I'm thinking there are at least two limits coming into play. First is
the efficiency/design of the regulator in the SM, and the second is the
current limit of the power supply.
bp
<part15sbs{at}gmail{dot}com>
On 3/11/2017 4:19 PM, Chuck McCown wrote:
Load impedance changes constantly while booting. It levels out after
booted up. The real data I posted yesterday I did exactly that. I
added 8 ohms and it quit working down in the 15 volt range. It was
below 11 volts on the SM.
So 6 watts/11 volts = .545 ohms.
*From:* Forrest Christian (List Account)
*Sent:* Saturday, March 11, 2017 4:50 PM
*To:* af
*Subject:* Re: [AFMUG] So anti-climatic
So, here's my question (which I would play with on the bench if I was
at home)....
Take a radio (such as a 100 series cambium) that is known to run down
to below 12V....
Put 29Vdc into the circuit. Add series resistance until the sqrt(4RP)
term reaches zero. This should be about 14.5vdc.
Now what happens? I mean in real life, not in the math world.
I'm almost 100% certain this has to do with the whole maximum power
transfer theorem which is discussed in electronics courses. That is,
the maximum power you can transfer is at the point where the impedance
of the power source matches the impedance of the load. Since the
math states that the point where the sqrt(4rp) term reaches zero
you're at 50% of the input voltage at the output, then it would lead
to the conclusion that the effective impedance of the power source
matches the impedance of the load since this would result in a 50/50
voltage division at this point.
Since increasing the impedance of the power source beyond this point
will result in a decrease in the power available to the load, it would
follow that a constant power load would no longer work beyond this
point - that is, below the 50/50 point you end up with the load not
being able to extract enough power from the line to work at all.
On Sat, Mar 11, 2017 at 12:35 PM, <[email protected]> wrote:
This is what I was looking for when I started this quest.
The formula is so simplistic, it almost seems like for the amount
of work I put into it it should be more complex.
If you cannot see the image, there is the bottom line.
If you have a device that uses a switch mode regulator, which is
many things these days because the are efficient, cheap and small,
you have a device that is essentially a constant power load.
If you have to power that constant power load, (like an SM or a
VDSL ethernet line extender) over wires that have a significant
amount of resistance, like 1000 feet of 24 gauge twisted pair,
there is a lower voltage beyond which they will not operate. (I
would say to not exceed their input voltage either without zener
protection).
That voltage is:
V(supply min) = 2 sqrt(PR)
Where P is the power of the load.
R is the power wire resistance.
In my case, I need 48.99 volts or more.
--
*Forrest Christian* /CEO//, PacketFlux Technologies, Inc./
Tel: 406-449-3345 | Address: 3577 Countryside Road, Helena, MT 59602
[email protected] | http://www.packetflux.com
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