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.
