Maker Pro
Maker Pro

low-cost 1800-amp heating source

G

Glen Walpert

Jan 1, 1970
0
Actually, why can't the two transformer primary windings be
in parallel? That'd reduce the copper resistance. And
L(leak) for that matter, right? Editing Tony's drawing:

AC high + Pri - Load current
--+-----------////////-------+--->---,
| T1 ======== | |
| DC --+---////////---, | |(
| | + Sec - | | |( L(leak)
| +_|_ | | |(
| --- | | |
| | + Sec - | | / Win's Load.
| DC --+---////////---' | \ Rload
| T2 ======== | /
'-----------////////-------' |
+ Pri - AC low

OK, now I see that Tony took advantage of a series primary
connection to use transformers with half the primary voltage
rating. Hah, it appears there's no free lunch.

It appears to me that the DC bias current can only saturate the core
for one half of the input cycle; the half where the magnetic fields
add rather than subtract, and that the original series configuration
had the fields add for both halves of the power cycle in one
transformer or the other, for symnetrical operation. The parallel
circuit would appear to deliver a bunch of DC to the load when
saturated. Good test of a Spice transformer model perhaps :).
 
G

Glen Walpert

Jan 1, 1970
0
It appears to me that the DC bias current can only saturate the core
for one half of the input cycle; the half where the magnetic fields
add rather than subtract, and that the original series configuration
had the fields add for both halves of the power cycle in one
transformer or the other, for symnetrical operation. The parallel
circuit would appear to deliver a bunch of DC to the load when
saturated. Good test of a Spice transformer model perhaps :).

Oops! Never mind, I looked at that drawing twice and still saw it
wrong :-(.
 
A

Archimedes' Lever

Jan 1, 1970
0
"Spurious Response" <[email protected]> skrev i en meddelelse




Well, since we are into plumbing supplies anyway, one might as well run
cooling water through the "wire" - i.e. make it from household copper tubing
;-)
1800 Amps will STILL likely destroy it in short order.

Ever see high speed micro-photography of a small fuse opening?
 
P

Phil Hobbs

Jan 1, 1970
0
Tony said:
I was interested in Ken's idea, so sketched a
possible circuit, as below.

+ Pri - Load current
230V high +------////////---------->-----+
T1 ======== |
DC-------////////-----+ |(
+ Sec - | |( L(leak)
| |(
.----------------' | Win's Load.
| \
| - Sec + /Rload
+---////////---------DC \
T2 ======== |
230V low +-------////////----------------+
- Pri +

Each transformer's primary winding has to be able to
take the final max load current. So if you had a
230V/2500VA load then you would possibly use a pair of
115V/1250VA transformers.

I suspect that if the transformers were designed for
this application then they would be smaller because
more window area could be allotted to the AC side.
Interesting. I'd have thought the opposite--for normal transformer use,
the core needs to avoid saturating on the current peaks, so to fully
saturate the core, the dc secondary current will need to be more than
the peak value in normal AC use, plus whatever safety factor was
designed into the transformer.

And Glen's point about the DC is important--Win's parallel arrangement
can be made to conduct heavily on both half cycles, but the series
arrangement can't. It's just like a full wave centre tapped rectifier.

Cheers,

Phil Hobbs
 
F

Frithiof Andreas Jensen

Jan 1, 1970
0
1800 Amps will STILL likely destroy it in short order.

A 1 Inch diameter water-cooled copper Pipe??
or a 1 inch bundle of the soft and pliant 3 mm stuff used for
refridgeration??

Nah!
 
T

Tony Williams

Jan 1, 1970
0
I don't think it is even as high as that Win. [snip]
3 to 5 times the secondary magnetising current peak
is probably much lower than 3 to 5 times the peak
secondary load current.

I breadboarded it and the control current turns out
to be far higher than I expected...... 60x the
secondary magnetising current peak at normal sec
voltage. Idcmax is practically equal to the normal
sec full load AC current.

However there is actually a good case for designing
for a controlled Vout that only goes up to about
90% of Vin..... that last 10% requires almost double
the control power. See below.


It needed 7.2W of control power for 80VA out (218Vrms
into a 590 ohm load, but only 3.2W for 70VA out
(203Vrms).

More later, probably after the F1 has finished.
 
W

Winfield

Jan 1, 1970
0
Tony said:
I breadboarded it and the control current turns out
to be far higher than I expected...... 60x the
secondary magnetising current peak at normal sec
voltage. Idcmax is practically equal to the normal
sec full load AC current.

