Revision 79adca8e-04b1-48dc-9285-f000a2a450df - Electrical Engineering Stack Exchange

Yes, there are an infinite number of ways to compute the finite value of \$R=300\:\Omega\$. However, you are actually computing \$R=\frac{V_\text{CC}-V_\text{LED}}{I_\text{LED}}\$. And there is only one way to calculate that in your example case:

$$R=\frac{V_\text{CC}-V_\text{LED}}{I_\text{LED}}=\frac{9\:\text{V}-3\:\text{V}}{20\:\text{mA}}=300\:\Omega\tag{0}$$

Note that for any given voltage source, \$V_\text{CC}\$, and for given values for the LED, you get exactly one way to calculate the resistor's magnitude.

The reality is a little more complex. LEDs vary, one from another, with no two of them exactly alike. The datasheets will specify a range of voltages that may be exhibited by any specific LED (of the same type and manufacture) when a certain current flows through it. This can be quite a range, too. So the value you use for \$V_\text{LED}\$ will only ever be an approximation/average value. You could use the equation with the upper and lower limits and get two different values for \$R\$ and then decide if you want to pick from one or more standard values within that range.


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The above formula calculates a resistor (or range of resistor values if you use the datasheet's full specification range of LED voltages.) And the resistor acts as a very rough current regulator regardless of the specific LED within a batch of LEDs from the same family used to calculate that resistor value.

I'm going to use a specific LED datasheet to illustrate how to estimate just how "good" a specific resistor might be in a specific situation, so that you can get a better sense of why a simple resistor works as well as they do and why the exact value of that resistor isn't as critical as you might think.

First, here's the LED: [Everlight 5mm white LED](http://www1.futureelectronics.com/doc/EVERLIGHT%C2%A0/334-15__T1C1-4WYA.pdf). We can see the following entry in that datasheet:

[![enter image description here][1]][1]

From the above and the formula I mentioned earlier, compute:

$$\begin{align*}
\left[\frac{9\:\text{V}-3.6\:\text{V}}{20\:\text{mA}}=270\:\Omega\right]\le R\le \left[\frac{9\:\text{V}-3\:\text{V}}{20\:\text{mA}}=300\:\Omega\right]
\end{align*}$$

It's very convenient to select the standard value of \$R=270\:\Omega\$. So let's do that, here.

To compute the quality of regulation using that resistor in this case, let's re-write the equation for the LED so that we can work out the current: \$I_\text{LED}=\frac{V_\text{CC}-V_\text{LED}}{R=270\:\Omega}\$. From that and your voltage source voltage, I think you can easily work out that the range of LED current will be: \$20\:\text{mA}\le I_\text{LED}\le 22.2\:\text{mA}\$.

As you can see, this is pretty good. Assuming the average value of \$V_\text{LED}\approx 3.3\:\text{V}\$, this is about \$21\pm 1\:\text{mA}\$ or about \$21\:\text{mA}\pm 5\%\$. So the resistor, given your supply voltage value, provides pretty good regulation. And it's why a resistor is often "good enough" for many uses with LEDs.

(Note: I didn't take into account here the % variation of the resistor, itself. But these days the values are pretty close to their nominal value. Why don't you do your own calculations using a 2% variation of the resistor value, as well, and see if it changes the results much.)


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You can analyze the regulation more generally. Here, you may want to know by what percent will the LED current vary if the power supply voltage varies by some percent value. Or, by what percent will the LED current vary if the current limit resistor itself varies by some percent value. Or, by what percent will the LED current vary if the LED operating voltage varies by some percent value. Those can be interesting questions, at times, depending on what matters.

Those factors are all called "sensitivity figures." How sensitive is one thing relative to another? Let's investigate that question.

