Electronics 1 — Filters & Op-Amps

Sizing a real part: the LED resistor

continues from lesson 3 — values defined earlier in the course stay live here

ELI5: an LED has a fixed voltage drop (~2 V) and will burn out if you let too much current through, so it needs a series resistor. The resistor "eats" the leftover voltage (supply − LED drop) at whatever current you pick: R = (V_cc − V_led)/I.

V_{cc} := 5 V

= 5 V supply

V_{led} := 2 V

= 2 V red LED forward drop

I_{led} := 15e-3 A

= 0.015 A bright but safe

R_{led} = \frac{V_{cc} - V_{led}}{I_{led}}

= 200 ohm = 200 Ω — use the next standard value up

✓ pass

(V_{cc} - V_{led}) * I_{led} < 0.25 W

a quarter-watt resistor survives

Try it yourself: size the resistor for a 2 V LED on a 9 V supply at 20 mA (20e-3 A). (R = (V_cc − V_led)/I.)

Rled_{you} := (9 V - 2 V) / 20e-3 A

= 350 ohm ✏️ Your turn: the resistor drops 9 V − 2 V = 7 V at 20 mA. Divide to get ohms.

✓ pass

abs(Rled_{you} - (9 V - 2 V) / 20e-3 A) < 0.1 ohm

green when your LED resistor is correct

Where next: AC Motors turns electricity into motion, then Power Systems scales it all up to kilovolts.