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If i remember correctly the input impedence of a BJT is r_pi=Vt/Ibq. I've read that using those two configurations I get a much higher input impedence. Can someone please tell me why? Maybe some formulas to prove the fact would be appreciated. Cheers

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    \$\begingroup\$ My "knee jerk" reaction:) is that you "get a much higher input impedence" since the input base current is beta times smaller than those of a single transistor... \$\endgroup\$ Jan 1 '20 at 17:05
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Simply follow the current path. Look at this example:

enter image description here

Where each transistor has a current gain equal to \$\beta = 99\$

As you can see the input base current is very small due to the BJT's current gain.

And the input signal source see the \$R_E\$ resistor as much large resistor

\$R \approx \beta_1 \times \beta_2 \times R_E \approx 100k\Omega \$

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    \$\begingroup\$ Shouldn't the collector current of T1 be 13.86 mA? \$\endgroup\$
    – simonov
    Jan 1 '20 at 19:54
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    \$\begingroup\$ @simonov Yes, you are right. I fix it. \$\endgroup\$
    – G36
    Jan 1 '20 at 19:58
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Here is my answer for the Darlington combination (input resistance hie,D):

  • hie,D=hie,3 + hfe,3*hie,1 (the input resistance of Q1 appears at the base of Q3 multiplied with the current gain)

  • We have hie,1=hfe,1/gm1 with gm1=gm,3*hfe,1 (because Ic1=hfe,1*Ic3).

  • From this: hie,D=hie,3 + hfe,3/gm,3= 2*hie,3

Hence, the input resistance of the darlington transistor is twice the input resistance of the first transistor (Q3)

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  • \$\begingroup\$ What do hie and hfe mean? \$\endgroup\$
    – Lucky-Luka
    Jan 4 '20 at 14:25
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    \$\begingroup\$ OK - I have used the so.called hybrid parameters which describe theBJT as a 4-pole. Physical explanation: hie is the small-signal base-emitter resistance rbe ...and hfe is the small-signal current ratio ic/ib=beta. \$\endgroup\$
    – LvW
    Jan 4 '20 at 14:41
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    \$\begingroup\$ Lucky-Luka...the first transistor (QA) has a resistance at its emitter that consists of the B-E path of the second transistor QB - similar to a simple ohmic resistor RE. Hence, we can use the classical feedback formula r1=hie,A + hfe,A*hie,B. \$\endgroup\$
    – LvW
    Jan 4 '20 at 15:00
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    \$\begingroup\$ I have deleted my comment - I have overlooked the 10 ohm resistance in the example. His answer is approximately correct, but he has neglected secondary terms. \$\endgroup\$
    – LvW
    Jan 4 '20 at 15:54
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    \$\begingroup\$ It is easy to see why his result is approximate only: Set RE=0 and the result would be zero, however, this cannot be true because we have still have the resistance of two B-E pathes ... \$\endgroup\$
    – LvW
    Jan 4 '20 at 15:59

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