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Freshman here, and I've been searching for an answer to this question, but I could not find any actual justifications yet. Basically, I am studying the concepts of a Darlington circuit, and saw that its current gain was really huge. My notion (and I hope I am correct) is that because the input current is passing through a transistor twice, that means that it gets amplified twice.

But what I also realized is that a Darlington circuit has a high input impedance! Now for that, I can't seem to find an answer to. Is it also because there are two resistors, so the total input impedance is the summation of transistor 1 and 2's input impedance?

Hope I can get some clarification. Thanks!

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    \$\begingroup\$ Are you familiar with the BJT's small signal model? Then draw the model for the Darlington transistor, it consists of two ordinary BJT models. Then it will become clear why the small signal input impedance is multiplied by \$\beta\$ (or \$h_{fe}\$ if you prefer). \$\endgroup\$ Sep 3 '20 at 11:19
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The first transistor in the darlington pair has a much lower base current than the second transistor .Remember that the emitter current in the first transistor is the base current of the second transistor when the effects of any base emitter resistors are small .This is usually the case when currents are high .The input current in the darlington is divided by the current gain of the first transistor .The input voltage is 2 VBE or roughly double that of the straight BJT .For example if the first transistor has a current gain of 25 the input impedance goes up by about 50 .

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    \$\begingroup\$ Also: the small signal base-emitter resistor (\$r_\pi = \beta/gm\$) that is connected to the Darlinton's base isn't connected to the emitter of the second transistor! Instead it is connected to the emitter of the first transistor, that emitter carries a signal. So "bootstrapping" occurs making the input impedance much smaller than \$r_\pi\$. \$\endgroup\$ Sep 3 '20 at 11:25

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