Question

R11 Voc Cc2 SRO2 Cc3 Roi R215 HE Ti Ca Rool BRg tvo R12 TO RI Ri >REL REZY I3 099 RE For the feedback amplifier above, whichConsider the circuit below, which has cocIng capacitors and parasitic capacitances. HE Rg Cel WH AY 1k0 FÅ 2uF v=Q cuir R; Rkos Consider the s-domain transfer function below. A(s) = (s +12)(8 +3 x103)-(5+5x10^)(s+9x10²) if the input is a symmetricalplease solve all of them :)

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Answer #1

1) d.Positive Feedback / Series Voltage feedback

In band-pass:

Name Vi the node in the Emitter of T1:

v (+)- v()- v. (4.) = 0

= VRF + VOTO R+rg

Volg = VRF R: +r, Retro +

There is positive feedback in the input because Vo is positive.

It's Voltage driven and the input is a sum of the generator and the output voltage, a series relation of voltages.

2) c.

1 fc 2 RC

Low Frequencies (small capacitors -> short-circuit)

The dominant pole in Low frequency is the one that gives the smallest frequency, the highest value capacitor and resistance in this case are RL and CC2:

1 fpled 1 = 4Hz 2nR_Cc 27 (20K)(1uF)

In High frequency the dominant pole is the one that yields the highest value frequency, the smallest capacitance and resistance:

First Calculate the miller effect capacitance.

-In the input:

Cm = Cp(1 - A) = 5pF(1 +60) = 305pF

This capacitance is connected in parallel with Cw1.

And the effctive capacitance in the input is:

C1 = Cw1+ Cm = 100pF + 305pF = 405pF

-In the output:

cm (Ay - 1) cp_-61 (5pF) = 5.1pF -60 771 Av

And the effctive capacitance in the output is:

C2 = Cw2+ Cm = 200pF + 5.1pF = 205pF

Therefore, the dominant is C2 and Ro:

грн,а 1 1 21RCw2 2л (250) (205pF) 2.8MHz

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