GENERATIONS

  This is a 3 VOLT SYSTEM. Using a (2N 5551 B011 NPN) transistor  Resistor 1 is 180 ohms and resistor 2 is 90 ohms The reason I used resistor 2 which is not needed, is that sometimes the bace(B) can get alot of volts which can cause  permanent damage to the transistor. When light is applied to the photoresistor, the base gets power hence creating a potential deference between resistor 1 and the collector (C) which causes a 0 or on power to the sound buzzer. But when the light is out, the output is 1 and the buzzer gets power so it produces sound I used this system on my door. So that when i open the door, the led moves with the door hence coursing the photoresistor to lose light and then it buzzers  This is the image of the sound buzzer

 This IC is an inverter logic gate ic. So if u can make an inverter gate then no need of buying this IC. Resistor R1 will control the blinking speed and resistor R2 will protect the LED from too much current. This type of blinking doesn't require a fixed type of voltage so any voltage values can work as long as they are not too small of a value.

 This is a simple blinking led using a bc547 transistor. As it is in the diagram. You will need 1 transistor , 1 LED , Capacitor and a Resistor. This will work well on a 12v battery. Other types of batteries may not do the trick. The value of the capacitor will determine the blinking speeds.

 The charger works like a 500 mA iPhone charger and will work with most modern phones. The main component in this circuit is the voltage regulator 7805 (U1). This is a component that takes an input voltage of 7V or above and gives out 5V. The diode (D1) placed in front of the input of the 7805 does two things: It makes sure you can't damage the circuit by connecting the plus and the minus of the battery the wrong way around. And it drops the voltage from the batteries 1v down, from 9V to 8V, which helps take some heat away from the voltage regulator. The Light-Emitting Diode (LED) is there to show you when the charger is working. The resistor R1 makes sure the LED doesn't have too much current flowing through it. The resistors (R2-R5) are there to tell your phone that this charger supports up to 500 mA (0.5 A) charging. If you don't know these components and their symbols, check out one of my most popular articles ever:  These are the components you'll need for this pro...

  In the usual circuits, the capacitor is not included in. So why did i add a capacitor. if u have ever built a touch switch using only a mosfet, u do know that when u touch on the off wires the mosfet will gain power over time and light up again. Which can be somewhat annoying. But when u add a capacitor as seen above. When u apply a plus or add the wires temporarily or use a push switch( this switch should not lock when pressed.), u charge the capacitor and when u release the capacitor still has charge hence it is still providing power and working as a line which joins both the gate(G) and the source(S) of the mosfet. REMEMBER; When u try to touch using this method it will not work coz the capacitor will need alot or current and voltage for it to charge at a small amount of time which ur fingers can't provide. So when u press the push switch on the gate(G) Drain(D) part. This will discharge the capacitor and provide power to the gate(G) at the same time 

  There are alot of circuits on how to do this. Most of them require you to place the photosensor between the base (B) of the transistor and the Emitter (E). But i have found this method not to be working well so i decided to place the photosensor between the resistor and the power supply (DC). So how is this method different. First the photosensor is already sending power even at low lighting. So if the source is 12v, that means that power though the photosensor under low light is approximately 8v. This means that 8v won't be able to course an effect on the relay since the relay is 12v based. When a bright light is put in a close distance with the photosensor, There will be low resistance on the photosensor hence the desired power which will follow to the relay The point where I have labeled as " AC power out" is the off point at default on the relay. What I mean is that the point with AC power out is the one is not connected inside of  the relay at first. I have used a ...