1 KVA 240 v transformer output 6v,8v,12 v AC how to test if

[rob]

Hi

I am a total novice in electrics and am not sure were to post this
question?.. i hope some one can help me...I need a transformer that
can supply aprrox 6 Volts Dc at 75 amps..
I have just purchased a 240 v ac input 1 KVA rated transformer its AC
output is 6v ,8v and 12 V..Now i understand that i need to rectify
the current and have just got off ebay a dozen MBRP40030CTL 400A 30V
schottky diodes with low Vf.. i assume that these will be able to
supply a 75 amp current.. at 6 v?..I got them because there rated at
400 amps..and were very cheap!
http://www.onsemi.com/pub/Collateral/MBRP40030CTL-D.PDF

My other question is that there is label on side of the transformer
which is hand written saying that it is rated at 83 amps?..
Label pic
http://www.saspyro.pwp.blueyonder.co.uk/label.JPG

yet when i looked inside i found a large fuse holder with 63 amps at
550v printed on it.?
http://www.saspyro.pwp.blueyonder.co.uk/fuse.JPG

yet the fuse thats attached to it says 40 amps at 550VAC..


http://www.saspyro.pwp.blueyonder.co.uk/Fuse2

Now i am confused ? if the fuse is 40 amps does that mean i have only
transformer that can supply 40 max?..

Inside the transformer there is a board with 1KVA printed on it

http://www.saspyro.pwp.blueyonder.co.uk/inside.JPG

i hope this makes some sense?

Is there any way i can test to see if it can supply 75amps ?

Thanks
Rob

 

[Rob]

rob <sasman@sasman.com> wrote in message news:<c3fg21ln394nvkaapunea212lnaus6kanv@4ax.com>...

Hi

I am a total novice in electrics and am not sure were to post this
question?.. i hope some one can help me...I need a transformer that
can supply aprrox 6 Volts Dc at 75 amps..
I have just purchased a 240 v ac input 1 KVA rated transformer its AC
output is 6v ,8v and 12 V..Now i understand that i need to rectify
the current and have just got off ebay a dozen MBRP40030CTL 400A 30V
schottky diodes with low Vf.. i assume that these will be able to
supply a 75 amp current.. at 6 v?..I got them because there rated at
400 amps..and were very cheap!
http://www.onsemi.com/pub/Collateral/MBRP40030CTL-D.PDF

My other question is that there is label on side of the transformer
which is hand written saying that it is rated at 83 amps?..
Label pic
http://www.saspyro.pwp.blueyonder.co.uk/label.JPG

yet when i looked inside i found a large fuse holder with 63 amps at
550v printed on it.?
http://www.saspyro.pwp.blueyonder.co.uk/fuse.JPG

yet the fuse thats attached to it says 40 amps at 550VAC..


http://www.saspyro.pwp.blueyonder.co.uk/Fuse2

Now i am confused ? if the fuse is 40 amps does that mean i have only
transformer that can supply 40 max?..

Inside the transformer there is a board with 1KVA printed on it

http://www.saspyro.pwp.blueyonder.co.uk/inside.JPG

i hope this makes some sense?

Is there any way i can test to see if it can supply 75amps ?

Thanks
Rob

Your transformer is rated at 1000kVA, so at 12V this gives
1000/12=83A, which fits with the label. As this transformer is second
hand, the original owner only needed 40A and so installed a 40A fuse.
After all, your AC mains outlet may be rated at 13A, but that doesn't
stop you fitting a 1A fuse in your table lamp. Remember to mount your
rectifier diodes on a heatsink.

 

[John Fields]

On 4 Mar 2005 12:28:24 -0800, rrjonasar@lineisp.co.uk (Rob) wrote:

rob <sasman@sasman.com> wrote in message news:<c3fg21ln394nvkaapunea212lnaus6kanv@4ax.com>...
Hi

I am a total novice in electrics and am not sure were to post this
question?.. i hope some one can help me...I need a transformer that
can supply aprrox 6 Volts Dc at 75 amps..
I have just purchased a 240 v ac input 1 KVA rated transformer its AC
output is 6v ,8v and 12 V..Now i understand that i need to rectify
the current and have just got off ebay a dozen MBRP40030CTL 400A 30V
schottky diodes with low Vf.. i assume that these will be able to
supply a 75 amp current.. at 6 v?..I got them because there rated at
400 amps..and were very cheap!
http://www.onsemi.com/pub/Collateral/MBRP40030CTL-D.PDF

My other question is that there is label on side of the transformer
which is hand written saying that it is rated at 83 amps?..
Label pic
http://www.saspyro.pwp.blueyonder.co.uk/label.JPG

yet when i looked inside i found a large fuse holder with 63 amps at
550v printed on it.?
http://www.saspyro.pwp.blueyonder.co.uk/fuse.JPG

yet the fuse thats attached to it says 40 amps at 550VAC..


http://www.saspyro.pwp.blueyonder.co.uk/Fuse2

Now i am confused ? if the fuse is 40 amps does that mean i have only
transformer that can supply 40 max?..

Inside the transformer there is a board with 1KVA printed on it

http://www.saspyro.pwp.blueyonder.co.uk/inside.JPG

i hope this makes some sense?

Is there any way i can test to see if it can supply 75amps ?

