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paralleing bridge rectifiers

J

Jon Slaughter

Jan 1, 1970
0
Anyone see any issues with paralleling two or more bridge rectifiers to
increase the total current capacity and reduce each components individual
heat signature?

Thanks,
Jon
 
H

Homer J Simpson

Jan 1, 1970
0
Anyone see any issues with paralleling two or more bridge rectifiers to
increase the total current capacity and reduce each components individual
heat signature?

How will you ensure equal sharing of the current?
 
J

Jon Slaughter

Jan 1, 1970
0
Homer J Simpson said:
How will you ensure equal sharing of the current?

I don't have to ensure equal sharing... I just need to ensure that the max
rating isn't exceeded. I would imagine that between two identical diodes
that they would share approximately the same current.
 
J

Jamie

Jan 1, 1970
0
Jon said:
Anyone see any issues with paralleling two or more bridge rectifiers to
increase the total current capacity and reduce each components individual
heat signature?

Thanks,
Jon
yes.
it won't work.
which ever bridge starts conducting first will be the one
doing most of the work.
it's not what you want to do.
diodes have a region of no current flow. When it starts flowing,
the resistance between the joints decreases as current increases.
so if one should start before the other, then it will be the
one doing the work.
For power systems, you need to have a single bridge large enough
for the job, or join them together via power resistors which most likely
is not what you want due to power loss.
 
H

Homer J Simpson

Jan 1, 1970
0
I don't have to ensure equal sharing... I just need to ensure that the max
rating isn't exceeded. I would imagine that between two identical diodes
that they would share approximately the same current.

I wouldn't assume that. If you run each via a one foot length of wire you'll
probably be closer to sharing as it will act as a resistor.
 
J

Jon Slaughter

Jan 1, 1970
0
Jamie said:
yes.
it won't work.
which ever bridge starts conducting first will be the one
doing most of the work.
it's not what you want to do.
diodes have a region of no current flow. When it starts flowing,
the resistance between the joints decreases as current increases.
so if one should start before the other, then it will be the
one doing the work.

This is not true. There is always current flowing. (diodes are not ideal
switches)

The point is not that one will start before the other but if they will
approximately equalize. I don't need to push each diode close to its max
but, say, if I had 100 diodes each with a max of 1A and I paralleled them
then I would expect I could run a few amps through them with no problem.
Now if one blew then maybe there would be a cascade effect and obviously the
more than blew the more that will blow.


e.g., it doesn't concern me if one diode uses 80% of its max and the other
50% as long as its well below 100%. Ofcourse if one diode went bad then you
would be running 100% of the current through just one diode. I could put a
fuse on each diode to prevent this(that is, let me know when one goes bad)
or even maybe add some current limiting device to each diode?

that is, maybe a foldback current limiter on each branch that is maybe 80%
of its max?


ultimately the best solution is to have the proper components but I don't. I
have many 4A bridge rectifiers but no 12A ones. Now I can use the 4A but
rather get about 8A out of it. Maybe paralleling 3 or 4 of them would be
safe? putting fuses on each one for 4A would be even safer.

Ofcourse if this runaway current draw that your talking actually happens
then it won't work. That is, if only one diode will conduct at the start
while all others will not then it will obviously draw all the current
through it. But if there is some sort of distribution that is approximately
average then I can deal with it and it will work for my circumstances.

Thanks,
Jon
 
J

Jon Slaughter

Jan 1, 1970
0
Homer J Simpson said:
I wouldn't assume that. If you run each via a one foot length of wire
you'll probably be closer to sharing as it will act as a resistor.

Maybe theres some way to use negative feedback between two bridges so that
the current between them will equalize? Something like a comparator that
adds resistance to one branch if it has more current than the other and vice
versa.

Of course it would be more work and money than just buying the proper bridge
but would be interesting to see.

Thanks,
Jon
 
J

john jardine

Jan 1, 1970
0
Jon Slaughter said:
Anyone see any issues with paralleling two or more bridge rectifiers to
increase the total current capacity and reduce each components individual
heat signature?

