Ok, let's try this solving for Vdrop instead of Vload... See if I come up
with the same solution as chuck:
For the Load:
Vload=Pload/I
For the resistive drop of the cable:
I=Vdrop/Rcable
Overall Voltage math:
Vin=Vdrop+Vload
Substituting Vload:
Vin=Vdrop+Pload/I
Substitute I:
Vin=Vdrop+Pload/(Vdrop/Rcable)
Simplify... And I get exactly the same formula as before....
Ahhh... the + term turns out to be the output voltage and the - term turns
out to be the vdrop.
Doesn't explain chuck's additional stuff at the bottom of the formula
though.
On Fri, Mar 10, 2017 at 5:24 PM, Forrest Christian (List Account) <
li...@packetflux.com> wrote:
> 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 <part15...@gmail.com> 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, ch...@wbmfg.com 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:* af@afmug.com
>> *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, ch...@wbmfg.com 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:* af@afmug.com
>> *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 <ch...@wbmfg.com> 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
> forre...@imach.com | http://www.packetflux.com
> <http://www.linkedin.com/in/fwchristian> <http://facebook.com/packetflux>
> <http://twitter.com/@packetflux>
>
>
--
*Forrest Christian* *CEO**, PacketFlux Technologies, Inc.*
Tel: 406-449-3345 | Address: 3577 Countryside Road, Helena, MT 59602
forre...@imach.com | http://www.packetflux.com
<http://www.linkedin.com/in/fwchristian> <http://facebook.com/packetflux>
<http://twitter.com/@packetflux>