Thanks
Rob

Your transformer is rated at 1000kVA, so at 12V this gives
1000/12=83A, which fits with the label. As this transformer is second
hand, the original owner only needed 40A and so installed a 40A fuse.
After all, your AC mains outlet may be rated at 13A, but that doesn't
stop you fitting a 1A fuse in your table lamp. Remember to mount your
rectifier diodes on a heatsink.

---
Why? take a look at the data sheet and find the curves for Vf VS If.
If he's trying to eke out that last little bit of voltage from the 6V
secondary tap he might want to let the diodes run warm.

BTW, if you had bothered to read the thread before you posted, you
might have noticed that you were only echoing what had been posted
previously.

--
John Fields

 

[Fritz Schlunder]

"John Fields" <jfields@austininstruments.com> wrote in message
news:5mmh219kovrpu4afn9j3mig5gs78ugmrpt@4ax.com...

On 4 Mar 2005 12:28:24 -0800, rrjonasar@lineisp.co.uk (Rob) wrote:
Your transformer is rated at 1000kVA, so at 12V this gives
1000/12=83A, which fits with the label. As this transformer is second
hand, the original owner only needed 40A and so installed a 40A fuse.
After all, your AC mains outlet may be rated at 13A, but that doesn't
stop you fitting a 1A fuse in your table lamp. Remember to mount your
rectifier diodes on a heatsink.

---
Why? take a look at the data sheet and find the curves for Vf VS If.
If he's trying to eke out that last little bit of voltage from the 6V
secondary tap he might want to let the diodes run warm.

Err... I know you are smarter than this Johnnie boy. At 75A DC output the
diodes will start smoking very quickly without adequate heatsinking. At
around 75A instantaneous forward current the voltage drop at maximum
junction temp. is somewhere around 0.25V, or around 19W per element. Add to
this something vaguely around say 2W for reverse leakage current, and a
heatsink is definitely required.

BTW, if you had bothered to read the thread before you posted, you
might have noticed that you were only echoing what had been posted
previously.

It appears Rob posted from Google Groups which has a large delay between
when people post and when messages appear. It is possible he never even saw
your post before he posted his own message. Cut him a little slack.

 

[Fritz Schlunder]

"rob" <sasman@sasman.com> wrote in message
news:c3fg21ln394nvkaapunea212lnaus6kanv@4ax.com...

Hi

I am a total novice in electrics and am not sure were to post this
question?.. i hope some one can help me...I need a transformer that
can supply aprrox 6 Volts Dc at 75 amps..
I have just purchased a 240 v ac input 1 KVA rated transformer its AC
output is 6v ,8v and 12 V..Now i understand that i need to rectify
the current and have just got off ebay a dozen MBRP40030CTL 400A 30V
schottky diodes with low Vf.. i assume that these will be able to
supply a 75 amp current.. at 6 v?..I got them because there rated at
400 amps..and were very cheap!
http://www.onsemi.com/pub/Collateral/MBRP40030CTL-D.PDF

These diodes are well suited to the task at hand.

Is there any way i can test to see if it can supply 75amps ?

If you don't need any kind of filtering and are perfectly content with
really lumpy DC power direct from the bridge rectifier, then yes this
transformer should be adequately capable of producing your 6V(ish) at 75A
DC. Naturally you would have to remove or replace the fuse with something
greater than 75A. Fusing on the primary side is somewhat easier and cheaper
since 250V fuses at a few amps are much cheaper and more common than giant
75A+ fuses (although at your low voltages 30V automotive blade fuses should
suffice). On the other hand a fuse on the primary side would need to have
greater overrating to avoid nuisance fusing since it must contend with
transformer saturation surge currents at initial power up.

If you need 6V DC at 75A DC well filtered, then the situation changes
significantly. The transformer outputs vaguely 6V AC, but when filtered
yields a peak voltage of roughly 8.5V less two schottky diode drops. When
well filtered with a large capacitor, the power factor will be substantially
reduced. Capacitive input filters fed from full wave rectified sinusoidal
sources can have typical power factors of around 0.6. In other words, if
your load still draws 75A DC (at around or a bit less than 8V DC with the 6V
AC tap) with the capacitive filter, the effective heating in the transformer
secondary will be more than what a 6V 75A RMS AC load would produce.

Suppose your load draws 75A DC at 8V but with a powerfactor of 0.6 (due to
input filter capacitor). This means your load uses 75*8 = 600W, but
600W/0.6 = 1000VA which is the maximum rating of your transformer. It isn't
totally clear if your transformer is rated for 1000VA load from any of the
taps, or if it is rated for 1000VA from the 12V tap only, but somewhat less
on the lower voltage taps. The 6V tap has half the turns of the 12V tap, so
if one assumes the length and size of the wire used for the 6V tap is half
that of the 12V tap, then 1000VA at half the voltage is twice the current.
However, P=I^2*R, so for half the resistance (half the length of wire for
the full 12V tap), but twice the current, the power dissipated is double.
On the other hand the 6V winding is probably located closer to the center of
the transformer where the diameter of each loop of wire is smaller, so the
6V tap probably has somewhat less than half the resistance of the 12V tap,
so the heating situation isn't quite so bad.

All and all, it is hard to say for certain if 75A DC is overdoing it a
little bit if you use a capacitive input filter. Without the filter a load
current of 75A DC is fine, but with the filter 75A might be pushing it
slightly over its limits. Long term reliability may be somewhat
compromised, so for peace of mind improved cooling using forced air and/or
reducing the load current to something a bit less than 75A might be
advisable.

