LED resistor calculator

A red LED on a 5 V supply, wired straight across, will pull as much current as the supply can give and burn out. The LED doesn’t limit its own current; a resistor in series does. This LED resistor calculator takes your supply voltage and the LED’s datasheet figures, picks the standard resistor, and shows the steps, the wattage and what your meter should read. The other tabs use the same reasoning on a long DC wire run and on a battery that sags under load.

Size the series resistor for an LED

Start from the supply and the LED datasheet. Presets come from named datasheets; pick custom to type your own part.

Your parts

LED

Red: Kingbright WP7113SRD/D (1.85 V typical at 20 mA)

Supply voltage

5 V

Forward voltage, typical

1.85 V

Forward voltage, maximum

2.5 V

Current you want

20 mA

Maximum DC current (datasheet)

30 mA

LEDs in series

1

Resistor series

E24 (5% parts)

Sketch

Answer

Resistor to fit

160 Ω

Exact value

157.5 Ω

Current with that resistor

19.7 mA

Current if Vf is at its maximum

15.6 mA

Power in the resistor

62 mW

Resistor rating (2× margin)

⅛ W

Power in the LEDs

36.4 mW

Checks

• Nothing to flag with these values.

In words

The 5 V supply shares its voltage between the LED and the resistor. The LED takes about 1.85 V, so 3.15 V is left across the resistor. Ohm’s law on the resistor gives 157.5 Ω for 20 mA. The next E24 value up is 160 Ω, which lets 19.7 mA flow.

Maths

1. Volts across the resistorV_R = Vs − n·Vf = 5 V − 1 × 1.85 V = 3.15 V
2. Resistance for the current you wantR = V_R / I = 3.15 V / 20 mA = 157.5 Ω → 160 Ω (E24)
3. Current with the standard partI = V_R / R = 3.15 V / 160 Ω = 19.7 mA
4. Current if Vf is at its maximumI = (5 V − 1 × 2.5 V) / 160 Ω = 15.6 mA
5. Resistor power, then the 2× marginP = V_R × I = 3.15 V × 19.7 mA = 62 mW; 2 × P = 124 mW → ⅛ W

Check it with a meter

With the circuit on, measure across the resistor: expect about 3.15 V. Divide that reading by 160 Ω to get the LED current (about 19.7 mA). Or put the meter in series on its mA range and read the current directly.

Voltage drop and heating along a DC wire run

Annealed copper, both conductors counted (out and back). This shows what Ohm’s law says about the run; it does not pick a conductor size.

Your parts

Supply voltage

12 V

Run length, one way

10 m

Conductor cross-section

2.5 mm²

Load current

10 A

Conductor temperature

20 °C

Sketch

Answer

Round-trip resistance

0.138 Ω

Volts lost in the wire

1.38 V

Share of the supply lost

11.5%

Heat in the wire

13.8 W

Voltage left at the load

10.6 V

Checks

• More than 10% of the supply is lost in the wire (11.5%). The load sees noticeably less than the supply.
• The wire is turning 13.8 W into heat along its 20 m of conductor.

In words

10 A flows out and back through 20 m of copper, which adds up to 0.138 Ω. Ohm’s law says that resistance takes 1.38 V, so the load gets 10.6 V of the 12 V. The same current warms the wire by 13.8 W.

Maths

1. Copper resistivity at this temperatureρ = 0.0172414 Ω·mm²/m × (1 + 0.00393 × (20 °C − 20 °C)) = 0.0172 Ω·mm²/m
2. Resistance of both conductorsR = ρ × 2L / A = 0.0172 Ω·mm²/m × 20 m / 2.5 mm² = 0.138 Ω
3. Volts lostV = I × R = 10 A × 0.138 Ω = 1.38 V
4. Heat in the wireP = I² × R = (10 A)² × 0.138 Ω = 13.8 W
5. Voltage at the load12 V − 1.38 V = 10.6 V

Check it with a meter

With the load running, measure the voltage at the supply terminals, then at the load terminals. The difference is the drop: expect about 1.38 V.

Battery or supply sag under load

A battery behaves like an ideal voltage (its EMF) behind a small internal resistance. Presets use the AA datasheet’s fresh-cell range.

Your parts

Battery

Energizer EN91 (AA alkaline), fresh: 1.5 V, 0.15 Ω internal

Open-circuit voltage (EMF)

1.5 V

Internal resistance

150 mΩ

Load resistance

1 Ω

Sketch

Answer

Voltage at the terminals

1.3 V

Current

1.3 A

Sag inside the battery

0.196 V

Power to the load

1.7 W

Power heating the battery

255 mW

Share of power lost inside

13%

Checks

• More than 10% of the power is heating the battery (13%) instead of the load.

