11. Amplifier Analysis. A two-stage amplifier is used between a voltage source and a load. The nominal values are: Rs = 1k92 R = 10 kQ2 R;1 = 10 k92 R;2 = 10 k92 Roj = 1 k92 Roz = 1 kQ Avoi = 10 Avo2= 10 The overall gain (A ) is the ratio of the voltage at the load over the voltage at the source, and can be expressed with the following equations: Vil = SR1 + RS Avol Riz viz = Vil' Rip + Rol Av Avo2-RL Vo = 12 Ry + Ro2 Note that A, is independent of the value of v. When you solve for A, V, will cancel. a) Assume that the amplifier parameters (Ri1, Ro1, Avo1, Ri2, Ro2 and AVo2) all have normal distributions with a standard deviation of 10% of the nominal values. Run a Monte Carlo simulation. Assume Rs and R, are the given values. Answer the following: a. What is the average (mean) value of the overall gain? b. What is the 25 percentile value? . What is the 75 percentile value? d. What percent of the values of the overall gain fall between +-5% of the mean? b) Repeat if the amplifier values have a 5% standard deviation. 12. LED Current. An LED with a current limiting resistor is attached to an output pin of a microcontroller. To be visible, the LED needs at least 7 mA of current going through it. The microcontroller can provide a maximum of 10 mA. Here is the data: Resistor (RD): a normal distribution with a nominal value of 360 Ohms and standard deviation of 10% of the nominal value. LED (VD): a normal distribution with a nominal value of 2 V and standard deviation of 10% of the nominal value. Output Pin (Vout): a uniform distribution from 4.9 to 5.1 V. The current provided to the LED is ID = (Vout-VD)/RD a. Running a Monte Carlo simulation, what is the expected value of the LED current