Paralleling Three Full Wave Rectifier Bridges

[mike]

wind-up

Thanks, that helps a lot.

 

[Anand P. Paralkar]

Hi everyone,

I am trying to build a relatively high voltage-high current DC source.
The scheme is simple and uses no regulation (therefore no feedback
control). The scheme is as follows:

Utility mains supply => Variac => Three full wave rectifier bridges in
parallel => Huge capacitor bank => Load.

Variac: something similar to this one:
http://orgchem.colorado.edu/Technique/Equipment/Communityequip/Variac.html

Bridge: KBPC3510

Capacitor bank: 6800uF, 400V

I could not find a full wave bridge rectifier with a sufficiently high
current rating and therefore, I thought of paralleling three that were
readily available.

I start (slowly) increasing the output AC voltage of variac so as to
increase the DC supply to the load. However, the variac fuse blows-up at
around 10V AC output!

Paralleling three bridges may not be the most elegant way to build a
high-current DC source, but I do not understand what could cause the
fuse in the variac to blow-up. (Please note, everything works fine with
a single bridge rectifier. This ofcourse limits the amount of load
current I can draw out from the source.)

Thank you for your help and greetings for festive season. Wish everyone
a new year full of good health and prosperity!

Regards,
Anand

Thank you everyone for your replies. My replies to some of the posts:

1. I did check each one of the bridges and the capacitors for faulty
devices, incorrect polarity, short circuits and open. Didn't find
anything there. Also, the three bridges are identical.

2. I would doubt that inrush current is an issue here. Each time that I
ran this experiment, I was careful enough to increase the variac output
voltage very slowly. (Ofcourse, connecting an uncharged capacitor bank
to full output voltage of the variac would surely trip the circuit
breakers on the utility mains line.)

3. Not that it matters, but we use a 230V/50Hz mains supply with the
installation capable of handling 15A of current. (The variac fuse would
blow somewhere around 10V AC output with no load connected to the output.)

4. I am not sure, but I get a feeling that some of us reading this post
have "registered" this as "bridge blowing-up". No, it is the variac
fuse that blows-up. Infact, there was no heating on any of the bridge
(or capacitors for that matter).

I didn't know about the balancing resistor scheme. As already
indicated, it may not be the solution for this problem, but, thanks for
the idea, simple way of controlling current hogging.

I guess, monitoring the variac current with a clamp meter (or may be
even an oscilloscope may be a good idea).

Thanks once again to everyone.

Regards,
Anand

 

[[email protected]]

Thank you everyone for your replies. My replies to some of the posts:

1. I did check each one of the bridges and the capacitors for faulty
devices, incorrect polarity, short circuits and open. Didn't find
anything there. Also, the three bridges are identical.

2. I would doubt that inrush current is an issue here. Each time that I
ran this experiment, I was careful enough to increase the variac output
voltage very slowly. (Ofcourse, connecting an uncharged capacitor bank
to full output voltage of the variac would surely trip the circuit
breakers on the utility mains line.)

3. Not that it matters, but we use a 230V/50Hz mains supply with the
installation capable of handling 15A of current. (The variac fuse would
blow somewhere around 10V AC output with no load connected to the output.)

Variacs are usually auto transformers and auto transformers work best
when the output is +/-30 % from the input (70 - 130 %) voltage. Trying
to get out only 5 % of the input voltage may cause some problems to
the autotransformer. If there are some advanced protection mechanism
on the variac, this might be triggered at such low setting, even if it
might OK around the input voltage.

A no load full wave rectified 230 Vac will generate about 325 Vdc and
your capacitor bank will be charged to that potential at the top of
each half cycle. As a back of the envelope calculation, assuming the
voltage is allowed to drop to 300 Vdc during 8 ms, until the diodes
starts to conduct again during the next half cycle, with 6.8 mF the
load can consume more than 20 Adc or 6.2 kWavg, quite lot for a single
phase feed.

If only possible, I would use a three phase 230/400 V feed and with
standard 6 pulse rectifier, 480 Vdc(avg) will be generated with 4.2 %
rms ripple _without_ any filtering capacitors. Reducing the ripple,
quite small capacitors would be sufficient. If the 480 Vdc is slightly
too much/too little, a small autotransformer will handle that change.

4. I am not sure, but I get a feeling that some of us reading this post
have "registered" this as "bridge blowing-up". No, it is the variac
fuse that blows-up. Infact, there was no heating on any of the bridge
(or capacitors for that matter).

I didn't know about the balancing resistor scheme. As already
indicated, it may not be the solution for this problem, but, thanks for
the idea, simple way of controlling current hogging.

Get three 12 V car headlights and connect each in series from a
rectifier to the variac output and increase the output voltage slowly
and monitor the illumination levels of each lamp. The illumination
level should be quite similar and they definitively will even out the
current in different bridges.

