v_c said:This looks like an emitter follower topology. The output resistance seen by R_L is the total resistance in the base divided by the total \[\beta\]. The total amplification factor \[\beta = \beta_{pnp} \beta_{npn}\]. The output resistance is \[10k\Omega/\beta\]. I agree that it is voltage feedback since the variations in the output voltage get reflected back to the base of the NPN transistor. Why? because in the large signal sense, the base voltage is with 0.7V of the emitter voltage. Note that due to this the phase of the base voltage is the same as the phase of the emitter voltage. In the small signal sense these signals are pretty much identical -- so the voltage feedback factor is nearly 1 (usually a little less since \[\beta\] is finite).
So this is pretty much an emitter follower but the output impedance is inversely proportional to \[1/\beta^2\] instead of \[1/\beta\].
Best regards,
\[v_c\]
yes, i'm sure.ptmsl said:Are you sure that the feedback factor is depenent only on the value of one resistor?
thank you for your info. but my question is about the Beta(GAIN) of feedback network.rkodaira said:The name of this transistor configuration is "SZIKLAI CONNECTION" or Complementary Feedback Pair (CFP), also named Complementary Darlington.
Used in output stage of power amplifiers. The advantage is that there is only one Vbe and not 2x Vbe as in the common Darlington connection. The gain is the product of the transistors gains. This was invented by George Sziklai.
already i said that i only draw the ac circuit.magnetra said:That is not a darlington pair, there is no DC biasing for the transistors used.
mtkee2003 said:hi
yes, i'm sure.ptmsl said:Are you sure that the feedback factor is depenent only on the value of one resistor?
and also i'm sure that this is a current-voltage feedback topology.
thank you my friend.
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