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I'm studying feedback amplifier and I stumbled upon these slides from Jaeger-Blalock book. enter image description here

enter image description here

However, I just cannot understand what happens in slide 2. Two R_F resistances appears: but why? I thought of Miller theorem, but the Z1,Z2 values aren't close to the original 100kOhm value. Could you explain me what is the circuit transformation?

EDIT: Just to clarify: this is how I'd draw the schematic in slide 2. Apparently it isn't equivalent to the one in the slides. Why Rf is being doubled and grounded? I don't fully understand the explanation "it needs to be considered both in input and output parts".

schematic

simulate this circuit – Schematic created using CircuitLab

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RF is used two times because it needs to be considered in both the calculation of the input resistance (Rin) and in the calculation of the output resistance (Rout). In the equivalent circuit (on slide two) RF is drawn two times to show exactly how it is being considered, (in parallel with the input, and in parallel with the output).

As for the transformation from the first circuit to the second, there is some inconsistency when the RI and RL are added in with no initial warning, but the use of RF on both sides of the second circuit is still correct.

In relation to the Miller Theorem, once all the equivalent resistances and the "K" parameter is known the miller theorem can be applied to find the systems overall input and output resistances, (as shown in the last two equations).

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  • \$\begingroup\$ thanks for the answer, but could you check the updated question? \$\endgroup\$ – Surfer on the fall Jan 22 '18 at 6:24
  • \$\begingroup\$ It could be claimed to some degree that RF is not directly grounded, however to a good approximation it is. Even in your redrawn circuit consider the affects of RC, RL, and ro being at the right side of RF. Similarly it might be claimed that RF is not needed on the output calculation (to some approximation) since its affect is minimal when in parallel with RC, RL, and ro. \$\endgroup\$ – Nedd Jan 22 '18 at 6:51
  • \$\begingroup\$ sorry, I don't understand why "to a good approximation it is".... \$\endgroup\$ – Surfer on the fall Jan 22 '18 at 6:54
  • \$\begingroup\$ In many calculations it may not be practical to find the absolute value of a result. For example if you were using 5% resistors would you calculate equivalent resistances to 0.1% ? Using a good approximation can be similar in some cases to rounding off a decimal number. To use an example with a 100k resistor in parallel or series with a 1k resistor, in parallel you might say the result is 0.99k, (or is it 0.990099k), or to a good approximation you might just say 1k. In series the result would be 101k, but to a good approximation you might still say it is 100k. \$\endgroup\$ – Nedd Jan 22 '18 at 7:30
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It looks like the slide 18-14 is missing essential details added to previous slide 18-13.

Allow me to speculate on behalf of authors;

  • Add Vin with series \$R_I=5k\$ to base , convert to Norton equiv.
  • Add Vout to collector ac-coupled (large C) with load \$R_L=5k\$

Since \$R_F \$ connects to both base and collector, this feedback is modeled as shown for Zin, Zout and Av.

Note that \$R_L\$ must be > \$R_C\$ when AC coupled. ( I think someone just forgot to include the schematic)


The value of Rf reduces the voltage gain by opposing the source current so it tends to follow Rf/Ri ratio rather than bias current and hFE.

Since Rf is opposing the input voltage by the ratio Ri/Rf, it also reduces the voltage gain to this ratio and it lowers the input impedance to Ri+Rf/β and it also shunts the output resistance with Rf. So this is modeled by shunting Rf to both input and output.


Now hopefully that makes sense.

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  • \$\begingroup\$ thanks for the answer, but could you check the updated question? \$\endgroup\$ – Surfer on the fall Jan 22 '18 at 6:24

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