What exactly is your application?

 

[John Fields]

On Sat, 5 Mar 2005 00:52:44 -0700, "Fritz Schlunder" <me@privacy.net>
wrote:

"rob" <sasman@sasman.com> wrote in message
news:c3fg21ln394nvkaapunea212lnaus6kanv@4ax.com...
Hi

I am a total novice in electrics and am not sure were to post this
question?.. i hope some one can help me...I need a transformer that
can supply aprrox 6 Volts Dc at 75 amps..
I have just purchased a 240 v ac input 1 KVA rated transformer its AC
output is 6v ,8v and 12 V..Now i understand that i need to rectify
the current and have just got off ebay a dozen MBRP40030CTL 400A 30V
schottky diodes with low Vf.. i assume that these will be able to
supply a 75 amp current.. at 6 v?..I got them because there rated at
400 amps..and were very cheap!
http://www.onsemi.com/pub/Collateral/MBRP40030CTL-D.PDF


These diodes are well suited to the task at hand.


Is there any way i can test to see if it can supply 75amps ?


If you don't need any kind of filtering and are perfectly content with
really lumpy DC power direct from the bridge rectifier, then yes this
transformer should be adequately capable of producing your 6V(ish) at 75A
DC.

---
Yes, but... I believe the question was if there was any way to _test_
the transformer to see whether it'll put out 75 amps.

--
John Fields

 

[Fritz Schlunder]

"Fritz Schlunder" <me@privacy.net> wrote in message
news:38t3q9F5logrpU1@individual.net...

At 75A DC output the
diodes will start smoking very quickly without adequate heatsinking. At
around 75A instantaneous forward current the voltage drop at maximum
junction temp. is somewhere around 0.25V, or around 19W per element. Add
to
this something vaguely around say 2W for reverse leakage current, and a
heatsink is definitely required.

Before someone calls me on this oversight, I should call myself on it. In a
full wave bridge rectifier each diode only conducts the load current part of
the time, and only blocks voltage part of the time, so the above theoretical
dissipation figures would probably be closer to half that stated or
somewhere around 10W total. 10W per device is still probably too much for
use without proper heatsinking.

 

[Lord Garth]

"rob" <sasman@sasman.com> wrote in message
news:71jj2156rh6tbuuhqpf89s40opa253auhr@4ax.com...

On Fri, 04 Mar 2005 10:43:17 -0600, John Fields
jfields@austininstruments.com> wrote:


From: John Fields <jfields@austininstruments.com


Thanks John for that fantastic reply..I am much more confident now
that i have more information on my PSU...

I am using the PSU to power an electrochemical cell for making
Chlorate/Perchlorate.. Unsmoothed Dc current will work ok for this
purpose....The voltage is not critical ..

snip?
Rob

Take care with this substance...it is a contact explosive.

 

[Rich Grise]

On Sat, 05 Mar 2005 04:42:45 -0600, John Fields wrote:

On Sat, 5 Mar 2005 00:52:44 -0700, "Fritz Schlunder" <me@privacy.net
wrote:


"rob" <sasman@sasman.com> wrote in message
news:c3fg21ln394nvkaapunea212lnaus6kanv@4ax.com...
Hi

I am a total novice in electrics and am not sure were to post this
question?.. i hope some one can help me...I need a transformer that
can supply aprrox 6 Volts Dc at 75 amps..
I have just purchased a 240 v ac input 1 KVA rated transformer its AC
output is 6v ,8v and 12 V..Now i understand that i need to rectify
the current and have just got off ebay a dozen MBRP40030CTL 400A 30V
schottky diodes with low Vf.. i assume that these will be able to
supply a 75 amp current.. at 6 v?..I got them because there rated at
400 amps..and were very cheap!
http://www.onsemi.com/pub/Collateral/MBRP40030CTL-D.PDF


These diodes are well suited to the task at hand.


Is there any way i can test to see if it can supply 75amps ?


If you don't need any kind of filtering and are perfectly content with
really lumpy DC power direct from the bridge rectifier, then yes this
transformer should be adequately capable of producing your 6V(ish) at 75A
DC.

---
Yes, but... I believe the question was if there was any way to _test_
the transformer to see whether it'll put out 75 amps.

Sure. Just put a 0.08 ohm, 450 watt resistor on the secondary, and meter
the volts and amps. Or 75 ea. 6V @ 6W light bulbs. ;-)

Fuse the primary at NO MORE THAN 4 amps. At 75 amps on the output, it
should only be drawing - oh, I'll leave that arithmetic as an exercise
for the reader. ;-)

Cheers!
Rich


Cheers!
Rich

 

[Rich Grise]

On Sat, 05 Mar 2005 00:01:52 -0700, Fritz Schlunder wrote:

It appears Rob posted from Google Groups which has a large delay between
when people post and when messages appear. It is possible he never even
saw your post before he posted his own message. Cut him a little slack.

People who post from google groups don't deserve slack.

The _only_ thing google beta has going for it is posts show up right away.
But it's still the person's responsibility to read the thread, and copy/
paste context material.

What they really _should_ do, of course, is use a real newsreader and
their ISP's news server. I've never heard of an ISP that doesn't have
a news server.

