levy@ttrdc.UUCP (Daniel R. Levy) (05/20/85)
My respected colleague, Mr. Drescher, writes: >You can build a voltage doubler by using a diode charge pump with the >555 as an oscillator. The diode harge pump is a standard voltage >doubling (ort tripling, quadrupling, ...) and recifying circuit. > ... > > This approach will not work with a 555, but there are many circuits > like this which will work with one side of the the transformer > grounded. Any of these will work with a 555. > 1. Set up your 555 as an astable multivibrator. > 2. This circuit will provide a doubler: > 555 D > output -------||----+--->|---+-----o > C | | > | --- > | --- C > | | > +---|<---+ > D | > | > +-----o > | > ----- > --- > - > >More circuits can be found in "The Art of Electronics" by Horowitz and Hill. > Trouble with the circuit given above, is it won't work! At least it won't work unless the 555 is being powered from a double ended power supply. E.g., if the 555 is being powered from a single ended power supply of +5 volts (say), the output of Mr. Drescher's circuit will also be 5 volts at the most. But if the 555 is being powered from a double ended power supply of +/-5 volts (say), then the circuit will put out +10 volts (neglecting diode drops, which would be about .7 volt for silicon diodes or .3 volt for germanium, multiplied by two diodes, and the limitations on the output swing of the 555 itself). The circuit given is an example of a peak-to-peak reading circuit; such are used in some VTVMs in the AC ranges (Heathkit comes to mind). Consider, however, this variation on the given circuit, which should act as the desired doubler within the limitations of the diode drops and the 555 output swing: - C1 + D1 (Signal from 555) ----------------|(----+---|<--- +Vdd | ~+Vdd ____ | D2 | | +--->|---+--O + 0v ____| |____ | C2 | ~2Vdd (-diode drops +---)|---+ and 555 swing | limitations) +-----------O - | | ----- --- - Assume the power is applied with the input signal at zero volts. Then C1 will charge up to Vdd less one diode drop through D1, with the polarity shown. Then let the input signal rise near to Vdd. It will now be additive with the voltage across C1, and will charge C2 to a voltage near 2Vdd, less two diode drops, through D2 (presuming the capacitance of C1 is great enough with respect to C2 that it is not substantially discharged in the process). Of course, one must consider the load to be applied when calculating the values of the capacitors to be used (which is also dependent on the output frequency of the 555, the higher the better for filtering purposes). The 555 itself, obviously, has output lim- itations which should be taken into consideration too. Dan Levy (sys$hacker) AT&T Teletype Corporation, Skokie, Ill.