68 Ω resistor color code

Where you'll meet it

An LED on two AA batteries

Two AA cells in a holder give 3 V, enough for a red LED with a little to spare. The 68 Ω resistor sits between the battery's plus and the LED, takes up that spare voltage and sets the current.

I = (3 V − 2.0 V) / 68 Ω ≈ 14.7 mA

That is bright, and under the 20 mA that most standard LEDs are rated for. With 47 Ω instead it would be (3 V − 2.0 V) / 47 Ω ≈ 21.3 mA, a little over, so 68 Ω is the safer pick.

A blue or white LED drops about 3.0 V, as much as the two cells give, so the formula leaves (3 V − 3.0 V) / 68 Ω = 0 mA. It may still light on fresh cells, because an LED drops less at small currents, but it fades fast as they run down.

The resistor turns (14.7 mA)² × 68 Ω ≈ 0.015 W into heat, so any ¼ W (0.25 W) part stays cool.

2.0 V and 20 mA are typical figures for a standard 5 mm red LED. Check your LED's datasheet. Fresh AA cells read a little over 1.5 V each and sag as they empty, and with only 1 V left across the resistor the current follows them closely. The LED lights only one way round: its longer leg goes toward the battery's plus. Wired backwards it just stays dark, because 3 V is less than the 5 V most LEDs are rated to block.

On SMD parts

Look for 680 or 68R0

• 680680 = 68 × 10⁰ = 68 Ω (3-digit code).
• 68R068R0 = 68 Ω (4-digit code, usually ±1% parts; the R sits where the decimal point would be, so it reads 68.0).

There is no EIA-96 code, because 68 Ω is not an E96 value (the nearest E96 values are 66.5 Ω and 68.1 Ω).

Check it with a meter

Read 64.6–71.4 Ω

1. Take the resistor out of the circuit.
2. Set the meter to Ω, range 200 Ω (or auto).
3. Touch one probe to each lead. Keep your fingers off the metal: your body is a resistor too.
4. Anything from 64.6 to 71.4 Ω is a good ±5% part. A 5-band ±1% part should read 67.32 to 68.68 Ω.

A meter has a small error of its own, so a reading a hair outside the range does not prove the part is bad. Measure again with firm contact before you swap it.