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):
https://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!
Michel Verdieu says
Curious to know how can I predict which LED starts blinking first after this circuit has been powered up ?
admin says
Hey Michel, I don’t know of any way of predicting that. But my first guess would be to check which resistor – R2 or R3 – has the lowest resistance. Assuming the capacitors are equal, the one with less resistance should charge faster than the other. And thereby turn on the opposite transistor and LED.
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.
john klicpera says
voltage is a source of power with added to amps
when forming a circuit with components create
different phase. to make components work on circuit board have frequency response .
amps is power force back in 70s when went to a
school need a refresher course p .s. p=e/I or something e=I/r n r I radio instate thanks
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:
https://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
Scott says
r1 and r4 also change the timing of the circuit. When led 2 is off, c2 is charging up to rail voltage using the time constant r4×c2. The amount of charge the capactor depends on this. Led 1 is on and the right side of c1 is charging up from a negative voltage value via r2. When it gets to 0.7v above ground, q2 switches on. The right side of c2 is referenced to ground (or just above) and the left side is below ground by the amount that c2 was able to charge to in the time given by r4 c2 and r2 c1.
Wilson Lalo says
Thank you for your explanation, it helps alot, I now understand the concept of charging and discharging the capacitor
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.
Passiday says
This is confusing: “The current from R2 finds easier way to mass flowing through the capacitor C1 and charging it in the meanwhile” – the polarity of the C1 does not allow it to be charged by the current flowing from R2, does it? Also in the simulator, the current from R2 flows through C1 only when it is discharging.
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.
Scott says
When the circuit is turned on both capacitors start to charge via r2 and r3, but not at exactly the same rate due to component tolerances and initial charge. The first capacitor to reach approximately 0.7 volts above ground will switch the opposite transistor on and start the process.
Vince says
Hi Oyvind,
Really enjoyed your explanation of the astable vibrater circuit. Very simple and easy to understand. I especially like the way you illustrated it and showed the relation of voltages between the various components. Thanks again for your self-sacrificing efforts to help us all. Vince ( Van. B.C. Canada)
admin says
Hi Vince, thanks a lot! Hearing that it helped you really means a lot to me.
Oyvind
Artem says
Thank you for the article. One question though. How do I choose the components to make the LEDs blink with the desired frequency?
Mercy says
Thanks for the explanation
Does capacitor rating affect the set up (uf by volt)
I used 10uf by 25v but it’s not working
admin says
25V should work fine.
neil says
do capactiors only charge through the lesser resistance (470) or is it because of the + on the cap is pointing to the lesser resistor? Why couldn’t they charge through the other ones?
admin says
It can charge both ways. But it will charge in the direction that current flows. And current always flows from a higher potential (Volt) to a lower potential (Volt).
alabima says
tanks so much av been studying it lng tym but with your explanation now I understand its workings but please what criteria to follow in choosing d capacitor
Electronics guy says
When you are explaining the voltages of the q2 transistor the c1 capacitor gets charged in reverse (the positive side with the plus sign has LOWER potential than the other side).Why doesn’t the capacitor blow up/break?Thanks in advance for the answer.
admin says
Good catch! You’re right – that happens. But it’s only 0.7V, so it’s not enough to damage the capacitor.
Electronics guy says
Thanks for the answer!And btw how do the capacitors discharge in the circuit?
Electronics guy says
Or do they discharge at all?
Electronics guy says
I am pretty sure that they discharge in some way,because it wouldn’t make sense for them to not to.
Electronics guy says
How do the capacitors discharge in the circuit?
admin says
Capacitors discharge when there is a path for the current to take from the side with the highest potential to the side with the lowest potential.
Passiday says
Thank you for the explanation! Was looking very hard to find one that goes into detail, and not trying to explain what happens “simple but wrong”.
Here are the things that feel very counter-intuitive for me:
– In the simulator, there are two voltage scopes attached to plain wires (the ones that are crossing in the middle). The idea that a plain wire has any voltage drop across its ends, feels weird. I mean, if resistance is zero, then any voltage results in infinite current. What, obviously, is not the case, in real life.
– The negative voltage at the minus end of the capacitor. I thought there could be no lower voltage anywhere in the circuit than ground. Apparently, just a wrong assumption, led by faulty intuition.
– In the simulator (your falstad.com link) one can see that the current flowing through LEDs is almost never zero. When its respective transistor is open, then the current is at its max, but when the transistor is closed, then it is charging its capacitor. So, I figure then that the switching is actually not immediate, there is some moment while the both LEDs are on, until the current that charges capacitor falls below a certain threshold level and LED goes out.
