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!
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.
Great addition, thanks for your comment!
The right side of C1 has to rise only 0.7V in relation to its left side ( ground) to turn Q2 on, while the left of C2 has to rise 8.7 V ( from -7.3V to +.7V) to turn Transistor 1 on. It seems that Q1 will turn on much later than Q2. I am assuming C2 is also getting charged by R3 as c1 is getting charged by R2.
There must be more to this.
Left of C1 so far has never reached 8V and its right side to .7V up there while right side of C2 reacahed 8V and its left side .7V.
Plus I like to see the right side of C1 reaching -7.3V just like C2 did on its left side. to see total picture. Otherwise it doesn’t seem even, unless I missed something.
Curious to know how can I predict which LED starts blinking first after this circuit has been powered up ?
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.
It’s generally the difference in the Vbe of the two transistors that determines which one starts first.
How would I modify this circuit to become a 3-way Multivibrator?
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.
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
Excellent explanation!
this is VERY good, clear and easy to understand.
at last, I get it !
thanks heaps and keep up the good work.
Glad you liked it!
Oyvind
i’m sorry but the negative voltage explanation is not clear at all. nor is it more understandable even if you know how a capacitor works. it’s an entirely weird phenomenon that deserves a paragraph and a couple of pictures. it’s a crucial part of how the turn-off happens. there are a few posts online of people being puzzled by this and no great answer is given besides that it’s just what happens.
it’s also not clear how the capacitors get +8V being positioned below two LEDs that are not conducting at the beginning at least.
I’ve written about negative voltage here: https://www.build-electronic-circuits.com/what-is-negative-voltage/
Perhaps it would help to look at it like this:
Take a 9v battery and connect an electrolytic capacitor in parallel with it. Then disconnect and reconnect with the positive end of the cap connected to the negative of the battery.
Now connect your volt meter – to battery – and measure the battery voltage at 9v+
Now move the positive lead to the negative side of the capacitor to see something like -9v.
This is what is happening over and over in the circuit.
Now if you connect a 47k resistor from capacitor- to battery + you will see how the capacitor discharges through the resistor each cycle.
This is a fantastic circuit to look at with an oscilloscope.
B- to either transistor base will nicely show the -7 volt phenomenon.
Collector to collector gives you a near perfect square wave.
Connect across either capacitor for a sawtooth wave.
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.
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
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.
Thank you for your explanation, it helps alot, I now understand the concept of charging and discharging the capacitor
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.
That’s really cool to hear!
Best,
Oyvind
Very good explanation of a basic concept. Not all in electronics are chips.
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
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.
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.
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 ?
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.
please add charging and discharging path of each capacitor at each step,it will be then more efficient to understand.
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 .
A capacitor doesn’t allow DC to flow when it is charged. That’s why no current flows through c2 when LED1 is on.
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
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
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
Both transistors start to turn on from the beginning through resistors R2 and R3.
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.
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)
Hi Vince, thanks a lot! Hearing that it helped you really means a lot to me.
Oyvind
Thank you for the article. One question though. How do I choose the components to make the LEDs blink with the desired frequency?
Thanks for the explanation
Does capacitor rating affect the set up (uf by volt)
I used 10uf by 25v but it’s not working
25V should work fine.
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?
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).
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
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.
Good catch! You’re right – that happens. But it’s only 0.7V, so it’s not enough to damage the capacitor.
Thanks for the answer!And btw how do the capacitors discharge in the circuit?
Or do they discharge at all?
I am pretty sure that they discharge in some way,because it wouldn’t make sense for them to not to.
How do the capacitors discharge in the circuit?
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.
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.
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.
Hi passiday. The voltage you are seeing on the wire is the voltage in relation to ground.
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
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).
The reverse voltage across the capacitors never goes above 0.7V. Usually polarized capacitors can withstand a reverse voltage of 1 to 1.5V.
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.
Hi & thanks
Do the capacitors have a conductor role, soon after getting charged or even while discharging?
Current flow through the capacitor when it is charging or discharging. When it’s not charging or discharging, no current flows through the capacitor.
Please the construction, operation and observation
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
Yes, that can work if you have an active buzzer that only need DC and not an audio signal.
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!!
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
What transistors do I use?
Pretty much any general purpose NPN transistors. For example BC546/547/548 or 2n3904
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.
