Welcome to Professor Mad.
Electronics for everyone.
Today we are going to learn about the
Blinking LED circuit
or also known as Astable multivibrator.
The blinking LED circuit is like the
electronics version of the “Hello World”-program.
It’s a simple electronic circuit
that gives you a visual cue if it works.
It was the first circuit most electronic enthusiasts ever built,
and it felt GREAT!
A multivibrator is an electronic circuit
used to implement a variety of
simple two-states.
Let's pay attention to one LED,
it oscillates between a “HIGH” state
and a “LOW” state
producing a continuous output.
Astable multivibrator,
means in which the circuit is not stable
in either state
It continually switches from one state to the other.
Lets see how this works.
For ease of understanding
let's use the schematic diagram
of the circuit.
This simple LED flasher blinker circuit
is made by using two NPN transistors
four resistors and two capacitors.
The speed of blinking can be changed
by changing the value of capacitors.
Let's forget about the capacitors for a moment.
At the beginning neither of the transistors are at "ON state".
The transistors act as off switches.
So as soon as we apply the supply voltage,
the voltage of both collector terminals becomes
equal to the supply voltage which is 9 V.
Both base terminals get 0.7 V through the resistors R2 and R3.
When a base terminal of a transistor gets the saturation voltage which is 0.7V,
the transistor acts as ON Switch.
Then the resistance between collector and emitter
terminal becomes 0 .
and collector voltage becomes equal to emitter voltage which is 0 V.
Then the current starts to flow between collector and emitter.
In this case both LEDs must light up together.
Most of the time we assume
the transistor instantly turns ON when it gets 0.7 V to its base terminal.
But in the real world it takes a few microseconds
to complete this turn ON action.
During this time the voltage of the collector gradually decreases
until it becomes equal to the emitter voltage which is 0 V.
But , how does the blinking happen?
That is where we need capacitors.
Let's again see what happens in the circuit when we connect the capacitors.
Again, At the beginning, neither of the transistors are “ON”.
So as soon as we apply the supply voltage,
the voltage of both collector terminals becomes equal to the supply voltage.
since the positive terminals of the capacitors
are connected to the collector terminals of the transistors,
both positive sides “P” and “S” also get the potential 9 V.
On the other hand both base junctions get 0.7 V through the resistors R2 and R3.
Since the negative sides of the capacitors “Q” and “R”
connected to the base of each transistor,
“Q” and “R” get the potential 0.7 V.
During this turning ON time duration,
voltage of Q and R sides of the capacitors which are connected
to the base terminals of the transistors are equal to 0.7 V,
but since the P and S sides are connected to the collectors of the transistors,
voltage of the P and S decrease gradually.
Let's pause here.
Another important thing that you must pay attention to is,
until the collector voltage becomes less than 0.7 V,
this whole time, the voltage of the capacitor's positive side is
higher than the negative side.
So the capacitors are charging.
But the capacitor charging time is much higher than the Transistor switching time.
So capacitors can't charge at the rate the transistor collector changes its voltage.
Let's pay attention to one side of the circuit at a moment
before the transistor turns ON.
While the collector pin potential was decreasing
and the capacitor was being charged,
let's assume at this moment,
the capacitor was able to achieve given potentials to its two sides.
That means the negative side of the transistor
has 0.3 V less potential than the positive side.
when the collector pin becomes 0 V,
the transistor turns ON
But keep in mind
that since the capacitor cannot discharge at that much higher speed as a transistor,
capacitor has to maintain the previous voltage difference
between the two sides for some time.
To maintain 0.3 V difference
while keeping the P side at 0 V,
the Q side becomes -0.3 V.
Let's go back again and see the full circuit.
There is a another important fact that we have to pay attention,
Even though both Transistor T1 and T2 are on a run to achieve “ON” state,
Both can’t achieve “ON” state simultaneously .
Always one will win the race by several microseconds.
The winner will be decided according to the resistance of the circuit paths
and several other unpredictable variables.
Let's assume transistor T1 turns ON first.
That means the voltage of the collector terminal will become zero.
Look closely, C1 has 0.7 V on its right side Q and 1 V on its left side P.
Or to say it in another way,
the left side potential was 0.3 V higher than the right side.
But now that T1 turns ON,
the voltage on the P side of C1 will be pulled down to 0 V through the transistor.
The internal charge of the capacitor does not change though,
so the Q side has to keep being 0.3 V lower than the right side.
So the Q side becomes -0.3 V
which is less than the knee voltage of the transistor T2.
So T2 is going back to turn OFF mode.
Now the circuit is turned ON,
only one led is turned ON.
Keep in mind that all of these initial steps only take a few milliseconds,
probably we cannot see these steps with the naked eye.
Let's again pay attention to the capacitor C1.
Since the transistor T1 is in its ON state,
now the side P which is connected to the collector pin is at 0 V.
This voltage will remain until the T1 becomes OFF state.
Side Q is charging through the resistor R2.
In another word we can say the capacitor C1 is discharging.
And on the other hand, look at the capacitor C2.
Since transistor T1 is in “ON state” ,
Now the R side, which is connected to the Base pin is at 0.7 V.
This voltage will remain until the T1 becomes “OFF state”.
At the same time, Side S is charging through the resistor R4.
Let's combine all these together.
Voltage of the “Q side” increases gradually from -0.3 V to 0.7 V .
And the “S side's” voltage gradually increases from 1 V to supply voltage 9 V.
When the plate Q becomes 0.7 V,
the Base of transistor T2 becomes 0.7 V and the T2 turns ON.
Now again we had that C2 at 0.7 V on its left side R and 9 V on its right side S .
Or to say it in another way,
the left side was 8.3 V lower than the right side.
But when the T2 turns ON,
the voltage on the S side of C2 is suddenly pulled down to 0 V
through the transistor.
The internal charge of the capacitor does not change though,
so the R side keeps being 8.3V lower than the S side.
Which is -8.3 V.
With -8.3 V on the R side of C2,
the base of transistor T1 also gets -8.3 V on its base,
which changes it to “OFF state”.
Now T1 is “OFF” and T2 is “ON”.
So the LED 1 will turn OFF and LED 2 will turn ON.
Let's see what's happening in capacitors.
since the transistor T2 is in its ON state,
now Capacitor C1 is charging through the resistor,
so its voltage increased gradually from -8.3 V.
On the other hand T2 keeps the voltage at 0.7 V at the base,
that means the plate Q of C1 is at +0.7 V the whole time T2 is ON.
But the R side of the C2 is discharging through the resistor furthermore.
Another thing that you must notice here is,
the resistance of resistor R1 and R3 is much less than the resistance of R2 and R4.
Hence the P side of C1 is charging faster than the R side of C2.
When the plate R becomes +0.7 V, the T1 turns ON again.
And It leas the T2 to turn off.
Again the C1 starts discharging and C2 is charging.
C2 is charging faster than the C1.
This happens over and over again, which creates the astable multivibrator.