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.
