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Grex > Science > #69: Anatomy and Behavior of a Wall Wart |  |
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drew
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Anatomy and Behavior of a Wall Wart
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Jan 4 18:22 UTC 2001 |
I have a number of wall warts, all of which are inadequate to a
task that I have for them. One of them that I can spare is rated at
13 VDC and 650 mA output.
With only a digital VOM plugged into the connector, the voltage
reads 19.3 to 19.4 volts. When switched to current measurement and
10 amp scale, a bit over 4 amps is indicated on the display. Yet
when I actually attempt to power something with this adaptor, with
the VOM in series to measure the current, only its rated amount (or
even less) gets through.
On opening the case, I found a transformer and what appears to
be a simple bridge rectifier with a 25 volt 2200 uF capacitor across
the DC output (presumably to smooth out the ripples). There doesn't
seem to be any provision for limiting current or voltage.
What would cause the limitation to available power in this
circuit? If I replaced this circuit with my own diodes, selected
for adequate current and voltage ratings, could I get better
performance? (I only need it for running a few tests. If they pan
out well, I can spring for a proper AC adaptor.)
(Please link where appropriate.)
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| 27 responses total. |
rcurl
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response 1 of 27:
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Jan 4 20:15 UTC 2001 |
The resistances of the diodes and transformer limit the current. When
you measured it shorted, of course the output voltage was essentially
zero. The current you get when you power something with it depends
on that something too (what was it?). You probably stressed those
diodes a bit when you measured the shorted current, but maybe they
survived... Higher current diodes would have lower resistance but
mainly they would dissipate the heat better. How much current do
do you need?
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drew
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response 2 of 27:
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Jan 4 22:03 UTC 2001 |
A couple of amps at around 15 volts.
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gull
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response 3 of 27:
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Jan 5 05:30 UTC 2001 |
You have a few options, depending on how 'clean' you need the power to be,
and how critical that voltage is, and how handy you are.
You can buy a ready-made supply at Radio Shack. Supplies made to power
things like CB radios and such generally put out 13-14 volts.
You can build your own supply. I believe the LM317 three-terminal regulator
will supply 2 amps if heatsinked. Radio Shack carries this chip, heat
sinks, bridge rectifiers, transformers, diodes, and resistors; this is
really all you need to make a supply at any voltage from 1.5 to 25 volts. I
think the data sheet for the LM317 has a sample circuit. You could also use
a 7812 and lift the ground pin a few volts above ground, to "fool" it into
putting out 15 volts instead of 12. An LED or two would probably provide a
voltage drop very close to what you'd need. I did something similar to this
with a 7805 to make a 7.5 volt supply.
You can recycle a computer power supply. 12 volts at a few amps is available
from the disk drive connector. This might be close enough to what you need.
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rcurl
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response 4 of 27:
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Jan 5 06:17 UTC 2001 |
Since your WW is rated 13 V @ 0.65 A, it will fall below 13 V @ 2 A
(more or less - those ratings are pretty approximate).
Put a 1 amp load on it, and measure the voltage across the transformer
secondary and the output voltage. That will indicate where the main
voltage loss is (from transformer impedance or rectifier resistance).
The *maximum* DC you can get out of the bridge rectifier is, of course,
1.4x the input voltage. You might also measure the open transformer
secondary voltage (no load) to check this. If the main voltage loss
is across the diode, you have the possibility of replacing the bridge
with diodes with a larger current rating. The transformer will get
hotter at 2 A, of course, but leave off the case (since this seems to
be a temporary application).
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bdh3
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response 5 of 27:
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Jan 5 08:26 UTC 2001 |
r u crazed? 13volts and 650milliamps rated and you want to get 'a
couple amps at 15 volts'? Now long? Like a couple minutes until it
melts down into a slag setting your whole house on fire? R U nuts?
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bdh3
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response 6 of 27:
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Jan 5 08:27 UTC 2001 |
U must be a 'renter'.
