LEDs & Light: Forward Voltage, Current & Arrays
Seven hands-on lessons that take you from what an LED actually is, through forward voltage and polarity, to calculating the right current-limiting resistor and building series and parallel arrays — with live interactive circuits, worked examples and troubleshooting. No prior electronics assumed.
Lesson 1 · Introduction to LEDs and How They Produce Light
Level: Beginner
- Explain what an LED is and how it's different from a regular diode.
- Describe, in plain terms, how a LED turns electricity into light.
- Preview why LEDs are controlled by current, not voltage.
An LED (light-emitting diode) is a tiny semiconductor component that emits light when electricity flows through it in the correct direction. You meet them everywhere: the indicator lights on a TV remote, the glow of a phone's charging LED, the strings of lights on a tree, and the bright panels of modern displays and car headlights. Electrically, an LED is a special kind of diode — a component that only conducts current one way. What makes an LED a diode with a superpower is that, when it conducts, it doesn't just let current pass, it gives off visible light as a side effect.
Here's the physics in everyday language. Inside the LED there's a junction where two types of semiconductor material meet. When you push current through in the right direction, electrons cross that junction and "fall" into lower-energy spots. The energy they lose in that drop is released as a packet of light called a photon. The colour of the light is set by how much energy each photon carries, which depends on the semiconductor material used. A smaller energy drop gives a redder, lower-energy photon; a bigger drop gives a bluer, higher-energy one. That's why a red LED and a blue LED are built from different materials — and, as you'll see next, why they need different voltages to light.
Three ideas from this lesson matter for everything that follows. First, LEDs are polarized: they only work one way round (anode to cathode). Second, an LED has a roughly fixed forward voltage it "uses up" once lit, and that value depends on its colour. Third — and this is the big one for design — an LED is a current device. You set how bright it is by choosing how much current flows, not by choosing a voltage. Left unchecked, an LED happily draws far too much current and destroys itself, which is exactly why every practical LED circuit includes a current-limiting resistor. That's the thread that runs through the whole module.
An LED produces light because…
LED light is electroluminescence: electrons give up energy as photons at the junction. The photon energy (and colour) is set by the material.
Lesson 2 · LED Forward Voltage and Polarity
Level: Beginner
- Identify an LED's anode (+) and cathode (−).
- Explain forward voltage (Vf) and why it depends on colour.
- Connect Vf to the idea of "leftover" voltage a resistor must drop.
Every LED has two leads and only one correct way to connect them. The anode is the positive side (current flows in here) and the cathode is the negative side (current flows out). On a typical through-hole LED you can tell them apart two ways: the shorter lead is the cathode, and the flat spot on the round case marks the cathode. If you connect the LED the wrong way round (reverse-biased), it simply won't light — and holding a large reverse voltage across it for long can damage it. Getting polarity right is the first thing to check whenever an LED refuses to work.
The second key number is the forward voltage, written Vf. Once an LED is lit, it "uses up" a fairly fixed amount of voltage across itself, and the rest of your supply voltage is left over. Think of the LED as a component with a price tag in volts: a red LED costs about 2 V, while a blue or white one costs about 3 V. That price depends on the semiconductor material (and therefore the colour), not on the particular part, so all red 5 mm LEDs will sit around 2 V. This matters because whatever your supply is not spent on the LED is exactly what your current-limiting resistor has to absorb — we turn that into a calculation in Lesson 4.
| Colour | Typical Vf | Notes |
|---|---|---|
| Red | 1.8 – 2.2 V (≈2.0) | Most common, easiest to start with |
| Orange | 2.0 – 2.2 V (≈2.1) | |
| Yellow | 1.9 – 2.2 V (≈2.0) | |
| Green | 2.0 – 2.2 V (≈2.1) | Some newer greens are ~3 V |
| Blue | 3.0 – 3.3 V (≈3.1) | Higher energy photons |
| White | 3.0 – 3.4 V (≈3.2) | Usually a blue chip + phosphor |
Try it — forward vs reverse
The forward voltage (Vf) of an LED is best described as…
Vf is the "price tag in volts" a lit LED uses up (~2 V red, ~3 V blue/white). The rest of the supply is what the resistor must drop.