Thanks,
Jon
For low powers they reckon on average you'll get a 30:70 balance on two
diodes running in parallel. This implies something like 3 bridges needed to
double the current. (you only see a 60mV change in a diodes ON voltage for
it to be passing 10x the original current)
But ... As as current goes up the diodes start generating noticable heat and
the loadsharing starts getting much worse. Resulting in the initial 'best
diode' in the batch hogging more and more of the current the hotter it
becomes and the hotter it becomes then the more current it hogs and on so
on.
That best diode is losing 2mV forward voltage drop for every degreeC
increase in it's temperature which means the other poorer diodes with their
higher mV on-voltages don't get a look in as they are running cool with low
currents and higher forward voltages and can only remain that way.
Low diode currents gives some semblance of order. Beyond a critical current
and a clear winner emerges by positive thermal feedback even though all the
diodes may be physically separate. (a chaotic function?)
 
J

John Popelish

Jan 1, 1970
0
Jon said:
Anyone see any issues with paralleling two or more bridge rectifiers to
increase the total current capacity and reduce each components individual
heat signature?

As replied, you can't rely on better than about 30% to 70%
sharing, if connected directly together, and mounted on a
common heat sink. If you can add a low resistance (one that
will drop about a tenth of a volt at average rated current,
which will drop more at instantaneous peak current) you can
force the sharing to be much closer. You can put the
resistor in either AC leg, but you need to put it in the
same leg of both bridges. Now you have the temperatures of
two more components to worry about. But there may be an up
side, besides the sharing. They will tend to slightly lower
the peak current in total, helping the transformer and
capacitors reduce their RMS current for a given DC average
output current. The other down side is the slightly lower
output voltage (that should be less than a volt).
 
J

jasen

Jan 1, 1970
0
Anyone see any issues with paralleling two or more bridge rectifiers to
increase the total current capacity and reduce each components individual
heat signature?

only thermal runaway and/or current hogging.

If you arrange some series resistance to balance the current
(winding resistance should be enough) all should be fine.

Bye.
Jasen
 
R

Rich Grise

Jan 1, 1970
0
Anyone see any issues with paralleling two or more bridge rectifiers to
increase the total current capacity and reduce each components individual
heat signature?

Buy the bigger rectifier. Under load, any amount of difference in their
forward voltage will cause positive feedback. As the diode heats up, its
forward voltage decreases, increasing its share of the current, and you
get thermal runaway. One hogs all of the current, melts, and then the
other gets stuck with all of the current, and melts.

Or, you could use current-balancing resistors, which will waste power,
but could protect the diodes; but bottom line, it'll be cheaper in the
long run to just spring for the bigger bridge.

Hope This Helps!
Rich
 
J

John Fields

Jan 1, 1970
0
Maybe theres some way to use negative feedback between two bridges so that
the current between them will equalize? Something like a comparator that
adds resistance to one branch if it has more current than the other and vice
versa.

Of course it would be more work and money than just buying the proper bridge
but would be interesting to see.

---
I think the best way would be to put Peltier junctions and
thermocouples on each of the diodes and then as the diodes started
warming up to cool them all off so they were all at the same
temperature as the one with the lowest temperature which was being
cooled the least.

Either that or to run them in a bath of circulating deionized
refrigerated water.
 
