Question

Find valt) in the ideal op-amp circuit below for t o. sketch Vodt), show t = Initial condifion 12 I1 V o)

Answer 1

" Since the op - amp is ideal, I_- = 0 and it is in negative feed back we can use virtual short at v_+ = v_- = DV by transforming the circuit into Laplace domain c rightarrow 1/sc R rightarrow R V_z (t greaterthan 0) rightarrow V_I/s Therefore c = 1 mu F 1/sc = 1/s (10^-6) I_1 (s) = 2/s/10 k Ohm = 2 times 10^-4/s Here 1/sc is in parallel with 1 k Ohm \r\n so resultant Resistance = (1k) (1/s (10^-6))/10^3/1 + s (10^-6) = 10^9/s times s (10^-6)/10^3 times s times 10^-6 + 1 = 10^3/1 + s (10^-6) times 10^3 = 10^3 times 10^3/s + 10^63 = 10^6/s + 10^3 Therefore v_0 (s) = -[I_1 (s) times 10^6/s + 10^3] = -[2/3 times 10^-4 times 10^6/s + 10^3] = -[2 times 10^2/s (s + 10^3)] by inverse Laplace: - 2 times 10^2/s (s + 10^3) = A/S + B/S + 10^3 A = S (2 times 10^2/s (s + 10^3))

Answer 2

" Since the op - amp is ideal, I_- = 0 and it is in negative feed back we can use virtual short at v_+ = v_- = DV by transforming the circuit into Laplace domain c rightarrow 1/sc R rightarrow R V_z (t greaterthan 0) rightarrow V_I/s Therefore c = 1 mu F 1/sc = 1/s (10^-6) I_1 (s) = 2/s/10 k Ohm = 2 times 10^-4/s Here 1/sc is in parallel with 1 k Ohm \r\n so resultant Resistance = (1k) (1/s (10^-6))/10^3/1 + s (10^-6) = 10^9/s times s (10^-6)/10^3 times s times 10^-6 + 1 = 10^3/1 + s (10^-6) times 10^3 = 10^3 times 10^3/s + 10^63 = 10^6/s + 10^3 Therefore v_0 (s) = -[I_1 (s) times 10^6/s + 10^3] = -[2/3 times 10^-4 times 10^6/s + 10^3] = -[2 times 10^2/s (s + 10^3)] by inverse Laplace: - 2 times 10^2/s (s + 10^3) = A/S + B/S + 10^3 A = S (2 times 10^2/s (s + 10^3))

Answer 3

" Since the op - amp is ideal, I_- = 0 and it is in negative feed back we can use virtual short at v_+ = v_- = DV by transforming the circuit into Laplace domain c rightarrow 1/sc R rightarrow R V_z (t greaterthan 0) rightarrow V_I/s Therefore c = 1 mu F 1/sc = 1/s (10^-6) I_1 (s) = 2/s/10 k Ohm = 2 times 10^-4/s Here 1/sc is in parallel with 1 k Ohm \r\n so resultant Resistance = (1k) (1/s (10^-6))/10^3/1 + s (10^-6) = 10^9/s times s (10^-6)/10^3 times s times 10^-6 + 1 = 10^3/1 + s (10^-6) times 10^3 = 10^3 times 10^3/s + 10^63 = 10^6/s + 10^3 Therefore v_0 (s) = -[I_1 (s) times 10^6/s + 10^3] = -[2/3 times 10^-4 times 10^6/s + 10^3] = -[2 times 10^2/s (s + 10^3)] by inverse Laplace: - 2 times 10^2/s (s + 10^3) = A/S + B/S + 10^3 A = S (2 times 10^2/s (s + 10^3))