Basic Astable Multivibrator Circuit
Assume that transistor, TR1 has just switched “OFF” (cut-off) and its collector voltage is rising towards Vcc, meanwhile transistor TR2 has just turned “ON”. Plate “A” of capacitor C1 is also rising towards the +6 volts supply rail of Vcc as it is connected to the collector of TR1 which is now cut-off. Since TR1 is in cut-off, it conducts no current so there is no volt drop across load resistor R1.
The other side of capacitor, C1, plate “B”, is connected to the base terminal of transistor TR2 and at 0.6v because transistor TR2 is conducting (saturation). Therefore, capacitor C1has a potential difference of +5.4 volts across its plates, (6.0 – 0.6v) from point A to point B.
Since TR2 is fully-on, capacitor C2 starts to charge up through resistor R2 towards Vcc. When the voltage across capacitor C2 rises to more than 0.6v, it biases transistor TR1 into conduction and into saturation.
The instant that transistor, TR1 switches “ON”, plate “A” of the capacitor which was originally at Vcc potential, immediately falls to 0.6 volts. This rapid fall of voltage on plate “A” causes an equal and instantaneous fall in voltage on plate “B” therefore plate “B” of C1is pulled down to -5.4v (a reverse charge) and this negative voltage swing is applied the base of TR2 turning it hard “OFF”. One unstable state.
Transistor TR2 is driven into cut-off so capacitor C1 now begins to charge in the opposite direction via resistor R3 which is also connected to the +6 volts supply rail, Vcc. Thus the base of transistor TR2 is now moving upwards in a positive direction towards Vcc with a time constant equal to the C1 x R3 combination.
However, it never reaches the value of Vcc because as soon as it gets to 0.6 volts positive, transistor TR2 turns fully “ON” into saturation. This action starts the whole process over again but now with capacitor C2 taking the base of transistor TR1 to -5.4v while charging up via resistor R2 and entering the second unstable state.
Then we can see that the circuit alternates between one unstable state in which transistor TR1 is “OFF” and transistor TR2 is “ON”, and a second unstable in which TR1 is “ON” and TR2 is “OFF” at a rate determined by the RC values. This process will repeat itself over and over again as long as the supply voltage is present.
The amplitude of the output waveform is approximately the same as the supply voltage, Vcc with the time period of each switching state determined by the time constant of the RC networks connected across the base terminals of the transistors. As the transistors are switching both “ON” and “OFF”, the output at either collector will be a square wave with slightly rounded corners because of the current which charges the capacitors. This could be corrected by using more components as we will discuss later.
If the two time constants produced by C2 x R2 and C1 x R3 in the base circuits are the same, the mark-to-space ratio ( t1/t2 ) will be equal to one-to-one making the output waveform symmetrical in shape. By varying the capacitors, C1, C2 or the resistors, R2, R3the mark-to-space ratio and therefore the frequency can be altered.
We saw in the RC Discharging tutorial that the time taken for the voltage across a capacitor to fall to half the supply voltage, 0.5Vcc is equal to 0.69 time constants of the capacitor and resistor combination. Then taking one side of the astable multivibrator, the length of time that transistor TR2 is “OFF” will be equal to 0.69T or 0.69 times the time constant of C1 x R3. Likewise, the length of time that transistor TR1 is “OFF” will be equal to 0.69T or 0.69 times the time constant of C2 x R2 and this is defined as.
Astable Multivibrators Periodic Time
Where, R is in Ω’s and C in Farads.
By altering the time constant of just one RC network the mark-to-space ratio and frequency of the output waveform can be changed but normally by changing both RC time constants together at the same time, the output frequency will be altered keeping the mark-to-space ratios the same at one-to-one.
If the value of the capacitor C1 equals the value of the capacitor, C2, C1 = C2 and also the value of the base resistor R2 equals the value of the base resistor, R3, R2 = R3 then the total length of time of the Multivibrators cycle is given below for a symmetrical output waveform.
Frequency of Oscillation
Where, R is in Ω’s, C is in Farads, T is in seconds and ƒ is in Hertz.
and this is known as the “Pulse Repetition Frequency”. So Astable Multivibrators can produce TWO very short square wave output waveforms from each transistor or a much longer rectangular shaped output either symmetrical or non-symmetrical depending upon the time constant of the RC network as shown below.
No comments:
Post a Comment