Voltage Multiplier

Posted on Aug 13, 2012

Figure 99-l(a)`s circuit exhibits a high-output impedance as a result of the small effective capacitance of the series-c

Voltage Multiplier
Click here to download the full size of the above Circuit.

onnected capacitors, and it exhibits considerable voltage loss due to all of the diode drops. Further, this circuit requires 2 diodes and 2 capacitors to produce a dc output voltage approximately times the rail voltage. Figure 99-1 (b)"s circuit multiplies more effectively using fewer diodes and capacitors. The parallel arrangement of the capacitors lets you use smaller capacitors than those required in Fig. 99-1(a). Alternatively, when using the same capacitor values of Fig. 99-1 (a), the output impedance will be lower. Whereas the clock source directly drives only one of the two strings of capacitors in Fig. 99-1(a), Fig. 99-l(b)"s clock drives both strings with opposite phases. This drive scheme doubles the voltage per stage of two diodes. A final diode is necessary to pick off the dc output voltage because both strings of capacitors now carry the - p ac input-voltage waveform. The ICL7667 dual-FET driver accepts a TTL drive swing and provides a low-impedance push-pull drive to the diode string. This low impedance is particularly helpful when using a long string to raise output voltage to more than 100 V, starting from a low rail voltage.

Leave Comment

characters left:

New Circuits



Popular Circuits

Audio Power Limiter
Sweep/Function Generator
Single-Chip Chime
Car battery monitor circuit diagram
Zero Crossing Detector (ZCD): Comparator circuit using 741 op amp
Build Your Own Motorcycle Alarm
Two electric motors has started manual and automatic control circuit