LED FLIP FLOP CIRCUIT ,USING BC547 TRANSISTOR |LED PROJECT
As a beginner in electronic projects, I loved making this simple LED Flip-Flop circuit! It's fun to build and also teaches you about basics of electronics. Now a days with Arduino and similar development boards, it has become rare to see such elegant basic electronic projects without any microcontroller. So let's use basic electronic components to build this magical circuit and learn how magically it works as well!
Let's keep the transistor as our starting point and start rigging up the circuit as shown in the figure.
Connect the emitter of transistor 1 (Let's call it Q1) to the emitter of
the other transistor (Q2).
Connect Q1's base to the negative of the capacitor (C2).
Connect Q2's base to the negative of the capacitor (C1).
Connect the collector of Q1 to positive of the capacitor (C1).
Connect the collector of Q2 to positive of the capacitor (C2).
Let's go ahead with the LED part. It's nothing new, by now you would know that a LED is almost always connected in series with a resistor. You might also know that this LED is called as 'current limiting resistor'.
Connect 470 ohm resistor to positive of LED 1
Connect the negative lead of the LED 1 to the positive of the capacitor C1 (to the same node where collector of Q1 is connected).
Connect 470 ohm resistor to positive of LED 2
Connect the negative lead of the LED 2 to the positive of the capacitor C2 (to the same node where collector of Q2 is connected).
By now you would have observed that the LED part of the circuit is exactly symmetrical with respect to right and left plane.
Now connect a lead of 10K resistor to the negative of capacitor C1 and similarly another resistor to negative of C2.
To drive an LED you'll need a 3.3 forward voltage minimum. So any DC power supply between 4.5V to 9V would do great!
Here, I've used a 9V battery for my convenience.
Connect Positive of the power source to all the other ends of the resistors that we've not connected yet.
Connect Negative of the power source to the emitter of both the transistor that we've shorted.
Although the circuit seems quite simple to build, it has quite complex explanation for the way it works. To understand it in the most simple terms, as you can see in the circuit, transistor Q1 is controlled by the resistor-capacitor pair (10K resistor) which acts as oscillator (with the help of transistor which works as stable multivibrator).
For simplicity, let's consider the transistors as a switch between the collector and the emitter. For the base of the transistor, let's represent it as a diode (because it has the same behavior).
Initially, when we power up the circuit, both the transistors act like open switch. The base of both transistors has a threshold voltage of 0.7 and below that voltage, the transistors will remain open.
Now, at the collector of the transistors, the voltage will be supply voltage. Now, since the middle resistors (10k) are of much greater value than the outer resistors (470 ohm) the capacitors will take some time to fully charge up.
As the capacitor starts to charge, because of component tolerances, one side of the circuit will be slightly faster in charging up the capacitor than the other. After a while, this capacitor reaches the threshold value of 0.7V and this causes the transistor to power up and closes the transistor. The LED of that side will light up now.
By this time, the other side of the circuit would have undergone the same process so before that part of the LED starts glowing, this first transistor closes and causes the LED to light up, the voltage at this end of the capacitor drops to zero And the voltage built up on the capacitor becomes negative and causes the other side of the transistor to open. The same process repeats on the other side as well and it keeps on repeating.
What is the very interesting thing that happens here is how when the left side capacitor (Q1) starts to charge and reaches the threshold voltage, the circuit closes the right transistor (Q2) and lights up the right side LED (LED 2). As soon as the LED is turned ON, the voltage at the capacitor becomes negative (because when the charging of the capacitor reaches the threshold, it starts to conduct and hence it drops to zero volts). This negative voltage at the base of the other transistor (Q1) and hence this transistor acts as open. And this cycle keeps repeating.
It is important that the internal resistors must be of the same value. Also both the capacitors must be of the same value.
The oscillation frequency is set by the internal Resistor-Capacitor network and it is given by f=1/(1.38RC)
Since I've chosen R=10k and C=100uf, I'll get frequency around 0.7 Hertz per second.
If you want to blink a single LED, simply remove the other LED and connect the resistor directly.
This circuit has many applications,
They can be used to get a square wave.
They are used as vehicle indicators.
They are used in toys to produce flashin
g lights.
They are also used as turning indicators.
