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Estimating transfomer current rating?

T

Terry Pinnell

Jan 1, 1970
0
I have a salvaged transformer and I'd like to estimate its max current
rating. I've just installed it in a home-brew power supply whose
transformer I burnt out the other day (after about 20 years usage).
This 0-30V supply includes a current limiter with various settings,
including 2A and 4A. So I need the estimate in order to be confident I
can retain one or both of those top end settings.

Physically the transformer is a bit smaller than its predecessor, with
dimensions of about 7 x 3.5 x 6 cm (LxWxH of metal outer) and weighing
about 1.0 kg.

Unloaded
Sec pair Ohms AC V
-------- ---- --------
Grey 8.0 9.6 ~ 10
Blue 2.8 17.6 ~ 18
Red 0.6 10.6 ~ 11
Orange 0.9 26.2 ~ 26

I assume that the grey winding is much thinner wire than the other
three pairs, and that the orange pair, with 0.03 ohms/V, is the
heaviest duty? That's the one I've used for the DC power supply.

I've also made the red and blue secondaries accessible, to give me
roughly 10, 18 and 28 V AC. The red pair gave me this data:

Load current A AC V
-------------- ----
0 10.6
1.45 9.3
1.60 9.1
1.77 9.0
2.58 8.1
2.7 8.0

So presumably I could also use the red secondary at a fairly high
current. But what - 1.5A, 2A, 3A...?

For the main orange secondary (disconnected from the bridge and all
subsequent DC circuitry), I measured these:

Load current A AC V
-------------- ----
0 25.9
0.84 24.8
0.88 24.6
1.76 23.0
2.52 22.2
3.70 20.1

Does that provide an estimate of the 'max safe current' I can use over
a long period (assuming the DC circuitry remains robust, which it has
appeared to be for the last couple of decades)?
 
J

John Larkin

Jan 1, 1970
0
I have a salvaged transformer and I'd like to estimate its max current
rating.

Figure about 15 watts per pound for conventional transformers, maybe
30 or so for torroids. That assumes all the windings are properly
loaded, and resistive loads.

John
 
J

John Woodgate

Jan 1, 1970
0
I read in sci.electronics.design that Terry Pinnell <[email protected]
Edial.pipex.com> wrote (in said:
Does that provide an estimate of the 'max safe current' I can use over a
long period (assuming the DC circuitry remains robust, which it has
appeared to be for the last couple of decades)?

As a rough guide, for that mass of transformer, you can take the load
regulation to be between 5% and 10%. Using just one secondary at once,
you can probably go for 10%. So the load current that causes the voltage
to drop by 10% from the no-load voltage is the maximum permissible
current.

But don't forget that the AC current from the transformer is much larger
than the DC output current - 1.6 to 1.8 times for a bridge rectifier.

The final test is 'how hot does it get?', which depends on where you
have put it as well as its own properties. It should not get hotter than
you can touch and stay touching, as, again, a rough guide.
 
T

Terry Pinnell

Jan 1, 1970
0
John Woodgate said:
I read in sci.electronics.design that Terry Pinnell <[email protected]


As a rough guide, for that mass of transformer, you can take the load
regulation to be between 5% and 10%. Using just one secondary at once,
you can probably go for 10%. So the load current that causes the voltage
to drop by 10% from the no-load voltage is the maximum permissible
current.

But don't forget that the AC current from the transformer is much larger
than the DC output current - 1.6 to 1.8 times for a bridge rectifier.

The final test is 'how hot does it get?', which depends on where you
have put it as well as its own properties. It should not get hotter than
you can touch and stay touching, as, again, a rough guide.


Thanks all. On the basis of that advice I'll be cautious about
anything over say 2.5A until I've used it enough to assess long term
temperature rise.
 
R

Robert Baer

Jan 1, 1970
0
Terry said:
I have a salvaged transformer and I'd like to estimate its max current
rating. I've just installed it in a home-brew power supply whose
transformer I burnt out the other day (after about 20 years usage).
This 0-30V supply includes a current limiter with various settings,
including 2A and 4A. So I need the estimate in order to be confident I
can retain one or both of those top end settings.