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.
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)?
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.
it’s my question too!
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?
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.
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
Thank you so much for this, it explains a lot to me and I’m gaining a bit more understanding about circuits !
Chris
Scotland
Glad to hear that, Chris!
Why is there a difference in on and off time of the LED?
Why is there a difference in time that the LED is on vs the time it is off?
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.
I have a similar circuit just without the L1 diode
good and a simple explanation. easy to understand. thank you
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.
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…..
Hi, how are you measuring the voltage? One of the sides to minus? Or across it?
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.
Hi passiday. The voltage you are seeing on the wire is the voltage in relation to ground.
That was great thank you so much!
How to know which LED starts blinking first after this circuit has been powered up? Is it randomly? thank you very much
It depends on microscopic differences between the components.
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
C2 is charged through R3, not R4.
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
Can we replace Polarized capacitors C1 and C2 with non-polarized capacitors? what will happen? Could you explain to me? thank you
Yes you can. Nothing special will happen.
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?
Thank you for your explanation. It was useful for me :)
Thank you for your great explanation. What I can’t understand is that the collector-emitter voltage can never be zero and is about 0.2 when the transistor is in saturation state (0.2 – 8 = -7.8 still negative). When the transistor is on but is not saturated, the voltage is determined by the resistance in the collector branch. On the other hand, the transistor can only be saturated if base voltage is +0.5v higher than the collector voltage. According to your explanations, turing a transistor from off to on state, does not necessarily result in turning the other transistor off, unless getting saturated itself. I can’t understand what’s the reason for getting the transistor saturated.
Bravo les bricoleurs, en 1978 je réalisais des décodeurs RTTY et j’utilisais déjà principe “”astable “” de type Flip Flop” pour visualisation par LED des signaux Mark & Space en façade de mes constructions terminées. -Porteuse : LED Verte , Mark ; Led Rouge, Space ; Led jaune. au moins on voyait vivre la construction. Amities.
So in this model, Which Capacitor stores electrical charge first?
Thanks, very helpful. Can you run this on 12Vdc?
Yes. Then you could also increase R1 and R4 from 470 ohm to 1k to decrease the current a bit.
Hi Oyvind,
Thank you for your emails. I am interested in electronics and especially in how circuits work. At the moment I am having trouble understanding one aspect of your explanation on how the Astable Multivibrator 2 LED circuit works
Your explanation states that when Q2 turns on, the voltage on the right hand side of C2 is suddenly pulled down to 0V through Q2.
Why isn’t C2 suddenly discharged to 0V as it is connected to ground? The left hand side of C2 could be discharged to <0.7V through Q1's base. Both left and right sides of C2 are + charged (0.7V and 8V respectively).
What is meant by "internal charge of C2" ? When is this internal charge discharged?
Thank you for your help. John Bathie
Hi John,
“Why isn’t C2 suddenly discharged to 0V as it is connected to ground?”
That’s because charging takes time, it doesn’t happen in an instant. Before Q2 was on, you had a capacitor with a voltage difference of 7.3V between the two sides. At the moment when the right hand side of C2 is connected down to ground, you still have a capacitor with a voltage difference of 7.3V. But from now on it will start to discharge.
I’m a total nube with no knowledge of electronics, and though I didn’t really understand the circuit, with the earlier explanation of the flow within a circuit, I do get a fundamental feel for the shifting from side to side and how voltages rising and falling are opening and closing “gates” which, in turn, effectively shift the voltage s back and forth.
Great! It’s not a simple circuit to understand if you’re not used to looking at circuits with currents and voltages going here and there. Come back to this circuit after looking at and understanding some simpler ones for a while and you’ll understand even more.
Is my understanding here correct please?
Bit confused
So, when we remove the capacitors in place of simple wires the lights stop flashing. So it appears the capacitors have the main role to have flashing lights. As the same manufactured capacitors are not always the same one will charge at a rate that is different to the other. When this charge meets the voltage requirement for the transistor the light flashes and the capacitor discharges, The same happens on the other side making the lights flash.