– So, in your section “When LED 1 is on”, you say “The right side of the capacitor C2 connects to 9V through R4 and L2, so it is charging and the voltage is rising.” But the L2 is off! So it’s not entirely clear how the C2 is charging is the L2 is off.
admin says
I say “The right side of the capacitor C2 connects to 9V through R4 and L2, so it is charging and the voltage is rising.” And I can see that it sounds strange. But at the moment in time when the transistor Q2 turns off, L2 is still on. And current is flowing through L2, charging C2 from the right side. But, the voltage on the right side of C2 reaches around 8V very quickly, turning the L2 off. There’s still a tiny bit of current flowing through R4 and L2, but not enough to light up the LED.
About the falstad simulator, I don’t see the voltage drop in wire that you’re seeing.
Voltage can be negative. And ground does not have to be the lowest voltage in the circuit. They are just labels.
When we call the plus of a 9V battery, “9 volts”, that just means we have defined the reference point to be the minus terminal, and the plus terminal is 9V higher.
But you can also say that you want your reference point (or your 0V) to be in the middle between the plus and the minus terminal. This would mean that the minus of the battery would be called -4.5V and the plus would be called +4.5V.
Scott says
Hi passiday. The voltage you are seeing on the wire is the voltage in relation to ground.
Yahaya hassan says
Please I wanto Cary out mine also
I have used the d components u uses in the circuit
I need a transistor I don’t know which value should I used at Q1 and Q2
And also capacitors have voltage and uf
I can only see uf in d capacitors u used, what should be d volts please
AL says
How is this possible without damaging the capacitors? As I understand it C1 and C2 are switching polarity and the capacitors are polarized (often electrolytic capacitors are utilized in Astable multivibrators).
admin says
The reverse voltage across the capacitors never goes above 0.7V. Usually polarized capacitors can withstand a reverse voltage of 1 to 1.5V.
Robert a Vernon says
It seems that the caps would charge and discharge equally. Otherwise this circuit wouldn’t work. The full time one is charging the other is discharging. And vise versa.
Bijan says
Hi & thanks
Do the capacitors have a conductor role, soon after getting charged or even while discharging?
admin says
Current flow through the capacitor when it is charging or discharging. When it’s not charging or discharging, no current flows through the capacitor.
Joseph says
Please the construction, operation and observation
Jatin says
I want to build the single buzzer (Piezo) circuit which on and off after 1 sec so If I replace LED so can I connect buzzer there. So can it work
admin says
Yes, that can work if you have an active buzzer that only need DC and not an audio signal.
salah alhaddabi says
Hello Oyvind,
Thank you very much for the detailed explanation really much appreciated!!
Just one question here: can both C1 & C2 be fully charged at the same time with positive side towards the LED to be 8V and the side towards the transistor to be 0.7V. Can that happen to both capacitors at the same time? If not, can you please elaborate why?
Thanks and I hope the next circuit “one led flasher” would be much simpler than this one as this one made me very scared to play with circuits!!
admin says
Hi Salah!
No, that can’t happen.
Let’s say C2 is 0.7V on the left side and 8V on the other side.
Since C2 is 0.7V, the transistor Q1 must be on. When transistor Q1 in on, the collector to emitter part of the transistor works more or less like a wire. That means the left side of C1 is connected to minus with a wire. So it must be 0V, it can’t be 8V.
The one led flasher is the same type of circuit.
I recommend you continue with this circuit:
https://ohmify.com/courses/how-to-make-a-light-blink/
Best,
Oyvind
Cal says
What transistors do I use?
admin says
Pretty much any general purpose NPN transistors. For example BC546/547/548 or 2n3904
Robert a Vernon says
No they can’t because at any time one of the caps is charging and the other is discharging. Look at the simulator. You will see that the power from both caps is always moving in the same direction. This is because they are reversed.
The .7 volts is irrelevant. The key factor is the state of the line segment below the led. When it’s on power is flowing so the state is low. Like a river that it’s damn is open. When the led is off then the state is high like a damned up river. When the state is high it forces the cap to charge, low it discharges. When the cap is charging it is pushing power into the base it connects to. The opposite when discharging.
Hope this helps.
Robert a Vernon says
Great explaination of the function of this circuit.
I do think that you have left out a key concept though. That is how a capacitor works in conjunction with the power state of segments 1 and 2 (the line segments between the leds, capacitors, and transistors). When the led is on then the line segment below it is in a low state and the capacitor is forced to discharge and the opposite when the led it off. When the capacitor is charging it’s pushing power into the base of it’s respective transistor. When the capacitor is discharging it’s pulling power from the base of it’s respective transistor.
