I'm doing a simple lab (I'm a hobby EE) to reinforce my ohm's law math and learn a little about how to do proper measurements with a multimeter.

I have simple circuit with a 2.2k ohm resistor connected in series with an LED. Everything works fine up to the point I go to calculate voltage drop across the resistor and LED.

My initial calculations only accounted for the 2.2k ohm resistor. As such I got the full voltage dropped across the resistor. However, when I measured the circuit for real I found the result to be nearly half of the input voltage, which would indicate to me

- My math is wrong
- There's resistance left unaccounted for

The only thing left to account for is the LED. What is the best method for determining the resistance of a simple LED? I tried doing what I do with resistors (hold it up to the probes with my fingers) but I don't get a proper reading. Is there a technique I'm missing here?

## Best Answer

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).