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.