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danr
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response 7 of 27:
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Jan 5 12:19 UTC 2001 |
I doubt it would set his house on fire, but I dont' think there's any way he's
going to get the current he wants from the wall wart. I think he'd be better
off buying a real supply from some surplus house. I doubt it would cost more
then ten bucks.
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rcurl
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response 8 of 27:
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Jan 5 18:33 UTC 2001 |
Not so fast. He wants 2 A at 15 V. The unit gives 19.3 V at 0 A, so where
are the 6.3 volts going at the nominal 13 V at 0.65 A? If they are mostly
lost across the bridge, then there is some hope. If they are mostly lost
in the secondary winding, there is no hope. He *can* draw 2 A (since
he can draw 4): maybe he should just let it draw 2 A and only replace
whatever smokes first... 8^}.
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drew
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response 9 of 27:
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Jan 5 20:43 UTC 2001 |
I think I see the problem, after converting said wall wart to an AC output
device by removing the board, and trying other AC only devices. (I have a
single-component bridge rectifier rated at 25 volts and 4 amps.)
DC resistance across the primaries generally measure on the order of 50 to
100 ohms, allowing 1 to 2 amps, or 100 to 250 watts - plenty of capacity.
Resistance across secondaries are around 2.5 ohms, causing a significant
voltage drop. Looks like a similar device rated at 20 VAC is indicated.
The one transformer I found in the scrap heap had a 125V output and a 6.5 volt
output, according to the label. I thought I had another, but must have tossed
it out last spring. I'll see what else I can dig up...
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mdw
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response 10 of 27:
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Jan 5 21:42 UTC 2001 |
A cheap power supply usually has:
transformer
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rectifier
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big capacitor
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resister
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current regulator
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electronic logic
This style of power supply is not very efficient - the voltage drop &
current through the resister is essentially constant, and the regulator
has to be capable of absorbing under "no load" the maximum power the
unit would produce at "full load". The current regulator should be
"close" to the logic. If it's not close, and the wires are too thin,
there will be a voltage drop between the regulator and the device, which
means any change in current draw will change the voltage. More
expensive power supplies avoid the big resistor loss with smarter logic.
An extreme in this is the "switcher" power supply, which replaces the
big transformer with a much smaller transformer running at a much higher
frequency. Most computers today use switching supplies.
A really cheap wall "wart" consists of just the transformer. This style
of device can often be recognized by the "hot spot" in the logic where
the onboard power supply dissipates its excess heat. A more expensive
style of power supply will produce rough DC at a higher voltage (say,
+15), then step it down to +12 for the actual onboard use. The power
supply will run hotter, and the logic cooler. An even more expensive
style of power supply has a "sense" line running back from the logic.
This allows the power supply to automatically compensate for any loss
between it & the logic, but requires a more expensive supply and a more
expensive connector and wire as well. S-100 machines used to use
+15V/+12V 2 step regulation. IBM think-pads used the external sense
line approach.
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n8nxf
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response 11 of 27:
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Jan 6 00:54 UTC 2001 |
The plate on the wart said 12v @ 650ma. Draw less current and the voltage
goes up. Draw more current and the voltage goes down. Rectifiers are not
resistors. They drop .07 v unless they short or open. i.e. the losses you
are seeing are transformer losses. Not only in the secondary but also in the
primary. The only way your going to get 15 volts @ 2 amps out of that wart
is if you boost the input voltage. Expect it to get quite warm, or should
I say hot? If you want a 12 volt, 2 amp supply buy of build one.
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mdw
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response 12 of 27:
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Jan 6 03:15 UTC 2001 |
I meant to say that it sounded like the "wall wart" might be using the
resistance of the transformer as part of the regulator - ie, replacing
it with a larger transformer might just blow the regulator in the
device.
Uh, what is this thing powering anyways?
