.
Forward voltage is entered, not looked up from color. A real LED’s drop depends on current and conditions, so a fixed-drop calculation is only one scenario. Tolerance, temperature, supply variation and data-sheet limits are not inferred by rounding a resistor value.
No positive headroom means the target calculation no longer applies
At or below the summed assumed forward drop, there is no positive headroom for a positive resistor to establish the target current in this simplified model. The result says so instead of returning a negative resistance or using a zero-ohm value to imply a determined current.
The Ohm’s law calculator can check the resistor’s voltage/current relationship once the appropriate resistor voltage is known. The battery nominal energy and runtime calculator can explore a separate supplied-load energy budget. Neither supplies the LED’s missing nonlinear I–V curve, a current-regulated driver model or a thermal derating choice.
A few questions, answered
Does the example forward voltage apply to every LED?
No. It is an editable fixed-drop assumption, not a color lookup or data-sheet guarantee.
Does the tool choose a standard resistor value?
No. It reports the exact modeled target resistance and evaluates the candidate value you separately enter.
Is computed resistor power its required marked rating?
No. Ratings and thermal derating are not selected by this arithmetic model.
Does insufficient headroom prove actual diode current is zero?
No. It only shows the positive-resistor target calculation is not available under the entered fixed-drop assumptions.