Lesson 3 · LED Current and Why Current Must Be Limited
Level: Beginner
- Explain that an LED's brightness is set by its current.
- Describe what happens if an LED is connected with no current limiting.
- State the role of the series current-limiting resistor.
With a resistor you set the voltage across it and the current follows; with an LED it's the other way round. An LED's brightness is controlled by the current flowing through it, measured in milliamps (mA) — a thousandth of an amp. A standard 5 mm LED is designed to run happily around 20 mA for full brightness, and it will be damaged if it's made to carry much more than that for any length of time. Below about 5 mA it gets noticeably dim. So "how bright do I want it?" is really "how many milliamps do I want to allow?"
This is where the danger hides. Once a LED is lit it has almost no internal resistance — it doesn't "push back" against the current the way a resistor does. So if you connect a 2 V red LED straight across a 5 V supply with nothing in between, the LED only "wants" 2 V, and the extra 3 V goes entirely into forcing current. The only thing left to limit that current is the thin resistance of the wires and the battery, which is tiny — the result is hundreds of milliamps, and the LED overheats and burns out in a few seconds. This is the single most common beginner failure, and the reason you never wire a bare LED directly to a supply.
The fix is a current-limiting resistor placed in series with the LED. The resistor's job is to absorb the "leftover" voltage — the part of your supply that the LED doesn't use — and, by doing so, to cap the current at the safe value you choose. A 5 V supply, a 2 V red LED, and a resistor sized for 20 mA: the resistor drops the 3 V it's left with and limits the flow to 20 mA. That single resistor is what keeps your LED bright and alive. The next lesson turns that idea into a formula you can use for any LED and any supply.
| Current | Result |
|---|---|
| 1 – 5 mA | Very dim — visible up close only |
| 10 – 20 mA | Comfortable brightness (20 mA is typical for a 5 mm LED) |
| ~20 – 25 mA | At / near the maximum continuous rating |
| > 25 mA | Overheating — shortened life or immediate failure |
Try it — why limit the current?
If you connect a 2 V LED directly across a 5 V supply with no resistor, it will…
A lit LED has almost no internal resistance, so the extra 3 V forces hundreds of milliamps through it and it fails. A series resistor is essential.
Lesson 4 · Determining the Right Current-Limiting Resistor
Level: Beginner
- To size a resistor, use the formula R = (Vsupply − Vf) / ILED.
- Go through the calculation step by step and choose a standard value.
- You can also use the interactive lab to test any LED, power supply, and current.
The resistor is an essential tool for your LED projects. It works in series and has two main functions: it drops the voltage that the LED does not use, and it limits the current to your desired level. The voltage across the resistor can be calculated by taking the supply voltage and subtracting the LED's forward voltage, which gives you VR = Vsupply − Vf. According to Ohm's law, R = V / I. Combining these gives you the formula for the LED resistor: R = (Vsupply − Vf) / ILED. The units match up perfectly: volts divided by amperes yields ohms (Ω). Remember, in this formula, the current must be expressed in amperes, so for example, 20 mA translates to 0.020 A.
- Identify the values: Vsupply = 5 V, Vf (red) = 2.0 V, I = 20 mA = 0.020 A.
- Voltage the resistor must drop: 5 − 2.0 = 3.0 V.
- Apply Ohm's law: R = 3.0 V / 0.020 A = 150 Ω.
- Pick a standard value: 150 Ω is an E12 value, so use it as-is.
- Check: I = 3.0 V / 150 Ω = 20 mA. ✓ Exactly the target.
Resistors come in standard values, specifically the E12 series, which includes 100, 120, 150, 180, 220, 270, 330, 390, 470, 560, 680, 820, and 1 kΩ, among others. You will likely round your calculated answer to the nearest standard value. If your answer is close to a standard value, it’s better to round up. A slightly larger resistor will lead to slightly less current, which is always safer; the LED will just be a bit dimmer.
