Darlington Voltage Follower

This article describes a simple transistor Darlington pair voltage follower circuit. A voltage follower can be made with just one transistor. However, the circuit presented in this article has a higher input resistance.

There are standard Darlington pair component transistors, two transistors (Darlington pair) in one component. A good example is the TIP122 power transistor.

I made this circuit by cutting a lid of a plastic container:

You can see my circuit working in this YouTube video:

The following video shows transistor working as a touch switch:

Designing the Circuit

I drawn the circuit via online https://easyeda.com software.

The input value should not be above:

VinMax = Vs + Vbe

If Vs = 3 V (shown in the circuit diagram) then:

VinMax = 3 V + 0. 7 V = 3.7 V

We can calculate minimum Rd value by deciding on the maximum LED current:

Rd = (Ve - Vled) / IledMax = (Vi - Vbe1 - Vbe2 - Vled) / IledMax

= (12 V - 0. 7 V - 0.7 V - 2 V) / 10 mA

= 8.6 V / 10 mA

= 860 ohms

We can easily calculate the circuit input resistance:

Ri = (Re || (Rd + Rled) || (Ro + Rl)) * (Beta + 1) * (Beta + 1)

Where: Beta is the transistor current gain.

Lets assume the Q2 Ve voltage is 12 V.

Then:

Rled = Vled / Iled = Vled / ((Ve - Vled) / Rd) = 2 V / ((12 V - 2 V) / 1000 ohms) = 2 V / 10 mA = 200 ohms

Therefore:

Ri = (100 kohms || (1 kohm + 200 ohm) || (1 kohm + 10 kohm))*(Beta + 1)*(Beta + 1)

= (1 / (1 / 100 kohms + 1 / (1 kohm + 200 ohm) + 1 / (1 kohm + 10 kohm)))*(20 + 1)*(20 + 1)

= (1 / (1 / 100,000 + 1 / 1,200 + 1 / 11,000))*21*21

= 1070.38598767*21*21

= 472,040.220564 ohms

= 472.04 kohms

Now we calculate the power dissipation to see why this circuit is not perfect:

Maximum power dissipation occurs at half supply voltage. The collector current is approximately equal to the emitter current. If supply voltage is 12 V, the output is shorted, the LED is taking 10 mA (assuming that we reduce the Rd value to just 400 ohms) then the emitter and collector current will equal to approximately:

Ie = Ve / Ro + Iled + Ve / Re = 6 V / 1000 ohms + 10 mA + 6 V / 100,000 ohms = 16.06 mA

Q2 Transistor Power Dissipation = Ie * Vce = 16.06 mA * 6 V = 0.09636 W = 96.36 mW

Q1 Transistor Power Dissipation = Ie * (Vce - Vbe) = 16.06 mA / (Beta + 1) * (6 V - 0.7 V) =

 = 16.06 mA / 21 * 5.3 V = 0.00405323809 W = 4.05323809 mW

For some general purpose transistors the maximum power rating is 100 mW and this is with the heat sink. Thus 96.36 mW power dissipation is just under the limit. However, if you reduce the supply voltage from 12 V to just 6 V, the power dissipation will be half that amount. Another option is to use a power transistor or put additional transistors in a parallel.