Bring up an LED and a 3.3 V input from a 5 V supply
Two small circuits that nearly every project needs: an LED with its series resistor, and a resistor divider that brings a 5 V signal down to what a 3.3 V input can take. This kit takes them in order, with each step ready-loaded with a worked example.
The tools use ideal components and steady DC. They tell you what to expect so you can compare it with your meter, not what a real part will do under every condition.
- 5 steps
- About 15 minutes
- Runs on your device; nothing is uploaded
Review status. Not yet reviewed by a specialist in this craft. The arithmetic is checked by tests against independent calculations, but the practical advice is a starting point: if you can improve it, a pull request is welcome (see docs/niche-kits.md in the Toolmoz repository).
Before you start
The boxes are for you while you work. They are not saved and leave this page when you do.
The steps
Size the LED resistor
The resistor sets the LED current: the supply voltage minus the LED’s forward voltage, divided by the current you want. The tool also gives the resistor’s dissipation, so you can check its power rating.
Take from the result: The ideal resistance and the dissipation. Choose the nearest standard value at or above the ideal one: a slightly higher resistance gives slightly less current.
Example loaded: Red LED on a 5 V supply, 20 mA. The classic first circuit: one red LED (about 2.0 V forward) from 5 V at 20 mA needs 150 Ω.
Read the bands on the part you picked
Confirm the value of the resistor in your hand before it goes on the board. A 150 Ω resistor with a 5 % tolerance reads brown, green, brown, gold.
Take from the result: The nominal value and the range its tolerance allows.
Example loaded: 150 Ω, four bands: brown green brown gold. Brown (1), green (5), brown (×10) and gold (±5 %) read 150 Ω ±5 %, the LED resistor for a red LED on 5 V at 20 mA.
Divide 5 V down to 3.3 V
Two resistors in series share the input voltage in proportion to their values: the output is the input times the lower resistor divided by the total. With 1 kΩ over 2 kΩ, 5 V becomes 3.33 V.
Take from the result: The unloaded output voltage, and the current the divider draws from the supply.
Example loaded: 5 V logic level down to 3.3 V (1 kΩ and 2 kΩ). Two resistors divide 5 V by two thirds, giving 3.33 V for a 3.3 V input, with the output unloaded.
Check what the input does to the divider
Whatever the divider feeds sits in parallel with the lower resistor and pulls the output down. Enter your input’s resistance as the load and compare the loaded and unloaded outputs.
Take from the result: How far the output drops. If it matters, use smaller resistors (which draw more current) or buffer the signal.
Example loaded: The same divider driving a 10 kΩ load. Shows why loading matters: a 10 kΩ load on the 2 kΩ leg pulls the output from 3.33 V down to 3.125 V.
Check a batch of resistors against their tolerance
Measure each resistor out of circuit and compare it with the limits its tolerance band promises. The tool marks each reading as inside or outside and shows the signed deviation.
Take from the result: Which readings fall outside the limits. Remember that your meter has its own error.
Example loaded: Check five 2 kΩ, 1 % resistors from a new reel. Five measured readings against 2 kΩ ±1 % (1980 to 2020 Ω): four are inside the limit and one reads 2024 Ω.
When you have the numbers
What this kit does not cover
- Ideal components and steady DC only. LED forward voltage changes with current and temperature, and resistors drift.
- A resistor divider is not a level shifter for fast or two-way signals, and it cannot supply current.
- Nothing here covers mains voltage, battery safety, static discharge or the maximum ratings of your parts: the datasheet comes first.
Sources and further reading
The tools’ own pages explain their methods and list their sources. Presets are worked examples from the Toolmoz repository, each checked against an independent calculation.