Thanks,
Rich

 

[Rich Grise]

On Sat, 05 Mar 2005 16:08:34 +0000, Lord Garth wrote:

"rob" <sasman@sasman.com> wrote in message
news:71jj2156rh6tbuuhqpf89s40opa253auhr@4ax.com...
On Fri, 04 Mar 2005 10:43:17 -0600, John Fields
jfields@austininstruments.com> wrote:


From: John Fields <jfields@austininstruments.com


Thanks John for that fantastic reply..I am much more confident now
that i have more information on my PSU...

I am using the PSU to power an electrochemical cell for making
Chlorate/Perchlorate.. Unsmoothed Dc current will work ok for this
purpose....The voltage is not critical ..

snip?
Rob

Take care with this substance...it is a contact explosive.

According to a current thread on news:rec.pyrotechnics, you don't even
need an electrolytic cell - with the right mixture of reagants, the KClO4
precipitates out of solution on its own.

And, on its own, it's not explosive. It's a very strong oxidizer, but it
still needs fuel. You might be thinking of ammonium perchlorate - that's
pretty nasty stuff.

Good Luck!
Rich

 

[John Fields]

On Sat, 05 Mar 2005 15:53:54 +0000, rob <sasman@sasman.com> wrote:

On Fri, 04 Mar 2005 10:43:17 -0600, John Fields
jfields@austininstruments.com> wrote:


From: John Fields <jfields@austininstruments.com


Thanks John for that fantastic reply..I am much more confident now
that i have more information on my PSU...

I am using the PSU to power an electrochemical cell for making
Chlorate/Perchlorate.. Unsmoothed Dc current will work ok for this
purpose....The voltage is not critical ..

You mention that i will need a 0.08 Ohms resistor to produce a 75 amp
load...
I have purchased some 18 gauge Nikrothal 80 wire with a .040"
diameter, 0.4062 ohms per foot..

Am i correct in thinking that i can divide 0.4062 (foot ) by 12 to
get the Ohms per inch?.. which = 0.03385 Ohms (per inch) so i would
need 2.37 inches ?...I intend to use this as a variable resistor to
increase/decrease the current flowing in to the cell..

---
I mentioned using a 0.08 ohm load to test your transformer because if
you place a 0.08 ohm load across a 6 volt source, from Ohm's law
you'll get:

E 6V
I = --- = ------- = 75 amperes
R 0.08R

flowing through the resistance if the source is capable of supplying
that current. Also, I mentioned that the resistor will be dissipating
450 watts, which is a HUGE amount of power for a 2.37 inch long piece
of Nichrome to handle.

There's also another problem, and that's that if you're planning on
using the Nichrome wire as a current limiting resistor, it'll be in
series with the cell, so if you want to get 75 amps into the cell the
sum of the resistances of the cell _and_ the resistor will have to be
equal to 0.08 ohms, which will make the resistance of the resistor
substantially less that 0.08 ohms.

Here's your proposed circuit:

XFMR BRIDGE CELL
+------+ +------+ +------+
MAINS-----|P S|--[FUSE]--|~ +|---[R1]<--|--+ |
| | | | | | |
| | | | | [R2] |
| | | | | | |
MAINS>----|P S|----------|~ -|----------|--+ |
+------+ +------+ +------+


Notice that if the output of the bridge is 6VRMS, in order for 75 amps
to flow through the cell the _sum_ of R1 (the nichrome resistor) and
R2 (the cell itself) must be 0.08 ohms. That means that you'll
probably be dealing with _very_ short lengths of white-hot Nichrome
wire to get 75A into the cell. Not a good idea, for two reasons: The
first is that white-hot Nichrome is dangerous, and the second is that
as Nichrome heats, its resistance increases. You could make things
better by getting longer lengths of Nichrome and connecting them in
parallel, but that's still a messy way way to do it, and if you need
to be changing resistances in the middle of a run it's going to turn
into a nightmare.

What I _strongly_ suggest you do is get a Variac (240V 50Hz in, 0 to
240V out at 5 or 10 amps. Five would work and would be less expensive
than ten, but ten wouldn't have to work as hard.) and use it to
control the voltage _into_ the transformer. Neat, clean, and you can
monitor the input current and get a pretty good idea of what's going
into the cell, although mneasuring the output current would still be
best. Also, using the Variac would allow you to use the 8V tap (or
even the 12V tap) and get away from too low a voltage out of the
bridge because of the diode drops and too low a voltage into the cell
because of wiring.

--
John Fields

 

[Roger Johansson]

Rich Grise <richgrise@example.net> wrote:

What they really _should_ do, of course, is use a real newsreader and
their ISP's news server. I've never heard of an ISP that doesn't have
a news server.

You would have been right 10 years ago, but today very few ISP's have a
newsserver, and even fewer run it in a good way, if they have one.

Today most participators in usenet have to find newsservers on their own.

Up till a few weeks ago the individual net university server in Berlin
was the best, but they decided to start taking a small fee, so a lot of
people have to find new free newsservers, or pay.

More and more people are forced to use some web interface, because they
do not know where to find a free newsserver. And many youngsters have no
idea that there is a thing called usenet to begin with. They think
newsgroups is the same as web forums.


--
Roger J.

 

[Fritz Schlunder]

"The Phantom" <phantom@aol.com> wrote in message
news:86rk21d3v53pp9928bmb9qto2so5jpl7mn@4ax.com...

