1. Generally, is there a correlation between a (DC) wall-wart's current rating and its internal resistance? (i.e. given 6V, 300 mA and a 6V, 700mA wall-warts, which would you expect to have the greater internal resistance?)
2. Suppose you have KVM switch which may be USB powered or externally powered through this DC jack:
What is your best guess as to what kind of DC power is expected - regulated 5V or unregulated 5V?
If they are unregulated, the higher current one will have lower resistance. The rated voltage is given at rated load, so if you use it with a lower current, expect the voltage to be higher.
Look up the model of the KVM switch, you might find reference to the power adapter. There are both types around now, no way to tell.
dtief wrote:If they are unregulated, the higher current one will have lower resistance. The rated voltage is given at rated load, so if you use it with a lower current, expect the voltage to be higher.
What's a good way to understand this? Would a higher current wall-wart have less turns on the secondary and thus less internal resistance?
Well actually a higher current wall wart would have a larger gauge
wire on the secondary and thus less resistance.
I dont know how much good it does to know this however,
as the wall wart secondary would be made whatever way the
manufacturer wants to make it. If they want slightly thinner
wire then they will use that, or if heavier, then that.
I've designed power transformers in the past and the secondary
wire gauge is partly determined by how much voltage drop
is acceptable, and i think this would vary a bit between manufacturers.
The apparent output resistance is also affected by the primary
winding, so the secondary gauge isnt the only thing that affects
the apparent output resistance.
One way to understand this better is to do some tests...
Measure secondary voltage with no load, half load, and full load.
Use that info to get a feel for what kind of equivalent internal
resistance you are dealing with. This test automatically takes
into account the primary resistance, which gets reflected to
the output so you can think of the transformer as having a
perfect primary winding with resistance only in the secondary.
Makes conceptualizing the workings of the device a little simpler
so you can think of the device as a battery of constant voltage
with a resistor in series with it so as the load increases, the
actual output voltage dips down.
Drawing any conclusions from a WWs spec just isn't that simple. I believe that for an unregulated WW that is rated for V voltage at I current, it should deliver V volts for a load that draws I amps. The measured open circuit voltage will be a fair amount higher. This confuses a lot of people. The number of turns on the secondary will be related to the number of turns on the primary. Al is right - the rated current would be primarily related to the secondary wire's gauge though there will be a weak relationship between number of turns and rated voltage. I believe you could deduce this from the loaded vs open circuit voltage but I'm too lazy to work out the relationship. I've seen cheap WWs that are considerably higher open circuit and are relying on droop to meet their "specs". I presume this is to reduce the amount of copper in the transformer.
nav wrote:What's a good way to understand this? Would a higher current wall-wart have less turns on the secondary and thus less internal resistance?
There's two parts to understanding this isssue.
Firstly, a plug in power adapter may be electronic
(switching regulator) or electrical (transformer and
rectifier). A lot depends upon the desired performance
and selling price.
Secondly, if the design is transformer based there
are many vairables that allow trade-offs for performance
that could have an effect on the "output resistance"
that you are investigating.
For your needs I would suggest that you treat the
power supply as a black box and from simple
testing create an equivalent circuit, showing a
lumped resistance for the "output resistance" parameter.
To do so connect the unit to a DMM and record the
voltage with no load and again with desired load.
Using simple math you can derive a numeric value
for equivalent "output resistance".
I guess i should have also mentioned that there is a sort of
'convention' used when designing power transformers.
That is, since the secondary number of turns is related to
the primary number of turns, and the secondary has to
put out a certain minimum current level (like 1 amp)
that nails the relationship between the primary and secondary
resistance to a constant that is also related to the turns
ratio.
For example, given a 120vac input and 12vac output, the
turns ratio (we'll call N) is 0.1 (not 10 really, unless you look
from the secondary to the primary)...
Vout=120*N=12 (simple multiplication)
Now since the output current is also dependent on N, we have
Iout=Iin/N (simple division).
This means if the output current is 1 amp then the input
current is 0.1 amp when N=1/10 as for our example.
