Scheme astable multivibrator is a classic circuit for flashing the two LEDs. Not necessarily two LEDs blink. only one LED can blink.
First, let me just show you an example of one of the schemes in action:
Do you want to know the theory of how the circuit works?
So, here is my modest attempt.
Basic circuit astable multivibrator
It's a classic scheme of unstable multivibrator.
LED L1 is lit, when the transistor Q1 is turned on. LED L2 is lit when the transistor Q2 is turned on the right side.
Resistors R1 and R4 are only for setting the current through the LEDs.
This means that the remaining six components constitute oscillator: Q1, Q2, C1, C2, R2 and R3.
Understanding the astable multivibrator
Tension on the left plate C2 controls transistor Q1.
The voltage on the right plate C1 controls the transistor Q2.
When transistor Q1 is turned on, it changes the voltage of C1, so that Q2 is turned off.
After some time, the voltage rises again and C1 includes a transistor Q2.
When transistor Q2 turns on, it changes the voltage of C2, so that Q1 is turned off.
This is repeated.
But this is a very superficial explanation.
What if you want to understand why this is happening?
If you want to really understand how the circuit works unstable multivibrator, you need to look in more detail the behavior of the voltage across the two capacitors.
What else do you need to know?
You need to know how transistors work.
It is important that you are well aware, behave as a voltage in the circuit and current flows as a.
Detailed explanation
A few things that will help you, before diving into an explanation ...
1. Voltage is always measured between two points
When we talk about the voltage in one particular point, which means the voltage measured from this point to the battery minus. (That's why we call the negative battery 0V)
2. Think of a transistor as a switch
To activate, 0.7 on average output (base). When it is on, its upper terminal (collector) connected to the lower terminal (emitter), so that through current can flow.
This also means that the upper pin has the same voltage as the lower output when the transistor is turned on. When the transistor is turned off, the contact between the upper and lower contact connections have, therefore, the current can not flow.
3. Use the simulator to see this
I recommend checking out the things that I write here, using a simulator. Here is a great example that you can use directly (without login or anything else):
http://www.falstad.com/circuit/e-multivib-a.html
When the LED 1 is turned on
Let's start by looking at the circuit, when the LED L1 is lit, and the other LED is off.
L1 is lit only when the transistor Q1 is turned on.
From how the transistors, we know that Q1 is turned on only if its base is 0.7 V. Since the left plate of C2 is connected to the base of Q1, this means that it is at 0.7 V.
The right plate of capacitor C2 is connected to 9B through R4 and L2, so it is charged and the voltage increases.
The capacitor is charged exponentially, which means that the voltage initially increases rapidly and then slows down more and more. Voltage quickly reaches 7-8V, but there is growing tension slowly.
Stresses around the transistor Q2
Since transistor Q2 is off, its base should be below 0.7 V.
The right plate of C1 is connected to the base of Q2, which means that it is also below 0.7 V.
But also the right plate of C1 is connected to a 9 V through resistor R2, which means that it is charging.
This means that a voltage below 0.7 V, but growing.
A turning point
Thus, the voltage on the right plate C1 increases.
And when it reaches 0.7 V, the real action begins!
When the right plate C1 reaches 0.7 V, this means that the base of transistor Q2 receives the 0.7 volts at its base and is included.
... That means that the LED on the right is also included.
But when switched on Q2, there is something interesting to stress that we have had over the capacitor C2 ...
Preparation of negative voltage
We did that at C2 was 0.7 on the left plate and 8 on the right.
Or, in other words, the left plate has a potential of 7.3 V lower than than the right.
But now, when Q2 is turned on, the voltage on the right plate C2 suddenly drops to 0 V through the transistor.
The internal capacitor charge does not change, so the potential at the plate is left at 7.3 In is lower than on the right.
But now, when the potential of the plate on the right is equal to 0, this means that the potential of the left plate becomes 7.3 V lower than 0!
Yes, it is -7.3 V.
Transistor Q1 receives negative at its base
At -7.3 in the left plate of C2 - Q1 transistor base and gets -7.3 in at its base, which eventually shuts it down.
So now the left LED and the transistor is turned off. And right LED and the transistor is turned on.
The left plate C2 begins to -7.3 V and charged through the resistor R3 and therefore the voltage rises. Since it is connected to the base of transistor Q1, when it reaches 0.7 V, Q1 turns on again.
And so it goes.
Two transistors constantly switch between on and off, allowing the two LEDs alternately switched on and off.
Questions?
I had so many problems with the understanding of the circuit unstable multivibrator when it starts. And it upset me, because I thought it was simple and straightforward scheme.
But the truth is that you need a good understanding of the basics of electronics before you can understand this circuit.