Physically the transformer is a bit smaller than its predecessor, with
dimensions of about 7 x 3.5 x 6 cm (LxWxH of metal outer) and weighing
about 1.0 kg.

Unloaded
Sec pair Ohms AC V
-------- ---- --------
Grey 8.0 9.6 ~ 10
Blue 2.8 17.6 ~ 18
Red 0.6 10.6 ~ 11
Orange 0.9 26.2 ~ 26

I assume that the grey winding is much thinner wire than the other
three pairs, and that the orange pair, with 0.03 ohms/V, is the
heaviest duty? That's the one I've used for the DC power supply.

I've also made the red and blue secondaries accessible, to give me
roughly 10, 18 and 28 V AC. The red pair gave me this data:

Load current A AC V
-------------- ----
0 10.6
1.45 9.3
1.60 9.1
1.77 9.0
2.58 8.1
2.7 8.0

So presumably I could also use the red secondary at a fairly high
current. But what - 1.5A, 2A, 3A...?

For the main orange secondary (disconnected from the bridge and all
subsequent DC circuitry), I measured these:

Load current A AC V
-------------- ----
0 25.9
0.84 24.8
0.88 24.6
1.76 23.0
2.52 22.2
3.70 20.1

Does that provide an estimate of the 'max safe current' I can use over
a long period (assuming the DC circuitry remains robust, which it has
appeared to be for the last couple of decades)?

A transformer is rated by power, which is directly related to the
cross-sectional area, core material, and to some degree, frequency.
In general, if your new line power transformer has about the same core
cross-sectional area, then the power rating is essentially the same.
Therefore, given a power rating (say of 100 watts), the current
capability can be calculated by a simple equation P = I * E (all RMS); a
30 V winding could be made to handle about 3.3 Amps - provided the wire
size used gives almost zero I*R loss.
Other secondaries can eat up so much space that a smaller than
adequate (for max current) wire size would be used, thereby limiting the
rating for that winding.
 
T

Tony Williams

Jan 1, 1970
0
Terry Pinnell said:
Physically the transformer is a bit smaller than its predecessor,
with dimensions of about 7 x 3.5 x 6 cm (LxWxH of metal outer)
and weighing about 1.0 kg.

Those gross figures suggest an overall output of
about 60VA. Stretch it to say 70VA tops, which
is probably pushing a bit.
Unloaded
Sec pair Ohms AC V
-------- ---- --------
Grey 8.0 9.6 ~ 10
Blue 2.8 17.6 ~ 18
Red 0.6 10.6 ~ 11
Orange 0.9 26.2 ~ 26

Assume that every winding was designed with the same
ratio of power output to copper loss. A big assumption
I know, but it allows a reasonable guesstimate of the
power output of each winding. As below.

Winding Current VA-out Copper Loss
~~~~~~~ ~~~~~~~ ~~~~~~ ~~~~~~~~~~~

Grey 0.08A 0.80W 0.05W

Blue 0.42 7.56 0.49

Red 1.19 13.00 0.84

Orange 1.87 48.60 3.15
------- ------
Totals= 70 4.53
------- ------

Those max output currents are for a resistive load,
and should be appropriately reduced for a rectifier
load and capacitive input filter.

That total of 4.5W copper loss on the secondary side
should be matched by another 4.5W in the primary.

4.5+4.5 plus say another 2W of core loss suggests that
the transformer would take about 81VA off the 240V mains
at full load, ie about 0.337A full load current.

You didn't give the primary resistance, but the sums
above suggests that it should be about 40 ohms. That
would be a useful cross check to see if the sums are
anywhere in the right ballpark.
 
T

Terry Pinnell

Jan 1, 1970
0
Tony Williams said:
Those gross figures suggest an overall output of
about 60VA. Stretch it to say 70VA tops, which
is probably pushing a bit.


Assume that every winding was designed with the same
ratio of power output to copper loss. A big assumption
I know, but it allows a reasonable guesstimate of the
power output of each winding. As below.

Winding Current VA-out Copper Loss
~~~~~~~ ~~~~~~~ ~~~~~~ ~~~~~~~~~~~

