Question

For the transistor feedback amplifier stage shown in figure, hfe-100, hie=1K, while hre and hoe are negligible. Determine RMf for Re=1K.
​

Answer 1

Answer:

$2 + 4 \div 8 \div$

Answer 2

Answer:

[$Rmf$ = -81.89kΩ ]

Explanation:

Given that:-

hfe= 100

hie= 1k

Re= 1k

To Determine:-

Rmf=?

Solution:-

when Rs=0

ac equivalent circuit

Is =$\frac{Vs}{Rs} +\frac{Vs}{hie} +\frac{Vs-V0}{Rs} i =\frac{Vs}{hie}$

$\frac{Vs}{Rs||hie||RB} = Is+\frac{Vs}{Rs}$

vs = $(Is +\frac{Vs}{RB} ) (Rs||hie||RB) ------------(1)$

Nodal at V0

$\frac{V0}{Rc} +Bi +\frac{V0-Vs}{RB} = 0\\\frac{Vo}{Rc||Rb} + \beta \frac{Vs}{hie} - \frac{Vs}{RB} = 0\\\\Vs = VS (\frac{-\beta }{hie} +\frac{1}{RB} ) (Rc||RB)$

From equation (1)

β= hfe = 100

$Vo = (Is +\frac{V0}{RB})(Rs||hie||Rb) (\frac{1}{Rs} -\frac{\beta }{hie} ) *(Rc||Rb)$

$V0 = Is(\frac{1}{Rb} -\frac{\beta }{hie} )(Rs||hie||RB)(RC||RB) +\frac{V0}{RB}( \frac{1}{RB} - \frac{\beta }{hie} ) (Rc||hie|\RB)(RC||Rb)$

$V0 (1-\frac{1}{RB}(\frac{1}{RB} -\frac{B}{hie} )(Rs||hie||RB)(RC||RB)\\ = Is (\frac{1}{RB} -\frac{\beta }{hie} )(RS||hie||RB)(RC||RB)\\\frac{Vo}{Is} = \frac{(\frac{1}{RB} -\frac{\beta }{hie} ) (Rs||hie||RB) (RC||RB)} {1 - \frac{1}{RB} (\frac{1}{RB} -\frac{\beta }{hie} ) (Rs||hie||RB)(RC||RB)} }$

$Rmf = \frac{\frac{1}{100k}-\frac{100k}{1k}(1k||1k||100k)(10k||100k) }{1-\frac{1}{100k}(\frac{1}{100k}-\frac{100k}{1k})(1k||1k||100k)(10k||100k) }$

 $= \frac{\frac{-9.999}{100k}*0.4975k*9.0909k }{1-\frac{1}{100k}-\frac{-9099}{100k}*0.4975k *9.0909k }$$\\= \frac{-452.22k}{1+4.522}$

[$Rmf$ = -81.89kΩ ]