What it shows
A capacitor charges slowly through a very large resistor. Only when its voltage reaches about 1.4 V does a pair of transistors, the Darlington pair, switch the LED on, giving a timed delay.
The science
The capacitor charges from 9 V through 1 MΩ, so its voltage rises slowly, with a time constant RC = 1 MΩ x 100 µF = 100 s. Two transistors with emitter of the first connected to the base of the second make a Darlington pair, whose gain is the product of the two, about 40000, and which needs about 1.4 V at its base to turn on. Reaching 1.4 V takes about RC x ln(9 / 7.6), which is 17 s; because the LED glows gradually and the base takes a little current as it turns on, the delay you see is about 15 to 25 s and varies with the transistors.
Materials
Cost: under Rs 200. Many items come from home or from waste; spare parts such as LEDs, motors and wire are in the store.
- Two BC547 transistors
- A 1 MΩ resistor and a 470 Ω resistor
- A 100 µF 16 V electrolytic capacitor
- A push button and an LED
- A 9 V battery with snap connector
- A breadboard and jumper wires, a watch
Safety
For this project: Battery power only. Fit the electrolytic capacitor with its negative leg to ground.
Never use mains electricity: power models with cells, batteries or a low-voltage adapter. Sharp and hot tools, flames and lab chemicals need an adult or your teacher beside you. Keep magnets, button cells and small parts away from young children, and wash your hands after making.
Step by step
8 steps. Do them in order, and test each part before you go on to the next. Write what you notice in a notebook as you go.
Place both BC547s side by side (flat face towards you, legs left to right are collector, base, emitter) and connect the emitter of the first to the base of the second.
Join the collectors of both transistors together and connect the second emitter to the negative rail.
Connect the 470 Ω resistor from 9 V to the LED long leg, and the LED short leg to the joined collectors.
Connect the 1 MΩ resistor from 9 V to the base of the first transistor.
Connect the positive leg of the capacitor to the base of the first transistor and the negative leg to the negative rail.
Connect a push button across the capacitor to discharge it when pressed.
Snap on the battery, press the button and release it, then start the watch; the LED switches on after about 20 s.
Change the capacitor to 47 µF and then 220 µF and record how the delay changes.
What you should see
The LED stays off for a while after the button is released, then switches on, with a longer delay for a larger capacitor.
Viva questions
A teacher or a judge will ask how it works. Try to answer each question before you open it, then compare.
01What controls the delay?
The values of the resistor and the capacitor, the RC time constant.
02What is a Darlington pair?
Two transistors joined so that the second amplifies the output of the first, giving very high gain.
03Why does the button reset the timer?
It discharges the capacitor so charging starts again from zero.
04Why is the delay not exact?
Transistor gain and capacitor tolerance vary, and the base draws current.
Take it further
Add a 2.2 MΩ trimmer in series with the 1 MΩ resistor to make an adjustable timer you can set from about 17 to 55 seconds.
Change one thing, measure what happens, and add a table or chart of your readings: that is what turns a model into your own investigation.
Free ebooks for this project
Robotics
Electronic Components Handbook
Resistors, capacitors, LEDs, diodes, transistors and more, with how to use each
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