However there is actually a good case for designing
for a controlled Vout that only goes up to about
90% of Vin..... that last 10% requires almost
double the control power. See below.


It needed 7.2W of control power for 80VA out (218Vrms
into a 590 ohm load, but only 3.2W for 70VA out
(203Vrms).

More later, probably after the F1 has finished.

That's pretty attractive, a 20:1 ratio, which would
translate to only 50W of DC to control 1kVA. Nice.
 
M

MooseFET

Jan 1, 1970
0
Winfield said:
For quick estimates, could we guess that the DC current
would be 3 to 5 times higher than the peak AC current at
the VA rating of transformer?
I don't think it is even as high as that Win. [snip]
3 to 5 times the secondary magnetising current peak
is probably much lower than 3 to 5 times the peak
secondary load current.

I breadboarded it and the control current turns out
to be far higher than I expected...... 60x the
secondary magnetising current peak at normal sec
voltage. Idcmax is practically equal to the normal
sec full load AC current.

However there is actually a good case for designing
for a controlled Vout that only goes up to about
90% of Vin..... that last 10% requires almost double
the control power. See below.

It needed 7.2W of control power for 80VA out (218Vrms
into a 590 ohm load, but only 3.2W for 70VA out
(203Vrms).

More later, probably after the F1 has finished.

Does the control circuit have a low AC impedance?

Without a low AC impedance, I suspect you will get a bunch of extra
power losses in the control circuit.
 
T

Tony Williams

Jan 1, 1970
0
Tony Williams said:
I breadboarded it.............

Low--------------------------------------+
/|\ |
230Vrms | /
Vout 590\ Load
120V 120V | /
+ - + - \|/ |
High-----//////---------//////-----------+
====== ======
+---//////---------//////---+
| + - - + |
| 18V, 0.4R 18V, 0.4R |
| 5R |
+-----------/\/\------------+
| + |
+------------||-------------+
Idc /|\ 10000uF |
| 11R |
+--/\/\--- + DC - ----------+
0-28V variable.

2x 50VA transformers, wired for 120V primaries and
18V secondaries. Total secondary resistance is 0.8R
and the max rated secondary current is 2.77Arms.

230Vrms supply and a 590R load.

I ignored Ken's nagging and just put the 5R in
first of all, to see what the circulating current
was....... it's large and there is a reason for
that. So Ken's cap is a neccessity. By using a
huge capacitor the AC peak-peak voltage is always
lower than Idc*0.8R, so a polarised electrolytic
can be used safely.

Quick first measurements, to be refined later.

Idc(amps) Vout(rms) dVout/dIdc

0.0 3.9 ______ 162.5 Volts/Amp
0.2 36.4 ______ 146
0.4 65 ______ 130
0.6 91 ______ 115
0.8 114 ______ 95
1.0 133 ______ 85
1.2 150 ______ 75
1.4 165 ______ 70
1.6 179 ______ 65
1.8 (187) 192 ______ 55
2.0 (203) 203 ______ 50
2.2 (212) 213 ______ 5
2.4 (209) 214 ______ 5
2.6 (210) 215 ______ 0
2.8 (215) 215

Those figures in brackets. Up in the 1.8 to 2A
dc control region there seems to be a control
hysteresis. If you casually fiddle the DC supply
up and down to get it spot on then the AC results
can be all over the place. The figures not in
brackets were taken by creeping the supply upwards
only.

You can see from the table that with a 230V supply
then the Load should be rated for about 200V max.
The control power increases unreasonably above
that figure.

Note that the sec windings are rated for 2.77Arms
maximum and at 2.8Adc they are carrying the dc,
plus whatever the circulating current is that point.

Note for Win..... I put a large low pass filter
across Vout and confirmed that there is Zero net
DC voltage there. This allowed a signal transformer
across Vout to allow for a scope to look at the
output waveform.

And why is there a large circulating current between
the two secs, that requires Ken's cap?

It's obvious when you look at the scope and see that
the circulating current is mainly peaks and at 100Hz.

The DC signal drives the two cores in opposite
directions, so one core handles +AC peaks and the
other -AC peaks. So the AC voltages across each
are mirror images, and that's where the circulating
current comes from when the secs are connected together.

More later when domestics permits........
 
M

MooseFET

Jan 1, 1970
0
Low--------------------------------------+
/|\ |
230Vrms | /
Vout 590\ Load
120V 120V | /
+ - + - \|/ |
High-----//////---------//////-----------+
====== ======
+---//////---------//////---+
| + - - + |
| 18V, 0.4R 18V, 0.4R |
| 5R |
+-----------/\/\------------+
| + |
+------------||-------------+
Idc /|\ 10000uF |
| 11R |
+--/\/\--- + DC - ----------+
0-28V variable.

