Ohm's Law Calculator

Ohm’s law is the equation everything else in electronics is built on. Rather than solving it on paper, drive a real circuit: change the supply, the series resistor or the LED itself, and watch the current, the heat and the brightness respond.

Push it too far and the LED blows — which is the point.

The equations

$V = I \times R$

$I = \frac{V}{R}$

$R = \frac{V}{I}$

$P = V \times I = I^2 \times R = \frac{V^2}{R}$

V — voltage, in volts
I — current, in amperes
R — resistance, in ohms
P — power, in watts

For an LED in series with a resistor, only what is left over after the LED’s forward voltage lands on the resistor:

$I = \frac{V_{supply} - V_f}{R}$

Forward voltage is the whole story

Every LED needs a minimum voltage across it before it conducts at all. Below that it is simply off — no dimming, no partial glow.

LED	Typical Vf	Headroom on 3.3 V
Red	1.8 V	1.5 V — comfortable
Yellow	2.1 V	1.2 V — fine
Green	2.2 V	1.1 V — fine
Blue / white	3.2 V	0.1 V — effectively nothing

This is why a blue LED on a 3.3 V tinyCore GPIO looks broken. It is not: there is no headroom left for the resistor to work with. Drive blue and white LEDs from 5 V, or accept that they will be dim. Switch the LED colour in the tool with 3.3 V selected and you can watch it happen.

Picking a resistor

$R = \frac{V_{supply} - V_f}{I_{target}}$

A red LED at 3.3 V, aimed at a comfortable 18 mA:

$R = \frac{3.3 - 1.8}{0.018} = 83\ \Omega \rightarrow \text{use } 100\ \Omega$

Always round up to the next standard value. Rounding down pushes current the wrong way. The tool suggests the E24 value that lands you at roughly 60% of the LED’s rating — bright, and with room for supply variation.

Worked examples

I²C pull-ups

I²C lines need pull-ups. At 3.3 V in fast mode (400 kHz), 4.7 kΩ is the usual pick:

$I_{pullup} = \frac{3.3}{4700} \approx 0.7\ mA$

Well inside GPIO drive capability, and it keeps bus capacitance manageable.

Voltage divider

To read a 5 V signal on a 3.3 V ADC:

$V_{out} = V_{in} \times \frac{R_2}{R_1 + R_2}$

With R1 = 10 kΩ and R2 = 22 kΩ: $V_{out} = 5 \times \frac{22}{32} \approx 3.4\ V$ — just at the edge, so leave yourself more margin than that in a real design.

Don’t cook the resistor

A standard through-hole resistor is rated at 1/4 W (250 mW). Check the dissipation whenever current gets serious:

$P = I^2 \times R$

100 mA through 10 Ω:

$P = (0.1)^2 \times 10 = 0.1\ W$

That is 100 mW — inside spec, but the part will be warm to the touch. The tool warns you as soon as the resistor crosses 250 mW.