Finding the time constant of a discharging capacitor.Estimating the Time constant by finding the time take for the Voltage to reduce decay by 63.2%From the equation Vc=V0e-t/rc we can work out the time constant, the time constant is when the time taken for the original voltage to reduce to approximately 37% e-1 of its value. This can only be achieved when the time is equal to the time constant (RC), This can be represented graphically by plotting a graph of the voltage across a discharging capacitor against time, and drawing a curve of best fit through the data points, and recording the results up until the point where there is no voltage across the capacitor nor current in the voltage. In the experiment given to us, the initial voltage across the capacitor was recorded to be 12 Volts, so to find the value of the time constant, we need to find the point in time at which the voltage is equal to 12e-1, Approximately 4.41 Volts. This can be achieved by drawing a straight horizontal line where the Voltage is equal to 4.41 and observing where it crosses the trendline we have drawn, at this point of intersection, the time constant can be estimated, by looking at the time at which this occurs.  From the graph above, instant at which the voltage is equal to 4.41 occurs at a time of 7 seconds, thus giving us the estimation that the time constant is equal to 7.Using this method in conjunction with software that is able to generate graphs we can increase the accuracy of our results further, as shown in the graph above, the equation of the trend line is shown, by making the equation equal to 4.41, and rearranging to find the value of “x” (In this case, x= time constant), by doing this method we are given a result of 6.995 seconds, and due to the results given only being precise to the degree of two significant figures we can round this value up to 7 seconds, so it is also  precise to two significant figures, this method serves to validate my findings where I used a purely graphical method to find the value of the time constant. Finding the time constant Using the gradient of the graph of Loge(12/v) against time.Much like the previous method, this method stems from the rearranging of the Vc=V0e-t/rc equation, the method revolves around rearranging the formula to the form of Ln(Vc/Vo) =-t/RC and from the results we are given the initial voltage is equal to 12V we can represent the formula as Ln(Vc/12)=-t/RC by plotting a graph of Ln(Vc/12) against time and drawing a linear line of best fit through all of the data points, we can find the value of the time constant, since the Gradient= Ln(Vc/12)/Time = -1/RC. From this rearrangement of the formula, the negative reciprocal of the Gradient must be equal to the Time Constant. By using Microsoft Excel and using it’s ability to generate a line of best it and display the equation for said line of best fit we can see that the gradient is equal to -0.1378, by taking the negative reciprocal of said gradient we can ascertain that the Time constant by using this method is equal to 7.3 seconds.Tangential MethodThe tangential method works by finding the point where a tangent on the curve at the point where voltage=12Volts and looking at where it crosses the x-intercept, the value at which it hits the y-intercept is taken to be the time constant, to increase your precision further when trying to discern the time constant graphically you can find the intercept at multiple points and take a mean average. From this graph, at the tangent of the curve at the point Where voltage=12Volts, the time constant can be observed to be approximately 7.25 seconds (although by differentiating the graph, and finding the equation of the tangent, the time constant Is calculated to be 7.24 seconds) and at the point where Voltage=6V the point where it crosses the x intercept is at 12.12 seconds meaning that the time constant for this value can be observed to be 7.12 seconds this value has a 1.7% discrepancy with the previous one calculated ((7.25-7.13)/7.13). The mean average these two results is 7.19, meaning that the average value for time constant is 7.19 seconds.Analysis and problems with the different methods   The main problem with the three methods above was mostly down to human error, for example when trying to find the time constant by looking at the point where the voltage is 37% of the original value, human error can occur when trying to look at the time where the V=4.41 line crosses with the trendline, I tried to alleviate as much human error as possible by changing the scale of the tick lines on the x-axis, this meant that I didn’t have to have to approximate the value as much if the time fell between two ticks marks.When applying the tangential method to the Voltage vs Time graphs, the largest error that could occur then would be drawing the tangent, since the software that I was using doesn’t come with the feature to calculate gradients at any point on the line, I was initially left to draw the tangent without any computational assistance, I opted to make the process more precise and accurate by differentiating the equation of the trendline (something that the software was able to display) and from their manually finding the value at which it crosses the x-axis, for this method the equation for any point on the curve was y= -207/125 e^(-69/50x), and since the graph crosses the y axis at the point where y=12, the general gradient of the curve is y= -207/125 e^(-69/500x)+12, at the point where y is equal to 0, we can work out the point where the tangent meets the X-intercept, in this case it was equal to 7.24 seconds, I applied this method to ensure that the tangent I had drawn was represented as accurately as possible on the graph, and I applied it for the tangents at points V=12V and where V=6V.Overall I would say that the method where one needs to plot the line of y=ln??v/12?  