2x 50VA transformers, wired for 120V primaries and
18V secondaries. Total secondary resistance is 0.8R
and the max rated secondary current is 2.77Arms.

230Vrms supply and a 590R load.

I ignored Ken's nagging and just put the 5R in
first of all, to see what the circulating current
was....... it's large and there is a reason for
that.

See what happens when you ignore my nagging :)

I admit that it is going on 40 years since I did this same sort of
experiment myself but it looks like the laws of physics haven't
changed too much in that time. It may be worth measuring the primary
side resistance so you can back that out of your figures and see how
much the controlled impedance is changing.


Also:

0.4R * 2 * (120/18)^2 = 35.5555 Ohms

When only one transformer is saturated, this will appear in the
primary side circuit.


230 * 590 /(590 + 36) = 217V

This combined with the primary resistance may explain some of the
knees in the curve.
Quick first measurements, to be refined later.

Idc(amps) Vout(rms) dVout/dIdc

0.0 3.9 ______ 162.5 Volts/Amp
0.2 36.4 ______ 146
0.4 65 ______ 130
0.6 91 ______ 115
0.8 114 ______ 95
1.0 133 ______ 85
1.2 150 ______ 75
1.4 165 ______ 70
1.6 179 ______ 65
1.8 (187) 192 ______ 55
2.0 (203) 203 ______ 50
2.2 (212) 213 ______ 5
2.4 (209) 214 ______ 5
2.6 (210) 215 ______ 0
2.8 (215) 215

Those figures in brackets. Up in the 1.8 to 2A
dc control region there seems to be a control
hysteresis. If you casually fiddle the DC supply
up and down to get it spot on then the AC results
can be all over the place. The figures not in
brackets were taken by creeping the supply upwards
only.

You can see from the table that with a 230V supply
then the Load should be rated for about 200V max.
The control power increases unreasonably above
that figure.

Note that the sec windings are rated for 2.77Arms
maximum and at 2.8Adc they are carrying the dc,
plus whatever the circulating current is that point.

Note for Win..... I put a large low pass filter
across Vout and confirmed that there is Zero net
DC voltage there. This allowed a signal transformer
across Vout to allow for a scope to look at the
output waveform.

And why is there a large circulating current between
the two secs, that requires Ken's cap?

When one core saturates, the other doesn't. One transformer still
functions as a transformer. Its secondary current is shorted out by
the one that is saturated. This is a disadvantage to this method vs a
real mag-amp.


It's obvious when you look at the scope and see that
the circulating current is mainly peaks and at 100Hz.

I assume you are in a 50Hz mains place. Here in the US, there would
be a lot of 120Hz on the secondaries. The difference int eh
secondaries only appears in that part of the cycle where the core is
saturated. This is late in the alternation.
 
T

Tony Williams

Jan 1, 1970
0
When one core saturates, the other doesn't. One transformer still
functions as a transformer. Its secondary current is shorted out
by the one that is saturated. This is a disadvantage to this
method vs a real mag-amp.

From what you say there it would be better to run the two
primaries in parallel so that the non-saturating primary
can directly pick up the voltage change that is happening
on the saturating primary. So I tried it, as below.

Primaries were relinked for 240Vrms (45 Ohms each)
and wired in parallel.

Low--------------------------------------+
|
230Vrms /
590\ Load
240V,45R 240V,45R /
+ - + - |
High--+--//////--+ +--//////---+-------+
| | | |
| +---|-----------+
+--------------+
====== ======
+---//////---------//////---+
| + - - + |
| 18V, 0.4R 18V, 0.4R |
| |
| | <- 5R removed.
| + |
+------------||-------------+
Idc /|\ 10000uF |
| 11R |
+--/\/\--- + DC - ----------+
0-28V variable.

It more or less works as before, but the big circulating
currents have disappeared so that there is far less AC
voltage across the series'd secondaries. The 10000uF
is still useful though because it keeps the (now 50Hz)
ripple lower than the DC voltage across the 0.4R+04R
winding resistances. Quick results tabled below.

