> From:         Matthew Brush[SMTP:matthewbrush@YAHOO.CA]
> Sent:         Tuesday, May 18, 2004 10:12 PM
> To:   PICLIST@MITVMA.MIT.EDU
> Subject:      [EE:] Simple Voltage Inverter

> Hello all,

> Not sure if voltage inverter is the proper name, it's seems to be
> used for a different type of circuit sometimes.  But does anyone
> have a super simple circuit to get a -5 vdc from a positive supply?
> I've seen them before with a 555, some diodes, and caps, but I
> can't find the circuit anymore.  I need the -5v for a MAX498 analog
> switch.  I don't know considerable current is needed or if it's just a
> reference.  Could I get away with just using a MAX232 to get my
> negative voltage then regulate it @ -5v ?

      The type of circuit you are referring to is called a "charge pump".

> I'm interested in knowing the theory of the voltage inverter so I
> could make one using the parts at hand rather than buying specific
> parts.  Any help/information would be greatly appreciated, seems
> like it would be handy to understand the theory behind this.  I have
> found some similar circuits on the net, but i didn't really understand them.

      . . .

If you are familiar with "voltage multiplier" circuits, you will understand how
charge pumps  work. The standard voltage doubler, for example, consists
of a series input capacitor C1 connected to the cathode of a diode, D1. The
anode of the diode is grounded. The cathode of D1 also goes to the anode
of a second diode D2; D2's cathode goes to a capacitor C2. The low end of the
C2 is grounded. The output voltage is taken from C2.

When the input goes negative, C1 charges through D1 to the peak negative
voltage of the input, making its input end negative. D2 is off since its anode is
at  - 0.7V. On the positive half of the cycle, the input is positive and is in
series with the charged C1. The peak output will then be twice the peak value
of the input voltage. C2 and D2 will act as a peak detector, charging C2 to
+ 2*Vpeak, where Vpeak is the peak value of the AC input voltage.

If the directions of the two diodes are reversed, the output will be minus 2*Vpeak.
Now note that the input of the circuit is AC coupled (in series with C1). The
presence of DC on the input has no effect on the output; the output is 2*Vpeak.
It is understood that Vpeak is the peak value of the AC component only. Any
DC input component will settle on C1, causing its cyclic value to alternate
around this voltage, effectively removing it.

The output of a 555 oscillator will switch between 0 and Vcc. If this output is
capacitively coupled, it appears between - Vcc/2 and + Vcc/2. The peak negative
value is then -Vcc/2, which has the desired polarity, but is only half of the required
amplitude. The voltage doubler circuit with the diodes oriented for negative output
will give you just what you want. The input is already capacitor coupled, and the
amplitude is doubled back to - Vcc.

The same can be done with a tripler. Higher order multipliers may have efficiency
problems.

John Power

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