Load- Line Analysis of Diode

[kenef]

Homework Statement

Determining the best resistance for my simple circuit consisting of a diode, battery, and resistor. Doing this using Load- Line Analysis

Homework Equations

I_D = V_D - V / R - Given by KVL[/B]

(I can disregard n-pretend )

The Attempt at a Solution

[/B]

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.

 

[Charles Link]

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.

 

[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!

 

[Charles Link]

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!

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.