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rcurl
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response 13 of 27:
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Jan 6 07:05 UTC 2001 |
Diodes do have resistance. Measure the curve sometime. There is a
small voltage drop at low current, but the voltage drop increases
as the current increases - not linearly, but very noticeably. Here
are some values I measured for rated 3 A 100 V diode (ER301):
Current (ma) Voltage drop
9 0.26
13 0.30
460 0.69
580 0.79
1000 1.04
A lot of WWs have no regulator - which seems to be the current
case.
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n8nxf
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response 14 of 27:
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Jan 7 12:31 UTC 2001 |
Alright. Rane is correct. Diodes are one way resistors. As a matter of
fact, all electrical devices have resistance, including wires and
superconductors. How about the magnetic portion of the transformer? The iron
in the transformer is responsible for most, but not all, or the coupling
between the primary and secondary windings. Can it handle the increased
magnetic field required for higher power transfer without going into
saturation? The Radio Amateurs Handbook has a section with a table to help
you determine the power handling capability of an iron core transformer by
looking at the cross sectional area of the iron passing through the center.
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n8nxf
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response 15 of 27:
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Jan 7 12:37 UTC 2001 |
Another mistake: Diodes have resistance in both directions. One way is just
less than the other.
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gull
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response 16 of 27:
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Jan 7 19:19 UTC 2001 |
Since iron costs money, and the goal of wall warts is to be as cheap as
possible, I'd be surprised if they could put out much more than rated
current without going into core saturation.
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russ
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response 17 of 27:
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Jan 8 01:00 UTC 2001 |
More to the point is whether the wall-wart could put out 30 watts.
I suspect not. The core of the transformer can only transmit a
certain amount of energy per cycle (this is why high-frequency
transformers can be so much smaller than low-frequency transformers).
A transformer core which is near its limits at 650 mA is not going
to be able to handle 2 A under similar conditions. You are going
to need a bigger transformer, or several wall-warts in parallel.
An alternative to parallel wall-warts is to borrow a power supply
from a defunct PC chassis. It shouldn't be terribly hard to adjust
the 12 V output up to 15 V. The big issue might be getting the
proper load on all the outputs so it will run. This is a more elegant
solution than multiple wall-warts, but it isn't as easy.
Speaking of the term "wall-wart", I just coined a term for a
transformer in the middle of a power cable: cord gall.
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russ
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response 18 of 27:
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Jan 8 01:01 UTC 2001 |
Winter Agora 56 <-> Science 69.
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gull
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response 19 of 27:
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Jan 8 02:14 UTC 2001 |
While technically you should put a load on the outputs of a computer power
supply, in reality most will run just fine with only one output loaded, or
no load at all -- the fan on most supplies runs off the 12V output, and
you'll notice the fan usually spins up then you turn on a supply with
mothing connected to it. Getting it to output voltages other than the +12,
+5, -5, and -12V it's intended to output would probably be tricky, though.
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drew
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response 20 of 27:
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Jan 8 04:06 UTC 2001 |
How exactly would the output voltage be "adjusted"? Is there a setscrew or
variable resistor that can be turned?
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rcurl
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response 21 of 27:
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Jan 8 07:06 UTC 2001 |
Another term for a power adapter in a power cable might be a "cordoma",
after "neuroma", which is a growth along a nerve (typically in the
nerves going to the toes).
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i
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response 22 of 27:
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Jan 9 03:20 UTC 2001 |
Re: #17, 21
No, no, no. Those bulges in cords are referred to as mice or rats
(depending on the size). In some circumstances they are referred to
as aneurysms.
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mdw
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response 23 of 27:
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Jan 9 03:46 UTC 2001 |
No, no, no. A rat is a computer input device that you roll around on
the floor with your feet instead of your hand.
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n8nxf
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response 24 of 27:
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Jan 10 01:40 UTC 2001 |
I've seen wall warts with switching power supplies in them capable of 30
watts. Not too common though. I've also seen humungo linear wall warts
capable of that kind of power. A little more common.
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