Idc Vout dVout/dIdc

0 2.6 ___ 192 Volts/Amp
0.2 41 ___ 170
0.4 75 ___ 145
0.6 104 ___ 120
0.8 128 ___ 105
1.0 149 ___ 90
1.2 167 ___ 85
1.4 184 ___ 75
1.6 199 ___ 60
1.8 211 ___ 40
2.0 219 ___ 0
2.2 220

At about 2Adc control the Vout waveshape is nearly
a sine (somewhat distorted) showing that both cores
are pushed so far into saturation that no further
Idc control is possible.

And you were right, the circuit reduces to two pri
winding resistances in parallel feeding the load.

45//45
231V ----/\/\-----+ <--- 222V in theory, but
| affected by distortion.
[590R]
|
low--------------+

So a form of saturable reactor using two transformers
does look viable (to knock something together quickly),
but it is probably better to use a three-winding
saturable reactor. Google suggests that they are
still available.
 
M

MooseFET

Jan 1, 1970
0
From what you say there it would be better to run the two
primaries in parallel so that the non-saturating primary
can directly pick up the voltage change that is happening
on the saturating primary. So I tried it, as below.

I'll read it after I say this:

With the promaries in series any inexactness in the matching of the
transformers has far less impact on the results. The two primaries
can settle on two different voltages as determined by there turns
ratios.

I mismatch in the open load inductance will cause an AC current in the
secondary that will force this into balance.

If somehow there is a difference in phase angle, this should also drop
out because the primaries can have slight differences in their phase
as well.

I had not though a great deal about nor actually tried the parallel
case because of the above thoughts. I now go on with interest to see
what happened.
Primaries were relinked for 240Vrms (45 Ohms each)
and wired in parallel.

Low--------------------------------------+
|
230Vrms /
590\ Load
240V,45R 240V,45R /
+ - + - |
High--+--//////--+ +--//////---+-------+
| | | |
| +---|-----------+
+--------------+
====== ======
+---//////---------//////---+
| + - - + |
| 18V, 0.4R 18V, 0.4R |
| |
| | <- 5R removed.
| + |
+------------||-------------+
Idc /|\ 10000uF |
| 11R |
+--/\/\--- + DC - ----------+
0-28V variable.

It more or less works as before, but the big circulating
currents have disappeared so that there is far less AC
voltage across the series'd secondaries. The 10000uF
is still useful though because it keeps the (now 50Hz)
ripple lower than the DC voltage across the 0.4R+04R
winding resistances. Quick results tabled below.

Idc Vout dVout/dIdc

0 2.6 ___ 192 Volts/Amp
0.2 41 ___ 170
0.4 75 ___ 145
0.6 104 ___ 120
0.8 128 ___ 105
1.0 149 ___ 90
1.2 167 ___ 85
1.4 184 ___ 75
1.6 199 ___ 60
1.8 211 ___ 40
2.0 219 ___ 0
2.2 220

At about 2Adc control the Vout waveshape is nearly
a sine (somewhat distorted) showing that both cores
are pushed so far into saturation that no further
Idc control is possible.

And you were right, the circuit reduces to two pri
winding resistances in parallel feeding the load.

45//45
231V ----/\/\-----+ <--- 222V in theory, but
| affected by distortion.
[590R]
|
low--------------+

So a form of saturable reactor using two transformers
does look viable (to knock something together quickly),
but it is probably better to use a three-winding
saturable reactor. Google suggests that they are
still available.

One very nice thing about the circuit is that you can get the
transformers from sites like Digikey or from the electronice surplus
shop. I suspect that for a given power, this will be a lower cost
system than one that uses a purpose made part. Normal transformers
are made in huge quantities.

Another nice thing is the fact that you are abusing a part in a clever
way. It adds to the mystery and magic of your designs. You always
want another engineer to feel "I would have never thought of that"
when he gets a look at your design.
 
H

Harry Dellamano

Jan 1, 1970
0
Tony Williams said:
When one core saturates, the other doesn't. One transformer still
functions as a transformer. Its secondary current is shorted out
by the one that is saturated. This is a disadvantage to this
method vs a real mag-amp.

From what you say there it would be better to run the two
primaries in parallel so that the non-saturating primary
can directly pick up the voltage change that is happening
on the saturating primary. So I tried it, as below.

Primaries were relinked for 240Vrms (45 Ohms each)
and wired in parallel.

Low--------------------------------------+
|
230Vrms /
590\ Load
240V,45R 240V,45R /
+ - + - |
High--+--//////--+ +--//////---+-------+
| | | |
| +---|-----------+
+--------------+
====== ======
+---//////---------//////---+
| + - - + |
| 18V, 0.4R 18V, 0.4R |
| |
| | <- 5R removed.
| + |
+------------||-------------+
Idc /|\ 10000uF |
| 11R |
+--/\/\--- + DC - ----------+
0-28V variable.

