# Resistor and LED - together and separated

Something very basic which I don't get no matter how much I read.

• circuit A: only a resistor and power supply - the resistor will burn.
• circuit B: only a LED and power supply - the LED will damaged.

As far as I understand these are the facts.

• in circuit B LED will draw as much current as we give it (we should limit it to about 20mA)
• in circuit A resistor will burn because it cannot dissipate the energy since the current is very high

If the LED is not limiting the resistor current, how come the mix of both is working, and the resistor does not burn?

Let me try to clarify. Let's take an example: in a series circuit of LED and resistor,

• $V_\text{in} = 12\text{ V}$
• LED $V_f = 2\text{ V}$
• LED $V_i = 20\text{ mA}$
• $R = \frac{10\text{ V}}{0.02} = 500\ \Omega$
• $P = 0.2\text{ W}$

The voltage across the resistor is $10\text{ V}$.

If you hook the same resistor to a $10\text{ V}$ power supply, wouldn't it burn?

For a resistor, the power is proportional to the square of the voltage across. Too much power and the resistor will 'burn':

$$p_R = \frac{V^2_R}{R} = \frac{V^2_{S}}{R}$$

where $V_S$ is the voltage from the power supply.

An LED has a nearly constant voltage across over a wide range of current through.

Thus, if the supply voltage is significantly larger than the nominal LED voltage, the current will be far larger than nominal current and the LED will likely be destroyed.

However, if a resistor is connected in series with the LED, the voltage across the resistor is reduced by the voltage across the LED:

$$V_R = V_S - V_{LED}$$

thus, the power dissipated by the resistor is reduced

$$\frac{V^2_R}{R}=\frac{(V_S - V_{LED})^2}{R} < \frac{V^2_{S}}{R}$$

Further, the current through the LED is fixed by the series resistor (series connected components have identical current):

$$I_{LED} = I_R = \frac{V_R}{R} = \frac{V_S - V_{LED}}{R}$$

For example, let the nominal LED voltage be $V_{LED} = 2V$ with nominal current $I_{LED} = 20mA$.

Let the supply voltage be $V_S = 2.5V$ and let the resistor be $R = 25 \Omega$ with a power rating of $0.125W$.

If the resistor is connected directly across the power supply, the power dissipated is:

$$p_R = \frac{(2.5V)^2}{25\Omega} = 0.25W$$

The resistor is dissipating twice as much power as it is rated for - the resistor will overheat.

However, if the resistor and LED are series connected across the power supply, the LED current is

$$I_{LED} = \frac{2.5V - 2V}{25 \Omega} = 20mA$$

and the resistor power is

$$p_R = \frac{(2.5V - 2V)^2}{25 \Omega} = 0.01W$$

which is much less than the rated power - the resistor will not overheat.

I assume you mean 'the mix' is a circuit where the resistor and the LED are connected in series.

To calculate the current flowing through both of them (when connected in series, the current through both of them must be the same, the current can't 'leak') one can see the LED as a device over which a constant voltage drops (see e.g. this graph on Wikipedia).

Subtract this voltage from the supply voltage and you get the voltage over the resistor, from which one can calculate (using Ohm's law) the current through the resistor and the LED.

From current through and voltage over the resistor one can calculate the power dissipated in the resistor. If it is within the maximum power rating, the resistor is normally not damaged.

For LEDs, typically a maximum current is specified. If the current (calculated above) is within this limit, there should be no damage to the LED either.

As you can see from the graph linked above does the current increase very quickly above the threshold voltage $V_d$, leading to huge currents very quickly as one increases the voltage. This current is essentially only limited by the very low internal resistance of the power supply and the wires.

To answer the original question, one can see the resistor as a current limiting component (thus limiting the current through the LED) and the LED as a component which reduces the voltage over the resistor. Or in other words, they 'share' the supply voltage.

If the power supply supplies a given voltage, and the resistor and LED are in series, then there with be a lesser voltage across the LED and a lesser voltage across the resistor. The sum of the two voltage drops with equal the power supply voltage.