One of the simplest project that can be done in Electronics to practically understand some of the basic components and the formulas involved is the LED Circuit. The objective is to power a LED which then emits light, however it is not typically possible to directly connect the LED to the battery because a resistor is required to regulate the current that flows in the circuit (and goes into the LED) and the voltage of the LED.

This small circuit assumes that the reader has some knowledge on resistors and LEDs. If this is not the case, the two links below provide an introduction on these components.

Intro on Resistors

Intro on LEDs

Figure 1 shows the schematics of the circuit used to power a LED: there is a battery with as certain voltage $V_{battery}$, then there is a resistor $R$ with a specific resistance and finally there is a LED which needs a specific voltage and current to work properly.

While the current of LED is typically the same for many types of LED, around 20mA, the voltage range is often dependent on the colour of the LED. Figure 2 shows typical voltage ranges for different colours.

So, given a certain LED, the objective is to find the correct resistor that is able to provide enough voltage and current. The voltage from the battery, $V_{battery}$ is the same voltage shared between the resistor and the LED: $V_{battery}=V_{R}+V_{LED}$. With the Ohm's Law, $V=RI$, we can rewrite the formula $V_{battery}=RI+V_{LED}$. In this equation the only unknown is the resistance $R$ because, once a LED is selected, this is associated to a specific current $I$ (again, typically 20mA) and $V_{LED}$. So, the formula to find the resistor is:

\begin{equation} R=\frac{V_{battery}-V_{LED}}{I} \end{equation}

For example, with a Red LED the voltage is around $2V$ (i.e. 2 Volts) and the current is $20mA$, assuming a battery of $9V$ then the resistance is $R=\frac{9-2}{0.02}=350\Omega$. Given that a single resistor of $330\Omega$ exists in the market, it can be used because close enough to the ideal value (otherwise several resistors in series or parallel are required to obtain the value). It is important to remember that the power dissipated by the resistor is $W=RI^2$, which in this case is $W=330 \cdot 0.02^2=0.132$, given that different resistors have a different Maximum Wattage (like $\frac{1}{8}W$, $\frac{1}{6}W$, $\frac{1}{4}W$, $\frac{1}{2}W$, $1W$ and so on). In this case a resistor with $0.25W$ is the safe option given that it is almost twice the Wattage effectively needed. Figure 3 shows the design of the circuit on a breadboard with the appropriate resistor and the Red LED.

The section below has a calculator which allows to find the value of the resistor given specific battery voltage, led voltage and current.

This calculator finds the resistor required in the LED Circuit and its dissipated power.

Battery Voltage (V):  LED Voltage (V):      LED Current (A):      Calculate Values Calculation Status: Resistor Value ($\Omega$): Resistor Dissipated Power (W):

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The LED Circuit with LED Circuit Calculator
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