# Load- Line Analysis of Diode

1. Sep 21, 2016

### kenef

1. The problem statement, all variables and given/known data
Determining the best resistance for my simple circuit consisting of a diode, battery, and resistor. Doing this using Load- Line Analysis

2. Relevant equations
ID = VD - V / R - Given by KVL

(I can disregard n-pretend )
3. The attempt at a solution

Here I have attempted using a 1KΩ resistor, obviously too small because I have a voltage drop of about 8.2V across the resistor so then there should be a drop of approx. 1.8 V across the diode. However using the diode equation this gives me way too large of a current.

2. Sep 21, 2016

I don't know that your sketch of $I=I_0 (exp^{eV/(n k_b T)}-1)$ is real accurate. Otherwise your load line looks close to correct. The $I_0 =1 E-12$ is only an estimate. Did you compute the exponential with $e=1.602 E-19$, $k_b=1.381 E-23$, and $T=300$ with $V=.7$? And n=2? The diode equation is only an estimate, but I think experimentally you are likely to get close to I=9 mA with $R=1 k \Omega$. I think the $V_{on}=.7 \, Volts$ for a silicon diode is a pretty good number.

3. Sep 21, 2016

### kenef

Dear Charles, yes I did compute it with all of those values, however the computation I perform with a 1KΩ gives me a current that is too high for my diode, thus I know that I need a resistor with higher resistance. I will attempt to use the online DESMO software and see what that will give me. Should be a tad bit more accurate with the graph. From my work though is it accurate the steps I took to determine the load line? I am simply asking because this is what I THINK is correct, but this is also my first type encountering such a load - line problem.

Thanks!

4. Sep 21, 2016

You did it correctly. The load line is very useful also in cases where the DC voltage varies for constant resistance. In that case the x-intercept (Voltage of the voltage source) will vary with the entire load line moving in the horizontal direction. Every so often I have encountered problems that use the load line method to solve them. $\\$ For your diode current, I do think if you use $V_{Diode}=.7$ and compute current $I=(V_0-.7)/R$ it will give you a fairly accurate answer without the load line method. I googled the Schockley diode equation, and they did not give a precise value for $I_0$ or n. I think you can get considerable variation in these parameters from one diode to another. $\\$ Just an additional item about the load line method: You begin with a graph of your working component, in this case $I=I(V_{Diode})$ i.e. of $I$ vs. $V_{Diode}$. In a series circuit(It only works for the series circuit) you have $V_0=I R +V_{diode}$. This equation can be written (solving for $I=I_{Diode} )$ as $I=-(1/R) V_{Diode}+(V_0/R)$ which is a straight line (with $I=y$ and $V_{Diode}=x$) that has slope $m=-1/R$ and the x-intercept is $V_0$. (Usually you use $\$ $y=mx+b$, and $b$ is the y-intercept), but here the x-intercept is important. The intersection of the load line with the graph of $I=I(V_{Diode})$ is the operating point.