Grey 0.08A 0.80W 0.05W

Blue 0.42 7.56 0.49

Red 1.19 13.00 0.84

Orange 1.87 48.60 3.15
------- ------
Totals= 70 4.53
------- ------

Those max output currents are for a resistive load,
and should be appropriately reduced for a rectifier
load and capacitive input filter.

That total of 4.5W copper loss on the secondary side
should be matched by another 4.5W in the primary.

4.5+4.5 plus say another 2W of core loss suggests that
the transformer would take about 81VA off the 240V mains
at full load, ie about 0.337A full load current.

You didn't give the primary resistance, but the sums
above suggests that it should be about 40 ohms. That
would be a useful cross check to see if the sums are
anywhere in the right ballpark.

Robert, Tony: Thanks, both. Reckon I'll downrate that 2.5A guess to
1.5A for time being then.

You're very close, Tony. The three primary windings measured 4.5, 42.5
and 46.9 ohms respectively. I used the last.

BTW, I was thinking of sticking my temperature sensor flush against
the transformer and measuring temp with my DVM while using heavyish
currents. But the heat dissipated from the 2N3716 mounted onto the
rear of the case would grossly distort the result. To use that
approach I'd have to disconnect the secondary, which of course means I
can't use the unit for normal bench work.
 
T

Tony Williams

Jan 1, 1970
0
Terry Pinnell said:
You're very close, Tony. The three primary windings measured 4.5,
42.5 and 46.9 ohms respectively. I used the last.

That doesn't sit well Terry..... the volume
allowed for the primary winding is the most
valuable real estate on the bobbin. Having
primary windings that do nothing doesn't sound
right.

Is that how they are marked, or could there be
some combination of windings that allow those
idlers to be used?
 
G

Gary Lecomte

Jan 1, 1970
0
Why not put some Thermal Switches in the Circuit. So if anything
overheats, It Shuts off. Much cheaper than New Transformers and other
parts. An Old Computer fan to blow air through it is also a good idea.

Take care.......Gary
 
R

Rich Grise

Jan 1, 1970
0
Terry said:
Physically the transformer is a bit smaller than its predecessor, with
dimensions of about 7 x 3.5 x 6 cm (LxWxH of metal outer) and weighing
about 1.0 kg.

Unloaded
Sec pair Ohms AC V
-------- ---- --------
Grey 8.0 9.6 ~ 10
Blue 2.8 17.6 ~ 18
Red 0.6 10.6 ~ 11
Orange 0.9 26.2 ~ 26

roughly 10, 18 and 28 V AC. The red pair gave me this data:

Load current A AC V
-------------- ----
0 10.6
1.45 9.3
1.60 9.1
1.77 9.0
2.58 8.1
2.7 8.0

So presumably I could also use the red secondary at a fairly high
current. But what - 1.5A, 2A, 3A...?

For the main orange secondary (disconnected from the bridge and all
subsequent DC circuitry), I measured these:

Load current A AC V
-------------- ----
0 25.9
0.84 24.8
0.88 24.6
1.76 23.0
2.52 22.2
3.70 20.1
These numbers indirectly tell you the internal resistance of the
supply. It's the slope of the graph, of course. Use that to figure
out the dissipation at various current levels for both of these
windings, add them up, and decide how much you want the transformer
to dissipate.

Or, just put a load on it and watch the temp. rise. I think one of
the gurus beat me to this one, however.

Have Fun!
Rich
 
R

Rich Grise

Jan 1, 1970
0
Tony said:
That doesn't sit well Terry..... the volume
allowed for the primary winding is the most
valuable real estate on the bobbin. Having
primary windings that do nothing doesn't sound
right.

