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How do you make charging time period equal to discharging time period in astable multivibrator using 555 timer?
Wiki User
∙ 11y ago
As the timing capacitor, C charges through resistors R1 and R2 but only discharges through resistor R2 the output duty cycle can be varied between 50 and 100% by changing the value of resistor R2. By decreasing the value of R2 the duty cycle increases towards 100% and by increasing R2 the duty cycle reduces towards 50%. If resistor, R2 is very large relative to resistor R1 the output frequency of the 555 astable circuit will determined by R2.C only. The problem with this basic astable 555 oscillator configuration is that the duty cycle, the "mark-to-space" ratio will never go below 50% as the presence of resistor R2 prevents this. In other words we cannot make the "ON" time shorter than the "OFF" time as (R1 + R2)C will always be greater than R1.C. One way to overcome this problem is to connect a signal bypassing diode in parallel with resistor R2 as shown in the below link.
Improved 555 Oscillator Duty CycleBy connecting this diode, D1 between the trigger input and the discharge input, the timing capacitor will now charge up directly through resistor R1 only, as resistor R2 is effectively shorted out by the diode. The capacitor discharges as normal through resistor, R2. Now the previous charging time of t1 = 0.693(R1 + R2)C is modified to take account of this new charging circuit and is given as: 0.693(R1.C). The duty cycle is therefore given as D = R1/(R1 + R2). Then to generate a duty cycle of less than 50%, resistor R1 needs to be less than resistor R2.
Wiki User
∙ 11y ago
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