This post is a re-write of an article I first published on this blog in 2008, but has been lost in the mists of time and maintenance. The original illustration has been used.

When a lot of gain is required in a single amplification stage within an electronic circuit; for instance when initially amplifying the tiny output of a piezo-crystal device, two transistors wired as a Darlington-pair becomes a useful commodity to use for the purpose.

Connecting two transistors as a Darlington pair by connecting the emitter of the first transistor directly to the base of the second transistor multiplies the beta of the first transistor by the beta of the second transistor to give an extremely high-gain device.

(Whilst it may also hold true that using a couple of series-connected pre-built operational amplifier components (741) wired as inverting amplifiers with high-impedance feedback resistors will serve the purpose, it may be better in the long run to design a circuit around a Darlington-pair in order that increased pre-set custom-attenuation-levels can be brought into play by adding a few extra discrete components here and there; thus producing exactly the desired signal – to be passed on to the subsequent amplification-stages.)

Imagine, for instance; two NPN devices, each with a maximum gain of 50, connected in such a way, giving a device with a maximum gain of 2500: This would be useful for boosting the output of a piezo-microphone for example; notorious for its low output.

(MOSFET devices would probably be the designers choice with regard to the Darlington-pair, in preference to the transistors used in this example. – However, for the purposes of simplicity, I’ll be using bipolar NPN transistors for the devices in my example.)

In the basic circuit-diagram of a Darlington amplifier below; the input is DC decoupled by C1, resistors R1 and 2 form a potential divider biasing the base of Q1 at exactly 0.7 Volts, R3 is the load resistor on the collector of Q1 – which drives the base of Q2; R4 restricting its collector load, and the R4/C2 combination decoupling its emitter to ground.

We’re using bipolar NPN transistors in this example; so since the base of Q1 is at 0.7 Volts (The transconductance threshold.), both Q1 and 2 will be in an “always on” state, and sensitive to any tiny ripple passing through the input capacitor C1.

Bearing in mind that the input ripple will probably be of only a few microamperes; the R1,2 pair should be selected with as high a resistance as possible – Within the megohm range to limit the current already present on the base of Q1 to a fraction of a microampere if at all possible.

Q1 should be chosen such that its base current need only be negligible for it to respond. With a beta of 50 the resistance of R3 should be within the range of around -2 to -40 times that of R1, so as not to drive the transistor into saturation.

Again, having a beta of 50; Q2 should be run ideally at between 2 and 40 gain.

Suggested component values to run the circuit at a voltage of 1.5 Volts are as follows:-

R1: 2M2, R2:(1M with 250K Lin. Preset in series.) R3: 100K, R4: 22K, R5: 110R

C1: 1uF 10V Elect., C2: 100uF 10V Elect., C3: 10uF 10V Elect.

Q1: BC107B, Q2: BC109C

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Note: I haven’t built this circuit myself; and it’s been drawn up for demonstration purposes only: It’s very basic and wouldn’t give brilliant sound quality anyway, but should nevertheless “work”.

Have you anything you’d like to add or comment upon?

P.S. Here is the text of the original article: -

Connecting two transistors as a Darlington pair by connecting the emitter of the first transistor directly to the base of the second transistor multiplies the beta of the first transistor by the beta of the second transistor to give an extremely high-gain device.

Imagine, for instance; two NPN devices, each with a maximum gain of 50, connected in such a way, giving a device with a maximum gain of 2500: This would be useful for boosting the output of a piezo-microphone for example; notorious for its low output.

In the basic circuit-diagram of a Darlington amplifier below; the input is DC decoupled by C1, resistors R1 and 2 form a potential divider biasing the base of Q1 at exactly 0.7 Volts, R3 is the load resistor on the collector of Q1 – which drives the base of Q2; R4 restricting its collector load, and the R4/C2 combination decoupling its emitter to ground.

Since the base of Q1 is at 0.7 Volts, both Q1 and 2 will be in an “always on” state, and sensitive to any tiny ripple passing through the input capacitor C1.

Bearing in mind that the input ripple will probably be of only a few microamperes; the R1,2 pair should be selected with as high a resistance as possible – Within the megohm range to limit the current already present on the base of Q1 to a fraction of a microampere if at all possible.

Q1 should be chosen such that its base current need only be negligible for it to respond. With a beta of 50 the resistance of R3 should be within the range of around -2 to -40 times that of R1, so as not to drive the transistor into saturation.

Again, having a beta of 50; Q2 should be run ideally at between 2 and 40 gain.

Suggested component values to run the circuit at a voltage of 1.5 Volts are as follows:-

R1: 2M2, R2:(1M with 250K Lin. Preset in series.) R3: 100K, R4: 22K, R5: 110R

C1: 1uF 10V Elect., C2: 100uF 10V Elect., C3: 10uF 10V Elect.

Q1: BC107B, Q2: BC109C

Note: I haven’t built this circuit myself; and it’s been drawn up for demonstration purposes only: It’s very basic and wouldn’t give brilliant sound quality anyway, but should nevertheless “work”.