1. Light the LED

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Problem 1

Close a 5 V loop through one resistor and a training LED so the LED current is between 8 mA and 12 mA.

The source and ground are already on the sheet and locked.
Place one resistor and one LED. Either order is fine.
The LED anode faces the source. Target current is 8–12 mA.

Context

A training LED holds about 2.0 V while it conducts. The resistor sees the rest of the 5.0 V. Current in a series loop is that leftover voltage divided by resistance. The anode is the pin current enters.

Open the lesson Component guide

Components on this challenge

DC source. An ideal 5.0 V source in this training model.
Resistor. Limits current. The voltage across it divided by its resistance is the current through it.
LED. A diode that emits light when current flows from anode to cathode.

Full brief

The 5.0 V source and GND are already on the sheet and locked. Add one resistor and the training LED (VF = 2.0 V) so they form a single series loop. Either order is fine. The LED anode must face the source. Choose the resistor that puts the LED current in the target window. The server simulates the circuit; the browser does not decide the score.

How PCBGrade measures this

The task above is the whole brief. The weights below are the published points.

connectivity · 25 pts
polarity · 20 pts
current/resistor choice · 35 pts
successful simulation · 20 pts
Cap 60: An open loop or a circuit the simulator rejects cannot pass.
Cap 60: A reversed LED cannot pass.

Reference

Training LED

Conducts at a fixed 2.0 V drop when forward-biased. Current is (Vsupply − 2.0 V) / R in a series circuit.

1. A. Anode
2. K. Cathode

Schematic · Drag a part. Drag empty sheet to pan. Scroll to zoom.

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