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So the problem goes as follows:

We have a simple circuit with a voltage source of Vs in series with a LED and a Resistor R_L. We are given that the barrier voltage of the LED is 1.5-2.2V and the the LED works best in the region of 20-30mA. Also the PIV is around -4.5-5V.

The Question is:

What are the upper and lower bounds of the voltage source so that we have a current of 20-30mA and the LED won't burn out if someone connects it to the circuit in the opposite way.

So in the solution Vs_max=4.5 for obvious reasons. But what troubles me is the way the load lines are crossing the graph of the LED.

Why is the 30mA current "matched" with the 1.5V of the LED. I mean why don't we consider it the other way around. Edit: PIV is the breakdown voltage. The (negative) voltage in which the diode, in this case the LED, will become useless as it will no longer block the currents in one direction.

why these two points, and not choose to intersect the Vs_max load line in the point of 20mA

Why these two points, and not choose to intersect the Vs_max load line in the point of 20mA?

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  • \$\begingroup\$ Use <Enter> x 2 for paragraph breaks. (I've fixed it for you.) You need a minor edit to your question to explain what PIV is. \$\endgroup\$
    – Transistor
    Apr 14, 2018 at 13:40

2 Answers 2

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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

  1. 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\$: $$\begin{align*}y-30\:\text{mA}&=\frac{0\:\text{mA}-30\:\text{mA}}{4.5\:\text{V}-1.5\:\text{V}}\cdot\left(x-1.5\:\text{V}\right)\\\\x&=4.5\:\text{V}-100\:\Omega\cdot y\end{align*}$$ and here you can see the value of \$V_{S_{MAX}}\$ and \$R\$.
  2. 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\$:$$\begin{align*}y-20\:\text{mA}&=-0.01\cdot\left(x-2.2\:\text{V}\right)\\\\x&=4.2\:\text{V}-100\:\Omega\cdot y\end{align*}$$ 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}}\$.

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  • \$\begingroup\$ Thanks for the exact explanation. But I want to know the reasoning behind this "coupling". I mean, the upper voltage (4.5V) must be matched with the upper current which is 30mA. That seems logical, the more we increase the voltage the more the current increases too(I presume that even though LEDs emit light instead of heat they still abide to something like ohm's law so that voltage is proportional to the current-correct me if wrong please). \$\endgroup\$
    – paris_th
    Apr 15, 2018 at 20:22
  • \$\begingroup\$ So what we do is consider that in the "worst-case senario" that if our LED gets the maximum current cause of the maximum voltage of the source, then the voltage drop of the LED must be the least possible. Why do we do that? Can't it be a case that it gets the maximum voltage drop? \$\endgroup\$
    – paris_th
    Apr 15, 2018 at 20:23
  • \$\begingroup\$ @paris_th Because the least voltage drop across the LED subtracts from the voltage source, leaving the most voltage drop possible for the resistor -- which means the maximum current in the resistor. Since you don't want to exceed this maximum, it's important that you work out the current in the resistor using the smallest possible LED voltage drop. Does that make sense? \$\endgroup\$
    – jonk
    Apr 15, 2018 at 20:40
  • \$\begingroup\$ But the LED and the resistor are connected in series, so the current of 20-30mA is the same for both of them. So by considering that when I apply the maximum voltage source I get the maximum current for my circuit, by considering the least voltage drop across the LED I can find the maximum voltage drop for that resistor. Because ultimately I want a resistor that can work in the worst-case senario, that is, it being exposed to the most voltage possible. I got that right? \$\endgroup\$
    – paris_th
    Apr 16, 2018 at 21:31
  • \$\begingroup\$ @paris_th You can only pick one R value. That's it. You have to support a range of 20-30mA with it. So the drop will be at least R*20mA and at most R*30mA. You would pair up the smallest possible source voltage with the largest possible voltage drop across R to see how bad the minimum remaining voltage left over for the LED would be. You would pair up the largest possible source voltage with the lowest possible voltage drop across R to see how bad the maximum remaining voltage left over for the LED would be. \$\endgroup\$
    – jonk
    Apr 17, 2018 at 1:51
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Why these two points, and not choose to intersect the Vs_max load line in the point of 20mA?

That problem is a bit odd to me. As I understand it, we're trying to figure out what upper and lower voltage limits are theoretically viable. So if we take the equation your provided, which is correct by Kirchoff's voltage Law, we can plot the graph you also provided. This graph exhibits four possible intersection points because of the current and voltage drop upper and lower limits. So now the question is between those four points, which ones are going to show the maximum and minimum voltage source? This is where I am quite puzzled. Indeed if we want to make sure we do not burn the LED, we will find the lowest possible upper limit and the highest lower limit to keep the LED lighted up. Hence the two points we should use to find out Vs_max and Vs_min are the two used in your graph to draw the load lines.

However from a practical approach, these two points are not realistic because of the diode's behavior. If you look it up, you will see graph where current is plotted against voltage drop showing that as current increases, so does te voltage drop. Hence the only possible combinations are (Imax,vmax) and (Imin,vmin) which imply:

\$ Vs_{max} = v_{max} + i_{max}.R \$ and \$ Vs_{min} = v_{min} + i_{min}.R \$

In your problem, they decided to consider the combinations (Imax,vmin) and (Imin,vmax) which output a lower Vsmax and higher Vsmin as if all combinations were allegedly possible, though this will probably not be the case with a practical circuit. Another possible explanation would be that the problem considers the barrier voltage to be already reached regardless of the current. What the graph here lacks is the actual behavior of the diode to intersect the current and voltage drop limits. I would then agree with you when you said:

I mean why don't we consider it the other way around

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