1. Why do we use Darlington pair?
  2. I've heard that we don't use it as common base. Why is that?
  3. Why don't we just develop a BJT with high beta value istead of using this structure?
  4. What are the different types of it?
  5. How can I recognize the emitter and collector of it (for example in the image below)?

enter image description here

  • \$\begingroup\$ They do make superbeta >1k, transistors but cost $$ \$\endgroup\$ Commented Jan 3, 2020 at 5:33
  • \$\begingroup\$ This is a quasi Darlington by another name. Can you find it? \$\endgroup\$ Commented Jan 3, 2020 at 5:54
  • 2
    \$\begingroup\$ Neither of your diagrams shows a Darlington Pair. \$\endgroup\$
    – Russell McMahon
    Commented Jan 3, 2020 at 6:22
  • \$\begingroup\$ en.wikipedia.org/wiki/Complementary_feedback_pair \$\endgroup\$
    – Janka
    Commented Jan 3, 2020 at 6:28
  • \$\begingroup\$ This is superbeta connection, not darlington. Hint: darlington contains 2 (or more) transistors of the same polarity, and all collectors are connected together. \$\endgroup\$ Commented Jan 3, 2020 at 6:44

3 Answers 3

  1. For high current loads. BJTs are a current-controlled device, so one might not provide enough current amplification. This is called the beta value and is normally around 100-300 for general purpose transistors
  2. The base current is being amplified and projected through the collector to the emitter (in NPN). If it's common base, the current is the same through both transistors. Stacking the base of the second transistor to the emitter of the first one provides amplification in the 1000s
  3. Good point, these are expensive! Alternatives are MOSFETs which are voltage controlled and can handle high currents.
  4. NPN, PNP and Szikial pair (push pull amplifier)
  5. These are generally labelled in Darlington ICs such as the ULN2003
  • \$\begingroup\$ ...."enough current amplification"??? Michael J,, please, can you explain WHY you think that "current amplification" is such a critical parameter? For my knowledge (as shown in my contribution), the only point is: Input resistance. Nothing else. Otherwise, a single high-current BJT could do the job. \$\endgroup\$
    – LvW
    Commented Jan 3, 2020 at 13:44
  • \$\begingroup\$ @LvW I have never been a fan of the term "dynamic input resistance" when talking about a BJT. It gives students the false impression that one can model the base behavior as a resistance - this is only true over a very narrow range of base voltages, since the I-V relation for the base current is exponential WRT to voltage (hence, "dynamic" resistance). It is much more useful to think of the BJT base as representing a forward biased diode connected to the emitter, which nicely captures the fact that above around 0.7 V the "resistance" is basically negligible. \$\endgroup\$ Commented Jan 3, 2020 at 14:49
  • \$\begingroup\$ @LvW As for why current gain is a figure of merit - simple, whatever is actually driving the transistor is going to have some limitation on its ability to source current. For example, a typical microcontroller GPIO pin is only going to be able to source a few milliamps maximum. If you can source 5 mA and you need to drive 2 A, you know that you need an "effective" h_FE of around 400. So, you just go and find two transistors such that their minimum h_FEs multiplied together is >400. That's a lot simpler than the dynamic resistance viewpoint. \$\endgroup\$ Commented Jan 3, 2020 at 14:57
  • \$\begingroup\$ @Peter..I cannot agree at all and I do not understand your reasoning.. As shown in my numeric example, the signal input resistance of a single transistor at the base may be in the range of some kohms only - and you suggest to neglect it? That means - to assume that the input resistance is infinite? What do you think is the reason the Darlington transistor was introduced? Is there any other motivation than the signal input resistance at the base node? \$\endgroup\$
    – LvW
    Commented Jan 3, 2020 at 15:01
  • \$\begingroup\$ @LvW I am suggesting that for intuition's sake you assume the input impedance at the base is essentially zero (not infinity) when V_be > 0.7V in forward active mode. This builds a more useful intuition of the BJT as a "current amplifier" rather than going through the rigmarole of trying to derive expressions that let us treat the BJT as a transimpedance amplifier. Under this viewpoint, it's easy to see that the purpose of the Darlington configuration is to increase current gain and thus allow a source with low current sourcing capability to drive large currents. \$\endgroup\$ Commented Jan 3, 2020 at 15:20

I think your question has too many questions. So I'll answer what I'm motivated to answer. Also, you don't write much at all about the context for these questions. And that's important in allowing me to focus what I say. So I will be somewhat terse, as well, in reply because you've not motivated more from me. I hope I strike a balance you find acceptable.

why do we use darlington pair?

