Voltage trippler ?

I need someone to explain how the following circuit works:
(I am referring to the voltage trippler, the doubler is easy
peasy)
http://hyperphysics.phy-astr.gsu.edu/hbase/electronic/voldoub.html#c3

Tia

 

[John Popelish]

beginner@btopenworld.com wrote:

I need someone to explain how the following circuit works:
(I am referring to the voltage trippler, the doubler is easy
peasy)
http://hyperphysics.phy-astr.gsu.edu/hbase/electronic/voldoub.html#c3

Lets call the left output terminal zero volts to simplify the
description of other voltages. And I will name the caps from left ot
right, C1, C2, C3, likewise the diodes, D1, D2, D3.

The right end of C1 is charged to 1X (one times) the negative half
cycle peak voltage of the AC by the conduction of the D1 when the
bottom of the transformer winding is negative (remember I have defined
the top end of the transformer winding as zero volts). It retains
most of this voltage throughout the rest of the AC cycle.

C2 sees the full AC voltage from the transformer. During the positive
half cycles, the D2 conducts, limiting the voltage at the right end of
C2 to be no more positive than the negative half cycle peak voltage on
the right end of C1. This charges C2 to the full peak to peak voltage
of the AC cycle (left end at positive peak while the right end is held
at the negative peak stored on C1) or 2X the half cycle peak voltage.

Then on the negative half cycles, the right end of C2 swings from 1X
the negative half cycle voltage (that it got from C1) to a full cycle
peak to peak voltage more negative than that (3X the half cycle
voltage) and that negative peak is pushed into the right end of C3
through D3. The left end of C3 is still at the 1X negative half cycle
voltage that was forced into C1.

So C1 stores a DC voltage of 1X, but C2 and C3 store 2X times the half
cycle peak voltage. C1 and C3 have their voltages stacked like
batteries to produce the output voltage of 3X, while C2 bobs up and
down on the AC waveform pumping charge out of C1 and into C3. C1
should be twice as big a capacitor as C2 and C3 (since it supplies not
only load current, but also the charging current for C3 via C2),
though it needs only half the voltage rating.

How was that?
--
John Popelish

 

http://hyperphysics.phy-astr.gsu.edu/hbase/electronic/voldoub.html#c3

Lets call the left output terminal zero volts to simplify the
description of other voltages. And I will name the caps from left ot
right, C1, C2, C3, likewise the diodes, D1, D2, D3.

The right end of C1 is charged to 1X (one times) the negative half
cycle peak voltage of the AC by the conduction of the D1 when the
bottom of the transformer winding is negative (remember I have defined
the top end of the transformer winding as zero volts). It retains
most of this voltage throughout the rest of the AC cycle.

C2 sees the full AC voltage from the transformer. During the positive
half cycles, the D2 conducts, limiting the voltage at the right end of

D2 is the middle one, it conducts during the -ve cycle

C2 to be no more positive than the negative half cycle peak voltage on
the right end of C1. This charges C2 to the full peak to peak voltage
of the AC cycle (left end at positive peak while the right end is held
at the negative peak stored on C1) or 2X the half cycle peak voltage.

Then on the negative half cycles, the right end of C2 swings from 1X
the negative half cycle voltage (that it got from C1) to a full cycle
peak to peak voltage more negative than that (3X the half cycle
voltage) and that negative peak is pushed into the right end of C3
through D3. The left end of C3 is still at the 1X negative half cycle
voltage that was forced into C1.

So C1 stores a DC voltage of 1X, but C2 and C3 store 2X times the half
cycle peak voltage. C1 and C3 have their voltages stacked like
batteries to produce the output voltage of 3X, while C2 bobs up and
down on the AC waveform pumping charge out of C1 and into C3. C1
should be twice as big a capacitor as C2 and C3 (since it supplies not
only load current, but also the charging current for C3 via C2),
though it needs only half the voltage rating.

How was that?
--
John Popelish

 

[John Popelish]

beginner@btopenworld.com wrote:

http://hyperphysics.phy-astr.gsu.edu/hbase/electronic/voldoub.html#c3

Lets call the left output terminal zero volts to simplify the
description of other voltages. And I will name the caps from left ot
right, C1, C2, C3, likewise the diodes, D1, D2, D3.

The right end of C1 is charged to 1X (one times) the negative half
cycle peak voltage of the AC by the conduction of the D1 when the
bottom of the transformer winding is negative (remember I have defined
the top end of the transformer winding as zero volts). It retains
most of this voltage throughout the rest of the AC cycle.

C2 sees the full AC voltage from the transformer. During the positive
half cycles, the D2 conducts, limiting the voltage at the right end of

D2 is the middle one, it conducts during the -ve cycle

The middle diode becomes forward biased when the bottom of the
secondary is most positive.
--
John Popelish

 

[Robert Monsen]

This is a version of a charge pump.

The easiest way to understand these things is to download a free spice
simulator and try it out, probing various places in the circuit. You could
also build one, of course, and probe it with an oscilloscope, which would be
even better.

They are confusing circuits, which depend on the one way conduction of the
diodes to push up the charge level of each stage, thus causing each
sucessive stage to oscillate at a higher voltage.

Note that the one is hyperphysics actually lowers the charge at each stage,
so the left side is higher voltage than the right side. They are usually
drawn the other way, so left is lower voltage than right.

As you probably know, the power isn't increased by this configuration (its
actually decreased by losses in the diodes) so you aren't getting something
for nothing. They have lousy regulation and current output, in general.

Regards,
Bob Monsen

<beginner@btopenworld.com> wrote in message
news:bgtmma$ki1$1@titan.btinternet.com...

I need someone to explain how the following circuit works:
(I am referring to the voltage trippler, the doubler is easy
peasy)
http://hyperphysics.phy-astr.gsu.edu/hbase/electronic/voldoub.html#c3

Tia

 

[John Popelish]

Robert Monsen wrote:
(snip)

Note that the one is hyperphysics actually lowers the charge at each stage,
so the left side is higher voltage than the right side. They are usually
drawn the other way, so left is lower voltage than right.

Sorry, but it is not correct to equate more positive with higher
voltage. These circuits are equally good at making the end that is
furthest from the transformer, more positive or more negative, just by
flipping each diode and capacitor. If you get zapped by a big
negative voltage you would never again think of it as a lower voltage.

--
John Popelish