E

Eric R Snow

Jan 1, 1970
0
As replied, you can't rely on better than about 30% to 70%
sharing, if connected directly together, and mounted on a
common heat sink. If you can add a low resistance (one that
will drop about a tenth of a volt at average rated current,
which will drop more at instantaneous peak current) you can
force the sharing to be much closer. You can put the
resistor in either AC leg, but you need to put it in the
same leg of both bridges. Now you have the temperatures of
two more components to worry about. But there may be an up
side, besides the sharing. They will tend to slightly lower
the peak current in total, helping the transformer and
capacitors reduce their RMS current for a given DC average
output current. The other down side is the slightly lower
output voltage (that should be less than a volt).
Greetings John,
I built a power supply for a CNC machine retrofit using the
original 3 phase xmfr and 5 full wave rectifiers. I used 4 1/2 of the
5 because the xmfr was actually 9 windings wound on 3 cores. Anyway,
the machine is only using the original contactors, spindle motor,
servos and the xmfr that the machine came with new . All the other
electric and electronic parts are new. I did not use any resistance
with the rectifiers to force the diodes to share the current more
equally because I didn't know I should. The machine runs fine and the
conversion has hundreds of hours on it. I think the rectifiers are
rated at 20 amps. The servos draw 15 amps max before the control shots
down. This has only happened once when I jammed one of the axes. The
xmfr is wired to the rectifiers with 10 gauge wires a little over 3
feet long each.
Some questions: How does the resistance help spread the load across
all the diodes?
Why should the resistance be located between the xmfr and the
diodes?
Why haven't the diodes in my setup been destroyed? Is this because
of the long 10 gauge wires?
Thank You,
Eric
 
J

John Popelish

Jan 1, 1970
0
Eric said:
Greetings John,
I built a power supply for a CNC machine retrofit using the
original 3 phase xmfr and 5 full wave rectifiers. I used 4 1/2 of the
5 because the xmfr was actually 9 windings wound on 3 cores.

Without drawing pictures, I am having a little difficulty
imaging that schematic. Is the secondary array essentially
3 Ys or 3 Deltas that all feed one DC supply? If this is
the case, and you use separate rectifiers for each
secondary, then the impedance of those secondaries acts as
the current balancing resistance (and inductance), to help
force all the rectifiers to share the load current.
Anyway,
the machine is only using the original contactors, spindle motor,
servos and the xmfr that the machine came with new . All the other
electric and electronic parts are new. I did not use any resistance
with the rectifiers to force the diodes to share the current more
equally because I didn't know I should. The machine runs fine and the
conversion has hundreds of hours on it. I think the rectifiers are
rated at 20 amps. The servos draw 15 amps max before the control shots
down. This has only happened once when I jammed one of the axes.

So the 20 amp rated rectifier is usually running well below
its rating. That's good.
The
xmfr is wired to the rectifiers with 10 gauge wires a little over 3
feet long each.
Some questions: How does the resistance help spread the load across
all the diodes?

It just adds additional voltage drop to he rectification
that is proportional to the instantaneous current, unlike
the very nonlinear and negative tempco drop of the diode
junctions. It helps to make the overall drop more
proportional to current and with a less negative tempco.
It doesn't force perfect sharing, but only makes the sharing
less unequal.
Why should the resistance be located between the xmfr and the
diodes?

I was afraid you would ask me that. Ideally, you would add
a bit of linear (and optimally, with a positive tempco)
resistance in series with each diode junction. I was
picturing the parallel rectifiers being fed from a single
secondary, and powering a single storage capacitor. So it
may work as well in series with any of the 4 legs of the
bridge, since each leg carries either all the AC current, or
all the DC current.

I was not taking the time to visualize all the permutations
and their relative performance, but just told you about the
first one I thought about. I stand by the requirement that
if there is one AC source, and one DC load, the sharing
resistors all need to be in the same led of the bridge.
Why haven't the diodes in my setup been destroyed? Is this because
of the long 10 gauge wires?

Perhaps it is because, most of the time, the average current
is pretty low, and the diodes have a fair thermal mass to
absorb the big thumps. Most rectifier diodes have
impressive surge current ratings.

And, yes, the wires are low value resistors with positive
temperature coefficient.
 
E

Eric R Snow

Jan 1, 1970
0
Without drawing pictures, I am having a little difficulty
imaging that schematic. Is the secondary array essentially
3 Ys or 3 Deltas that all feed one DC supply? If this is
the case, and you use separate rectifiers for each
secondary, then the impedance of those secondaries acts as
the current balancing resistance (and inductance), to help
force all the rectifiers to share the load current.