Additionally, make sure to check the resistor's power rating using the formula P = I²R. For example, with 20 mA going through a 150 Ω resistor, the power would be 0.02² × 150, which equals approximately 0.06 W—well below the ¼ W rating of a small through-hole resistor. Therefore, power is not typically a concern when using a single LED. You can experiment with different combinations in the interactive lab below to see how the current changes.
Try it — single-LED current limiter
| Supply | LED (Vf) | Target I | R = (Vs−Vf)/I | Standard | Actual I |
|---|---|---|---|---|---|
| 5 V | Red (2.0) | 20 mA | 150 Ω | 150 Ω | 20.0 mA |
| 9 V | Red (2.0) | 20 mA | 350 Ω | 330 Ω | 21.2 mA |
| 3.3 V | Red (2.0) | 10 mA | 130 Ω | 130 Ω | 10.0 mA |
| 5 V | Blue (3.1) | 15 mA | 127 Ω | 130 Ω | 14.6 mA |
Check A white LED (Vf ≈ 3.2 V) runs from a 9 V supply at 15 mA. What resistor do you calculate, and what's the nearest standard value?
R = (9 − 3.2) / 0.015 = 5.8 / 0.015 = 386.7 Ω. The nearest E12 value is 390 Ω. With 390 Ω the actual current is (9 − 3.2) / 390 ≈ 14.9 mA — a touch under 15 mA, which is fine (slightly dimmer, slightly safer).
Rounding the resistor up a little reduces current and is always the safer choice.
Lesson 5 · Building LEDs in Series
Level: Beginner
- Explain that the same current flows through every LED in a series string.
- Compute the string's total forward voltage and its single resistor.
- Recognise the constraint that n × Vf must be below the supply.
Connecting LEDs in series means chaining them end to end in a single line: supply → LED → LED → LED → ground. The defining property of a series circuit is that the same current flows through every component. So all the LEDs in a string glow with the same brightness, and you only need to pick one current for the whole thing. The voltages, however, add up: if each LED uses 2 V, three in series use 6 V. That total is what the supply has to beat, and what the (single) resistor must be sized around.
There is just one resistor for the entire string, and the formula generalises nicely: R = (Vsupply − n·Vf) / I, where n is the number of LEDs. Two important consequences follow. First, there's a hard constraint: n·Vf must be less than Vsupply, or there isn't enough voltage to light them at all. Second, adding more LEDs eats up more of the supply voltage, so the leftover voltage for the resistor shrinks — the string needs a smaller resistor (or none, if the voltages almost exactly match). Series is great when you have a fairly high supply and you want to keep total current low, because the current is just I, not n×I.
Try it — a series string
| LEDs | Colour (Vf) | Supply | Target I | Total Vf | R | Standard |
|---|---|---|---|---|---|---|
| 2 | Red (2.0) | 5 V | 15 mA | 4.0 V | 67 Ω | 68 Ω |
| 3 | Red (2.0) | 9 V | 20 mA | 6.0 V | 150 Ω | 150 Ω |
| 2 | Blue (3.1) | 9 V | 15 mA | 6.2 V | 187 Ω | 180 Ω |
Series is efficient on current (the supply only provides I, not n×I), but it needs a high enough supply voltage, and if one LED fails open the entire string goes dark — there's only one path for the current. That's a real reliability consideration for long strings.
In a series string of 3 identical LEDs, the current through each LED is…
Current is the same everywhere in a series loop; it's the voltages that add up (n × Vf). One resistor sizes the whole string.
Lesson 6 · Connecting LEDs in Parallel
Level: Beginner
- Explain that each LED in parallel sees the full supply voltage.
- State why each parallel LED needs its own resistor.
- Compute the total current drawn from the supply.
