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 maintainV = Vd
).If the voltage is less than
Vd
but greater than the breakdown voltageVbr
, 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 maintainV = 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 acrossR
isVR = 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
, andVd = 2V
?This time,
VS < Vd
, and the diode doesn't conduct. There's no current flowing throughR
, soVR = 0V
. That meansVD = VS = 1V
(here,VD
is the actual voltage acrossD
, where-asVd
is the saturation voltage drop of the diode).