# Basic Electronics on a Breadboard

How a breadboard, resistor, LED, capacitor, diode, transistor and 555 timer combine into working circuits, one part added at a time\.

By PETE\!

Topic: Computing, AI & Electronics

[Open the interactive Jig](https://ask.jig.com/j/hCzw9npbtWO1)

## Transcript

### 1. Meet the breadboard

A breadboard lets you build a circuit without solder\. Under each column of five holes, a to e or f to j, a metal clip joins the holes, so two leads pushed into the same strip are connected\. The gap down the middle keeps the two halves apart\. The long rows along each edge are power rails: every hole along the red \+ line is joined, and so is every hole along the blue − line\.

Drag empty space to turn the board\. The gold bands show the hidden metal clips\.

### 2. Power on the rails

The supply on the left gives 5 volts\. Its red lead goes into the \+ rail and its black lead into the − rail, so the whole length of both near rails is now live\. Nothing happens yet: current only flows when there is a complete loop from \+ back to −\.

The glowing rails show which holes are connected to the supply\.

### 3. A resistor limits current

A red jumper brings 5 V from the \+ rail to column 12\. A short yellow jumper carries it across to column 14, where the resistor starts\. The resistor resists the flow of current: the more ohms, the less current\. This one is 330 Ω; its bands read orange, orange, brown \(3, 3, ×10\) and gold for ±5 %\. The loop is still open, so no current flows\.

Each lead sits in its own column: the resistor joins column 14 to column 22\.

### 4. The LED lights

An LED gives light when current flows through it in one direction only\. Its long leg, the anode, goes toward \+ in column 22; the short leg by the flat rim, the cathode, goes toward − in column 23, and a black jumper closes the loop to the − rail\. A red LED drops about 2 V, so 330 Ω sets the current: \(5 − 2\) V ÷ 330 Ω ≈ 9 mA\. Try other resistors\.

Gold dots show conventional current, flowing from \+ through the parts to −\. 100 Ω gives 30 mA, more than a typical 20 mA LED should carry\.

### 5. A switch opens and closes the loop

The yellow jumper is replaced by a pushbutton straddling the centre gap\. While it is pressed, it joins its left legs to its right legs and carries current from column 12 to column 14\. Release it and the loop opens: the current stops and the LED goes out at once\.

Press and hold the button on the board, or use the guide button\.

### 6. A capacitor stores charge

A 1000 µF capacitor joins column 14 to the − rail\. Its striped side marks the negative lead\. While the button is pressed it charges to 5 V\. When you let go, it keeps pushing current through the resistor and LED, and the light fades instead of snapping off\. With 330 Ω the time constant is 330 Ω × 1000 µF ≈ 0\.33 s\.

Press briefly, then watch the fade\. A bigger resistor makes the fade slower and dimmer\.

### 7. A diode guards an LED

A second LED circuit sits on the right half\. A red jumper brings 5 V to column 33, and a 330 Ω resistor carries it to column 36\. The red LED's anode is in column 36 and its cathode in column 37\. From there a 1N4007 diode runs down into the − rail, its silver band, the cathode, toward −\. A diode passes current only from anode to cathode and drops about 0\.7 V, so \(5 − 0\.7 − 2\) V ÷ 330 Ω ≈ 7 mA lights the LED\.

Turn the diode round\. It now blocks the current, just as it would if a battery were connected backwards, and the LED stays safely dark\.

### 8. A potentiometer dims an LED

Two long jumpers at the right end carry \+ and − across to the far rails, so the top half has power too\. A red jumper feeds 5 V to the potentiometer's end leg in column 33\. Its middle leg, the wiper, sends the current on through a 330 Ω resistor to a yellow LED and back to the − rail\. Turning the shaft changes how much of the 10 kΩ track lies between the end leg and the wiper, so the current changes from about 9 mA down to 0\.3 mA\.

Drag the slider to turn the knob\. The 330 Ω resistor keeps the current safe even with the knob turned fully up\.

### 9. A transistor switches an LED

The NPN transistor's legs, emitter, base and collector, sit in columns 42, 43 and 44 with its flat face toward you\. A black jumper takes the emitter to −\. A 330 Ω resistor from the \+ rail and a green LED lead down into the collector\. A 10 kΩ resistor feeds the base from column 39, which a yellow jumper links to \+\. The small base current, \(5 − 0\.7\) V ÷ 10 kΩ ≈ 0\.4 mA, switches on about 8 mA through the LED\.

Pull out the yellow jumper: with no base current the transistor turns off and so does the LED\. E B C is the 2N3904 order; other transistors differ, so check the datasheet\.

### 10. A 555 timer makes a clock

The 555 straddles the gap with pin 1, by the dimple, at front left in column 46\. Pin 1 goes to −, and pins 4 and 8 go to \+\. On the far half, a 1 kΩ resistor stands from pin 8 to pin 7 and a 68 kΩ resistor from pin 7 to pin 6\. A blue jumper over the chip joins pin 6 to pin 2, and a 10 µF capacitor joins them to −\. The capacitor charges through both resistors to two thirds of 5 V, then pin 7 drains it through the 68 kΩ down to one third, and it starts again\.

Frequency ≈ 1\.44 ÷ \(\(1 kΩ \+ 2 × 68 kΩ\) × 10 µF\) ≈ 1 Hz\. The orange 100 nF capacitor across the far rails, beside pin 8's supply jumper, keeps the chip's supply steady\.

### 11. The output flashes and beeps

Pin 3, the output, is high while the capacitor charges and low while it drains\. A 330 Ω resistor takes it to a red LED, and a yellow jumper takes it to an active buzzer whose − leg sits in the − rail\. Each time pin 3 goes high, about 5 mA lights the LED and the buzzer's built\-in oscillator makes it beep\. When pin 3 goes low, both stop\. The cycle repeats about once a second, for as long as the power is on\.

An active buzzer beeps on a steady voltage\. A passive piezo disc would need the 555 to switch hundreds of times a second to make a tone\.