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Image text transcribed for accessibility: A Miller integrator is shown in Figure 2 below, with R = 10 k ohm and C= 100 nF Assuming that the op-amp is ideal, derive an explicit expression for the output V 0 and show that the circuit does indeed act as an integrator. In practice, this circuit will not operate as desired because dc imperfections in the op-amp will give rise to an output offset current that will charge the capacitor. The dc parameters of the op-amp are as follows: input bias current lB = 100 nA (directed into the input terminals) input offset current lio = plusminus 10 nA input offset voltage V i0 = plusminus 2 mV Calculate the worst-case output offset current lc for the case where the polarity of the dc imperfections lead to positive limiting (i.e., lc is positive). The input V1 should be connected to ground for this calculation. Calculate the worst-case output offset current lc for the negative limiting case (i.e., lc is negative), again with V1 = 0 V. Thus, even with no input (V1 = 0), the op-amp's dc imperfections lead to saturation at plusminus Vcc (the supply voltages), depending on the polarity of the imperfections. Find the worst-case (i.e., minimum) time to saturation if the op-amp output saturates at Vcc = plusminus 10 V and the capacitor is initially discharged. Figure 2