Sol says
What are the specs for Q1, Q2? It’s the only component without a spec. Is there a reason for that? Does it not matter? I’m assuming a NPN transistor but there are many to choose from. I looked at your “how do transistors work” link and there’s also no mention of the values and what they mean, and which one you would need for this circuit. I tried different values of Beta/hFE on the linked simulation and only saw a difference when I made the value too small. I’m assuming the Beta/hFE simply define a maximum voltage throughput it can handle? Is the “on” voltage the same for all (about 0.7v)?
admin says
Most (or all) general purpose NPN transistors will work. For example BC546/547/548 or 2N3904.
The beta/Hfe value is the current gain. Ex if it’s 100, and you have 0.01 mA flowing through the base, you’ll get 100 times that current (= 1mA) flowing from collector to emittor.
Jordi says
Hi, this article is great, it has helped to understand many thing Thank you.
I can see that as current goes everywhere at the same time, if we were in a perfect world it will stay all the time on. But as the transistors and capacitors are not perfect, in just some small time the blinking we see happen, because one charges faster than the other.
What if I want to make it blink faster or slower? I can replace the central resistances R2 and R3. But, as they are the same resistance, can I replace both of them with a single potentiometer that then goes to both? Or those resistances also collaborate to create the imperfection of the circuit that creates the blink?
admin says
You can’t combine them into a single potentiometer, no, because they are not connected in parallel. But you can use a dual potentiometer to change the speed of both at the same time.
Nick Betts says
Thanks, I’m no electronics genius and have read several descriptions regarding the operation of this circuit. Your description is great and easy to understand. Top marks, cheers
Chris says
Thank you so much for this, it explains a lot to me and I’m gaining a bit more understanding about circuits !
Chris
Scotland
admin says
Glad to hear that, Chris!
Luke says
Why is there a difference in on and off time of the LED?
Luke says
Why is there a difference in time that the LED is on vs the time it is off?
admin says
There’s no difference in my circuit (check video). If you have a difference, it’s because C1 is not equal to C2. Or because R2 is not equal to R3.
Luke says
I have a similar circuit just without the L1 diode
thasneem says
good and a simple explanation. easy to understand. thank you
doesntmatter says
The charging characteristics of the capacitor is not exponentially, it is quite the opposite.
Exponentially means slow in the beginning and than even faster in the end.
Hebe says
Thanks a lot for your explanation. I don’t understand the part when right side LED lit. I measured C2 using a oscilloscope and the lowest voltage C2 get is -0.5V. Different from -7.3 V as you said. The circuit works well. So it frustrate me…..
admin says
Hi, how are you measuring the voltage? One of the sides to minus? Or across it?
Scott says
When the circuit is turned on both capacitors start to charge via r2 and r3, but not at exactly the same rate due to component tolerances and initial charge. The first capacitor to reach approximately 0.7 volts above ground will switch the opposite transistor on and start the process.
Scott says
Hi passiday. The voltage you are seeing on the wire is the voltage in relation to ground.
Saumya says
That was great thank you so much!
Sherlock says
How to know which LED starts blinking first after this circuit has been powered up? Is it randomly? thank you very much
admin says
It depends on microscopic differences between the components.
Sherlock says
When Q1 is on, the L1 is lit, the right side of the capacitor C2 connects to 9V through R4 and L2, so it is charging and the voltage is rising. Then why is the L2 not lit while there’s the current flow through it (R4->L2->C2->Q1)? Please help me explain why. Thank you
admin says
C2 is charged through R3, not R4.
Sherlock says
Sorry but i don’t still understand yet. As you said above (in the stage of L1 is lit and L2 is turn off): “The right side of the capacitor C2 connects to 9V through R4 and L2, so it is charging and the voltage is rising”, You said through R4 not through R3.
And if C2 is charged through R3 how could it happen? I mean how can C2 capacitor be charged if It’s right side is not connected to the lower voltage point for the current flowing through it and charge it during Q2 is turned off (the collector and emitter of Q2 are not joined together) Could you show me the path that make C2 be charged. Thank you very much
Sherlock says
Can we replace Polarized capacitors C1 and C2 with non-polarized capacitors? what will happen? Could you explain to me? thank you
admin says
Yes you can. Nothing special will happen.
Utkarsh Pathak says
That was a great explanation and probably the only place it’s available. Could you let us know how we can increase the blinking time and what would help us create the circuit more powerful for a series of more than 10 LED?