# Confusion with voltage drop equation

I had never really thought of resistors causing a voltage drop, I have only used them to limit current to say an LED.

Using something such as E = IR I can use a single resistor as an example:

5v / 100ohm = 0.05A * 100Ohm = 5V


Now I can maybe assume that is 5 volt drop, leaving 0 after it. How does this allow an LED that requires, say, 1.8 volts that it drops if it leaves 0V at the other side of the resistor?

Using a simple circuit simulator..

5v ------- 220Ohm resistor resistor -----1.8v-----led-------0v


Why is 1.8V "left" that is required by the 1.8 drop led?

It is apparent that the resistor just "consumes" 5 - 1.8 volts magically.

I am confused by this, as you can calculate voltage dividers much more simply. Although, does a voltage divider "split" some of the voltage to ground? it is parallel however, so I am unsure why one branch (with resistor) affects the second branch (the output)

The 1.8V is a given for the LED; if current flows through a LED it will show a more or less constant voltage across it, just like a common silicon diode will drop about 0.7V at low currents. The anode will always be 1.8V higher than the cathode for your type of LED (the actual voltage mainly depends on the LED's color). The fact that this voltage is constant allows you to calculate the appropriate current limiting series resistor:

$R = \dfrac{V_+ - V_{LED}}{I_{LED}}$

or the current for a given resistor value:

$I_{LED} = \dfrac{V_+ - V_{LED}}{R} = \dfrac{5V -1.8V}{220 \Omega} = 15mA$

To answer we need to examine the I-V relationships of both the resistor and LED.

The resistor is easy as it has a linear I-V relationship:

So we know that the current is always linearly proportional to the voltage applied - if the current is 1mA at 1V, we know it will be 2mA at 2V and so on.

The diode is a bit more complicated, here is the I-V curve:

The relationship is exponential rather than linear - here is the "ideal" Shockley equation from Wiki:

Shockley diode equation The Shockley ideal diode equation or the diode law (named after transistor co-inventor William Bradford Shockley, not to be confused with tetrode inventor Walter H. Schottky) gives the I–V characteristic of an ideal diode in either forward or reverse bias (or no bias). The equation is:

where I is the diode current, IS is the reverse bias saturation current (or scale current), VD is the voltage across the diode, VT is the thermal voltage, and n is the ideality factor, also known as the quality factor or sometimes emission coefficient. The ideality factor n varies from 1 to 2 depending on the fabrication process and semiconductor material and > in many cases is assumed to be approximately equal to 1 (thus the notation n is omitted).

So what does this mean? It means at a certain voltage, the current will rise sharply (or we can see it as the resistance drops sharply) effectively "clamping" the voltage to a certain point (the physics is beyond the scope here, but worth reading about or asking on the physics stack) This point is dependent on the diode type, but will be around 1.8V for a typical red LED.
So if we have a resistor in series with a red LED, and we want to limit the current to 20mA (common operational current) we trace the I-V point on the curve. On the curve above we see that at 20mA the voltage drop is approximately 1.8V for the red LED, so we subtract this from the supply voltage and use Ohms law to calculate the resistor needed:

SO for a red LED running at 20mA from a 5V supply the equation is:

(5V - 1.8V) / 0.02A = 160 ohms.

Hopefully this clarifies things a bit.

• While what you tell is true, I'm not sure the Shockley equation is really useful here. For one thing Shockley applies to the theoretical diode model. And no design engineer will ever use the equation to calculate the series resistor. – stevenvh Feb 4 '12 at 9:03
• And I'm not even sure that LED follow exactly the Shockley's law, but I may be wrong. – clabacchio Feb 4 '12 at 9:16
• No diode follows exactly Shockley's equation, but it gives a basic shape of the I-V curve. Not essential to the answer maybe, but the OP may find it useful to follow up on. Of course I wasn't suggesting you would ever use this to calculate the series resistor, the last part of the answer covers that. – Oli Glaser Feb 4 '12 at 9:41
• @OliGlaser, so whatever voltage the LED is supplied, it still is 1.8V over it? That really makes sense why people do 5-1.8/0.020 or whatnot, as you're limiting the current to it and not the circuit. I think I am really getting to understand this. – Kim N. Feb 5 '12 at 21:04