Great circuit! Now my steampunk themed Halloween mask can have a real working circuit on it instead of something just for looks! I played around with the capacitor values to get it to blink how I think is the coolest, and I’m using a 3v watch battery for power. I noticed 3v is dimmer and blinks faster where with 9v the LED’s are brighter (which it to be expected), but blinks slower?
great explanation. I could never really understand the astable multivibrator circuit before. I would be interested in the mathematics behind such a circuit. What differential equations you can set up to figure out the frequency of the switching depending on the resistance and the capacitance of the components?
Can you use capacitors that are not electrolytic? I have a space limited application for this and even the 8.5mm SMD ECaps stick out too far from the board. A chip type capacitor would be ideal with the limited space I have to work with.
Yes, you can use any capacitor type. It’s the capacitance value that matters.
The transistors change state due to CURRENT. You do not have to mention voltage AT ALL.
The transistors change state due to the the CHARGING OF THE CAPACITORS. The base resistors are merely included to DISCHARGE the capacitors. And finally, the base resistors must be a high value so they have very little effect on being able to turn the transistors ON. In other words, the transistors need a high base current for the circuit to work.
The circuit does not change state when one transistor sees a rise of voltage on its base, but the other transistor turns OFF when the base current falls to a low value because the capacitor on its base is nearing fully charged. At this point the transistor starts to turn OFF and it pulls the capacitor on its collector HIGH and this causes the capacitor on its collector to start to charge via the opposite base and create the rapid change of state.
Thank God someone asked that question!!! Why does it not get mentioned right away?
Hi. Why do these projects work for everyone else but not for me? I have just built this circuit but the LED’s are permanently on. I have checked, checked and rechecked the connections and cannot find any problems. The only difference is that I am using 2.2uf / 50 capacitors. Surely this would only affect the timing, not the operation. Thank you in advance.
Hi! If that is the only difference, then yes, the timing should be the only difference. One possibility is that it’s blinking too fast for you to see that it’s blinking.
Another possibility is that you haven’t used NPN transistors, but PNP instead. A third option is that you’ve used the wrong resistors values.
Thank you for your reply. Apologies, I used 470R resistors instead of 470K. It is now working okay. Another issue, I need one side to have a delay of 10 to 15 minutes. Through experimentation, I have found that a 100uf capacitor to be the best out of the ones that I have. For the resistance, I started with a 1R but shut it down as it was taking too long. I then changed this for a 2R, a 2.2R then a 7.5R and after a couple of seconds I had to pull them out because they were over heating. My question is, why did the 1R work but not the three higher value ones. I checked the colour coding on the internet against mine and its value and it is correct. Thanks again.
Sir, thank you. This is by far the best explanation that I could find for how Astable Multivibrator Circuits work. I didn’t get that the transistor brought the + side of the Capacitor down to 0 this forcing the other side to -7.3. Literally a light bulb went on when I read your description.
I can’t tell you how grateful I am to come across this description. If it was in my power I’d give you a Knighthood. It’s not so please accept my thanks instead.
That’s great! And too bad about the knighthood, that would have been cool, haha. But your thanks is more than enough!
hi my friend
thanks to you finally i got the idea of astable vibrator.
hats off to you.
Hi, I’m curious about the length of time each LED is to be on for at each time. I would estimate it to be a second, is it on for a lesser period or longer period?
Many Thanks
Hello, I would like to ask, why is it that at R2 and R3, the voltage goes to the capacitor rather than to the base pins of Q2 and Q1? In addition, are the 47K resistors there in order to drop the voltage enough so that the capacitors only charge to 0.7V on their negative pins?
The voltage doesn’t go anywhere. It stays there and can be measured both at the capacitor pin and at the transistor pin. The resistor are there to reduce the current to slow down the charging of the capacitors. The more you slow down the charging of the capacitors, the slower the LEDs blink.
Thank you very much for the lesson you did so well
There is a lot more to the operation to this circuit. You have not mentioned anything about current. The circuit changes due to CURRENT not voltage. You have not mentioned anything about SATURATION or REGENERATION.
That’s why the circuit is much more complex than your description and NO-ONE has fully mentioned how the circuit ACTUALLY works. .
What I don’t get is the negative voltage that turns off the transistor. U have two different R connected from Vcc to the base so both transistors have to be turned on.and I noticed that the cap that are used in this circuit are polarised so they will never charge from the right side . This circuit is really confusing, I still don’t quit understand what is going on
A polarized capacitor will also charge from both sides. It’s just that it’s not recommended to do it…