How Ohm’s law works, using your pedal’s status light

The rule itself. Voltage is the push, resistance holds the current back, and current is what actually flows. Ohm’s law ties them together: current equals voltage divided by resistance. Set the LED to No LED so the loop is just a supply and a resistor. At 9 V with 1 kΩ, 9 mA flows. Double the supply to 18 V and the current doubles to 18 mA. Double the resistor instead and the current drops by half. Watch the dots: they speed up and slow down with the current.

Why an LED changes the math. An LED takes a fixed bite of the voltage before it lets any current through, about 2 V for a red one and about 3.2 V for blue or white. Only what is left over sits across the resistor, and that leftover is what Ohm’s law works on. On 9 V with 1 kΩ, a red LED gets (9 − 2) ÷ 1000 = 7 mA. Swap in a blue LED and it drops to 5.8 mA, with nothing else changed. Below its drop an LED stays dark: a blue LED on 3 V never lights. Use the multimeter to see it. Across the resistor it reads 7 V, across the red LED 2 V, and across the whole supply 9 V. The pieces always add back up to the supply.

Picking the resistor for your pedal’s LED. Decide the current you want, subtract the LED’s drop from the supply, and divide. For a red LED at 10 mA on 9 V that is (9 − 2) ÷ 0.010 = 700 Ω, and the nearest standard value, 680 Ω, gives about 10.3 mA. Many pedal builders go much higher, like 4.7 kΩ, because a modern LED is plenty bright at a milliamp or two: 4.7 kΩ with a red LED on 9 V runs about 1.5 mA, and it saves battery life.

The limits. A typical LED is rated for 20 mA, and common datasheets list about 30 mA as the most it can take. Past 30 mA this page burns it out, and it stays burnt until you replace it, just like the real thing. Put a red LED straight across 9 V through a 10 Ω resistor and the math asks for 700 mA: it pops instantly. The resistor has a limit too. It turns current into heat, and the heat is current times current times resistance. A red LED on 9 V through 470 Ω runs 14.9 mA and puts 104 mW into the resistor, which is under half of a 1/4 W part’s rating, so it stays cool. Switch to a tiny 1/8 W resistor and that same 104 mW is 83% of its rating, so it runs warm. A common rule is to keep a resistor under half its rating. With no LED, 10 Ω across 9 V makes 8.1 W of heat, more than 30 times what a 1/4 W resistor can take.

What this page assumes. The supply is treated as perfect. A real 9 V battery sags when you pull a lot of current from it, so the huge currents in the burnout cases are more than a real battery could actually deliver, though the LED still dies. The LED is treated as a fixed drop, which is the math builders use. A real LED’s drop moves a little with current, so real currents land close to these numbers, not exactly on them. The dots move on a stretched scale so tiny currents still creep along and big ones race. The graph shows the current at every supply voltage with your resistor: the solid line is your circuit, and the dashed line is the same resistor with no LED. The gap between them is the LED’s drop. The build list rounds your resistor to the nearest standard E12 value and works out what that real part gives you.

I = V ÷ R  ·  with an LED: I = (V_supply − V_LED) ÷ R
resistor heat: P = I × I × R