Lecture - 10: Voltage Shunt Feedback

Example - 1

(a).An inverting amplifier is implemented with R1 = 1K and Rf = 100 K. Find the percentge change in the closed loop gain A is the open       loop gain a changes from 2 x 105 V / V to 5 x 104 V/V.
(b) Repeat, but for a non-inverting amplifier with R1 = 1K at Rf = 99 K.

Solution: (a). Inverting amplifier

Here      Rf = 100 K
               R1 = 1K
When,

(b) Non-inverting amplifier

Here      Rf = 99 K
               R1 = 1K
              

Example - 2

An inverting amplifier shown in fig. 4 with R1 = 10Ω and R2 = 1MΩ is driven by a source v1 = 0.1 V. Find the closed loop gain A, the percentage division of A from the ideal value - R2 / R1, and the inverting input voltage VN for the cases A = 100 V/V, 105 and 105 V/V.

Solution:

we have
when A = 103,


                  

Fig. 4

Example - 3

Find VN, V1 and VO for the circuit shown in fig. 5.

Solution:

Applying KCL at N
                 
or                2VN + VN = VO.

                       
Now         VO - Vi = 6 as point A and N are virtually shorted.
                   VO - VN = 6 V
Therefore,  VO = VN + 6 V

           
Therefore, VN = Vi = 3 V.

Fig. 5

GOTO >> 1 || 2 || 3 || Home