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How A Darlington Pair Transistor Works

A Darlington pair is two transistors that act as a single transistor but with a much higher current gain. This mean that a tiny amount of current from a sensor, micro-controller or similar can be used to drive a larger load. An example circuit is shown below:


The Darlington Pair can be made from two transistors as shown in the diagram or Darlington Pair transistors are available where the two transistors are contained within the same package.

What is current gain?

Transistors have a characteristic called current gain. This is referred to as its hFE.

The amount of current that can pass through the load in the circuit above when the transistor is turned on is:

Load current = input current x transistor gain (hFE)

The current gain varies for different transistors and can be looked up in the data sheet for the device. For a normal transistor this would typically be about 100. This would mean that the current available to drive the load would be 100 times larger than the input to the transistor.

Why use a Darlington Pair?

In some applications the amount of input current available to switch on a transistor is very low. This may mean that a single transistor may not be able to pass sufficient current required by the load.

As stated earlier this equals the input current x the gain of the transistor (hFE). If it is not possible to increase the input current then the gain of the transistor will need to be increased. This can be achieved by using a Darlington Pair.

A Darlington Pair acts as one transistor but with a current gain that equals:

Total current gain (hFE total) = current gain of transistor 1 (hFE t1) x current gain of transistor 2 (hFE t2)

So for example if you had two transistors with a current gain (hFE) = 100:

(hFE total) = 100 x 100

(hFE total) = 10,000

You can see that this gives a vastly increased current gain when compared to a single transistor. Therefore this will allow a very low input current to switch a much bigger load current.

Base Activation Voltage

Normally to turn on a transistor the base input voltage of the transistor will need to be greater than 0.7V. As two transistors are used in a Darlington Pair this value is doubled. Therefore the base voltage will need to be greater than 0.7V x 2 = 1.4V.

It is also worth noting that the voltage drop across collector and emitter pins of the Darlington Pair when the turn on will be around 0.9V Therefore if the supply voltage is 5V (as above) the voltage across the load will be will be around 4.1V (5V – 0.9V)

Other resources

This article is listed in the resource section of the web site within the understanding electronics section.

You may be interested in other similar resources:

See it in a real project

We have two electronic kits which utilise a Darlington pair to allow an output current of up to half an Amp to be controlled. We also sell individual Darlington pair transistors, see below for more details.

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8 thoughts on “How A Darlington Pair Transistor Works”

  • what would happen to input voltage applied to base?

    • It remains the same when the transistor is “on”. If it drops low then the transistor would turn off again.

  • I have been trying to build a power bank using Darlington pair transistors, my aim is to increase a supply voltage of 4.7 to at least 5volts, but I keep on getting less voltage. How do I achieve that?

    • A Darlington Pair is actually to increase current not voltage, a voltage drop across a Darlington Pair is normal of around 0.9V. To increase voltage you would need a step up regulator which we don't sell. To research more I suggest you look up buck boost converter.


  • I am glad that someone is keeping up with the basics in electronics

  • bobby shmurda June 5, 2019 at 11:20 am

    this is the best dar

  • I’m trying to figure out how to replace a specific ‘no-name’ transistor. No doubt it’s a Darlington seated in a TO220 housing. Measuring with one of these cheap digi testers also do say Darlington, BUT returns an hFE of 18 and uF of 995mv. This doesn’t make sense...does it? This tester is driven by a small 9 volt battery, so am I being fooled by not supplying enough ‘juice’ to really open it?
    The transistor is located in an old type electronic ignition module, that is not produced anymore. I have two in working order....and one dead. Disabling both revealed a BC337 and this mystery Darlington. 2SD1071 seems to be ‘close’...but then still not a match. Any good advice, ideas how I can map out the characteristics of my ‘no namer’ Darlington.

    • Hi Christian, without the circuit in front of us it's really difficult for us to advise. The problems you've encountered when using a tester are quite common as the numbers they return can often be inaccurate. If it's not possible to find an old diagram of the module via google then I would say more trial and error might be your only option. Not an ideal answer I know and I'm sorry I couldn't be more help with this issue. The only Darlington Pair Transistor we currently carry is (which may or not be of use in this instance):

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