Q7) a) Peak-to-peak ripple 1.01V.

b) From part (a) we see that the maximum value of the differential output is around 3.6V and the output discharges to 2.57V over a discharge time, t, which can be calculated as follows:

3.6 �1 − 𝑒𝑒−𝑡𝑡

𝑅𝑅𝐿𝐿𝐶𝐶� � = 1.01

And therefore:

𝑡𝑡 = 0.33𝑅𝑅𝐿𝐿𝐶𝐶

Assuming the discharge time, and maximum voltages stay almost the same, we can calculate the new time constant to achieve 0.5V from:

3.6 �1 − 𝑒𝑒−0.33𝑅𝑅𝐿𝐿𝐶𝐶

𝑅𝑅𝐿𝐿𝐶𝐶𝑛𝑛𝑛𝑛𝑛𝑛� � = 0.5

𝐶𝐶𝑛𝑛𝑛𝑛𝑛𝑛 = 𝐶𝐶0.33

0.149= 2.2 𝜇𝜇𝐹𝐹

Simulating with the new capacitance value results in a ripple around 525mV. By tweaking the cap value to 2.27uF the desired 0.5V ripple can be achieved.