How to Hook Up an LED Light to a Battery
Led Circuit used as a Mini Lamp
Led Circuit is Used as a Source of Light
I´ve created this little mini lamp. This is a lamp that can be used as a source of illumination either when writing or reading or other kind of night time activity that requires a little light in order to be performed. There are many other applications for this simple circuit. For example, it can be used as a piece of home interior decoration or as the source of light for the outside or interior of your home, or it can be used as the front light for a bicycle. Other uses depend on your imagination.
Hooking a led (light emitting diode) to a battery is a very simple and easy electronic kind of process. LEDs are some of the most used light emitting devices for an ample variety of purposes, such as advertising, signaling, and for capturing the attention of people for commercial and public purposes. They´re also used in many home appliances, such as TVs, DVD players, remote controls and other electronic devices.
LEDs can be used to create a number of attractive circuits. In this hub I will show you how to connect a led to a battery. This simple led circuit can then be used as a lighting device for practical purposes. There is some basic tools and components that you´ll need to connect a led to a battery. Most of these items can be found in an electronic supply store such as radio shack. To hook up an led to a battery you´ll need some basic tools, such as pliers, voltmeter, and a few components, including black and red wiring, a 9 volt battery and a variable resistance.
Components Needed to Hook up an Led to a Battery
- One Led
- One 9 V battery
- One battery clip
- 330 Omh resistor
- 22 gauge black and red wire
- One variable resistance
- Wire cutters
- Wire strippers
Led Circuit Diagram
With a Variable Resistance
Formula for the Resistance of the Circuit
9 Vs - 3.2 V Led / 0.020 amperes
= 290 Ohms
Hooking an Led to a Battery
For this circuit, I have used an ultra-bright white led with a voltage drop of 3.2 volts; a 9 V battery; a 330 Ohm resistor; insulating red and black wire gauge 22. I have also used a variable resistance so that I can dim the light of the led. On the right, there is a list of all the components used in this circuit.
In order to light up an led, you need to supply the voltage or power. In this diagram, I´m using a 9 volt battery, which is connected to the wire end of a 330 Ohms resistor. The other end of the resistor is connected to the positive end of the led (anode), which corresponds to the long leg of the led. The short leg of the led (cathode) is the negative side and is connected to the negative side of the battery, which is where the circuit is closed.
The led in this circuit needs to be wired correctly with the longer pin (the anode) connected in orientation to the positive side of the battery and the shorter pin of the led (the cathode) connected to the negative side of the battery. The led is a polarized component. If the pins are inverted, it won´t conduct a current.
Also you´ll have to supply the correct voltage, as the led has a voltage threshold of 3.2 volts. When this threshold is not exceeded, the led will not conduct a current, whereas if the voltage needed is supplied, the led will conduct a current and light up; however if the voltage is exceeded, you risk burning the led.
The LED in this circuit has a voltage drop of 3.2 volts. When this voltage is reached, the LED will conduct a current and emit electromagnetic energy. To calculate the resistor needed to attain the desired current, you can use the formula shown to the right, where the LED´s voltage, in this case 3.2 volts, is substracted from the main power supply voltage, which is 9 volts.
The result is then divided by the desired operating current, which in this circuit is 20 mA. The final result should is 290 Ohms. Since there are no resistors with that resistance number, I used a 330 ohm resistor with a 20% resistance tolerance, which gives me 66% tolerance resistance value either up or down from the 290 Omhs I need.
So the 330 ohm resistor can work well in this circuit to lower the voltage to the desired level to conduct a current.
Led Circuit on a Breadboard
In the circuit, the potential difference or voltage in the 9 V battery. produces a current, which is in turn limited by the resistor. The needed voltage and current are then supplied to the desired amounts to the led. The resistor is used to to drop the voltage from the 9 V battery to a voltage that can be used by the led.
on the picture to the right, I have wired the circuit on a breadboard, which is a tool to design electronic circuits before building them on permanent printed circuit board. You can buy one at any electronics store for a modest price. A breadboard is designed to allow you to insert the circuitry of your designs so that you can build any circuit without the need to solder a single component.
On this circuit, I have wired the led circuit on a breadboard with all the components. the black test lead of the voltmeter corresponds to the negative side of the battery and the red test lead corresponds to the positive side of the battery. The variable resistance, which is the knob behind the red lead of the voltmeter, can be used to dim the led higher or lower. Notice the order in which the circuit is wired on the breadboard and compare it to the above diagrams.
Ohm´s Law Triangle
Voltage over current voltage over resistance (Vertically)
current multiplied by resistance (Horizontally)
With a voltmeter, you can use Ohm´s law to measure voltage, current and resistance throughout the circuit. The voltage drop of any part across the circuit is equal to the current multiplied by the resistance, thus, according to Omh´s law:
V = I x R; I = V / R; R = V / I
In the picture to the right, there is a useful way to remember the formula for voltage, current and resistance, and on this site, there is an explanation on how to use Ohm´s law to check these values on a circuit.
How Does an Led Emit Light?
LEDs are made of semiconductor materials. When a led is polarized directly, a current flows and causes the electrons in the semi conductive material to recombine with the electron holes. This produces a release of energy in the form of photons. The energy released is determined by the type of semiconductor material. Gallium Arsenide Phosphide emits wavelengths of red energy, whereas LEDs made from Silicon carbide emit blue photons.
All LEDs emit electromagnetic radiation when they´re polarized directly. There are a variety of types of LEDs that emit in an ample wavelength of colors, including the invisible infrared, visible light and ultraviolet. While some LEDs can emit light of only one color, there are others that emit in multicolored wavelengths. There are LEDs that can flash and others are able to emit two colors.
LEDs and Semiconductor Materials
Gallium arsenide phosphide
Aluminium gallium indium phosphide
585-595nm Yellow 2.2v
Aluminium gallium indium phosphide
Indium gallium nitride
Blue/UV diode with yellow phosphor