 

[Spehro Pefhany]

I guess, monitoring the variac current with a clamp meter (or may be
even an oscilloscope may be a good idea).

Use an RMS-reading meter to see what the fuse is "measuring".


Best regards,
Spehro Pefhany

 

[[email protected]]

Variacs are usually auto transformers and auto transformers work best
when the output is +/-30 % from the input (70 - 130 %) voltage. Trying
to get out only 5 % of the input voltage may cause some problems to
the autotransformer. If there are some advanced protection mechanism
on the variac, this might be triggered at such low setting, even if it
might OK around the input voltage.

To understand why auto transformers are attractive with small up/down
voltage conversion (say +/- 10 % to +/- 30 %) , one should remember
that in an ordinary transformer, all the power is transferred through
the iron core, while in an auto transformer, only the power related to
the voltage _difference_ goes through the iron core, in this case only
10 % to 30 %.

You can think about the auto transformer as an ordinary transformer
with the primary connected to the input voltage and secondary with 10
to 30 % of the input voltage. connect the secondary in series with the
input voltage and you get 10-30 % boost (in phase) or 10-30 % drop
(connect in antiphase). Only a small amount of the power goes through
the iron core.

Trying to run an auto transformer (variac) at only 5 % of input,
practically all power flows through the iron core of the transformer,
might cause saturation and similar problems.

 

[[email protected]]

To the OP
I think you mentioned that the application is to charge a battery.
Is this true?

Also, the others are trying to tell you that the reason the variac is blowing the fuse may be that one of the diodes is now defective.
Just because the fuse is in the variac and the diodes don't get hot does not the diodes are good.

Do you know how to check the diodes with an Ohm meter?

Happy new year
Mark

 

[Fred Abse]

<LTspice listing snipped.>

For reality's sake, you need some series resistance in the source.

Why hide V1 parameters?

 

[Fred Abse]

Thank you everyone for your replies. My replies to some of the posts:

1. I did check each one of the bridges and the capacitors for faulty
devices, incorrect polarity, short circuits and open. Didn't find
anything there. Also, the three bridges are identical.

2. I would doubt that inrush current is an issue here. Each time that I
ran this experiment, I was careful enough to increase the variac output
voltage very slowly. (Ofcourse, connecting an uncharged capacitor bank to
full output voltage of the variac would surely trip the circuit breakers
on the utility mains line.)

3. Not that it matters, but we use a 230V/50Hz mains supply with the
installation capable of handling 15A of current. (The variac fuse would
blow somewhere around 10V AC output with no load connected to the output.)

4. I am not sure, but I get a feeling that some of us reading this post
have "registered" this as "bridge blowing-up". No, it is the variac fuse
that blows-up. Infact, there was no heating on any of the bridge (or
capacitors for that matter).

I didn't know about the balancing resistor scheme. As already indicated,
it may not be the solution for this problem, but, thanks for the idea,
simple way of controlling current hogging.

I guess, monitoring the variac current with a clamp meter (or may be even
an oscilloscope may be a good idea).

You still haven't told us what the variac fuse is rated at.

 

[Fred Abse]

To understand why auto transformers are attractive with small up/down
voltage conversion (say +/- 10 % to +/- 30 %) , one should remember that
in an ordinary transformer, all the power is transferred through the iron
core, while in an auto transformer, only the power related to the voltage
_difference_ goes through the iron core, in this case only 10 % to 30 %.

Huh?
In *any* transformer, all the primary current goes through the primary,
and all the secondary current through the secondary. The secondary amp
turns oppose the primary amp turns, resulting (in a perfect transformer
having no leakage inductance) in zero net amp turns hence zero flux.
Connecting windings in series makes no difference.

You can think about the auto transformer as an ordinary transformer with
the primary connected to the input voltage and secondary with 10 to 30 %
of the input voltage. connect the secondary in series with the input
voltage and you get 10-30 % boost (in phase) or 10-30 % drop (connect in
antiphase). Only a small amount of the power goes through the iron core.

Where does the rest go?

Trying to run an auto transformer (variac) at only 5 % of input,
practically all power flows through the iron core of the transformer,
might cause saturation and similar problems.

You cannot saturate a transformer core with secondary amp turns. Even a
shorted secondary won't do it.
Saturation occurs where the magnetizing inductance (inductance of primary
with open secondary) allows enough current to flow to saturate the core.
That's why big current transformers, such as use in distribution equipment
buzz like hell if the secondary is O/C. That's a good danger warning.

COTS transformers are generally designed to run as close to saturation as
possible, at rated voltage and frequency, to economize on iron.

