Work out the series resistor an LED needs on a given supply, rounded up to a value you can actually buy, with the power it has to dissipate.
Exact resistor
150 Ω
Nearest E12 above it
150 Ω
The resistor dissipates
0.06 W
R = (supply − forward voltage) ÷ current
An LED is not a resistor: its voltage barely moves while the current climbs, so connecting one straight across a supply either does nothing or destroys it. The series resistor exists to take the difference between the supply and the diode's forward voltage, and its value follows from Ohm's law applied to that difference — not to the whole supply, which is the mistake that halves the brightness or doubles it.
R = (supply − forward voltage) ÷ current
Only the leftover voltage appears across the resistor, so only that leftover goes into the division. The answer is then rounded up to the next value in the E12 series, because rounding down means more current than you asked for, and the last figure is the power the resistor turns into heat.
Source: NIST SP 811 (2008), Appendix B.8 — ohm (Ω), the SI unit of resistance: 1 Ω = 1 V/A
A number that disagrees with its source is a defect, not a rounding preference.
What did you enter, what did the tool show, and what did you expect instead? If you have a source that disagrees with ours, a link to it is the most useful thing you can send.
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