Transcription
From the blinking light of a car’s turn indicator to the precise timing of a computer, the square wave is common. A square wave generated inside a computer, also referred to as a clock signal, synchronizes all the operations within a computer. When a low-frequency square wave is connected across an LED, it blinks, and it has numerous applications. An astable multivibrator is a circuit that can produce a square wave. The name “astable” means that the output is not stable in any state. There are different ways to build an astable multivibrator circuit, but in this video, I will focus on a transistor-based astable multivibrator circuit. Without further delay, let’s get [Music] started.
This is the circuit diagram of an astable multivibrator. Please don’t be scared if you don’t have any idea of the purpose of its components; we will understand it step by step in a minute. From observation, we see that the circuit consists of four resistors, two capacitors, and two transistors. Let’s understand the purpose of the capacitor. If I connect it across a voltage source, such as a 9V battery, the capacitor charges. But what does that mean? It means that the voltage across it increases until it reaches the source voltage, which is 9 volts in my case. Here is a demonstration. As I connect the capacitor across the voltage source, the voltage across the capacitor, V<sub>C</sub>, rises from 0 volts to 9 volts. We can say that the capacitor is charged. Now, even if the voltage source is removed, the voltage across the capacitor remains at the same level until you short the two leads of the capacitor using a conductor, which immediately discharges the capacitor.
Next, let’s try connecting a resistor in series with the capacitor and then connect it to the voltage source. This time, guess what? The voltage across the capacitor does not increase instantly but takes time. After some time, the voltage across it reaches the supplied voltage, which is 9 volts. It takes time because the resistor resists the flow of current and slows down the voltage buildup across the capacitor. If you increase the resistance, the voltage buildup across the capacitor will be even slower. You can also increase the capacitance of the capacitor to increase charging time. This time required by the capacitor to charge is essential for the working of the astable multivibrator circuit. The capacitor can also be discharged by using the same resistor; just connect the resistor in parallel by shorting these two points, and as you can see, the voltage across the capacitor decreases slowly.
One more thing you need to know is that a capacitor will charge up to the difference in voltage supplied between these two points. Let’s say at point A, the voltage given is 5 volts, and at point B, 3 volts. So the capacitor will charge up to the difference in voltage, that is 5 - 3, which is equal to 2 volts. So the voltage across the capacitor will increase exponentially from 0 volts up to 2 volts over time.
Next up, let’s understand the purpose of the transistor. It is employed as a switch in the circuit. How it works is that when a voltage is applied across the base and emitter, the transistor acts as a closed switch between the collector and emitter. This base-to-emitter voltage must be around 0.7 volts in order to activate the transistor. But if the voltage given is lower than 0.7 volts, the transistor turns off. So 0.7 volts is the threshold voltage for a bipolar junction transistor. Now you can connect a load or a resistor at the collector side, then power it up. This way, you can control the flow of current through the collector by using the base voltage. If 0.7 volts is applied, the transistor turns on, and current flows to the collector. But if lower than 0.7 volts is applied or even removed, the transistor turns off, and no current flows to the collector.
One more thing about base voltage: the base-to-emitter junction is just like a PN junction diode, so the base voltage should not be more than 0.7 volts. For that reason, to control the transistor using a higher voltage, such as 9 volts, the resistor must be connected in series to limit the current. This way, the voltage at the base with respect to the emitter will be 0.7 volts. Now, can you tell me what will be the voltage at the collector point with respect to the emitter when the transistor is at a closed position or at an open position? Think about it. At the open position, since no current is flowing, hence no voltage drop across the collector resistor, therefore the whole supplied voltage appears across the collector and emitter. But when the base voltage is given, the collector point is shorted to the emitter, and that’s why the voltage across the collector and emitter at the closed position is 0 volts. So remember, closed means 0 volts, and open means the supplied voltage; if 9 volts is given, 9 volts will appear across the collector and emitter.
Okay, now we can combine the capacitor and transistor. Let’s connect the capacitor in the circuit like this. You can notice that it forms an RC circuit that we previously discussed. The transistor immediately turns off after the capacitor is connected. Can you tell why? Yes, as you have guessed, the voltage across the capacitor slowly increases from 0 volts, and the transistor turns off because the voltage between the base and emitter must be around 0.7 volts to turn it on. But as the capacitor charges and the voltage across it reaches 0.7 volts, the transistor turns on, and that’s how a simple timer can be made.
