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We propose to build at a cost insignificant (a few euros) an oscillator using the integrated timer 555. This proposal is a type of relaxation oscillator for generating output square wave of fixed frequency up to 500kHz with the possibility of different Duty Cycle 50-100% Unlike the use of the 555 timer in a monostable circuit which produces a single pulse output (one-shot), when activated on its input pin 2 (trigger), in order to obtain the oscillator from 555 to astable Multivibrator, You must continuously re-activate the IC 555, after each cycle timing.This is largely possible by connecting the trigger input (pin 2) and the threshold input (pin 6) together, there by allowing the device to act as an astable oscillator. The oscillator 555 thus moves continuously from one state to another. The determination of the duration of singoloimpulso and Duty cycle will be possible to size the two resistors R1 and R2, as shown below.



The oscillator 555 described above, the pin 2 and pin 6 are connected together, allowing the circuit re-activate itself on every cycle that enables it to operate as an oscillator of Free Running. During each cycle, the capacitor C is charged through both resistors R1 and R2, but download only through the resistor R2, which has another terminalesu to discharge (pin 7).

Then the capacitor charges up 2/3Vcc (the upper limit comparator), which is determined by 0.693*(R1 + R2)*C and discharged down to 1/3Vcc (the lower limit of comparison), determined by 0.693*(R2*C) This results in a waveform output whose voltage level is approximately equal to Vcc – 1.5 V and whose output “ON” and “off” periods was determined by the capacitor and the combinations resistance. The time required to complete an individual charge and discharge cycle is expressed as:


Where, R is in Ω and C in Farad.

When you are logged on as astable Multivibrator, the output from ‘oscillator 555 will continue indefinitely, alternating between charging and discharging 2/3Vcc and 1/3Vcc until power is removed. As with the monostable multivibrator these times of loading and unloading, then the frequency, are independent of supply voltage. The duration of a complete cycle is then equal to the sum of the two times the individual from office and discharge the capacitor, thus given as:


The output frequency of the oscillations can be found by inverting the equation above for the entire duration of the cycle to obtain a ‘final equation for the output frequency for an astable oscillator 555:


By altering the time constant of a single RC combination, the duty cycle can be set with precision and is calculated as the ratio between the total resistance (R1 + R2) the sum (R1 +2 * R2). The work cycle for the oscillator 555, which is the ratio of the time “ON” time and “OFF”, is given by:




Consider the following astable oscillator 555 constructed using the following components, R1 = 1kΩ, R2 = 2kΩ and capacitor C = 10 uF.Calcolare the output frequency from the oscillator 555 and the duty cycle of the waveform output.

t1 – charging time ( “ON”) is calculated as follows:


t2 – discharge time ( “OFF”) is calculated as follows:


The total will then:


And the resulting wave frequency output will be:


Which corresponds to a duty cycle of output:




Since the timing capacitor C is charged through resistors R1 and R2, but you download only through the resistor R2, the duty cycle output can be varied only between 50 and 100%, changing the resistance value R2. Reducing the value of R2 increases the value of the duty cycle of around 100% while increasing the duty cycle is reduced R2 around 50%. If the resistor, R2 is very large compared to the resistance R1 the output frequency of the circuit ‘s oscillator 555 will be determined by R2 * C only. The problem with this basic astable circuit 555 is that the duty cycle, the “mark-to-Space” will never fall below 50% due to the presence of the resistor R2 to prevent it. In other words, we can not do the time “ON” shortest time “off”, since (R1 + R2) * C will always be greater than R1 * C.

One way to overcome this problem is to connect a signal diode in parallel with the resistor R2, as shown below:


By connecting this diode D1 between the trigger input (pin2) and the entrance of discharge (pin7), the timing capacitor charging time can only directly through the resistor R1, since in that case R2 resistance short-circuited by the diode. The discharge of the capacitor as usual will be through the resistor R2

So the previous charging time t1 = 0.693*(R1 + R2)*C is amended to t1 = 0.693*(R1*C), and the duty cycle is given as D = R1 / (R1 + R2)

So to generate a duty cycle of less than 50% will be enough to take a lower resistance R1 resistance R2.


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