Now since the output current is 1 amp, that requires a certain
size wire, we'll call its cross sectional area Wg. We know that Wg
supports 1 amp because it is chosen to meet that spec.
Interestingly, since the input current is 1/10 times the output
current, the wire size only has to have a cross sectional area
that is 1/10 of the secondary, so the primary cross sectional area
is Wg/10.
Also interesting, since we know the wire sizes of both primary and
secondary we can estimate the resistance of either winding knowing
only one windings resistance. Since the wire size is 1/10 the
secondary size, that makes its resistance 10 times more than
the secondary for the same number of turns. For the actual
number of turns (10 times more for this example) the resistance
would be 10 more times that, which is 100 times that of the
secondary (which is 1/N squared).
Since the secondary is often inaccessible because of output
rectifiers, lets measure the input resistance.
Ok, so say we measure 50 ohms. Since we have 120v in and
12v out that makes N=0.1 and (1/N)^2=100 so the resistance
of the primary is 100 times that of the secondary, or in reverse,
the secondary resistance is 1/100 times the primary resistance.
That means the secondary is 0.5 ohms.
Neat huh?
So all we have to do is measure the primary resistance, divide
by one over the turns ratio squared (1/N)^2, and we have an
estimate for the secondary resistance. Nice.
Now all we need to do is reflect the primary resistance to the
output resistance and we will have the required output resistance.
Since we measured 50 ohms the reflected output resistance
is 50/100 (again using 1/N squared) so we get 0.5 again.
Adding the secondary and reflected primary R's, we get 1 ohms
total.
Now we can ask the question, how come we dont get 12 amps
short circuit current, because an ideal source of 12 volts
divided by 1 ohm equals 12 amps.
The answer is, leakage inductance. Leakage inductance plays
a part in the overall current draw and that's a little harder to
measure, but even so, applying it to the calculation is a little
harder to do too.
This is why i suggested making three measurements, because
the curve may not follow that obtained with a pure resistance.
Part of the reason there is leakage inductance in the first place
is because the two coils are often not placed one on top of
the other, but side by side in order to meet the line isolation
standard (double insulated), but it also helps with filtering.
Also note that this discussion does not take into account the
excitation current required for the transformer core. The
effect this has on the primary wire size is to raise it slightly
so the primary can support both the anticipated load current
and this excitation current. This is usually a fraction of the
primary current required for the load, but might still change
the wire size. The secondary wire size, however, would
still be chosen based on the primary wire size without the
extra excitation current, so there would be some error
introduced in the procedure outlined above. This makes
calculating the secondary resistance just an estimate,
and a normal error could be 10 percent or higher, but
probably not more than 20 percent.
One last note, about the fact that wire sizes are mainly available
in integer sizes. This could skew the calculation slightly too,
depending on the error. If for a given design we needed the
primary to have a wire gauge that was 3.1 or 2.9 gauges less than
the secondary we would most likely choose 3 as that is the closest
integer size available.
If a given transfomer holds a certain voltage and the current demands need to be larger, then the VA rating must also be larger. And regardless of wire impedance, this requires IRON. That is the iron of its core. More VA - more iron and a heavier and larger unit. If you have seen enough transformers over the years, you almost get a feel for VA. You can also compare the unit to other listed units for VA (even by weight alone). True, some cores are of a more efficient design and material, but for most garden variety transfomers, they will be very similar.For any given voltage, you can get a fair assumption of the output current based on the amount of iron.
No that's my fault. I didnt word it right in my haste to get the
post posted. It should have read "10 times more", not
"10 times less".
I also didnt word the discussion about the wire
gauge correctly, where the primary gauge is 1/10 the secondary,
because most of us think of 'gauge' as AWG or similar, and
then that isnt correct. What i meant there was wire cross sectional area.
Im going to make the corrections now.
I also added a brief discussion about the excitation current and
its effect on the primary wire size, and how it might slightly skew the
calculation of the secondary wire resistance.
LATER:
Ok, two corrections made,
"less" to "more" and "gauge" to "cross sectional area".
Also added excitation current discussion and integer wire gauge
discussion.
If anyone spots any other typos like this please let me know as i
dont want to confuse the reader. Thanks.
