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Sunday, October 25, 2009

Oscillator Circuit with 50% Duty Cycle

This is a circuit f oscillator that can be configured to give symmetric oscillation (50% duty cycle). This circuit is using 555 timer IC for produces the 50% duty cycle. Other circuit uses diodes to split the charging and the discharging paths through different resistors, but here we need no such diodes. This is the figure of the circuit.


The time period for the output high is the same as regular configuration;
t1 = 0.693 RA C,
But for the output low, the period is;
t2 = [(RA.RB/(RA+RB)] x C x Ln[(RB-2RA)/(2RB-RA)]
Thus the oscillator frequency is equal to 1/(t1+t2).

Please remember this oscillator circuit will not oscillate if RB is greater than 1/2 RA because the junction of RA and RB cannot bring pin 2 down to 1/3 VCC and trigger the lower comparator.

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