The advantage of the short circuit test method is that you only
dissipate a power equal to the losses in the transformer which should
be under 100 watts for a transformer of this size. (light bulbs or
electric heaters extra) It does not include the core loss, but that
shoud be a negligible part of total losses at nearly full power out of
the transformer.

Are you sure? When designing a power transformer you get allot of variables
to play with to try to optimize the design. It has been my understanding
that often the most optimum design (for size/weight) makes the core losses
approximately equal to the copper losses while operated under full load.

 

[The Phantom]

On Sun, 6 Mar 2005 07:10:55 -0700, "Fritz Schlunder" <me@privacy.net> wrote:

"The Phantom" <phantom@aol.com> wrote in message
news:86rk21d3v53pp9928bmb9qto2so5jpl7mn@4ax.com...
The advantage of the short circuit test method is that you only
dissipate a power equal to the losses in the transformer which should
be under 100 watts for a transformer of this size. (light bulbs or
electric heaters extra) It does not include the core loss, but that
shoud be a negligible part of total losses at nearly full power out of
the transformer.

Are you sure? When designing a power transformer you get allot of variables
to play with to try to optimize the design. It has been my understanding
that often the most optimum design (for size/weight) makes the core losses
approximately equal to the copper losses while operated under full load.

It is quite true that the optimum design for a transformer under full load is to have

approximately equal core and copper losses. The operative phrase here is "under full
load". The transformers handling the output of Grand Coulee dam are probably operated
near full load a lot of the time, as are the transformers in major substations, and are
designed for nearly equal core and copper losses. But small transformers like the OP
described, and many others, are not usually designed that way.

Consider the so-called "pole pig" transformer; the ones you see on telephone poles. The
residential loads are rarely taking such a transformer to its full load capability for
extended periods of time. In fact, they spend most of their time lightly loaded, and
their core losses are a substantial quantity which the power companies try to minimize. I
remember when Metglas was first developed, there was talk of using it in the cores of pole
pigs because of its very low losses. There is enough total core loss in all the pole pigs
in a major utility district to make it worth while (maybe. They haven't done it yet).

There is a classic old text, "Magnetic Circuits and Transformers" by the staff of MIT,
published in 1943. It has many examples and descriptions of testing methods and modelling
of transformers. One example, on page 377, analyzes a 15 kva, 2400:240 volt, 60 Hz
distribution transformer. The copper loss at full load is 276 watts, the core loss is 84
watts, for an efficiency (at full load) of 97.09%. The book then goes on to describe how
to compute the "energy efficiency" (which is not "full load" efficiency), considering the
daily load curves. The core and copper loss can be apportioned for lowest cost of use and
ownership only by considering how the transformer is loaded. And if it isn't operated at
full load nearly all the time, the best apportionment is NOT to have equal core and copper
losses. At anything less than full load nearly all the time, you want to have lower core
loss.

 

[The Phantom]

On Sun, 06 Mar 2005 12:09:30 -0600, John Fields <jfields@austininstruments.com> wrote:

On 5 Mar 2005 21:29:04 -0600, The Phantom <phantom@aol.com> wrote:

On Fri, 04 Mar 2005 10:47:06 +0000, rob <sasman@sasman.com> wrote:

Hi

I am a total novice in electrics and am not sure were to post this
question?.. i hope some one can help me...I need a transformer that
can supply aprrox 6 Volts Dc at 75 amps..

Rob,
Do you have any instruments that can measure currents of ~75 amps
AC? Of 5 amps AC?

Do you have an AC voltmeter?

The tests that others have proposed so far require you to dissipate
power on the order of .5 kw.

The standard method for testing the capability of a transformer
without actually dissipating a power equivalent to the rating of the
transformer is the short circuit test.

---
AFAIK, the short test is used to measure the copper losses, not the
"capability" of a transformer, whatever that means.

I think, as I explained to Fritz Schlunder, that for small transformers like this, the
copper loss *is* what, to a great extent, determines the (power handling) "capability" of
the transformer.

---

You need a variac for this. You dead short the 6 volt output of the
transfomrer.

---
NO!!! If you do you will exceed the current rating of the secondary
and, possibly, damage the transformer. The transformer is rated for
1kVA out of the _entire_ secondary which, at 12VRMS out comes to the
83.3 amps noted on the transformer's faceplate. That is, the
transformer secondary is wound with wire which is designed to carry
83.3 amps no matter which voltage tap is used.

I think you are quite right. I didn't read your first response to Rob where you
interpreted the label, which I also didn't bother to look at. :-( I was thinking that
the secondary had multiple windings which could be paralleled, which, after looking at the
label, I would agree that it doesn't. The 83 amps times 12 volts gives 996 VA, the rating
of the transformer. Mea Culpa. He should limit the primary current to 1.875 amps as I
mention later.

---

You connect a variac capable of about 5 amps output to
the power line. Turn the output of the variac all the way down to
zero output (THIS IS VERY IMPORTANT). Connect the variac output to
the 240 volt winding of the transformer with an ammeter in series to
monitor the current. SLOWLY turn up the variac until the ammeter
reads the rated current for the 240 volt winding (about 4 amps AC in
this case). You may need to put a (power) resistor of a few ohms in
series with the 240 volt winding of the transformer if the current
comes up too fast (which it undoubtedly will) when you SLOWLY turn up
the variac. (If you don't have any suitable power resistors, use an
electric heater or some light bulbs as resistors). (An even better
alternative would be to follow the variac with a small step-down
transformer. Something with about 5 amps output at 6 or 12 volts.
The variac would power the line-voltage winding and the low voltage
winding would power the 240 volt winding of the 1kva transformer you
are testing.)

