The astable multivibrator circuit is a classic circuit for flashing two LEDs. It doesn’t have to flash two LEDs though. It can blink just one LED. Or it can create a tone to play on a speaker.
First, let me show you the circuit in action:
Want to know the theory behind how the circuit works?
There have been many attempts to explain this circuit. Most have failed to explain it to other than those already beyond the beginner level.
So here’s my humble attempt.
But let’s keep this interactive – ask me about the things you don’t understand in the comments section at the bottom. And I’ll update the article as I discover what is missing.
The Astable Multivibrator Circuit Diagram
This is a classic oscillator circuit.
The Light-Emitting Diode (LED) on the left side is lit when the transistor on the left side (Q1) is ON. The LED on the right side is lit when the transistor on the right side (Q2) is ON.
Resistors R1 and R4 are only there to set the current through the LEDs.
Which means the remaining six components make up the oscillator: Q1, Q2, C1, C2, R2, and R3.
Understanding the Astable Multivibrator
The voltage on the left side of C2 controls transistor Q1.
The voltage on the right side of C1 controls transistor Q2.
When transistor Q1 turns ON, it changes the voltage of C1 so that Q2 turns off.
After a short while, the voltage of C1 rises back up and turns on the transistor Q2.
When transistor Q2 turns on, it changes the voltage of C2 so that Q1 turns off.
This keeps repeating.
But that’s a very superficial explanation.
What if you want to understand why this happens?
Before we jump in
If you want to really understand how the astable multivibrator circuit works, you have to look more detailed at how the voltages over the two capacitors behave.
What do you need to know?
You need to know how transistors work.
And it is important that you have a good understanding of how voltages behave in a circuit, and how current flows.
These are all subjects you’ll learn when you join Ohmify, my online school for learning electronics.
The Detailed Explanation
A couple of things to help you before diving into the explanation…
1. Voltage is always measured between two points.
When we talk about the voltage at one specific point, it means the voltage measured from that point to the minus of the battery. (That’s why we call the minus of the battery 0V)
2. Think about the transistor as a switch.
It needs 0.7V on the middle pin (base) to turn ON. When it is ON, its top pin (collector) connects down to its bottom pin (emitter) so that current can flow through it.
This also means that the top pin has the same voltage as the bottom pin when the transistor is on. When the transistor is OFF, there are no connections between the top pin and the bottom pin, so no current can flow.
3. Use this simulator to see for yourself
I recommend verifying the things I am writing here by using a simulator. Here’s a great one that you can use right away (no login or anything needed):
http://www.falstad.com/circuit/e-multivib-a.html
When LED 1 is on
Let’s start by looking at the circuit when the LED L1 is lit and the other LED is off.
L1 is only lit when transistor Q1 is ON.
We know from how transistors work that Q1 is only turned ON if it has 0.7V on its base. Since the left side of C2 connects to the base of Q1, that means it’s at 0.7V.
The right side of the capacitor C2 connects to 9V through R4 and L2, so it is charging and the voltage is rising.
A capacitor charges exponentially, which means the voltage rises quickly in the beginning, then slows down more and more. The voltage reaches 7-8V quickly, but from there the voltage rises slowly.
The voltages around transistor Q2
Since the transistor Q2 is off, its base must be lower than 0.7V.
The right side of C1 connects to the base of Q2, so that means this is also lower than 0.7V.
But the right side of C1 is also connected to 9V through the resistor R2, which means it is being charged.
That means the voltage is below 0.7V but rising.
The Turning Point
So, the voltage on the right side of C1 is rising.
And when it reaches 0.7V, the action starts!
When the right side of C1 reaches 0.7V, that means the base of transistor Q2 gets 0.7V on its base and turns on.
…which means the LED on the right also turns on.
But when Q2 turns on, something interesting happens with the voltages we had over the capacitor C2…
Getting a negative voltage
We had that C2 had 0.7V on its left side and 8V on its right side.
Or to say it in another way, the left side was 7.3V lower than the right side.
But now that Q2 turns on, the voltage on the right side of C2 is suddenly pulled down to 0V through the transistor.
The internal charge of the capacitor does not change though, so the left side keeps being 7.3V lower than the right side.
But now that the right side is 0V, that means the left side becomes 7.3V below 0!
Yes, that’s -7.3V.
Transistor Q1 gets minus on its base
With -7.3V on the left side of C2, the base of transistor Q1 also gets -7.3V on its base, which turns it off.
So now, the left LED and transistor have turned off. And the right LED and transistor have turned on.
The left side of C2 starts at -7.3V and is charged through resistor R3 and therefore rising. Since it connects to the base of transistor Q1, when it reaches 0.7V, Q1 turns on again.
And so it continues.
The two transistors keep alternating between on and off, which makes the two LEDs alternate between on and off.
Questions?
I had so much trouble understanding the astable multivibrator circuit when starting out. And it frustrated me because I was thinking it was a simple and easy-to-understand circuit.
But the truth is that you need to have a good understanding of the basics of electronics before you will be able to understand it.
And you should have looked at several much simpler circuits first. Something I take all my Ohmify students through. Curious? Click to learn more about Ohmify.