Now that we have learned the fundamentals, let’s jump to the astable multivibrator circuit. The circuit consists of two NPN transistors with collector resistors. To turn the transistors on, a base resistor is connected for each transistor like this. Finally, connect the capacitors: one terminal at the collector of the first transistor and the other terminal at the base of the second transistor. Similarly, connect the second capacitor: one terminal at the collector and the other terminal at the base. Now power it up using a voltage source, such as a 9V battery. A square wave voltage will appear across the collector and emitter of each transistor.
Let’s make the circuit on a breadboard. Connect each and every component as per the circuit diagram. Finally, I have connected a 9V battery and an LED between the collector and emitter of each transistor to visualize the square wave. You can see that the square waves are exactly opposite to each other, but how does the circuit work? First, let’s label each and every component. As you connect a voltage source like this, both transistors will try to activate through their base resistors, RB1 and RB2. But no two transistors are exactly the same; one transistor will always activate faster than the other one. Let’s assume Q1 gets activated first. This forms an RC circuit, and the capacitor charges. In this part, the voltage across the capacitor appears between the base and emitter of the second transistor. Since the voltage across a capacitor slowly rises from 0 volts, that’s why Q2 remains open because the voltage is below 0.7 volts. While C1 is charging in this part, C2 also gets charged through a different path. This capacitor will charge up to the supplied voltage, V<sub>CC</sub> - 0.7 volts. That’s because one terminal of the capacitor is connected to V<sub>CC</sub> because Q2 is open, and the other terminal of the capacitor is connected to the base of Q1, which is 0.7 volts as Q1 is closed. That’s why the capacitor C2 will charge up to the difference of voltage between these two points, that is V<sub>CC</sub> - 0.7 volts.
One more thing to mention is that capacitor C2 charges much faster than C1 because the base resistors are of much higher resistance in comparison to the collector resistors. As I have already told, Q2 remains off because C1 charges from 0 volts, and as it increases and reaches 0.7 volts, which is enough to activate the transistor, the transistor Q2 turns on. After that, notice that the voltage across capacitor C2 appears between the base and emitter of transistor Q1. I have used NPN transistors, which require a positive voltage at the base with respect to the emitter. But notice that the negative polarity of C2 is at the base and positive at the emitter. This high negative voltage strongly turns the transistor Q1 off. As you can see, the state has been changed: Q1, which was previously on, is now off, and Q2, which was previously off, is now on. Capacitor C2 now charges in the opposite direction through this part, and the voltage across it appears between the base and emitter of Q1. The voltage across C2 increases from the previously charged voltage towards V<sub>CC</sub>. The previously charged voltage is V<sub>CC</sub> - 0.7 volts, but there is this negative sign because we are measuring the voltage at the base with respect to the emitter, which is negative. The voltage across the capacitor C2 increases towards V<sub>CC</sub>, and as it reaches 0.7 volts, the transistor Q1 turns on. But while that was happening, C1 charges in this part; it charges up to V<sub>CC</sub> - 0.7 volts, and it charges much faster than C2 because of low collector resistance. As the voltage across C2 reaches 0.7 volts, Q1 turns on, and due to the high negative voltage of C1, Q2 turns off immediately. As you can see, the state has changed again: Q1 is on, Q2 is off. We have come to the beginning state of the circuit. The same process then gets repeated again and again, and that’s how we get a square wave output between the collector and emitter of each transistor that oscillates between 0 volts and V<sub>CC</sub>.
This is how the waveform looks on the oscilloscope screen. The frequency of the square wave output of the astable multivibrator can be found by using this equation. What we can observe is that the frequency depends on the resistance of its base resistor and the capacitance of its capacitor. Higher the value of the base resistance or capacitance, lower will be the frequency of the square wave output, and vice versa. You can try different values of resistor or capacitor to get different frequencies. Last but not least, the resistance of each collector resistor should be sufficient that it can supply appropriate current to the load, such as an LED. The first step is to find out how much current the load takes and at what voltage. The blue LED I’m using takes around 20 mA at 3.2 volts. Next, use this formula to find the collector resistance: the supplied voltage, V<sub>CC</sub>, I am using is 9 volts, the load voltage is 3.2 volts, and the load current is 20 mA. By solving this, the collector resistance equals 290 ohms. The closest resistance value I got is 330 ohms. This way, the LEDs glow perfectly.
Videos like this take a lot of effort to make. Please support my channel by sharing this video with your friends and so like and subscribe to my channel. Thank you so much for watching.