This is where the infinitesimal precision of calculus comes into its own. A tiny % variation in current is \$\% I=\frac{\text{d}I}{I}\$ (from a calculus point of view.) Let's start out by applying the derivative operator to the LED current calculation formula from above:

$$\begin{align*}D\left[\: I_\text{LED}\:\right]&=D\left[\:\frac{V_\text{CC}-V_\text{LED}}{R}\:\right]\\\\\text{d}\,I_\text{LED}&=\frac{1}{R}\,\text{d}\,V_\text{CC}-\frac{1}{R}\,\text{d}\,V_\text{LED}-\frac{V_\text{CC}-V_\text{LED}}{R}\,\frac{\text{d}\,R}{R}\end{align*}$$

If we choose to look at the partials (holding the other variations as constant for the purpose), then we find the following three approximations:

$$\begin{align*}
\frac{\%\,I_\text{LED}}{\%\,V_\text{CC}}=\frac{\frac{\text{d}\,I_\text{LED}}{I_\text{LED}}}{\frac{\text{d}\,V_\text{CC}}{V_\text{CC}}}&=\frac{V_\text{CC}}{I_\text{LED}\,R}=\frac{1}{1-\frac{V_\text{LED}}{V_\text{CC}}}\tag{1}\\\\
\frac{\%\,I_\text{LED}}{\%\,V_\text{LED}}=\frac{\frac{\text{d}\,I_\text{LED}}{I_\text{LED}}}{\frac{\text{d}\,V_\text{LED}}{V_\text{LED}}}&=-\frac{V_\text{LED}}{I_\text{LED}\,R}=-\frac{1}{\frac{V_\text{CC}}{V_\text{LED}}-1}\tag{2}\\\\
\frac{\%\,I_\text{LED}}{\%\,R}=\frac{\frac{\text{d}\,I_\text{LED}}{I_\text{LED}}}{\frac{\text{d}\,R}{R}}&=-\frac{V_\text{CC}-V_\text{LED}}{I_\text{LED}\,R}=-1\tag{3}
\end{align*}$$

Both \$V_\text{CC}\$ and \$V_\text{LED}\$ are positive values (or, at least, the same sign) and also that in order to operate the LED it must be that \$V_\text{CC} \gt V_\text{LED}\$, the following conclusions can be made:

 1. Equation 1 says that regulation vs changes in \$V_\text{CC}\$ is better when \$V_\text{CC}\gg V_\text{LED}\$ and that increases in \$V_\text{CC}\$ will lead to increases in \$I_\text{LED}\$.
 2. Equation 2 says that regulation vs changes in \$V_\text{LED}\$ is better when \$V_\text{CC}\gg V_\text{LED}\$ and that increases in \$V_\text{LED}\$ will lead to decreases in \$I_\text{LED}\$.
 3. Equation 3 says that regulation vs changes in \$R\$ is fixed at 1:1. So a +1% change in the resistor value will correspond to a -1% change in the current. This is simply because \$R\$ is in the divisor (and that we are talking about small changes in \$R\$.)

Also note that the sensitivity equations can be used without knowing the value of \$R\$. The only thing that matters is the ratio of \$V_\text{CC}\$ and \$V_\text{LED}\$. This is an important observation for resistor regulation: regulation is better when the supply voltage is very much larger than the required load voltage. (Better regulation implies wasting more power. One of the reasons why active linear regulators were designed, which can provide good regulation without requiring a lot of overhead voltage to get it.)

In your case, \$\frac{\%\,I_\text{LED}}{\%\,V_\text{CC}}=1.58\$ and \$\frac{\%\,I_\text{LED}}{\%\,V_\text{LED}}=-0.58\$. Given the LED datasheet I'd provided, the LEDs are \$3.3\:\text{V}\pm 9\%\$ and so we can compute that a 9% change in \$V_\text{LED}\$ would lead to a \$-0.58\,\cdot\,\pm 9\%= \mp 5.22\,\%\$ change in the LED current. Which is very close to what was observed in earlier calculations above.

  [1]: https://i.sstatic.net/8zp06.png