The \$\beta\$ relationship between collector current and the required recombination base current can present an unacceptable load on the driver. In some of these situations, it's acceptable to simply use a second BJT to reduce that load. For example:


simulate this circuit – Schematic created using CircuitLab

On the left, the circuit is attempting to use just a single BJT for a switch. But to achieve that, it needs to guarantee about \$80\:\text{mA}\$ as the base current to make sure that \$Q_1\$ is saturated. (\$\beta_1\approx 10\$.) But most I/O pins from an MCU cannot handle that much current. So the left circuit simply won't work.

On the right, \$Q_2\$ has been added in order to provide additional support. Here, \$Q_1\$ is not saturated (it's collector is about \$600\:\text{mV}\$ higher than before, which may be an issue but probably isn't), so it's \$\beta\$ value is likely much higher. Let's call it \$\beta_1=80\$, as it is in active mode and we are using a higher current BJT here. But \$Q_2\$ here is highly saturated, so again it's \$\beta_2\approx 10\$. But \$Q_2\$ only needs to supply about \$\frac{800\:\text{mA}}{\beta_1=80}\approx 10\:\text{mA}\$ through its collector to the base of \$Q_1\$. So \$Q_2\$'s base (it's saturated, so \$\beta_2\approx 10\$) requires only \$\frac{80\:\text{mA}}{\beta_2=10}\approx 1\:\text{mA}\$. This much current can be supplied by most MCU I/O pins. So the right side circuit may very likely work, satisfactorily. The only question is that the combined collector voltage is a little higher, so there may (or may not) be a new problem. (It depends on the type of the load and what it actually requires.)

and I've heard that we don't use it as common base. why is that?

The common base configuration has low input impedance and the base is usually connected to a voltage source or else a capacitor of sufficient size that it acts sufficiently close to voltage source. Reducing base recombination current just doesn't arise as an important factor in these cases.

There are other downsides (and also cases where a Darlington actually is used in the common base configuration.) But the bottom line is that common base configurations have priorities that aren't solved by a Darlington arrangement and, in fact, are often more complicated by using one.

why don't we just develop a bjt with high beta value instead of using this structure?

There are high beta value BJTs. For example, the DSC2A01 is a single BJT with a fairly high \$\beta\$ value.

There are practical limitations. For example, making the base narrower increases \$\beta\$ because there is a smaller chance for recombination during transit from collector to emitter. But doing so also worsens the Early Effect.

I'm stopping at this point. As I said, you've piled a lot of questions together and I've decided to take as many that are directly related to each other as I feel I could. The remaining ones are really part of a different topic.


1. Why do we use Darlington pair?


simulate this circuit – Schematic created using CircuitLab

The Darlington pair can be seen as a new single transistor with modified parameters.

  • The dynamic input resistance is rather high: hie,D=2*hie,1

  • The "beta-factor" (hfe) is large: hfe,D=hfe,1*hfe,2

  • The gain (transconductance) is reduced: gm,D=0.5*gm,2 .


Common emitter gain stage with gain A=-100 (collector resistance Rc=1k, no negative signal feedback )

Transistors: hfe,1=100 and hfe,2=50 ;

Reqired transconductance: gm,D=0.1 A/V (A=-gm,D*Rc=-0.1*1000=-100) ;

Q2: gm,2=2*gm,D=0.2A/V and Ic,2=VT*gm,2=5mA;

Q1: Ic1=Ic,2/hfe,2=0.1mA and gm,1=4mA/V and hie,1=hfe,1/gm,1=25 kOhm :

Darlington input resistance : hie,D=2*hie,1=50 kOhm

Comparison with a single transistor stage (hfe=100; same gain A=-100):

Required transconductance (as before): gm=0.1 A/V

Input resistance: hie=hfe/gm=100/0.1=1 kOhm .

Result: For the same gain (A=-100) the input resistance at the base node of the Darlington combination is much larger if compared with a single transistor (Example: Factor 50). This the most important property of the Darlington compound transistor.


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