Question

5.6 Using the same parameters for the 741 op amp in Example 5.1, find v0 in the op amp circuit of Fig. 5.45

Figure 5.44 For Prob. 5.5. 6 Using the same parameters for the 741 op amp in Example 5.1, find v, in the op amp circuit of Fig. 5.45

Answer 1

Using the given model of op-amp we can draw as

Using the given model of op - amp we can draw as Using KVL in the loop we get Av_d - 50 times i + 10^-3 - 2 times 10^6 times i = 0 where v_d = 2 times 10^6 times i A times 2 times 10^6 times i - 50 times i + 10^- 3 - 2 times 10^6 times i = 0 where v_d = 2 times 10^6 times i A times 2 times 10^6 times i - 50 times i + 10^-3 - 2 times 10^6 times i = 0 (A times 2 times 10^6 - 50 - 2 times 10^6) times i = -10^- 3 Given open loop again A = 2 times 10^5 hence i almostequalto - 10^-3/2 times 10^5 times 2 times 10^6 = -0.25 times 10^-

Using KVL in the loop we get

$Av_d - 50\times i + 10^{-3} - 2\times 10^6\times i = 0$

where v_d = 2 x 10⁶ x i

$A\times 2\times 10^6\times i - 50\times i + 10^{-3} - 2\times 10^6\times i = 0$

$\left ( A\times 2\times 10^6 - 50 - 2\times 10^6 \right ) \times i = - 10^{-3}$

Given open loop gain A = 2 x 10⁵ hence

$i \approx \frac{ - 10^{-3}}{2\times 10^5 \times 2\times 10^6} = -0.25\times 10^{-14}\; A$

Thus V_d = -2 x 10⁶ x 0.25 x 10^-14 = -0.5 x 10^-8 A

Thus Vo is given by

$V_o = AV_d - 50\times i \approx AV_d = -2\times 10^5 \times 0.5\times 10^{-8} = -10^{-3}\; V$