35.3 ohms is what the formula says you will need to have it zero out at 29 volts on the power supply and 6 watts on the SM. It will be drawing .41 amps. 9/.41=14.63 volts on the SM. = 35.67 ohms.
Pretty danged close to the line resistance. I think your maximum power transfer theorem idea is what is happening here. Burning half the power in the load and half in the line. So I guess another way to calculate it is ask if the constant power load can work at a point where the load impedance can equal the line resistance. Can a canopy work at 100 ohms? Well, 6/100= .06 which would be the I squared part. Square root of .06 =.244 amps. 6/.244=24.5 volts on the canopy and 24.4 volts on the resistor. So rounding errors aside, it proves out. But this requires 48.9 volts on the power supply. I think the 2* sqrt (PR) is the easiest method to find the lowest Voltage power supply for a given line resistance. So you can go lower with resistance or higher with power supply voltage. (Another brain fart below 6 watts and 11 volts = 121/6= 20 ohms. The .545 was amps not ohms. ) From: Chuck McCown Sent: Saturday, March 11, 2017 5:19 PM To: af Subject: Re: [AFMUG] So anti-climatic 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
