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Electronics Engineering DISCUSSION
I have four white LEDs (about 3.2 V at 20 mA) running from 5 V. To save parts I connected them in parallel with a single 22 Ω resistor sized for the total of 80 mA. They light, although one looks slightly brighter than the others.
Everything I read says each LED needs its own resistor. If they are the same type from the same bag, why is sharing one resistor a problem, and what is the correct way to wire them?
An LED is not a resistor. Its current rises exponentially with voltage, so near the operating point a change of a tenth of a volt changes the current by a large fraction. At the same time the forward voltage at a given current varies from part to part; datasheets for white LEDs commonly give a range several hundred millivolts wide at 20 mA.
In parallel, all four are forced to the same voltage. The one with the lowest forward voltage takes the most current and the one with the highest takes the least. The resistor only fixes the total at about 80 mA; it has no say in how that total is divided. The brighter LED you noticed is the one taking more than its share.
Temperature makes the imbalance worse over time. The forward voltage of an LED falls by a few millivolts per degree as it warms. The LED carrying the most current runs hottest, its forward voltage drops further, and it takes an even larger share.
The failure case is the real danger. If one LED fails open circuit, the resistor still sets about 80 mA, now divided among three: roughly 27 mA each, above the 20 mA rating. The next weakest fails sooner, and the rest follow at increasing current.
The standard fix is one resistor per LED: R = (5 V - 3.2 V) / 20 mA = 90 Ω, so use 100 Ω for about 18 mA. Now each resistor absorbs its own LED's variation. If the forward voltages range from 3.0 V to 3.4 V, the currents range from 20 mA to 16 mA, a difference the eye barely notices, and a failed LED has no effect on the others.
Each resistor dissipates 1.8 V × 18 mA, about 32 mW, so any small resistor size is adequate.
If you have a higher supply available, a series string is better still. Three LEDs in series from 12 V need 9.6 V, and one resistor of (12 - 9.6) / 0.02 = 120 Ω sets the current. Series connection guarantees identical current in every LED, and 80 percent of the supply power goes into the LEDs, against 64 percent with 3.2 V LEDs on 5 V.
For more LEDs, use several strings with a resistor per string. For power LEDs of 350 mA and above, replace the resistor with a constant-current driver, since a resistor wastes too much power and regulates poorly as the supply varies.
To be fair to what you observed: parallel LEDs on a shared resistor do often work for a while, especially parts from one reel run well below their rating, and commercial LED modules do parallel dice that have been sorted into tight voltage bins and share one heat spreader. Neither condition holds for loose parts on a breadboard.
If you want to see the imbalance, measure it: put a 1 Ω resistor in series with each LED in your parallel circuit and read the millivolts across each. One millivolt is one milliamp.