In a parallel connection, each LED is linked across the same two points of the power supply. This means every LED experiences the full supply voltage. This setup differs significantly from a series connection: you only need a voltage sufficient for one LED, not for each LED multiplied by the number of LEDs, and if one LED fails, the others continue to function since each has its own separate path. However, an important consideration is current. Each branch draws its individual current, so the total current supplied must equal the number of LEDs multiplied by the current each one requires. For instance, if three LEDs each need 15 mA, they will draw a total of 45 mA together, which is three times what a single LED would use. While this amount is acceptable for a USB power source, it can drain a small battery quickly.
A common point of confusion is the need for each parallel LED to have its own current-limiting resistor. Using a single resistor for all the LEDs won’t work effectively because practical LEDs have slightly different forward voltages. The LED with the lowest forward voltage tends to take most of the current, which can cause it to overheat and potentially fail. By assigning a separate resistor to each LED, you ensure that each branch receives its intended current, regardless of the others. The formula for calculating the resistor for each branch is the same as for a single LED: R = (Vsupply − Vf) / I.
Try it — a parallel array
| Series | Parallel | |
|---|---|---|
| Supply voltage needed | High (n × Vf) | Low (just above one Vf) |
| Current from supply | Low (one × I) | High (n × I) |
| Resistors needed | One for the string | One per LED |
| If one LED fails | Whole string goes dark | Others keep working |
| Best when | Supply is plentiful / want low current | Supply is limited / want redundancy |
When wiring 3 LEDs in parallel from one supply, you should…
A shared resistor can't balance the branches; a slightly-lower-Vf LED would hog current. One resistor per LED keeps each branch safe. Total current = n × I.
Lesson 7 · Designing and Troubleshooting Practical LED Arrays
Level: Beginner → Intermediate — capstone
- Choose between series and parallel for a given supply and goal.
- Apply practical design considerations (power, tolerance, big arrays).
- Diagnose common LED faults from their symptoms.
Putting it together: choosing series or parallel is mostly a question of what your supply can do and what you want. If you have a high-voltage supply (say 9 or 12 V) and you'd like to keep current low, a series string is elegant — one resistor, low total current, but you must fit enough LEDs that n×Vf stays under the supply. If you're on a low supply (3.3 V from a microcontroller, or a coin cell) you can't stack many LEDs in series, so parallel is the natural choice — at the cost of higher total current and one resistor per LED. Many real designs mix the two: a few parallel "rows," each a small series string, to balance voltage, current and redundancy. Run your scenario through the checker below and compare.
A few practical notes separate a textbook answer from a robust build. Vf varies part to part and with temperature, so aim your current a little under the maximum rather than right at it. For a single or a few LEDs a hand-calculated resistor is perfect, but for large arrays the total current can get big and uneven — that's when a proper current-regulating supply or LED driver beats a pile of resistors. And always respect the resistor's power rating for big currents (P = I²R), though for typical LED work a ¼ W part is plenty.
Try it — design a scenario
| Symptom | Likely cause | Fix |
|---|---|---|
| LED dark | Reversed polarity / open in string / Vsupply ≤ total Vf / R too big | Check orientation, continuity, supply vs Vf, reduce R |
| LED dim | R too large / low supply / higher-than-expected Vf | Reduce R, raise supply, confirm the colour's Vf |
| Very bright / hot | R too small — too much current | Increase R to bring I down to ~20 mA |
| One branch dead (parallel) | That LED or its resistor faulty/open | Check that branch's LED and resistor |
| Whole string dead (series) | One open or reversed LED breaks the path | Check each LED in the string |
| Flickers / unstable | Marginal contact or weak supply | Firm up connections, check the supply can deliver the current |
LED design checklist
A whole series string of LEDs is suddenly dark. The most likely first thing to check is…
Series has one path for current, so a single open or reversed LED stops them all. Check each LED's continuity and orientation.
Now that LEDs are second nature, the current and voltage ideas generalise to every component. Continue to Ohm's Law in Practice for the math behind these resistors, First Circuits for more hands-on fundamentals, or DC Circuit Analysis to understand multi-component circuits. For the physical build side, see Breadboarding & Soldering.