Voltage shunt Feedback

Voltage shunt Feedback:

Fig. 1, shows the voltage shunt feedback amplifier using OPAMP.

Fig. 1

The input voltage drives the inverting terminal, and the amplified as well as inverted output signal is also applied to the inverting input via the feedback resistor R_f. This arrangement forms a negative feedback because any increase in the output signal results in a feedback signal into the inverting input signal causing a decrease in the output signal. The non-inverting terminal is grounded. Resistor R₁ is connected in series with the source.

The closed loop voltage gain can be obtained by, writing Kirchoff's current equation at the input node V₂.

The negative sign in equation indicates that the input and output signals are out of phase by 180. Therefore it is called inverting amplifier. The gain can be selected by selecting R_f and R₁ (even < 1).

Inverting Input at Virtual Ground:

In the fig. 1, shown earlier, the noninverting terminal is grounded and the- input signal is applied to the inverting terminal via resistor R₁. The difference input voltage v_d is ideally zero, (v_d= v_O/ A) is the voltage at the inverting terminals (v₂) is approximately equal to that of the noninverting terminal (v₁). In other words, the inverting terminal voltage (v₁) is approximately at ground potential. Therefore, it is said to be at virtual ground.

 Input Resistance with Feedback:

To find the input resistance Miller equivalent of the feedback resistor Rf, is obtained, i.e. Rf is splitted into its two Miller components as shown in fig. 2. Therefore, input resistance with feedback Rif is then

Fig. 2

Output Resistance with Feedback:

The output resistance with feedback Rof is the resistance measured at the output terminal of the feedback amplifier. The output resistance can be obtained using Thevenin's equivalent circuit,shown in fig. 3. iO = ia + ib Since RO is very small as compared to Rf +(R1 || R2 ) Therefore,i.e. iO= ia vO = RO iO + A vd. vd= vi – v2 = 0 - B vO Similarly, the bandwidth increases by (1+ AB) and total output offset voltage reduces by (1+AB).

Fig. 3

 Example - 1

(a).An inverting amplifier is implemented with R₁ = 1K and R_f = 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 R_f = 99 K.

Solution: (a). Inverting amplifier

Here      R_f = 100 K
              R₁ = 1K
When, 

(b) Non-inverting amplifier

Here      R_f = 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 V_N for the cases A = 100 V/V, 10⁵ and 10⁵ V/V.

Solution: we have  when A = 103,

Fig. 4

Example - 3

Find V_N, V₁ and V_O 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