When you have the current into the 240 volt winding set to 4 amps AC,
you just wait around and see how hot the transfomer gets.

---
That's not really a valid criterion, since the core losses are only
going to be a fraction of what they would normally be with 240VAC on
the primary. Moreover, if you're going to do it properly you need to
monitor the temperature rise of the transformer over ambient and make
sure it doesn't exceed the spec.

I'm not sure what instrumentation Rob has access to; that's why I asked about meters.
If you're really going to do it right, monitoring the temperature rise of the
"transformer" over ambient isn't enough. You need to know the temperature of the
windings, and the rating of the insulation. Since the transformer doesn't appear to have
thermocouples buried in the windings, what is what I usually do with a new design, one
would have to use the method of measuring the resistance of the windings at room temp, and
then applying current for several hours, followed by a measurement of the winding
resistances when hot. From this the temperature rise of the windings can be inferred.
One starts out with less than rated current and makes a measurement, then if temperatures
are under allowable values, increase the current and take some more measurements.

Given a starting temp (room temp), designated Tcold, and measured resistance of a winding
at that temp, designated Rcold, and a measured resistance Rhot after allowing the
transformer to reach equilibrium with heating from the current in the windings; the temp,
Thot, is given by:

Rhot
Thot = ------ (234.5 + Tcold) - 234.5
Rcold

in degrees centigrade.

Since we don't know whether his transformer has Class C, Class H, or whatever, insulation,
we can't really tell what temperature rise is acceptable. One way to find out would be to
assume (there I go again) that the transformer was designed properly, and will exhibit a
temp rise at full load of just about the rating of the insulation system. Rob could short
the 12 volt winding, and apply rated current to the primary, wait a few hours and measure
the temperature rises of the windings. This would presumably be what the transformer was
designed for. He could then measure the temp rises with his rectifier load and see if
they are <= to what he got with the full load test I just described.

I'm sure you know all this, John, but I'm going into detail for Rob's benefit.

---

The current in the shorted 6 volt winding should be about 160 amps.

This is too much, as John points out. Don't do this.

Ideally you should short the 6 volt winding into a suitably
rated shunt and then you could directly measure the current there.

---
The shunt's an OK idea, but it needs to short out the _entire_
secondary, for the reason given earlier.

Or limit the current to less than 83 amps in the shunt; this would be acceptable for the
6 volt tap. The shunt could also be used to short the rectifier output so that the actual
current there could be measured. To assess the heating in the 6 volt winding, a true RMS
meter would need to be used. The electrochemical effect of the current is proportional to
coulombs/sec, so an average responding meter would be appropriate to assess this.

Rob's original question was whether the transformer could supply 75 amps at 6 volts. It
would be reasonable to assume that the transformer can meet its nameplate rating. So if
he has access to a high current shunt, he can just put it in series with the 6 volt output
tap and, with a true RMS meter, verify that the current is less than 83 amps, and feel
safe without all this fooling around we've all been telling him to do!

I somehow doubt that Rob has any high current shunts around. Maybe he has access to some,
but I got the impression that he might be short on instrumentation.

So, what we all need to help him with is the measurement of the secondary current without
having to buy a high current shunt. One possibility is to make a current transformer
somehow. Or, how about this? Ten guage copper wire (here in the states, anyway) has very
close to .001 oms/ft. If he were to cut a 1.5-foot length of 10 guage wire, and solder on
a couple of smaller wires exactly 1 foot apart, he would have a shunt of 1 milliohm, and
with a current of 75 amps, he would need to measure 75 millivolts AC, and even the
cheapest Radio Shack DVM can do that. The problem is that the piece of wire would be
dissipating 5.6 watts and would get hot. The resistance would change enough to inspire a
vote of no confidence. He could parallel two such wires and have only 2.8 watts
dissipation, and finally, he could put the put the homemade shunt in a plastic tray of
water (at 20 degrees C), which would probably limit the temperature rise to an acceptable
value. (He can find out the resistance/foot of available wire of about this size and
figure out exactly how far apart to put the meter taps.) (And, he will need a true RMS
meter)

---


Without knowing the details of the transformer insulation system, I
can't say what the maximum allowable temperature rise is, but if the
transformer doesn't start stinking too much, you're probably ok. So
far, we would be testing the transformer at its full rated power.

---
No, you'd only be exciting the copper, not the core. To determine the
core losses, you'd need to do the open circuit test.

I'm aware of this, but as I explained in another reply to Fritz Schlunder, typically
small transformers like this aren't optimized for continuous operation at full load. It
could be, but I think it unlikely. Anyway, to do it right (measure the core loss) would
require a wattmeter designed for measurements under low power factor conditions.

But, he could get an idea of the relative magnitude of the core loss by measuring the room
temp resistances of the windings as mentioned earlier and then applying 240 volts to the
primary with the secondaries unloaded, waiting several hours, measuring the warm winding
resistances and using the formula given earlier.

---

Since you only want 75 amps from the 6 volt winding, you could set the
primary current in this test to 1.875 amps AC. This would correspond
to 75 amps in the 6 volt secondary. Just let it run for a few hours
while monitoring the temperature rise. This will provide a test of
the transformer at the actual current which you will be using.