LED FLIP FLOP CIRCUIT ,USING BC547 TRANSISTOR |LED PROJECT
As a beginner in electronic projects, I loved making this simple LED Flip-Flop circuit! It's fun to build and also teaches you about basics of electronics. Now a days with Arduino and similar development boards, it has become rare to see such elegant basic electronic projects without any microcontroller. So let's use basic electronic components to build this magical circuit and learn how magically it works as well!
Let's keep the transistor as our starting point and start rigging up the circuit as shown in the figure.
Connect the emitter of transistor 1 (Let's call it Q1) to the emitter of
the other transistor (Q2).
Connect Q1's base to the negative of the capacitor (C2).
Connect Q2's base to the negative of the capacitor (C1).
Connect the collector of Q1 to positive of the capacitor (C1).
Connect the collector of Q2 to positive of the capacitor (C2).
Let's go ahead with the LED part. It's nothing new, by now you would know that a LED is almost always connected in series with a resistor. You might also know that this LED is called as 'current limiting resistor'.
Connect 470 ohm resistor to positive of LED 1
Connect the negative lead of the LED 1 to the positive of the capacitor C1 (to the same node where collector of Q1 is connected).
Connect 470 ohm resistor to positive of LED 2
Connect the negative lead of the LED 2 to the positive of the capacitor C2 (to the same node where collector of Q2 is connected).
By now you would have observed that the LED part of the circuit is exactly symmetrical with respect to right and left plane.
Now connect a lead of 10K resistor to the negative of capacitor C1 and similarly another resistor to negative of C2.
To drive an LED you'll need a 3.3 forward voltage minimum. So any DC power supply between 4.5V to 9V would do great!
Here, I've used a 9V battery for my convenience.
Connect Positive of the power source to all the other ends of the resistors that we've not connected yet.
Connect Negative of the power source to the emitter of both the transistor that we've shorted.
Although the circuit seems quite simple to build, it has quite complex explanation for the way it works. To understand it in the most simple terms, as you can see in the circuit, transistor Q1 is controlled by the resistor-capacitor pair (10K resistor) which acts as oscillator (with the help of transistor which works as stable multivibrator).
For simplicity, let's consider the transistors as a switch between the collector and the emitter. For the base of the transistor, let's represent it as a diode (because it has the same behavior).
Initially, when we power up the circuit, both the transistors act like open switch. The base of both transistors has a threshold voltage of 0.7 and below that voltage, the transistors will remain open.
Now, at the collector of the transistors, the voltage will be supply voltage. Now, since the middle resistors (10k) are of much greater value than the outer resistors (470 ohm) the capacitors will take some time to fully charge up.
As the capacitor starts to charge, because of component tolerances, one side of the circuit will be slightly faster in charging up the capacitor than the other. After a while, this capacitor reaches the threshold value of 0.7V and this causes the transistor to power up and closes the transistor. The LED of that side will light up now.
By this time, the other side of the circuit would have undergone the same process so before that part of the LED starts glowing, this first transistor closes and causes the LED to light up, the voltage at this end of the capacitor drops to zero And the voltage built up on the capacitor becomes negative and causes the other side of the transistor to open. The same process repeats on the other side as well and it keeps on repeating.
What is the very interesting thing that happens here is how when the left side capacitor (Q1) starts to charge and reaches the threshold voltage, the circuit closes the right transistor (Q2) and lights up the right side LED (LED 2). As soon as the LED is turned ON, the voltage at the capacitor becomes negative (because when the charging of the capacitor reaches the threshold, it starts to conduct and hence it drops to zero volts). This negative voltage at the base of the other transistor (Q1) and hence this transistor acts as open. And this cycle keeps repeating.
It is important that the internal resistors must be of the same value. Also both the capacitors must be of the same value.
The oscillation frequency is set by the internal Resistor-Capacitor network and it is given by f=1/(1.38RC)
Since I've chosen R=10k and C=100uf, I'll get frequency around 0.7 Hertz per second.
If you want to blink a single LED, simply remove the other LED and connect the resistor directly.
This circuit has many applications,
They can be used to get a square wave.
They are used as vehicle indicators.
They are used in toys to produce flashin
g lights.
They are also used as turning indicators.
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