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 Mondaykids.com

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