The generalized solution: For the Load: I= Pload/Vload
For the resistive drop of the cable: Vdrop=Rcable*I Overall Voltage math: Vin=Vdrop+Vload Substituting Vdrop: Vin=Rcable*I+Vload Substitute I: Vin=Rcable*(Pload/Vload)+Vload Now is where my math gets messy... Enter the above into microsoft math, solve for vload, you get: [image: Inline image 1] and: [image: Inline image 2] Now some funny math to see if one or the other makes sense. Assume Rcable to be so close to zero to be zero, what is the result? On the top one, setting rcable to zero, results in that term being zero, which leaves Vin^2 alone under the square root which further simplifies to vin. So you end up with Vload = (Vin-Vin/2). Wrong. The second one, same procedure results in Vin+Vin/2, or simplifies to Vload =Vin. So the second one is correct for zero ohm cable drop. I'm going to guess it's valid for all cases. Just as a simple test: Vload=sqrt((48*48-4(1)(1))+48)/2) (48V in, 1 watt load, 1 ohm cable) Vload=(sqrt(48*48-4)+48)/2 Doing some math, results in a 47.97. Seems pretty accurate...... On Fri, Mar 10, 2017 at 2:08 PM, Bill Prince <[email protected]> wrote: > Then here is what I'd do. First, assume the wire has zero resistance. > Figure out what the dynamic resistance of the SM is at that condition. > > In the latest example (first example was 6 watts, now we're at 9 watts), > the SM pulls .1875 amps. In this condition, the SM presents a 256 ohm load. > > Then add in the resistance of the line, and see if it is a significant > percentage of that load. 100 ohms would be more than 1/3 of the load. > > If it was only 10 ohms, then no big deal. Given a big voltage drop across > the line, it would be reasonable to assume that the SM would draw more > current when given a lower voltage. > > My inclination would be to measure it rather than try to predict it when > it's a black box. > > > bp > <part15sbs{at}gmail{dot}com> > > > On 3/10/2017 11:43 AM, [email protected] wrote: > > But when you add a significant series loop resistance you cannot say > that. The canopy SM is more of a constant power load, like the formula I > am trying to solve. > > If the SM draws a constant 9 watts, I use a 24 volt power supply and I > insert 15 ohms of cat 5 resistance, how much current will there be? > > *From:* Bill Prince > *Sent:* Friday, March 10, 2017 12:41 PM > *To:* [email protected] > *Subject:* Re: [AFMUG] Fw: the solution > > > It would help a lot to see what the voltage drop across the SM is. In the > case of a Canopy SM, it's going to be about 29 volts. In which case, the > rest of it is easy. I'm sure each type of SM or CPE has it's own power > characteristics, so this would depend on that more than anything. > > > > bp > <part15sbs{at}gmail{dot}com> > > > On 3/10/2017 11:34 AM, [email protected] wrote: > > The load settles in at some resistance depending on what voltage it sees. > Canopy FSK SM did the exact same thing. It would run form 10.5 to 24 volts > and burn about 7 watts irrespective of what voltage it was seeing. > > So this calculation is important for stuff we all use. However we are > usually not trying to put the limits of a loop resistance or voltage. I > really don’t want to have to put 240 VDC on a twisted pair but I will if I > have to. > > A general solution where you input the load power and loop resistance and > it returns the minimum voltage is what I am trying to develop here. > > *From:* Bill Prince > *Sent:* Friday, March 10, 2017 11:34 AM > *To:* [email protected] > *Subject:* Re: [AFMUG] Fw: the solution > > > P=I**R > > (power equals current squared times resistance) > > The issue for this problem is that we know neither the current nor the > "resistance" of the load, nor the voltage drop over the 100 ohm part of the > circuit. > > > > bp > <part15sbs{at}gmail{dot}com> > > > On 3/9/2017 9:07 PM, Chuck Macenski wrote: > > Ok, doing software apparently has erased some important stuff from my > brain. Hard to know what else I lost. Having said that, why is Vr = I**2 * > R? Wouldn't Vr = I * R? > > On Thu, Mar 9, 2017 at 9:59 PM, Chuck McCown <[email protected]> wrote: > >> >> As you can see, I actually arrived at the solution early on, but then >> stumbled around searching for the linear solution which does not exist. > > > > > > > -- *Forrest Christian* *CEO**, PacketFlux Technologies, Inc.* Tel: 406-449-3345 | Address: 3577 Countryside Road, Helena, MT 59602 [email protected] | http://www.packetflux.com <http://www.linkedin.com/in/fwchristian> <http://facebook.com/packetflux> <http://twitter.com/@packetflux>
