Low frequency divider


Posted on Jul 19, 2012

The ratio of capacitors Cl and C2 determines division. With a positive pulse applied to the base of Ql, assume that Cl = C2 and that Cl and C2 are discharged. When {31 turns off, both Cl and C2 charge to 10 volts each through R3. On the next pulse to the base of Ql, Cl is again discharged but C2 remains charged to 10 volts. As Ql turns off this time, Cl and C2 again charge. This time C2 charges to the peak point firing voltage of the PUT causing it to fire. This discharges capacitor C2 and allows capacitor Cl to charge to the line voltage.


Low frequency divider
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The input and output frequency can be approximated by the equation (Cl = C2) Cl For a 10 kHz input frequency with an amplitude of 3 volts, the table shows the values for Cl and C2 needed to divide by 2 to 11. As soon as C2 discharges and Cl charges, the PUT turns off. The next cycle begins with another positive pulse on the base of Ql which again discharges Cl.




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