**1 Answer**

written 2.1 years ago by |

A = $\frac{vo}{vi}$ gain of ampli without f/b

Af = $\frac{vo}{vs}$ gain of ampli with f/b

$\beta$ = $\frac{vf}{vo}$ is feedback factor.

voltage gain with negative f/b

vs = vi + vf

but vf = $\beta$vo

vs = vi + $\beta$vo

"' y Vi = $\frac{vo}{A}$

vs = $\frac{vo}{A}$ + $\beta$ vo

vs = vo ( $\frac{1+a\beta}{A})$

$\therefore$ Af = $\frac{vo}{vs}$ = $\frac{A}{(1+A\beta})$

This indicates that in voltage series negative f/b overall voltage gain of ampli decreases by factor ( 1 + A$\beta$)

i/p Resistance with negative f/b Let Ri = $\frac{vi}{Ii}$ is i/p Resistance

without f/b

Rif = $\frac{vs}{Ii}$ is i/p Resistance with f/b

Vs = Vi + Vf

Vs = Ii Ri + $\beta$ Vo

Vs = Ii Ri + $\beta$ AVi

Vvs = Ii Ri + $\beta$ AIi Ri

Vs = Ii Ri (1+A$\beta$)

$\frac{vs}{Ii}$ = Ri (1+ A $\beta$)

$\therefore$ Rif = Ri ( 1 + A$\beta$)

This indicates that in voltage series negative f/b overall voltage gain increases by factor (1 + A $\beta$ )

o/p Resi with f/b

In order to find o/p Resi with f/b making the i/p VS = 0 and connecting a fictitious voltage source of value vx at the o/p then

Rof = $\frac{vx}{Ix}$

Where Ix is current drawn from $v_x$

Ix = $\frac{vo-Avi}{Ro}$

But Vi = - vf = - $\beta$ vo

Ix = $\frac{vo+A\beta vo}{Ro}$

$\therefore$ Ro Ix = Vo ( 1 + A$\beta$)

$\therefore$ Rof = $\frac{vo}{Ix}$ = $\frac{Ro}{(1+A\beta)}$

Which indicates that in voltage series negative f/b overall o/p Resi decreases by factor (1+A$\beta$)