Electronic – How deos this clock generator work


This circuit shows a clock generator, where a capacitor is charged to a certain voltage for half a period (phi opening switches S1 and S2). The same voltage but with negative polarity will be shown to the comparator input V_IN in the second half (phi bar) and slowly charged up back to ground. The comparator will create a clock signal if V_0 = 0 (shorted to ground), where the output is high in the first half and low in the second half.

Clock generator circuit

From what I understand, the D flip flop will change the output at the rising edge of the clock signal generated at the output of the comparator with frequency f. Now here is the problem: isn't the output of the D flip flop going to have half the frequency (f/2) of its input? Doesn't this create a loop in which the output signal frequency (frequency of phi) is divided by two at every cycle?

Best Answer

When the output flips, it flips the capacitor. So now you are looking at the capacitor the other way, and instead of \$V_0\$ volts, it's \$-V_0\$ volts.

You are thinking that the capacitor charges, output flips, capacitor discharges, output flips. If it were that way, the circuit wouldn't work because the output would only flip on the charging cycle.

However, because the capacitor is "flipped" along with the output, the circuit always acts like it's charging the capacitor, never discharging it. Capacitor charges, output flips, capacitor charges again, output flips again.