Did this explanation help you understand the circuit? Or are you just more confused than when starting out? Let me know your comments and questions in the comment field below!
cannonball says
I like your explanation, but to make it easier for a beginner to understand, the switching of these transistors is very fast.
The sole purpose of the capacitors is to turn off the transistor to which the – lead is connected and stays off until the capacitor discharges.
To prove that this happens using a breadboard, pull out either capacitor and both LEDs will light up.
admin says
Great addition, thanks for your comment!
Clayton Benignus says
How would I modify this circuit to become a 3-way Multivibrator?
admin says
You mean 1 out of 3 LEDs on at a time? Then I’d introduce a counter, like the 4017, and have it reset itself after 3 counts.
phil says
Excellent explanation!
this is VERY good, clear and easy to understand.
at last, I get it !
thanks heaps and keep up the good work.
admin says
Glad you liked it!
Oyvind
Rudra says
What an excellent explanation for astable multivibrator. Just a question though, is there is a rule of thumb in deciding what resistors to use for R1/R2 and R3/R4 as shown in the accompanying diagram in the above article? I see that both R1 and R4 are 470 ohms and R2 and R3 are 47K ohms.
admin says
Hey Rudra,
R1 and R4 control the brightness of the LEDs. You can use this article to decide on their values:
http://build-electronic-circuits.com/current-limiting-resistor
R2 and R3 control the current the charges the capacitors. The more current that flows, the faster the capacitors charge, and the faster the LEDs blink. This means that lower resistance equals faster blinking and vice-versa.
Best,
Oyvind
Preshila says
I understand electronics, but i have never been confident of making an electronic device; especially those having TRANSISTORS and ICs, but this multivibrator worked, despite i used two different transistors (13001 & 8050 ).
Though it did not work instantly because :
I used an already used PCB
and
I used different values of components; which caused the the LED to blink very fast.
But after reading how the circuit works; i was then able to find a suitable resistor that would give me the desired frequency at which the LED would blink.
admin says
That’s really cool to hear!
Best,
Oyvind
Tony says
Very good explanation of a basic concept. Not all in electronics are chips.
Alexander says
Excellent explanation…but Im struggling with one tiny concept around the voltages on the base of the transistors.
As the base of both transistors has a hard connection to +9v, why do they ever actually turn off? The way my mind keeps seeing it is they are essentially in parallel with their controlling capacitor..so how come the bases dont just get a hard 9v on them?
Does the act of charging the capacitor rob the transistor base of the applied voltage, until the capacitor is charged?
If so, what are the mechanics of that- current flowing in the capacitor or some such?
Thanks in advance
Tomasz says
Hi Alexander.
Not very precise but mostly true is the statement “Current takes the path of least resistance”. Thus remembering that capacitor is conducting while being charged(regards RC time constant) the current from R2 finds easier way to mass flowing through the capacitor C1 and charging it in the meanwhile, followed by the Q1 collector-emitter than flowing directly to the transistor base Q2.
Respectively with R3, C2 and so on.
Bob says
Do not understand this:
The internal charge of the capacitor does not change though, so the left side keeps being 7.3V lower than the right side.
If you click on the capacitor C2 on http://www.falstad.com/circuit/e-multivib-a.html , Vd is changing on it, why ?
admin says
Are you looking at C2 when it is charging? And comparing the voltage just before and just after the transistors switch? I see about 3.7V just before, and maybe 3.5V just after. That tiny difference is just because I’m not able to stop the simulation at the exact switching point.
Sagor Hazra says
please add charging and discharging path of each capacitor at each step,it will be then more efficient to understand.
Vojtech Panek says
Please help me.
When LED 1 is on why Led 2 isnt lightining ,current can go through r4, LED2 , avoids closed q2 and go through capacitor c2 to ground through q1 .
admin says
A capacitor doesn’t allow DC to flow when it is charged. That’s why no current flows through c2 when LED1 is on.
Sebastian Reddy says
If the positive side of the battery attracts electrons from the negative side of the battery how is it possible to charge the capacitors when the negative side of the battery is blocked by the switched off transistors. Consequently how will the base of the transistors switch on the collector to emitter circuit within the transistors. Please assist with this crucial understanding of this oscillation circuit. How is it initialized when positive and negative of battery is open circuit in the transistors, ie switched off
admin says
In the beginning, both transistors will start to turn on as there is a path for the current to take through R2 and R3 to the bases of each transistor.
Tiny differences in resistance and capacitance of the components make one of the transistors turn on a tiny bit faster than the other. When that happens, the one that turn on will pull down the voltage of the capacitor (and thereby the voltage to the base of the other transistor) so that the other one gets turned “more off”.
And they continue turning each other on and off.
I highly recommend you checking out this simulation:
http://www.falstad.com/circuit/e-multivib-a.html
There you can really study what is going on in the circuit.
But there is a lot of things going on in this circuit, so if you don’t “see” how it works right now, don’t worry. With time it becomes more clear.
Oyvind
Sebastian Reddy says
So to initialize the electron flow between positive and negative of the battery through the circuit what causes it to happen since transistors are off at that initial stage when power is connected to the circuit
admin says
Both transistors start to turn on from the beginning through resistors R2 and R3.