Make a Darlington Pair Using Two Npn BJTs

This instructable, will be all about Darlington Pair and it's applications. I will be going through in detail in terms of construction based on both NPN and PNP type (coming soon! - stay tuned). So, let's begin.

Darlington Pair ; is a name given to a cascaded set of transistors both being of the same type, say NPN or PNP. This configuration basically amplifies the input at the 'base of configuration' and that results in a very large current in the emitter collector circuit. This is a very useful property, as the amplification factor or Current gain of the setup is very large. It is given as follows

β (net) = β1 . β2 + β1 + β2

where , β1 is the current gain of one of the transistors. β2 is the gain of other transistor. β(net) is the current gain of the whole setup. β is the ratio of colector current to base current.

β = Ic/Ib

or

Ic = β . Ib

where , 'Ic' is the collector current and 'Ib' is the base current.

it means that for a small base current the collector current would be β times the base current. but for a darlington pair the β is β(net) which is effectively large (see step 3) so a small base current results in a very large collector current.

Don't get bored. Now let's see it in work.

1 - Breadboard
2 - NPN transistors, 547B (x2)
3 - 10kΩ resistor
4 - 100Ω resistor
5 - LED ( for seeing it in effect)
6 - Power Supply ( 5V or 3V would suffice / you could use a cell combination).
7 - don't forget the jumper wires
8 - Multimeter with transistor testing (hFE)

Let's begin by making the darlington pair.
Put one of the transistors in the breadboard. Now place the second transistor in the breadboard, such that the collector terminals of both transistors are connected. and the emmiter terminal of second transistor is connected to the base of first transistor. Also, the base of second transistor is not connected to anything and the emmiter of first transistor is not connected to anything.

That is the darlington pair. The core of our circuit here.

Fact - 547B transistors have a β of around 350 more or less.
which means β(net) will be

=350 . 350 + 350 + 350
=123,200

which means for a base current of about 1μA the collector will be 123,200 times 1μA which is about 123mA more or less (it depends on efficiency). So you can see, how much is the amplification factor.

Now let's proceed to connecting the LED.

connect the LED on the collector side of the darlington setup.

To be safe connect a 100Ω resistor in series with the LED, it'll keep the LED safe from sudden surges in voltage.

Connect the cathode of LED to the collector of the setup. and connect the 100Ω resistor to the anode of LED.

Now the LED setup is done.

Let's move towards the base of the setup. it is a good practice to install a base resistor here 10kΩ to protect the transistor (even though we are willfully giving weak input on base.)
!!Don't Forget This!!

Connect the Positive terminal of power supply to the other end of 100Ω resistor. and the negative terminal of power supply to the emitter of transistor 1.



It's Done yay!

Let's test it,
touch the unconnected end of 10kΩ resistor, if all went well. The LED should light up.

But why? why does it light up because of a simple touch

The simple answer is touching the resistor lead causes a very very very minute discharge from hand to lead, of the order of nano Amperes to micro Amperes. and then that is amplified by the darlington pair, resulting in a huge current in collector emitter circuit, huge enough to drive an LED or something else, it depends on the power supply and it's delivering capability.

This specific circuit is sensitive enough to even detect the electromagnetic noise just conne t a long enough wire to the 10kΩ resistors other end. The LED should glow.

Use the multimeter transistor tester to check and identify the terminals of the transistor. To avoid blowing the transistor.

- if the LED isn't glowing as bright (good enough) as I said, and you are using a cell or battery. Then there is a good chance that your battery might be running low.