Diode voltage doubling circuit

The first circuit ：

Input VG1 by 1kHZ The square wave ,, The main principle is to use C1 The voltage at both ends of the shall not change suddenly , The working process is as follows ：

The first stage ： When VG1 For the high time , Give the capacitor first C4 Charge , Fill to 5V.

The second stage ： When VG1 For the low ,C1,C4 Divide equally 5V voltage ,VF2 Output 2.5V.

The third stage ： When VG1 For the high time , capacitance C4 Charged to 5V, because C1 The left terminal voltage becomes 5V, however C1 The voltage at both ends should be maintained 2.5v, So at this time, the output terminal VF2 The voltage of is 7.5V.

The fourth stage ： When VG1 For the low ,C1,C4 One 5V, One 2.5V, After sharing, it becomes 3.75V,VF2 Output 3.75V.

The fifth stage ： When VG1 For the high time , diode SD2 It can also be conducted ,C4 Charged to 5V, because C1 The left terminal voltage becomes 5V, however C1 The voltage at both ends should be maintained 3.75v, So at this time, the output terminal VF2 The voltage of is 8.75V.

Repeat the process above .... until VF2 Output close to 10V.

The waveform is as follows ：

Second circuit ：

The analysis process is similar to , The waveform is as follows ：