 

[[email protected]]

Huh?
In *any* transformer, all the primary current goes through the primary,
and all the secondary current through the secondary. The secondary amp
turns oppose the primary amp turns, resulting (in a perfect transformer
having no leakage inductance) in zero net amp turns hence zero flux.
Connecting windings in series makes no difference.

True for ordinary transformers with separate primary and secondary
windings.

Where does the rest go?

It goes through the galvanic connection.

You cannot saturate a transformer core with secondary amp turns. Even a
shorted secondary won't do it.
Saturation occurs where the magnetizing inductance (inductance of primary
with open secondary) allows enough current to flow to saturate the core.
That's why big current transformers, such as use in distribution equipment
buzz like hell if the secondary is O/C. That's a good danger warning.

COTS transformers are generally designed to run as close to saturation as
possible, at rated voltage and frequency, to economize on iron.

Depending on the variac setting, you can run hundreds of kW at 1:1
setting (limited only by the contact ratings), at 90 .. 110 % we are
talking about 10 % of the nominal power goes through the core, the
rest goes through the galvanic connection.

At 5 %, most goes through the core. possibly 1/20 of the nominal power
can be transferred through the iron core.

 

[Fred Abse]

True for ordinary transformers with separate primary and secondary
windings.


It goes through the galvanic connection.

If, by that, you mean two inductors in series, how do you square that
with; voltage ratio=turns ratio, inductance ratio=turns ratio squared?

Depending on the variac setting, you can run hundreds of kW at 1:1
setting (limited only by the contact ratings), at 90 .. 110 % we are
talking about 10 % of the nominal power goes through the core, the rest
goes through the galvanic connection.

At 5 %, most goes through the core. possibly 1/20 of the nominal power
can be transferred through the iron core.

That's just plain wrong.

 

[[email protected]]

OP is that same eternal-september troll sicko harassing sed with stupid ignorant posts for quite some time now. Don't waste your time on "it."

 

[[email protected]]

If, by that, you mean two inductors in series, how do you square that
with; voltage ratio=turns ratio, inductance ratio=turns ratio squared?

Assume you have some AC point to point connection to a load.

Then connect a variac at 1:1 settings to the line. What happens ?

The actual power is still flowing to the load.

There might be some reactive (inductive) power flowing through the
auto transformer.

Changing the tap settings and more and more power will flow through
the magnetic core.

 

[[email protected]]

On Sun, 29 Dec 2013 08:59:09 -0800 (PST),

OP is that same eternal-september troll sicko harassing sed with stupid ignorant posts for quite some time now. Don't waste your time on "it."

I would disagree with you.

The OP is definitively not a "please do my homework for me" type.

Discussing this topic is definitively appropriate for this newsgroup.

 

[Fred Abse]

Assume you have some AC point to point connection to a load.

Then connect a variac at 1:1 settings to the line. What happens ?

The actual power is still flowing to the load.

There might be some reactive (inductive) power flowing through the auto
transformer.

Changing the tap settings and more and more power will flow through the
magnetic core.

Actually less power, for a constant load, since the voltage reduces.

Derive an expression for the voltages and currents in an autotransformer
tapped at, say 30%, hence the flux. Neglect winding resistance.

 

[Maynard A. Philbrook Jr.]

4. I am not sure, but I get a feeling that some of us reading this post
have "registered" this as "bridge blowing-up". No, it is the variac
fuse that blows-up. Infact, there was no heating on any of the bridge
(or capacitors for that matter).

Sounds like you have it connected backwards!
Are you first connected to the AC, then output of the
variac to the bridges?

Jamie

 

[John S]

<LTspice listing snipped.>

For reality's sake, you need some series resistance in the source.

Yes. What value would you recommend for this simulation?

Why hide V1 parameters?

As JF stated, it is easy to see the parameters by simply right-clicking
on the source. If you want to see them on the schematic permanently,
simply click the box in the lower left corner "Make this information
visible on the schematic" after right-clicking the source. Easy.

Cheers,
John S

 

[John S]

OP is that same eternal-september troll sicko harassing sed with stupid ignorant posts for quite some time now. Don't waste your time on "it."

Fred, please do not sully this thread. At this point it doesn't matter
what the OP had in mind. There is a dialogue going on that is
instructional to many and pleasing to others. Please leave us to enjoy
ourselves.

Thanks,
John S

 

[Fred Abse]

Yes. What value would you recommend for this simulation?


As JF stated, it is easy to see the parameters by simply right-clicking on
the source. If you want to see them on the schematic permanently, simply
click the box in the lower left corner "Make this information visible on
the schematic" after right-clicking the source. Easy.

I knew that already.
Shift-alt-ctrl-H will unhide it as well.

I was interested as to why John thought it necessary to hide the
parameters, especially in view of the fact that supply impedance has an
effect on peak rectifier current.

 

[John S]

You omitted what you though might be an appropriate value for the
simulation.

Thanks,
John S