Is that how they are marked, or could there be
some combination of windings that allow those
idlers to be used?
42.5
4.5
-----
47.0, which is pretty flippin' close to 46.9. I'd say, put the
two small ones in series, and that string in parallel with the
other one.

Maybe for a first smoke test, put a, say, .1R resistor in
series with each to see how well they current-share.

Have Fun!
Rich
 
H

Henry Kolesnik

Jan 1, 1970
0
Get a variable load like headlight bulbs and keep loading it up till it gets
hot after several hours so the temp will stabilize.. If you can comfortably
hold your hand on it you've found a load that it'll be happy with and should
work for years. Remember that half wave vs full wave recitifiers will have
double utilization factor.
 
R

Robert Baer

Jan 1, 1970
0
Tony said:
Those gross figures suggest an overall output of
about 60VA. Stretch it to say 70VA tops, which
is probably pushing a bit.


Assume that every winding was designed with the same
ratio of power output to copper loss. A big assumption
I know, but it allows a reasonable guesstimate of the
power output of each winding. As below.

Winding Current VA-out Copper Loss
~~~~~~~ ~~~~~~~ ~~~~~~ ~~~~~~~~~~~

Grey 0.08A 0.80W 0.05W

Blue 0.42 7.56 0.49

Red 1.19 13.00 0.84

Orange 1.87 48.60 3.15
------- ------
Totals= 70 4.53
------- ------

Those max output currents are for a resistive load,
and should be appropriately reduced for a rectifier
load and capacitive input filter.

That total of 4.5W copper loss on the secondary side
should be matched by another 4.5W in the primary.

4.5+4.5 plus say another 2W of core loss suggests that
the transformer would take about 81VA off the 240V mains
at full load, ie about 0.337A full load current.

You didn't give the primary resistance, but the sums
above suggests that it should be about 40 ohms. That
would be a useful cross check to see if the sums are
anywhere in the right ballpark.

I have hand-wound hundreds of power transformers for personal use, and
*NEVER* had to worry or compensate for "copper loss". Never ran out of
window space, either.
Now core loss is someting that cannot be avoided, but can be safely
ignored for 50W and higher power transformers (small percentage of core
rating).
 
T

Terry Pinnell

Jan 1, 1970
0
Tony Williams said:
That doesn't sit well Terry..... the volume
allowed for the primary winding is the most
valuable real estate on the bobbin. Having
primary windings that do nothing doesn't sound
right.

Is that how they are marked, or could there be
some combination of windings that allow those
idlers to be used?

Sorry, that's my careless description. There's only *one* primary,
tapped in the familiar way (presumably to accommodate mains in the
range 220-240V, although there are no markings). I simply measured the
three combinations, and used the full winding.

orange
o---------o
|
C|
C| 4.5 ohms
brown C|
o---------o
C| Total winding 47 ohms
C|
C|
C|42.5 ohms
C|
C|
C|
white |
o---------o

created by Andy´s ASCII-Circuit v1.23.080803 Beta www.tech-chat.de
 
T

Terry Pinnell

Jan 1, 1970
0
Henry Kolesnik said:
Get a variable load like headlight bulbs and keep loading it up till it gets
hot after several hours so the temp will stabilize.. If you can comfortably
hold your hand on it you've found a load that it'll be happy with and should
work for years. Remember that half wave vs full wave recitifiers will have
double utilization factor.

Thanks, but I mentioned what IMO is the snag with that up-thread in
Message-ID: <[email protected]>

I've since implemented the alternative I suggested. Currently (sorry
<g>) I have the secondary delivering 2.2A (an arbitrary first choice),
while I monitor transformer case temperature with my DVM.

But I still don't see how even a protracted series of such tests is
going to tell me with any accuracy what I can expect using DC loads at
various voltages in the range of my supply.
 
T

Tony Williams

Jan 1, 1970
0
Terry Pinnell said:
Sorry, that's my careless description. There's only *one*
primary, tapped in the familiar way (presumably to accommodate
mains in the range 220-240V, although there are no markings). I
