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- Photoresistor (LDR)<\/li>\n
- 100K\u2126 Resistor<\/li>\n
- OPAM<\/li>\n
- LED<\/li>\n
- 2 AA batteries or DC Power Supply<\/li>\n<\/ul>\n
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<\/p>\nA photoresistor is a light-sensitive resistor that changes resistance in response to light exposure. Photoresistors are used in a variety of applications, such as light dimmers and camera light meters. When light falls on a photoresistor, its resistance decreases. This decrease in resistance can be used to control the flow of electricity in a circuit. It is a light-sensitive resistor that can be used to measure light intensity. Photoresistors are made of semiconductor materials, such as cadmium sulfide (CdS) or cadmium selenide (CdSe). When light shines on a photoresistor, the material’s resistance decreases. The amount of resistance depends on the intensity of the light. Photoresistors are used in a variety of applications, such as light-sensitive switches, security systems, and light meters. They are also used in some types of optical sensors.<\/p>\n
A photoresistor is a resistor whose resistance changes according to the amount of light that it is exposed to. When exposed to total darkness, the photoresistor’s resistance is very high. This is called its rated dark resistance. For example, in our case, we are using a 2M\u2126 photoresistor. This means that when exposed to total darkness, its resistance will be around 2M\u2126. As the photoresistor is exposed to increasing amounts of light, its resistance begins to drop significantly. There is another rating on the datasheet labeled cell resistance @ illuminance. This is the resistance that the photoresistor will drop to when exposed to bright light, typically 10 lux. This is also called light resistance (since it is the resistance the photoresistor has when exposed to bright light). For our photoresistor, in particular, the cell resistance @ illuminance is 20-30k\u2126. This means the resistance of the photoresistor will drop to 20-30K\u2126 when exposed to a 10 lux level of light.<\/p>\n
So a photoresistor is basically a device that gives off very high resistance at dark light levels and low resistance at high light levels. Being that it does this, it can act as a sensor for light or a photosensor. With the photoresistor acting as the light sensor, the other significant component is the 2N3904 transistor. The transistor in this circuit will act as an amplifier, amplifying current so that sufficient current is produced to light the lamp. Any type of LED can be used for this circuit. The power we will use for this circuit is 3 volts, which can be obtained from either 2 AA batteries in series or from a DC power supply set to this voltage.<\/p>\n
The procedure of Dark Activated Circuit Design:<\/h3>\n
So, for this circuit, 3 volts are powering the circuit. These 3 volts are in parallel to a 100K\u2126 resistor and the photoresistor.<\/p>\n
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<\/p>\nWhen exposed to bright light, the photoresistor’s resistance is very low. It drops to around 20-30K\u2126. Current travels through the 100K\u2126 resistor and then has 2 paths- it can either go through the base of the transistor or go through the photoresistor. The base of the resistor to the collector has a resistance of around 400K\u2126. Current always takes the path of least resistance. When the photoresistor is exposed to bright light, its resistance is about 20-30K\u2126, which is significantly less than the 400K\u2126 of resistance the base of the transistor has. Therefore, most of the current will go through the photoresistor and very little will go to the base of the transistor. So the base of the transistor is bypassed. Thus, the transistor does not receive enough current to turn on and power the LED. Thus, the LED is off when there is a lot of light in the surroundings.<\/p>\n
However, when it begins to get dark, the photoresistor’s resistance becomes very high. Its resistance goes up to over 2M\u2126 of resistance. This creates a very high-resistance path. Being that 2M\u2126 is significantly greater than the 400K\u2126 of resistance that the base of the transistor offers, most of the current will go through the base of the transistor. This means that current does not go through the photoresistor when it is dark, due to this high resistance. Instead current goes through the 100K\u2126 resistor and through the base of the transistor. The transistor receives enough current to power on and turns on the LED connected to the collector terminal.<\/p>\n
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<\/p>\nSo this is how a dark-activated light circuit can work. Again, as always, variations of this circuit can be done. Instead of using an LED, we can use any other type of lighting fixture such as a lamp. You may want to use a bright LED. Or you can use any other various lighting source. You may want to use multiple lights, so you can place different lights in parallel with one another. All that would be needed is an adjustment in the voltage and current of the circuit. Customize the circuit according to your needs and preferences.<\/p>\n
Live Simulation: https:\/\/dcaclab.com\/en\/experiments\/61569-dark-activated-led-light<\/a><\/p>\n