is the most accurate method, since the gradient is already calculated in the software, and there are no observations that need to be done that can introduce an aspect of human error into the analysis (for example: finding the point where the tangent intercepts the X-axis.)Percentage Discrepancies between my results  This table that I have constructed in Microsoft excel serves to show the percentage discrepancy the results of each method have with each other, for example if you look in column 1 row 2, you can see that the result for the ln(v/12) method is 2%  larger than the result for when you need to find the point where the voltage is 37% the initial voltage. By analysing the table we can see which methods brought about the most similar results and we can attempt to pinpoint which method is the most accurate by looking at the magnitude of the percentage discrepancies, from the table we can see that the method that deviated the most from the other methods is the method where you find the point at time at which the voltage is 37% of the original voltage, with a total of 7 percentage point difference when compared to the other two methods, this is the largest percentage point difference across all the results gathered, conversely the method that deviated the least was surprisingly the tangential method, with only a 5% percentage point difference across when compared to the other two methods, This was surprising due to the amount of human error that could have arisen when trying to find the value of the time constant, although my method of finding the point where the tangent crosses the x-axis using mathematics served to increase it’s accuracy, and I can say in full confidence that without that process, the tangential method would have been the most inaccurate method of all three.Estimating model parameters for a diode The constant voltage modelThe constant voltage model is the most commonly used voltage model when modelling a diode and works on the assumption that up until a certain voltage (forward voltage), the current in the diode is equal to zero, the voltage in the diode then reamains constant, this model works on the fact that the I-V characteristic of a diode represents that of an exponential growth graph. So as the current increases, the gradient of the I-V characteristic approaches one, the constant votlage model works by finding the value of the current where the gradient of the characteristic is approximatley equal to one.  Using the graph of raw data obtained from the experiment I have chosen to set the forward voltage as being equal to 0.772, for it is at this point in the graph where the gradient starts to become constant. This can be determined simply by observing the gradient of the characteristic and seeing that it has started to become linear beyond this point. Orange: Constant Voltage ModelBlue: Raw dataThe Constant Voltage models assumes that the resistance of the diode is infinite before a voltage equal to the forward voltage is being induced through the diode, and is equal to zero whence the voltage induced is greater than or equal to the forward voltage.From the superimposed graph we can see that the any points prior to 0.772V have been reduced to 0V, if we were using the constant voltage model graph for any analysis, any voltage running through the diode lower than 0.772V wouldn’t allow a current to flow. The constant voltage model also removes large parts of the graph, more specifically within the range of 0.6V and 0.772V, meaning that the model in this specific case is only optimised for voltages larger than 0.772V.Piecewise linear modelThe Piecewise linear model is a model that servers to break up an I-V characteristic into manageable linear parts so that it can be more easily used in the application in mathematical modelling, the Piecewise linear model has it’s advantages over the Constant Voltage model with the fact that it is a closer approximation of the curve, in layman’s terms it removes less of the curve. Much like the constant voltage model the Piecewise linear model works on the assumption that the resistance in the diode switches between two values when a certain voltage is achieved (forward voltage) but unlike the constant voltage model, the values for resistance past the forward voltage is not limited to zero (although prior to the forward voltage, the resistance is assumed to be infinite)The Piecewise linear model works, by drawing connecting to parts of the I-V characteristic curve where the gradient is relatively constant, and extending that down to the X-axis, this allows you to get a value for the resistance that is in the diode when the voltage across it is larger than the forward voltage by finding the inverse of the gradient.      The gradient of the curve past the forward voltage is equal to: y=((450-170)*?10?^(-3))/(0.81-0.765) =56/9, thus the resistance inside the diode must be equal to 9/56?.The points of intersection with the Piecewise linear model and the curve is at the points (0.765,170) and (0.81,450)for when extending this line down to the x-axis and extending it to the final point on the graph , it provides us with the best approximation of the curve when it has progressed past the “knee” (the point at which the y-values start to increase rapidly