LED's aren't best modeled as a pure resistor. As noted in some other answers, real LED's do have resistance, but often that's not the primary concern when modeling a diode. An LED's current/voltage relationship graph:

Now this behavior is quite difficult to calculate by hand (especially for complicated circuits), but there is a good "approximation" which splits the diode into 3 discrete modes of operation:

If the voltage across the diode is greater than `Vd`

, the diode behaves like a constant voltage drop (i.e. it will allow whatever current through to maintain `V = Vd`

).

If the voltage is less than `Vd`

but greater than the breakdown voltage `Vbr`

, the diode doesn't conduct.

If the reverse bias voltage is above the breakdown voltage `Vbr`

, the diode again becomes conducting, and will allow whatever current through to maintain `V = Vbr`

.

So let's suppose we have some circuit:

^{simulate this circuit – Schematic created using CircuitLab}

First, we're going to assume that `VS > Vd`

. That means the voltage across `R`

is `VR = VS - Vd`

.

Using Ohm's law, we can tell that the current flowing through R (and thus D) is:

\begin{equation}
I = \frac{V_R}{R}
\end{equation}

Let's plug some numbers in. Say `VS = 5V`

, R=`2.2k`

, `Vd=2V`

(a typical red LED).

\begin{equation}
V_R = 5V - 2V = 3V\\
I = \frac{3V}{2.2k\Omega} = 1.36 mA
\end{equation}

Ok, what if `VS = 1V`

, R = `2.2k`

, and `Vd = 2V`

?

This time, `VS < Vd`

, and the diode doesn't conduct. There's no current flowing through `R`

, so `VR = 0V`

. That means `VD = VS = 1V`

(here, `VD`

is the actual voltage across `D`

, where-as `Vd`

is the saturation voltage drop of the diode).

You can't improve the expected accuracy by combining resistors with the same accuracy. But you will improve the standard deviation of the result. What this means is that if you had 10 individual 1 ohm, 5% resistors, the standard deviation of the 10 resistors would be close to the standard deviation of the lot they were manufactured from. However, if you combine 9 of them as you did to form a 1 ohm resistor, and did this 10 times with different resistors from the same lot, the standard deviation of these ten 1 ohm resistor networks would be smaller than the first set by a factor of 3 (the square root of 9). What this means is that you have improved the probability of being closer to 1 ohm by using a network of 9 resistors because you have narrowed the width of the probability distribution around 1 ohm.

## Best Answer

There's a tool solving exactly this problem:

http://kirr.homeunix.org/electronics/resistor-network-finder/

It constructs a network with your desired equivalent resistance, out of the values that you have in stock.