---
This isn't a good test either, because the core losses will be even
less than they were previously and won't reflect the conditions under
which the transformer will be expected to function in real life.
---

Rectifying the output of the transformer will cause the current to be
non-sinusoidal which will increase the copper (I^2*R) losses.

---
Yes. Conversion efficiency is about 81% for a full-wave bridge.

And the distortion of the current waveform will make the the copper losses even more
significant compared to the core loss than with a pure resistive load.

---

You could also connect your rectifiers in whatever configuration you
will use, and short the output from the rectifiers. This will give
you a method of testing the heatsinking of the rectifiers. As before,
turn the variac which is supplying the 240 volt winding ALL THE WAY
DOWN before you start the test. Turn up the variac SLOWLY while
monitoring the current in the 240 volt winding until you reach 1.875
amps AC. You should use a true RMS ammeter when the rectifiers are in
circuit for best results. I wouldn't turn the current up to 1.875
amps right away with the rectifiers in circuit; turn it up gradually
while monitoring the temperature of the rectifiers. If the rectifiers
sizzle when touched with a wet finger, you probably should use some
more heat sinking.

The advantage of the short circuit test method is that you only
dissipate a power equal to the losses in the transformer which should
be under 100 watts for a transformer of this size. (light bulbs or
electric heaters extra) It does not include the core loss, but that
shoud be a negligible part of total losses at nearly full power out of
the transformer.

---
As Fritz Schundler pointed out earlier, an optimum transformer design
will yield core losses equal to copper losses.

As I explained to Fritz, this is only true for a transformer used nearly all the time at
full rated load.

In some special applications the disparity between core and copper loss is even greater
than the example I cited to Fritz. Inverters for off grid use are designed for very low
core loss because the customer wants to be able to leave the inverter running for extended
periods without a large drain on the battery. A transformer I designed for a 2400 watt
inverter had about 12 watts of core loss and 347 watts of copper loss (full load) on EI200
laminations, 2.75" stack height. I think inverter transformers represent the extreme of
this disparity.

A good example of a transformer where the core loss is prominent is the transformer in a
"wall wart". Those transformers are usually blazing hot even without a load. They use
cheap iron and run 'em hot. These transformers are very small, and the resistance of the
primary is so high that the volt-seconds seen by the core is greatly reduced by the IR
loss in the primary when the transformer is loaded. This reduction in core loss
compensates for the increase in copper loss due to load so that the transformer is within
ratings (barely, I think).

 

[rob]

Well thanks for all the help and advice..I have took Johns advice and
this morning the post man delivered a Claude Lyons Regavolt 715-E 240V
15Amp Variac...
http://www.saspyro.pwp.blueyonder.co.uk/Variac.jpg

http://www.saspyro.pwp.blueyonder.co.uk/Variacrating.jpg


Will this variac work as a variable resistor? so that i can control
the current going in to my electrochemical cell?.. or will it just
control the voltage?..

Phantom I have a Digital Multimeter a 2 amp panel meter and a 25 amp
panel meter that is all the measuring equipment i have .. i am looking
for a 0 -100 amp DC panel meter + shunt..I cant do much more until my
Diodes arrive..i will let you know how i progress ..



Thanks

Rob

Ps Im posting this Via Giganews Newserver ...Altho i do you use
Google groups to read this and other newsgroups..

 

[Larry Brasfield]

"rob" <sasman@sasman.com> wrote in message
news:t7bu21142eldr84dnj0uvvrstmib0nfd9j@4ax.com...

Well thanks for all the help and advice..I have took Johns advice and
this morning the post man delivered a Claude Lyons Regavolt 715-E 240V
15Amp Variac...
http://www.saspyro.pwp.blueyonder.co.uk/Variac.jpg

http://www.saspyro.pwp.blueyonder.co.uk/Variacrating.jpg

You should be aware that this device very likely does not
have an isolated secondary. This means that your output
circuit will have an Ohmic connection to the power line.
Depending on the plug polarity and the correctness of
your outlet wiring, this can result in a personnel hazard.

You should take precautions to ensure that people are
not able to inadvertanly touch parts of your setup that
are connected to the secondary.

Will this variac work as a variable resistor? so that i can control
the current going in to my electrochemical cell?.. or will it just
control the voltage?..

The resistance in your circuit, both in the secondary
and what you intentionally introduce, together with
electrode/solution resistances, will set the current as
a function of the voltage you dial in with the variac.

--
--Larry Brasfield
email: donotspam_larry_brasfield@hotmail.com
Above views may belong only to me.

 

[The Phantom]

On Wed, 9 Mar 2005 14:52:20 -0800, "Larry Brasfield"
<donotspam_larry_brasfield@hotmail.com> wrote:

"rob" <sasman@sasman.com> wrote in message
news:t7bu21142eldr84dnj0uvvrstmib0nfd9j@4ax.com...

Well thanks for all the help and advice..I have took Johns advice and
this morning the post man delivered a Claude Lyons Regavolt 715-E 240V
15Amp Variac...
http://www.saspyro.pwp.blueyonder.co.uk/Variac.jpg

http://www.saspyro.pwp.blueyonder.co.uk/Variacrating.jpg

You should be aware that this device very likely does not
have an isolated secondary.

If when you say "secondary", you are referring to the "secondary" of the variac, then of
course he should make sure that all connections from the "secondary" of the variac (which
is really an autotransformer) are safely insulated.

