The Verification Venue · pointed at a thing everyone gets wrong
The Resistor That Saves the Light
It's the first circuit anyone builds, and the place the most parts die: an LED, a battery, and — if you're lucky — a resistor. Leave the resistor out and the LED flares and dies. Why?
“An LED has no resistance — zero ohms — so by Ohm's law it draws infinite current and burns out.” This is repeated everywhere, and it is false. A diode is not a zero-resistance wire. It is a device whose current climbs exponentially with voltage. That single fact — not a missing resistance — is the whole story, and you can watch it below.
A real diode obeys the Shockley equation, I = Is(eV/(n·Vt) − 1): current rises not in proportion to voltage but as its exponential. At room temperature the curve is so steep that about +60 mV multiplies the current by ten. Voltage is the wrong knob — nudge it a hair and the current explodes. So you don't set the voltage across an LED; you set the current through it, and the resistor is what does that. The operating point is where the resistor's straight load line crosses the LED's exponential curve. Drag the resistor and watch the crossing move.
Current through the LED
20.0 mA
rating: 20 mA · safe
Voltage across the LED
2.00 V
Slide R toward zero and the load line tips up toward vertical; the operating dot rockets up the exponential and the current readout turns red. Slide it back to 150 Ω — the value the load line picks out for a safe 20 mA.
The load line's x-intercept is the supply voltage; raising it slides the whole line right and steepens the demand on the resistor.
Notice what the curve does near the top of the axis: it stands almost straight up. Past the forward voltage, an enormous range of current is packed into a sliver of voltage. That is why “just connect it to the right voltage” is hopeless — the voltage you'd need is known to a millivolt, drifts with temperature, and varies part to part. The resistor sidesteps all of it: it converts your sloppy supply voltage into a current you actually control. Every value on screen is solved live; here is the arithmetic.
The check — every number recomputed in front of you
The operating point is the solution of the circuit's loop equation V = I·R + VLED with I = Is(eVLED/(n·Vt) − 1). That's transcendental — no algebra isolates I. The page solves it two independent ways every frame and shows they agree:
The textbook design check, and the catastrophe as R shrinks — each row is the live Lambert-W solution at that R:
| R (Ω) | I (mA) | V_LED (V) | vs 20 mA | verdict |
|---|
And the constants the whole thing rests on, recomputed from CODATA:
Run it yourself: node research/why-led-needs-a-resistor/verify-why-led-needs-a-resistor.mjs — it recomputes Vt from k and q, the 150 Ω design from Ohm's law, and checks the Lambert-W operating current against an independent bisection solve to under 1e-9 at every R.
What's exactly true here, what's a free choice, and where the model breaks
Exactly true. Diode current follows the Shockley law I = Is(eV/(nVt) − 1) in the normal forward region; the thermal voltage Vt = kT/q ≈ 25.85 mV at 300 K is computed from the exact 2019-SI values of Boltzmann's constant and the elementary charge. The operating point really is the intersection of the load line with the curve, and it has a closed form via the Lambert W function (Banwell & Jayakumar 2000); the page solves it that way and cross-checks against Newton/bisection. The decade-per-~60 mV steepness is a direct consequence of Vt: one factor of ten in current per n·Vt·ln10 ≈ 59.5 mV (n=1); a round 60 mV is actually ≈×10.19. R = (V−Vf)/I = (5−2)/0.02 = 150 Ω is just Ohm's law on the resistor's share of the voltage.
Free choices (named, because they aren't universal). A real LED's parameters vary by part, colour and temperature. We model a representative red LED: forward drop Vf = 2.0 V at the rated 20 mA (datasheet specs are always quoted at a test current — there is no single “the” forward voltage), and ideality factor n = 1.8 (LEDs run high, ~1–2). The saturation current Is is then pinned so the curve passes through that rated point — it is not a tabulated constant. Change the part and the numbers shift; the shape of the argument does not.
Where the model breaks — and it breaks exactly in the destructive regime. The ideal Shockley curve has no series resistance and no high-injection roll-off, so at very small R it overstates the current wildly (an ideal source through R≈0 gives a physically absurd current). A real battery and wires have internal resistance, and the LED has its own bulk series resistance, so the true “no-resistor” current is large-but-finite — enough to destroy the part, which is the honest point, but not the millions of milliamps the bare equation prints. The verifier shows this: adding ~0.5 Ω of parasitic resistance drops the ideal R≈0 current from millions of mA to a few thousand. What is robust is the conclusion: the current is brutally sensitive to R, blows past the rating as R falls, and the misconception (“zero resistance”) has the mechanism backwards.