4 years ago
Tuesday, December 16, 2008
having fun with leds is majaa you can do many things like a flip flop.its a simple circuit but it is using world wide in many big projects. what you need is:-
2-Leds (Light Emitting Diode)
2-Resistors (470 Ohms---Yellow, vilote, Brown)
2-Resistors(10k Ohms--- Brown, Black, Orange)
2-Capacitors (100 microfarads "µf" )
2-Transistors (Bc 547)
HOW THE CIRCUIT WORKS
When the power is applied, the slight difference in characteristics between the two transistors and electrolytics causes one transistor to turn on faster than the other. Suppose Q1 turns on faster viathe uncharged 100µf electrolytic C1, LED2 and the 470R resistor.
The voltage on the collector of Q1 will drop to about 0.35v and LED1 will light up. The positive lead of capacitor C2 will have 0.35v on it and this voltage will also be on the baseof Q2. Transistor Q2 will be turned off by this action but LED2 will come on for a short time while C1 charges.C2 begins to charge in the reverse direction (electrolytics can do this provided the voltage is not too high) and as the voltage rises above .6v, Q2 begins to turn on. This lowers the voltage on its collector and begins to turn on LED2.
The positive end of C1 is also connected to the collector and as the voltage drops, this effect is transferred to the base of Q1 via C1. This action begins to turn off Q1 and its collector voltage rises.Since C2 is connected to this point, the base of Q2 will see a rising voltage and it will turn on harder. In a very short time the two transistors have changed state.There's a little more concerning C1.
An electrolytic can be considered to be a rechargeable battery and when C1 is charged at the beginning of the cycle, it will have about 5v across it (for a 9v supply).
If we change this to a 5v rechargeable battery the explanation will be easier. The positive terminal of the battery will be connected to the collector of Q2 and when the transistor turns ON, the collector will be .35 above the negative rail. (the zero rail).
This means the negative terminal of the battery will be 4.85v BELOW the zero rail. In other words the base of Q1 will see a negative voltage of 4.85v.
And this is exactly what happens. The energy in the electrolytic will now be removed by the 10k resistor and after a short time the base will see a positive voltage of .6v and Q1 will begin to turn on and change the state of the circuit.
This is how the delay is created for each of the cycles.
Before we leave the multivibrator there's an important concept that should be explained.
Since each transistor is either ON or OFF, the circuit is classified as DIGITAL ELECTRONICS
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---------urs maroonrose------------
aslu udhaguleh nuvaane
ok i will make one 4 u maroon rose ok ennu dhoa
and butter fly
its really very simple
its look like lil difficalt dhoa
Mysterious!!
but Mysterious u know Physics is cool subject with that u can open ur mind to this tech world. And this curcuit is simple one and easy to follow steps
and Once u done it..
it will going make u happy