This means that your output
circuit

If by "output circuit", you mean that connection from the output of the variac to the 240
volt winding (primary) of his 1 kva transformer, then yes; but I got the impression that
his 1 kva transformer had a secondary without a galvanic connection to the primary. If
this is so, then touching parts of the setup which are connected to the 6 volt secondary
shouldn't be a problem.

will have an Ohmic connection to the power line.
Depending on the plug polarity and the correctness of
your outlet wiring, this can result in a personnel hazard.

You should take precautions to ensure that people are
not able to inadvertanly touch parts of your setup that
are connected to the secondary.

I got the impression that his 1 kva transformer had a secondary without a galvanic
connection to the primary. If this is so, then touching parts of the setup which are
connected to the 6 volt secondary shouldn't be a problem. If on the other hand, when you
say "secondary", you are referring to the "secondary" of the variac, then of course he
should make sure that all connections from the "secondary" of the variac (which is really
an autotransformer) are safely insulated.

Will this variac work as a variable resistor? so that i can control
the current going in to my electrochemical cell?.. or will it just
control the voltage?..

The resistance in your circuit, both in the secondary
and what you intentionally introduce, together with
electrode/solution resistances, will set the current as
a function of the voltage you dial in with the variac.

 

[John Fields]

On Mon, 11 Apr 2005 21:10:28 +0100, Rob <Rob@rob.co.uk> wrote:

Hi
well since last time i asked for advice i have had a few
problems...First problem was i had ordered some 400A 30V low Vf
Schottky diode rectifiers i won them on ebay
http://cgi.ebay.co.uk/ws/eBayISAPI.dll?ViewItem&rd=1&item=3877178672&ssPageName=STRK:MEWN:IT
...But never got them through the post so that delayed me a week or
so..Then i Bought from a Surplus shop 8 x 110 amp Rectifiers..they
were only a about 3$ each ...used but untested..I got them from Cpi
surplus in the USA ..very fast service...Unfortunatley the diodes i
bought were Silicon controlled rectifiers...Which my Radio Ham friend
told me they are no use for what i wanted...So i was realy pissed
off!!...So i decided to spend even more money on another PSU rated at
5 Volts at 325 amps..I won it on ebay a week orso ago..
http://cgi.ebay.co.uk/ws/eBayISAPI.dll?ViewItem&rd=1&item=7501227612&ssPageName=STRK:MEWN:IT..

http://www.abex.co.uk/sales/electronic/psu/rack/p10/p_series.pdf

Its made by Power Ten INC.. its a 19 inch rack mounted PSU with
variable Voltage and Current control..First impressions were WOW!!
when i turned it on it sounded like a jet taking of lots of noise from
the internal fans...However my Smile was soon wiped of my face when i
tried to test the power output...

I connected it up to a short length of nicrome wire and set the
voltage to 5 Volts and slowly turned up the current .. all i could get
was 20 amps the wire glowed white hot ..Now i have done this test
before on my old 400watt computer PSU and it went off scale on my 25
amp analoge meter.. and melted the nicrome wire...

I then tried a very crude electrolyisis of my Lead Nitrate solution,
using my old 400 watt PSU i can get 15 amps thru it ..But with this
PSU i only managed a poxy 2 amps... i am now totaly pissed off!!!?

Have i wasted even more money?...How can i test my PSU?.. will it
damage the unit if i make a dead short on the connections.. and slowly
turn up the current ?...The control for the current works by setting a
limit on the power supply...I hope some one can put a smile back on my
face...

Dont forget i want to use the power supply for electrolysing Sodium
Chloride solution..I have read several patents and the cells run at
about 4.75 volts at several hundred amps....

So i assumed that a 5V supply should be able to do the job... I have
got several amp meters to read the current..

http://cgi.ebay.com/ebaymotors/ws/eBayISAPI.dll?ViewItem&rd=1&item=4534893020&sspagename=STRK%3AMEWN%3AIT&rd=1

http://cgi.ebay.co.uk/ws/eBayISAPI.dll?ViewItem&rd=1&item=7500828509&ssPageName=STRK:MEWN:IT

But at the moment i have no power supply that can give me the current
i require..

Is the PSU faulty?...or is it something else i am overlooking?..

Hope some one can help ....

---
Sorry to hear about all your problems.

Since nichrome has a fairly high resistance to start with, (about an
ohm per foot for #22 AWG) and a positive tempco, I suspect that what's
happening is that when the wire you're using gets white hot its
resistance is rising to about 0.25 ohms, since by Ohm's law:

E 5V
R = --- = ----- = 0.25 ohms
I 20A

If you want to know for sure, you can measure the resistance of the
piece of wire when it's at room temperature and then apply:


Rt = Ro ( 1 + (0.4E-3 (T - To)))


Where Rt is the hot resistance of the wire,
T is the hot temperature, and
To is the ambient temperature.

Conservatively estimating white heat to be 1000°C, and solving for Ro:


Rt 0.25
Ro = ----------------------- = ---------------------- ~ 0.18 R
1 + (0.4E-3 (T - To)) 1 + 0.4E-3 (1000-20)

So, if you were using #22 wire at 1 ohm per foot, the length of wire
you used to test the PSU would have been:

12" * .18R
l = ------------- = 2.16"
1R

Close?

--
John Fields
Professional Circuit Designer