It more or less works as before, but the big circulating
currents have disappeared so that there is far less AC
voltage across the series'd secondaries. The 10000uF
is still useful though because it keeps the (now 50Hz)
ripple lower than the DC voltage across the 0.4R+04R
winding resistances. Quick results tabled below.

Idc Vout dVout/dIdc

0 2.6 ___ 192 Volts/Amp
0.2 41 ___ 170
0.4 75 ___ 145
0.6 104 ___ 120
0.8 128 ___ 105
1.0 149 ___ 90
1.2 167 ___ 85
1.4 184 ___ 75
1.6 199 ___ 60
1.8 211 ___ 40
2.0 219 ___ 0
2.2 220

At about 2Adc control the Vout waveshape is nearly
a sine (somewhat distorted) showing that both cores
are pushed so far into saturation that no further
Idc control is possible.

And you were right, the circuit reduces to two pri
winding resistances in parallel feeding the load.

45//45
231V ----/\/\-----+ <--- 222V in theory, but
| affected by distortion.
[590R]
|
low--------------+

So a form of saturable reactor using two transformers
does look viable (to knock something together quickly),
but it is probably better to use a three-winding
saturable reactor. Google suggests that they are
still available.

Great job in finding a viable solution to replace the hard to find 3
winding sat. reactor. Looking at your schematic of the primaries in parallel
and trying to reduce DC control power loss, is there any reason for not
using lower secondary voltage transformers with their lower DC resistance?
Also dose the 11R serve any purpose other than match to 28VDC? Looks like
removing the 11R and using a 0 to 2.0V DC supply would save power and maybe
reduce the 10mF cap.
Thanks,
Harry
 
W

Winfield

Jan 1, 1970
0
Harry said:
Tony Williams wrote ...


Great job in finding a viable solution to replace
the hard to find 3 winding sat. reactor.

Hear, hear!!!!! Thanks Tony!
Looking at your schematic of the primaries in parallel
and trying to reduce DC control power loss, is there
any reason for not using lower secondary voltage
transformers with their lower DC resistance?

It's the amp-turns that counts. so one would not
expect much improvement.
 
T

Tony Williams

Jan 1, 1970
0
MooseFET said:
I'll read it after I say this:
With the promaries in series any inexactness in the matching of
the transformers has far less impact on the results. The two
primaries can settle on two different voltages as determined by
there turns ratios.
I mismatch in the open load inductance will cause an AC current
in the secondary that will force this into balance.
If somehow there is a difference in phase angle, this should also
drop out because the primaries can have slight differences in
their phase as well.

Playing with the breadboard has given an opinion which
may be interesting Ken.

If the primaries are wired in series, then it is highly
desirable to allow an AC circulating current to flow in
the series'd control winding. Indeed, that balancing AC
current is a neccessity. It does mean though that the
control winding has to be able to carry both the DC
control and AC circulating currents.

There has to be a large capacitor across the control
winding, which the DC control circuit must be able
to drive.

However, if the primaries are in parallel then it is
not really desirable to have an AC circulating current
flowing in the series'd control winding. Any AC
circulating current is due to transformer mismatch, does
not aid in the saturable reactor operation, and causes a
small (and unneccessary) power loss in the transformers.

If possible the DC control driver should have a high
output impedance and able to run with the peak-peak
ripple voltage due to transformer mismatch. A capacitor
may be used to reduce the ripple voltage but only as
large as neccessary.
I had not though a great deal about nor actually tried the
parallel case because of the above thoughts.

Having played with it I prefer the paralleled primaries,
(mainly because I can better imagine what is happening),
but also because of a dislike of additional circulating
currents flowing in the control windings.
 
T

Tony Williams

Jan 1, 1970
0
Harry Dellamano said:
Great job in finding a viable solution to replace the hard to
find 3 winding sat. reactor.

Hello Harry. Ken's idea, not mine.
Looking at your schematic of the primaries in parallel and trying
to reduce DC control power loss, is there any reason for not
using lower secondary voltage transformers with their lower DC
resistance?

I haven't thought about optimising the control power
Harry but get the feeling that Ampere-Turns is Ampere-
Turns.... reduce the Turns and the Amps have to go up.
Also dose the 11R serve any purpose other than match to 28VDC?
Looks like removing the 11R and using a 0 to 2.0V DC supply would
save power and maybe reduce the 10mF cap.

