#### Re: Understanding Voltage Multipliers

Don Black <donald_black@...>

Nice description. I would point out that it doesn't matter what the DC bias is on the input square wave provided it doesn't exceed C1 voltage rating. C1 blocks any DC and it's only the peak to peak voltage which matters. Not trying to pick nits but since you specified the input at DC it might confuse some trying to understand it.

Don Black.

On 06-Oct-13 11:01 AM, Reed Dickinson wrote:
With all the current interest in voltage multipliers, including theory and construction, I feel it is about time to talk technically about them.
I am inserting a schematic of a voltage tripler with an input square wave of amplitude AB biased around 0 Volts.  AB would be the peak-to-peak voltage applied to the tripler.

Lets assume the tripler is working properly and analyze what waveforms and voltages can be found at each of the circled node points 1 through 6.

Node 1.
C1 charges to the voltage B through D1 so we will see a square wave of amplitude AB with the lower portion of the square wave at zero volts.

Node 4.
C4 will charge to voltage level A through D2 plus the charge on C1 so we will see a DC level of AB volts at point 4.

Node 2.
C2 charges to voltage level B through D3 plus the DC charge on C1 so we will see a square wave of amplitude AB with the lower portion of the square wave at AB volts at point 2.

Node 5.
C5 charges to voltage level A through D4 plus the DC charge on C4 so we will see a DC voltage of 2AB at point 5.

Node 3.
C3 charges to voltage level B through D5 plus the bias of AB from C2 so we will see a square wave of amplitude AB with the lower portion of the square wave at 2AB volts at point 3.

Node 6.
C6 charges to voltage A through D6 plus the voltage of 2AB from C5 so we will see a DC voltage of 3AB at point 6.

As you can see, each section composed of two capacitors and two diodes adds the value AB to the DC level found on the bottom side of the nodes 4,5 and 6 on the drawing.
A rule of thumb: the number of sections is the multiplier value.  So, multiply the square wave AB peak-to-peak value by the number of sections to get the output high voltage.
Each capacitor must have a working voltage of at least 1.5AB and each diode should have a peak inverse voltage of 2AB.  I specify 2AB because diodes are more prone to failure than capacitors.

If you wanted to make a voltage 6X multiplier you would just add three more sections to the drawing above.  The input current is the required output current multiplier by the number of sections.
To construct a negative voltage multiplier reverse each diode.

Reed Dickinson

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