# P2

## Introduction

The purpose of this lab was to familiarize ourselved with transistors and amplifiers by running an experiment collecting data on the current in a circut and estimating β in the equation Ice = βIbe. Additionally we prefomed some coding with different probability distributions.

## Description

### Experiments

In lab experiment we set up the following circut with a transistor to find beta in the equation described in the abstract.

The values of R2 an R3 could be calculated by the fact that the max current through BE was .1 mA and through CE was 3 mA.

In the second experiment we were tasked with creating a circut with an amplifier to produce a gain of 2. The circut diagram is below.

The values of R2 and R3 could be calculated using Ohm's Law as follows.

The gain of 2 that we are trying to acheive can be accomplished when R3 is twice the resistance of R2.

### Exercises

Exercise 1: We were given a four sided, six sided, and eight sided die and told to pick one at random and roll. We were tasked with findind the probability we picked each die given we rolled a 1 - 4 - 2 - 5.

Exercise 2: Given a defect distribution, we were tasked with estimating the probability there would be a defect at t = 4.5 min and t = 5.5 min after the last defect was observed.

Exercise 3: We had to use calculus to show the expectation value of t is tau.

## Results

### Experiment 1

### Error Discussion

Unfortunately, our team was unable to collect viable data from the Arduino. This makes solving for beta impossible. If we were to have viable data we would solve for beta by graphing the current through the base over the current through the collector and finding the slope.

### Experiment 2

The circut was built as shown below and the desired gain of 2 was shown in the oscilliscope when a sawtooth waveform was input.

### Exercise 1

We find that is impossible for use to have a die with 4 sides (which makes sense because we were required to roll numbers higher than 4) while the 6-sided die is 75.9% and the eight sided is 24%.

### Exercise 2

We were given the following distribution.

We can find the probability of the defect occuring after the two times by taking the integral of the CDF at either time and finding the difference between the two. This results in the following equation.

We then simply plug in 10 for tau and our times of t=4.5 and t=5.5 and we get .423 and .362 respectively. The difference tells us that there is a .061 chance of a defect occuring between the two times.