simply measured the three combinations, and used the full winding.
orange
o---------o
|
C|
C| 4.5 ohms
brown C|
o---------o
C| Total winding 47 ohms
C|
C|
C|42.5 ohms
C|
C|
C|
white |
o---------o
created by Andy´s ASCII-Circuit v1.23.080803 Beta
www.tech-chat.de

Ah! Ok now, thank you.
 
J

John Woodgate

Jan 1, 1970
0
I read in sci.electronics.design that Robert Baer
I have hand-wound hundreds of power transformers for personal use, and
*NEVER* had to worry or compensate for "copper loss". Never ran out of
window space, either.

Your windings have zero resistance? Maybe you are using a different
definition of copper loss than I^2R?
Now core loss is someting that cannot be avoided, but can be safely
ignored for 50W and higher power transformers (small percentage of core
rating).

This depends on the ratio of window height to limb width. For scrapless
and even semi-scrapless laminations, the copper loss considerably
exceeds the iron loss if proper allowance on maximum induction is made
fro high mains voltage. With modern silicon iron, an iron (hysteresis)
loss of 5 W/kg at 1.5 T is typical.
 
J

John Woodgate

Jan 1, 1970
0
I read in sci.electronics.design that Terry Pinnell <[email protected]
But I still don't see how even a protracted series of such tests is
going to tell me with any accuracy what I can expect using DC loads at
various voltages in the range of my supply.

If you are using a series regulator following a bridge rectifier with a
large filter/reservoir capacitor, the a.c. secondary current will be
between 1.6 and 1.8 times the d.c. load current. It has a peaky
waveform, so you need a true r.m.s. meter to measure it, or use your
scope. The waveform is close to repeated half-cycles of a higher
frequency than 50 Hz, maybe 150 Hz, interleaved by zero-current periods.
So you can take the r.m.s. value as the peak value divided by sqrt(2)
and then divided by the ratio of the duration of the pulse to the whole
half-period, i.e. 3, if the pulse looks like 150 Hz.

The load voltage is irrelevant, because the rectifier always produces
the full voltage across the capacitor.

All you need to check is that the transformer doesn't get too hot with
the maximum current you want to draw from it.
 
L

legg

Jan 1, 1970
0
I've since implemented the alternative I suggested. Currently (sorry
<g>) I have the secondary delivering 2.2A (an arbitrary first choice),
while I monitor transformer case temperature with my DVM.

But I still don't see how even a protracted series of such tests is
going to tell me with any accuracy what I can expect using DC loads at
various voltages in the range of my supply.

The rms current values in the secondary winding are directly
substitutable. The 1.8 relationship of RMSin/DCout is typical of
capacitive rectifier filters with 15% ripple.

Theres no reason why you couldn't bodge a capacitive rectifier-filter
onto your winding as a test load.

When shopping for suitable junk transformers, it's often wise to take
a meter or two and scrap paper with you.

L values give turns ratios.
R values give first-order current-generated rises.

Otherwise, why not think of other uses for those extra windings? While
neccessity may be the mother of invention, opportunity or incident is
it's father.

RL
 
T

Terry Pinnell

Jan 1, 1970
0
BTW, I was thinking of sticking my temperature sensor flush against
the transformer and measuring temp with my DVM while using heavyish
currents. But the heat dissipated from the 2N3716 mounted onto the
rear of the case would grossly distort the result. To use that
approach I'd have to disconnect the secondary, which of course means I
can't use the unit for normal bench work.

I've had time for testing the isolated secondary at only two loads so
far, but here are the results. The temperature measurement was with
the miniature sensor that comes with my DVM, taped to top of
transformer. Case was open though.

Time Load (A) Temp (C) V (AC RMS)
---- -------- -------- ----------
11:03 0 25 25.4

11:03 2.20 25 22.0
12:00 2.18 42 21.9 OK to hold.

12:03 2.92 42 20.6
12:50 2.87 53 20.4 Too hot to hold
 
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