DIY an Astable Multivibrator and Explain How It Works

Introduction: DIY an Astable Multivibrator and Explain How It Works

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Astable Multivibrator is a circuit which has no stable states and its output signal oscillates continuously between the two unstable states, high level and low level, without any external triggering.

The necessary materials:

2 x 68k resistors

2 x 100μF electrolytic capacitors

2 x red LED

2 x NPN transistors

Step 1: Step One: Solder the Resistors and LEDs and NPN Transistors Into the PCB

Please note that the long leg of the LED should be inserted into the hole with ‘+’ symbol on the PCB. The flat side of the transistor should be on the same side of the diameter of the semicircle on the PCB.

Step 2: Step Two: Solder the Electrolytic Capacitors Into the PCB

Electrolytic Capacitors have polarity that the long leg is anode while the short leg is cathode. This Astable Multivibrator circuit is quite simple that it is the best DIY kits for you to learn the knowledge of capacitors charging and discharging. Up to this step the DIY is finished. The most important part of this instructable is analysis.

Step 3: Explain How Astable Multivibrator Works

The power voltage of this circuit is recommended at the range of 2V to 15V, mine is 2.7V. You are free to select the supplied voltage from 2V to 15V as you want. When connect the power source with this circuit, in reality, both of the capacitors C1 and C2 start charging and it is hard to say which capacitor will get about +0.7V at its cathode side that will turn the base of NPN transistor on firstly even they are marked by the same value of capacitance. Because all the components would have tolerance, they are not 100% ideal components. Generally, when the voltage of base of transistor reaches 0.7V the transistor will be conducted and it becomes active.

(1)Let’s say Q1 is conducting heavily and Q2 is at off state and LED1 is light and LED2 is off. The collector of Q1 will be low output as will the left side of C1. In this project low output doesn’t mean 0V, it is about 2.1V, this is determined by the supply voltage you applied to the circuit. And now C1 begins to charge via R1 and its right side becomes increasingly positive until it reaches a voltage of about +0.7V. We can see from the circuit diagram that the right side of C1 is also connected to the base of transistor, Q2. (2) At this time the Q2 is conducting heavily. The rapidly increasing collector current through Q2 now causes a voltage drop across LED2, and Q2 collector voltage falls, causing the right side of C2 to fall rapidly in potential. It is the attribute of a capacitor that when the voltage on one side changes rapidly, the other side also undergoes a similar continuous change, therefore as the right side of C2 falls rapidly from supply voltage to low output (2.1V), the left side must fall in voltage by a similar amount. With Q1 conducting, its base would have been about 0.7V, so as Q2 conducts, the base of Q1 falls to 0.7-(2.7-2.1) = 0.1V. Then LED1 is off and LED2 is light. However, the LED2 doesn’t last long. C2 now begins to charge through R2, and once the voltage on the left side (Q1 base) reaches about +0.7V another rapid change of state takes place, Q1 is active, LED1 is being light, so as Q1 conducts, the base of Q2 falls to 0.1V, Q2 becomes inactive, LED2 is off. The on and off of the Q1 and Q2 are repeated from time to time, the duty cycle, T is determined by the time constant RC, T=0.7(R1.C1+R2.C2).

Step 4: Waveforms Show

The vertical offset of my oscilloscope is 0V, and I've marked the explanation text on each waveform image. This part is the supplement to step three. To get the material for learning please go to

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    Question 10 months ago on Step 1

    I understand that the C1 and C2 take turns in turning on the cross connected transistors. Ie, C1 will turn on Q2 (eventually causing a turn-off of Q1) and C2 will turn on Q1 (causing a turn-off Q2). My main difficulty is in understanding the charging process of these capacitors. The way they are connected in this circuit doesn't look like anything I've seen before. Usually, I've seen the plus of the battery connected to the plus of the capacitor and the negative of the battery connected to the negative of the capacitor. The capacitor can be connected in series with a load or in parallel with a load. Regardless, its positive terminal should lead to the plus of the battery and the negative to the negative.

    Here we have something unusual looking. First of all, the circuits use specifically polarized capacitors. So, looking at C2 for example, how come its negative side is connected to the positive terminal of the battery via R3? Its official right side is also connected to the positive (I guess?) when Q2 is turned off or negative when Q2 is turned on. That's so weird and so never mentioned in any descriptions of this circuit I've read. Yet you also read to always observe the polarity of the polarized capacitor. So yeah, for a newb like me, it's tough.

    Next thing that is odd to me is that when we just turned on the circuit, I'm told that the voltage on both C1's and C2's negative side begins to positively charge through R2 and R3 respectively until one of them reaches the BE threshold voltage of 0.7V thus causing the ensuing events (let's forget about those for now). Ok, so as I mentioned before, it's hard to imagine the negative side of the capacitor being charged. But somehow it does work?

    On top of it, the positive side of those capacitors (before any of them reach 0.7V on the negative side) reaches about the positive voltage of the battery (roughly). But how is that possible if the LEDs are not letting any current through yet and the positive sides of the capacitors are connected below the LEDs?