Basic current question

[Jonathan Kirwan]

Jonathan Kirwan wrote:


Losing 50% of your power at every connection? Commercial transmitters
would all be on fire.

My thinking, when saying what I said, came from something along these
lines:

http://www.trendcomms.com/multimedi...Copper/Theme/Chapter2/Impedance Matching.html

I apologize if I misrepresented that idea.

15 volt drop at 10 amps is WAY HIGH. Hope the insurance guy doesn't
find out about it after the house burns down. My house wiring is crappy
and it loses less than 1/3 of that.

I measured roughly a 15V drop, running a power saw in a room at the
far end of a home. The run of wire to that end of the home, including
various loops around the walls and via various connectors, appears to
have been around 100 feet each way. The wiring in the walls was 14
gauge aluminum. Not sure what the actual draw was from the saw, but
it was enough to drop about 15V. The reason I bothered with any
measurements at all was because the lights dimmed quite substantially.
Bugged the hell out of me.

I disagree about treating the wires as perfect conductors -- even for a
newbie. That is how the dreaded ground loops get started. It can make a
big difference even on signals as 'trivial' as 1V P-P video into a 75
ohm load. Most newbies treat 'ground' as an afterthought while to me,
it is the most important thing to get right from the beginning. Power
regulation/distribution comes next.

I can't imagine trying to understand a schematic by first replacing
every wire with some complex representation of its impedance and then
trying to fathom it. We will just have to disagree with each other on
this point.

By the way, I took courses on the Tektronix campus back in the 1970's
and this is how they also taught me to examine schematics. It's a
thinking method not of my own concoction.

Jon

 

[ehsjr]

I measured roughly a 15V drop, running a power saw in a room at the
far end of a home. The run of wire to that end of the home, including
various loops around the walls and via various connectors, appears to
have been around 100 feet each way. The wiring in the walls was 14
gauge aluminum. Not sure what the actual draw was from the saw, but
it was enough to drop about 15V. The reason I bothered with any
measurements at all was because the lights dimmed quite substantially.
Bugged the hell out of me.

From the wire table in the NEC, #14 Al is about .51 ohms/100 feet.
You know the rest, given that spec, but I'll spell it out in case
it is helpful to somebody.

200 feet wire totals ~ 1 ohm, so a 15V drop equates to 15 amps
drawn. Seems about right (mathematically) - I suppose it depends
on your saw. But electrically - you want to keep the drop a lot
smaller, and using a 15 amp saw on a 15 amp circuit will give you
dimming and is not too kind to the saw. #12 copper runs about .2
ohms per 100 feet, so you would have about 6 volt drop on a 100'
circuit wired with it. (200 feet total wire). That's acceptable
(barely) whereas 15 v drop is not. 20 amp circuits call for #12.
So it is better to use a 20 amp circuit for the saw. Actually,
one could pick a nit about it being acceptable - but set that
aside. The idea is to plug the power saw in to a 20 amp circuit
instead of a 15 amp circuit, when possible.

Ed

 

[Jonathan Kirwan]

From the wire table in the NEC, #14 Al is about .51 ohms/100 feet.
You know the rest, given that spec, but I'll spell it out in case
it is helpful to somebody.

200 feet wire totals ~ 1 ohm, so a 15V drop equates to 15 amps
drawn. Seems about right (mathematically) - I suppose it depends
on your saw. But electrically - you want to keep the drop a lot
smaller, and using a 15 amp saw on a 15 amp circuit will give you
dimming and is not too kind to the saw. #12 copper runs about .2
ohms per 100 feet, so you would have about 6 volt drop on a 100'
circuit wired with it. (200 feet total wire). That's acceptable
(barely) whereas 15 v drop is not. 20 amp circuits call for #12.
So it is better to use a 20 amp circuit for the saw. Actually,
one could pick a nit about it being acceptable - but set that
aside. The idea is to plug the power saw in to a 20 amp circuit
instead of a 15 amp circuit, when possible.

Ed

Agreed. However... there was only _one_ circuit to that room. Which
is the usual case.

Jon

 

[Alan B]

My thinking, when saying what I said, came from something along these
lines:

http://www.trendcomms.com/multimedi...Copper/Theme/Chapter2/Impedance Matching.html

I apologize if I misrepresented that idea.

Yes, you misrepresented the idea. Maximum power transfer matches load to
source, so that the source impedance and the load impedance are equal.
This load impedance is of the entire transmission system to which the
source is attached. Theoretically the connections used in a transmission
line will be perfect, and thus lossless. In reality, of course, there will
be some loss, but it will likely be measured in the tenths or hundredths of
dB.

Thus power is not halved at every connection; however power will be halved
if 1:2 branching is done, e.g. using a coax line splitter. In that case,
The signal will be reduced by about 3dB for each leg. The more branches,
the more loss when splitting. This is because each load cannot take 100%
of the power the source is delivering; rather each of two loads, if matched
properly, will get 50% of the source.

 

[Alan B]

You know, I often treat conductors exactly the opposite way ( as resistors ).
Many poor designs could be avoided by doing this.

I agree absolutely. It's a nice shortcut to simplify the math in
engineering classes, but really, every engineer should understand that
wires and connectors are components also, and have electrical
characteristics that may have critical effect on a circuit design. I could
tell a hundred stories....

I can't recall how many times I've said to the layout guy ' always think of a
track as a resistor ' ( and inductor too btw ) or they'll make them long and
stringy.

It's not just the "layout guys." A *lot* of engineers, left without adult
supervision, would run wild with poor wire design.

 

[Jonathan Kirwan]

Yes, you misrepresented the idea. Maximum power transfer matches load to
source, so that the source impedance and the load impedance are equal.
This load impedance is of the entire transmission system to which the
source is attached. Theoretically the connections used in a transmission
line will be perfect, and thus lossless. In reality, of course, there will
be some loss, but it will likely be measured in the tenths or hundredths of
dB.

Thus power is not halved at every connection; however power will be halved
if 1:2 branching is done, e.g. using a coax line splitter. In that case,
The signal will be reduced by about 3dB for each leg. The more branches,
the more loss when splitting. This is because each load cannot take 100%
of the power the source is delivering; rather each of two loads, if matched
properly, will get 50% of the source.

Thanks. That clears up my muddled thinking and verbiage.

Jon