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
Simply follow the current path. Look at this example:
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 \$
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)