Greetings John,
I don't know what the configuration of the secondarys is. When
connected to the original three servo amps the 9 taps from the xmfr
were connected such that each winding had one wire going to each servo
amp. I think. Anyway, what I did was connect those 9 taps to the
rectifiers so each winding does have it's own rectifier. And the
outputs from the rectifiers are all tied together. This is because the
new servo amp configuration has all three amps on one board and they
share the DC supply.
So the 20 amp rated rectifier is usually running well below
its rating. That's good.


It just adds additional voltage drop to he rectification
that is proportional to the instantaneous current, unlike
the very nonlinear and negative tempco drop of the diode
junctions. It helps to make the overall drop more
proportional to current and with a less negative tempco.
It doesn't force perfect sharing, but only makes the sharing
less unequal.


I was afraid you would ask me that. Ideally, you would add
a bit of linear (and optimally, with a positive tempco)
resistance in series with each diode junction. I was
picturing the parallel rectifiers being fed from a single
secondary, and powering a single storage capacitor. So it
may work as well in series with any of the 4 legs of the
bridge, since each leg carries either all the AC current, or
all the DC current.

I was not taking the time to visualize all the permutations
and their relative performance, but just told you about the
first one I thought about. I stand by the requirement that
if there is one AC source, and one DC load, the sharing
resistors all need to be in the same led of the bridge.


Perhaps it is because, most of the time, the average current
is pretty low, and the diodes have a fair thermal mass to
absorb the big thumps. Most rectifier diodes have
impressive surge current ratings.

And, yes, the wires are low value resistors with positive
temperature coefficient.
Thanks for the explanations John. I will keep them in mind next time I
need to build a similar power supply. And I'm sure there will be a
next time.
Cheers,
Eric
 
J

jasen

Jan 1, 1970
0
Greetings John,
I don't know what the configuration of the secondarys is. When
connected to the original three servo amps the 9 taps from the xmfr
were connected such that each winding had one wire going to each servo
amp. I think. Anyway, what I did was connect those 9 taps to the
rectifiers so each winding does have it's own rectifier. And the
outputs from the rectifiers are all tied together. This is because the
new servo amp configuration has all three amps on one board and they
share the DC supply.

If the three secondaries all produce the same voltage you should be safe.

Bye.
Jasen
 
A

Andy Dingley

Jan 1, 1970
0
I think the best way would be to put Peltier junctions and
thermocouples on each of the diodes and then as the diodes started
warming up to cool them all off so they were all at the same
temperature

If you're going to that sort of length, isn't there already some
commercial technology for 100A+ rectifiers (from International
Rectifier or the Hexfet people?) that uses a funny geometry so as to
gain the right sort of temperature coefficient inherently. You can
just parallel these things up as much as you like, they sort it out
amongst themselves.

As to paralleling a pair of rectifiers to share current on a
breadboard lash-up, then this is an old hack and works fine -- just
use a matching pair (same part, same manu.) and don't expect that the
total rating will be twice that of a single.

As to why you'd do it for production, then I don't know. High power
rectifiers are cheaper than low power ($/A)
 
J

John Fields

Jan 1, 1970
0
If you're going to that sort of length,

....

---
I guess I should have punctuated it like this:

I think the best way would be to put Peltier junctions and
thermocouples on each of the diodes and then as the diodes started
warming up to cool them all off so they were all at the same
temperature as the one with the lowest temperature which was being
cooled the least. :^)
 
R

redbelly

Jan 1, 1970
0
I think the best way would be to put Peltier junctions and
thermocouples on each of the diodes and then as the diodes started
warming up to cool them all off so they were all at the same
temperature as the one with the lowest temperature which was being
cooled the least.

Either that or to run them in a bath of circulating deionized
refrigerated water.

Uh, you're joking, right?

Mark
 
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