Diode (LED) example problem question
Just to clear away for others, this point about why \$V_{S_{MAX}}=4.5\:\text{V}\$, you've been provided the statement that someone might plug in the LED in the wrong direction and you are told that you cannot count on the LED standing off a reverse voltage of more than \$-4.5\:\text{V}\$. Since in reverse-biased arrangement, the assumption is that the current is zero or near that and therefore the voltage drop across \$R\$ will be zero or near that and therefore all of \$V_S\$ will appear across the LED. We're assuming reverse-biased so this \$V_S\$ cannot be allowed to exceed \$4.5\:\text{V}\$ to avoid exceeding the minimum reverse voltage specification. So that part is obvious and you say you understand that part.



At this point you can work out the resistor value. The worst case allowed current is \$30\:\text{mA}\$ and you already know that \$V_{S_{MAX}}=4.5\:\text{V}\$. The only remaining thing is to make the assumption that the LED drops the least voltage across itself, leaving the maximum remaining voltage across the resistor \$R\$. To avoid exceeding the maximum specification of \$30\:\text{mA}\$ in the LED, then:

$$R=\frac{4.5\:\text{V}-1.5\:\text{V}}{30\:\text{mA}}=100\:\Omega$$

The only remaining detail is to work out the value of \$V_{S_{MIN}}\$. Here, you assume the smallest current, or \$20\:\text{mA}\$, and also the largest possible voltage drop across the LED at this current, so:

$$V_{S_{MIN}}=20\:\text{mA}\cdot 100\:\Omega + 2.2\:\text{V} = 4.2\:\text{V}$$

The first equation uses two points, \$p_1=\left(1.5\:\text{V}, 30\:\text{mA}\right)\$ and \$p_2=\left(4.5\:\text{V}, 0\:\text{mA}\right)\$, to define the load line and work out the slope (\$R\$.) The second equation uses a new point \$p_1=\left(2.2\:\text{V}, 20\:\text{mA}\right)\$ and the just-determined slope (determined by \$R\$) to define a new load line and work out the \$x\$-axis intercept (which is \$V_{S_{MIN}}\$.)



Line Equations


The two-point equation of a line is: $$y-y_1=\frac{y_2-y_1}{x_2-x_1}\cdot\left(x-x_1\right)$$ Plugging in \$p_1=\left(1.5\:\text{V}, 30\:\text{mA}\right)\$ and \$p_2=\left(4.5\:\text{V}, 0\:\text{mA}\right)\$ and solving for \$x\$: $$\begin{align*}y-30\:\text{mA}&amp;=\frac{0\:\text{mA}-30\:\text{mA}}{4.5\:\text{V}-1.5\:\text{V}}\cdot\left(x-1.5\:\text{V}\right)\\\\x&amp;=4.5\:\text{V}-100\:\Omega\cdot y\end{align*}$$ and here you can see the value of \$V_{S_{MAX}}\$ and \$R\$.
The point and slope equation of a line is: $$y-y_1=m\cdot\left(x-x_1\right)$$ Plugging in \$p_1=\left(2.2\:\text{V}, 20\:\text{mA}\right)\$ and \$m=\frac{-1}{R}\$ and solving for \$x\$:$$\begin{align*}y-20\:\text{mA}&amp;=-0.01\cdot\left(x-2.2\:\text{V}\right)\\\\x&amp;=4.2\:\text{V}-100\:\Omega\cdot y\end{align*}$$ and here you can see the value of \$V_{S_{MIN}}\$ and \$R\$.




So the reasoning is pretty simple.

The outer (upper) line is determined by \$V_{S_{MAX}}\$ on the \$x\$-axis and the worst-case allowable current (\$30\:\text{mA}\$) through \$R\$ assuming that the LED drops the least-possible voltage (\$1.5\:\text{V}\$.) The slope is computed and you have \$R\$.

The inner (lower) line must retain that slope (be parallel to the line above it) but must now intersect through a point determined by the lowest allowable current (\$20\:\text{mA}\$) assuming that the LED drops the greatest-possible voltage (\$2.2\:\text{V}\$), leaving the least-possible voltage for \$R\$ to develop that minimum current. The \$x\$-axis intercept is computed and you have \$V_{S_{MIN}}\$.