Consider the shunt-shunt feedback amplifier shown in the figure below
The transistor is biased at Ic=3.94 mA, then the value of output resistance R0 is equal to
A
68.9Ω
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B
560Ω
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C
1kΩ
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D
1.1kΩ
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Solution
The correct option is A68.9Ω .
v0=(ib+rπibR2).R1+rπibR2×R2=ib(1+rπR2).R1+ibrπ
Applying KCL at node A is equal to I0=βib+ib+rπibR2 I0=ib(β+1+rπR2) v0I0=(1+rπR2)R1+rπβ+1+rπR2 =(1+657700×103)×6.3×103+6571+100+657700×103=68.9Ω