Electrical Engineering ⇒ Topic : Practical Inverting Op Amp
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Gaurav
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PRACTICAL INVERTING OP AMP In this case, Ri ≠ ∞, Av ≠ ∞, and R0 ≠ 0 The practical OP Amp is shown in Figure (a) Figure (a) Practical equivalent circuit. The equivalent circuit of Figure (a) is shown in Figure 1 (a) and its Thevenin's equivalent is shown in Figure 1 (b). Figure (1) Practical OP Amp. V'S and R'1 are given as: ............ (1) .............. (2) Since Ri ≠ ∞ , Ri. connected between inverting and non-inverting terminals. Since R0 ≠ 0,R0 is connected in series with the voltage source (AvVi). The equivalent circuit of Figure (a) is redrawn in Figure 1 (a). Let us replace the components to the left of the terminals 2 and 4 of Figure 1 (a) by the Thevenin's voltage V's and Thevenin's equivalent resistance R'1 as shown in Figure 1 (b). V'S and R'1 are given by ................. (3) ........... (4) Closed loop voltage gain AV The voltage between terminals 2 and 4 is V24. Let V24 = V. Therefore, it can be written as: Vi = IR2 + V0 ................ (5) Also V0 = AvVi + IR0, ............... (6) Applying KVL to the loop of Figure 1 (b), it can be written as: -V'S; + IR'1 + R2 + R0) AvVi = 0 ............ (7) Using Eqs. (5) and (6), we obtain ................ (8) Using Eq. (7), Eq. (6) can be written as: [Using Eq. (8)] ............. (9) Since Ri ≥ R1, the following approximations can be made: With approximations, Eq. (9) can be written as: ............ (10) where A, is gain of circuit with feedback or closed loop gain. If Av ≥1 and R1 ≥(R2 + R0),Av → R2/R1 Therefore, AV, has the same value as in ideal case. | |
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