In words

With no load the battery reads its 1.5 V. Connect the load and 1.3 A flows through the battery’s own 0.15 Ω too, which takes 0.196 V. The load is left with 1.3 V, and it receives 1.7 W.

Maths

1. Current round the loopI = E / (r + R_L) = 1.5 V / (0.15 Ω + 1 Ω) = 1.3 A
2. Volts lost insideI × r = 1.3 A × 0.15 Ω = 0.196 V
3. Terminal voltageV = E − I·r = 1.5 V − 0.196 V = 1.3 V
4. Power to the load and insideI² × R_L = 1.7 W; I² × r = 255 mW

Check it with a meter

Measure the open-circuit voltage (about 1.5 V), then the voltage with the load on (about 1.3 V) and the current (about 1.3 A). The internal resistance is roughly the voltage change divided by the current: Energizer’s pulse method works the same way.

!Theory model only: it shows circuit relationships, it does not size, approve or certify electrical work. An original Fieldbook tool.

How the numbers are worked out

The trick is to apply Ohm’s law to the resistor, not to the LED. An LED holds a roughly fixed voltage across itself, its forward voltage (Vf), so whatever the supply has left over sits across the resistor. That leftover voltage divided by the current you want gives the resistance.

Resistors only come in certain values. The calculator takes the next value up in the E24 (or E12) series, so the real current lands at or just under your target. It then works out the current again with that part, because that is the number your LED actually sees.

Datasheets give a typical and a maximum Vf. The calculator runs both: an LED from the high end of the spread draws less current with the same resistor, which is why two “identical” LEDs can glow differently. When the supply is only a little above the LED’s Vf, that spread swings the current a lot, and the tool says so.

The resistor’s power is its voltage times its current. The tool suggests a part rated for at least twice that, so it runs comfortably cool. That 2× margin is Fieldbook’s own rule of thumb, not a manufacturer’s figure.

What Ohm’s law says about voltage drop on a 12 V wire run

A wire is a resistor too, just a small one. Copper’s resistance grows with length, shrinks with cross-section and rises a little as it warms. On a long low-voltage run the current goes out and back, so both conductors count. Multiply that resistance by the load current and you get the volts the load never sees; multiply by the current again and you get the heat spread along the wire. The wire-run tab shows those numbers for annealed copper. It is theory only: conductor sizing follows the applicable electrical code edition and a qualified person, and the tool never recommends a size.

Why a battery’s voltage drops under load

A battery behaves like a perfect voltage source with a small resistance hidden inside it. With nothing connected you read the full voltage. Draw current and part of the voltage is lost across that internal resistance, so the terminals read less, and that part of the power warms the battery instead of reaching the device. A gadget that resets when a motor starts is often just this: the terminal voltage dips below what the device needs. The battery tab uses the fresh-cell internal resistance range from Energizer’s AA alkaline datasheet.

Model assumptions

• DC only, steady state. For AC circuits with inductance and capacitance, see the series RLC circuit calculator.
• The LED’s forward voltage is fixed at the datasheet value you enter; real LEDs shift it with current and temperature.
• Resistors and wires are treated as constant resistances, and the supply as ideal unless you give an internal resistance.
• Wire resistance uses annealed copper at 20 °C with its temperature coefficient (NBS Circular 31); other temperatures are a linear correction.
• The battery’s nominal voltage stands in for its open-circuit voltage, and its internal resistance is held constant, though both change with charge, temperature and load.
• The resistor rating margin, the “little headroom” check (current at maximum Vf below three quarters of typical) and the 10% loss checks are Fieldbook’s own rules of thumb.

Theory model only: it does not size conductors, select protection or replace a code or a qualified person.

Try these theory checks

• Set the LED preset to blue and lower the supply toward 3.3 V. Watch the gap between the typical and maximum-Vf current, then explain it in words.
• With a red LED on 12 V, add LEDs in series one at a time. Predict what happens to the resistor’s power before you look.
• On the wire-run tab, double the length, then double the cross-section. Which change undoes the other?
• On the battery tab, lower the load resistance step by step. At what point does more than a tenth of the power stay inside the battery?

These are original study prompts written for this site.

Sources

• LED presets: Kingbright datasheets for the WP7113SRD/D red, WP7113YD yellow, WP7113QBC/D blue and WP7113QWC/D white LEDs.
• Standard resistor values: IEC 60063:2015, as tabulated by Vishay.
• Resistor power ratings: Stackpole CF series datasheet.
• Copper resistivity and temperature coefficient: NBS Circular 31, Copper wire tables.
• AA alkaline voltage and internal resistance: Energizer EN91 datasheet; measuring method: Energizer, Battery internal resistance.