See the other post replying to Ken. The paralleled
primaries probably requires a high impedance drive.

However, with the series'd primaries you could possibly
make that required 10000uF the output capacitor of a low
voltage variable DC supply.
 
T

Tony Williams

Jan 1, 1970
0
Hear, hear!!!!! Thanks Tony!

It was an interesting exercise prompted by Ken's
suggestion.

I do prefer the paralleled primaries. Note that a
pair of 240V primaries were used on a 231V supply,
which obtained full control from near-zero output
upwards. If you don't need to go down to zero
then there could be some mileage in using lower
voltage primaries, that saturate early without any
Idc control current flowing.

I used potted toroids Win and there was a useful
experiment that was not possible to do.

This would have been to wind some turns on, and
get an estimate of the secondary turns-count.
Guess the le of the core and use H = 4.pi.N.I/10.le
to see if the maximum polarising field strength
comes anywhere near the suggested 4 to 5 Oersteds.
This would allow a pre-design estimate of the final
Idc(max) control current.
 
G

Glen Walpert

Jan 1, 1970
0
Hello Harry. Ken's idea, not mine.


I haven't thought about optimising the control power
Harry but get the feeling that Ampere-Turns is Ampere-
Turns.... reduce the Turns and the Amps have to go up.


See the other post replying to Ken. The paralleled
primaries probably requires a high impedance drive.

However, with the series'd primaries you could possibly
make that required 10000uF the output capacitor of a low
voltage variable DC supply.

Interesting to see this discussion migrate to the design of magnetic
amplifiers, or mag-amps. In the applications where I have seen
mag-amps used they were phased out a long time ago, alas due to just
plain silly classification of certain programs I can't discuss those
applications. But I found a manufacturers description at
http://www.butlerwinding.com/elelectronic-transformer/mag-amp.html
which shows both the parallel and series configurations you have been
discussing here as the two standard configurations. There are/were
other configurations in use with for example multiple control windings
for control from multiple independent inputs, and the rectifier
version already mentioned.

Typical 1940's application ran the mag-amp control winding at a fairly
high voltage from a tube amp, then ran the DC output from the mag-amp
to the generator field of a big motor-generator set which provided
power to the load. State of the art for fast, accurate positioning of
large objects at the time.
 
M

MooseFET

Jan 1, 1970
0
[....]
However, if the primaries are in parallel then it is
not really desirable to have an AC circulating current
flowing in the series'd control winding. Any AC
circulating current is due to transformer mismatch, does
not aid in the saturable reactor operation, and causes a
small (and unneccessary) power loss in the transformers.

If possible the DC control driver should have a high
output impedance and able to run with the peak-peak
ripple voltage due to transformer mismatch. A capacitor
may be used to reduce the ripple voltage but only as
large as neccessary.


Having played with it I prefer the paralleled primaries,
(mainly because I can better imagine what is happening),
but also because of a dislike of additional circulating
currents flowing in the control windings.

It looks like my series case has enough faults to lose out tot he
parallel case. You could feed the control current through an inductor
so that the ripple voltage doesn't get to the control circuit.

I have been thinking about using a "light dimmer" style circuit to
make the control current. Since the ripple from noe hits its maximum
just at the zero crossing, it will change the timing of when the core
saturates. In the current design, the cores saturate late in the
alternation. This would shift it even later making the power factor
worse. It would be nice if the control circuit instead made the power
factor better.

Just for fun I was also thinking about placing the tranformer that
drives the control circuit in parallel with the saturable reactor.
Unfortunately this runs into trouble at the full blast end. It does
have a nice local feedback effect that would make it easier to control
the voltage at the load more accurately. Perhaps a small transformer
at the load to sense the load voltage would be a better idea.

This saturable reactor idea has the nice feature of good isolation
between the control side and the power side.
 
H

Harry Dellamano

Jan 1, 1970
0
Tony Williams said:
Hello Harry. Ken's idea, not mine.


I haven't thought about optimising the control power
Harry but get the feeling that Ampere-Turns is Ampere-
Turns.... reduce the Turns and the Amps have to go up.


See the other post replying to Ken. The paralleled
primaries probably requires a high impedance drive.

However, with the series'd primaries you could possibly
make that required 10000uF the output capacitor of a low
voltage variable DC supply.
Ok Tony, agree that amp*turns must be satisfied. So if we had the luxury to
choose the ideal core materials we would choose a material with high flux
density (B) to reduce N and low coercive force (H) to reduce I with no air
gap. I am going with Supermenour.
Thanks,
Harry
 
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