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This circuit is two BJT transistors in so called 'darlington configuration'. I want to analyse this circuit for AC signals (where the capacitors are shorts).

enter image description here

The question is to calculate Delta_V_L/Delta_V_G (sorry I don't know how to use subscripts here)

This is what I think it should be but unfortunately when I plug in the numbers it's not correct. Solution should be Delta_V_L/Delta_V_G= -50.

enter image description here Any ideas on how to solve this and is this smallsignal equivalent circuit correct? Thanks!

Edit: h_ie1 and h_ie2 are the smallsignal model resistances, hence why they aren't on the original circuit

Edit2: I realised that there are different models for small signal analysis, I'm using this one: enter image description here

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  • \$\begingroup\$ Try analysing the circuit from a DC (biasing) perpective first \$\endgroup\$ Jun 1, 2016 at 15:07
  • \$\begingroup\$ Have you try to write a equation? For example for Vin we have \$ Vin = Ib1 *hie1 + (\beta 1 + 1)*Ib1*hie2 \$ and for Vout \$ Vout = (\beta 1 + 1)*Ib1 * \beta 2 * Rc|R_L \$ \$\endgroup\$
    – G36
    Jun 1, 2016 at 15:10
  • \$\begingroup\$ You should use the transconductance gm. Also your first stage equation is completly false, the first stage got a feedback effect from the second input base resistance. \$\endgroup\$
    – MathieuL
    Jun 1, 2016 at 15:10
  • \$\begingroup\$ @JImDearden I already did DC analysis, that was no problem; AC analysis is completely different though \$\endgroup\$
    – Arthur VP
    Jun 1, 2016 at 15:23
  • \$\begingroup\$ @Arthur VP I already give you the solution. You do not see it ? \$\endgroup\$
    – G36
    Jun 1, 2016 at 15:41

1 Answer 1

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Without using any small-signal equivalent diagram I consider the circuit as a two-stage transistor amplifier: (1) The first stage is in common-collector configuration and (2) the 2nd stage is a clasical common-emitter stage.

  • Assuming a DC voltage of 0.6 volts across both B-E diodes it is easy to find the base current IB1 for the first transistor and, hence, all the other three currents (IC1, IB2, IC2) because the current gains are given.

  • Applying the relation for the transconductance gm=Ic/VT and rbe=beta/gm we easily can find the gain values for both stages: G1=0.5 and G2=-100. Hence, the total gain, indeed, is G=G1*G2=-50.

(By the way: In a typical/classical Darlington configuration both collector nodes are connected).

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  • \$\begingroup\$ For this circuit it is better to assume 0.7 volts across two B-E junctions. Therefore Ic = 4.95mA; hie1 = 500 Ohms ; Ic2 = 100mA; hie2 = 5 ohm. Also for this type of a connection the voltage gain will be equal 0.5 times the common-emitter stage gain (the second stage). Av = - 0.5 * gm2 * Rc = 0.5 * 4 Siemens * 25 Ohms = -50. Because Vin splits two Vbe. \$\endgroup\$
    – G36
    Jun 1, 2016 at 19:58
  • \$\begingroup\$ Your result is Ic1=4.95mA and my result was Ic1=5.05 mA. Therefore, the "best" value, perhaps, would be VBE=0.65 volts. In any case, it is an estimate only (and, certainly, both B-E voltages will not be equal.) \$\endgroup\$
    – LvW
    Jun 1, 2016 at 20:04
  • \$\begingroup\$ But notice that no matter what what the re'b1 = hie2 where re'b = Vt/Ie. In this case re'b1 = Vt/Ie1 and it is also equal to hie2 = Vt/Ie1 and this is why the first stage gain (emitter-follower) is always equal to 0.5 (G1 = hie2/(re'b1 + hie2)). Therefore vbe1 = vbe2 (from small signal point of view). And this also explain why -0.5*gm2*Rc is true. \$\endgroup\$
    – G36
    Jun 1, 2016 at 20:34
  • \$\begingroup\$ Yes - of course. However, my point was only to mention that both Vbe values (DC) will not be equal. Nevertheless, without any additional information we must assume Vbe1=Vbe2. \$\endgroup\$
    – LvW
    Jun 2, 2016 at 8:11

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