Author Topic: xantrex c35 question  (Read 4177 times)

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cdog

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xantrex c35 question
« on: March 03, 2012, 04:35:00 PM »
Hello i an going to build a 13' mill to replace my destroyed 10'.
I have a c35 running two 2 ohm 300 watt resistors im parallel for a dump load....
In the past i tried hooking another one up but got the orange flashing light....
Is this because the ohms are too low? Its a24volt setup and if ive done my math correctly.....900 watts should still be in the safe zone for the controller??
I need the extra capacity for when i get my new mill going.....any advice?

cdog

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Re: xantrex c35 question
« Reply #1 on: March 03, 2012, 07:34:40 PM »
I should also ask why not use a voltage controlled switch running a solid state realay to run the resistors?????

cdog

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Re: xantrex c35 question
« Reply #2 on: March 04, 2012, 02:42:06 PM »
Does anybody run a morningstar relay driver running a large solid state relay?
Seems this would be a simple and robust method?

mab

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Re: xantrex c35 question
« Reply #3 on: March 04, 2012, 03:06:06 PM »
28v/2ohms = 14A per resistor so if you add a third 2 ohm resistor then you're running at 42A.

If you add a 4 ohm resistor that will take you to 35A if you want to push your C35 to its limit.

I should also ask why not use a voltage controlled switch running a solid state realay to run the resistors?????

I guess you could use that for the additional 2 ohm resistor, but I'm not sure how you're connecting it so I'm not sure how it'll interact with the xantrex.

mab

cdog

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Re: xantrex c35 question
« Reply #4 on: March 04, 2012, 03:13:04 PM »
What if i put the 3rd resistor in series with the previous two parraled ones?
I was thinking of a volage controlled swithch to run a 100 amp ssr.....that would run all six resistors i have
..

mab

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Re: xantrex c35 question
« Reply #5 on: March 04, 2012, 03:43:56 PM »
Quote
What if i put the 3rd resistor in series with the previous two parraled ones?

No - with your existing resistors you've got 1 ohm total = 28A at 28V. If you put 2 ohms in series that gives you 3 ohms total or just 9.3A at 28v. If you put two 2 ohms in series with each other and put that series pair in parallel with your existing two parallel 2 ohms resistors, that'll give you 0.8 ohm for the Xantrex (35A at 28v).

If you can set your voltage controlled switch to cut in and out at the right voltages then I suppose it would work - it could handle six  2 ohm resistor in parallel (84A) - I'm a bit reluctant to say for certain 'cos I've never used off the shelf VCS or an SSR so I don't know how well they work.

mab

cdog

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Re: xantrex c35 question
« Reply #6 on: March 04, 2012, 04:03:10 PM »
I understand what you are saying....but can you explain the math as to how you arrived at .8 ohm?

mab

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Re: xantrex c35 question
« Reply #7 on: March 04, 2012, 05:41:54 PM »
for adding resistors in parallel:
Rtotal = 1/ (1/R1 + 1/R2)

so two 2 ohms parallel = 1/(1/2+ 1/2) = 1/1 = 1 ohm
and: 1 ohm and 4 ohm p'llel = 1/ (1/1 + 1/4) = 1/ (1 1/4) = 1/1.25 = 0.8 ohm

or: two 2ohms and 4 ohms in p'llel =  1/(1/2 + 1/2 + 1/4) = 1/1.25 = 0.8 ohm

for adding resistors in series:
Rtotal= R1+R2

so: 1ohm + 2ohm = 3 ohm
and: 2 2ohms in series = 4 ohms

For working out amps through a resistor:

amps = Volts/ohms
eg. 28volts/2ohms = 14amps

the same eqn can be used 2 other ways:

ohms=volts/amps    eg: 28v/35amps = 0.8ohms
volts=amps*ohms   eg: 35amps*0.8ohms = 28volts




cdog

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Re: xantrex c35 question
« Reply #8 on: March 04, 2012, 05:49:42 PM »
All the resistors are two ohm...but how do i put the two resistors that would be in series to the two in parrallel?

mab

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Re: xantrex c35 question
« Reply #9 on: March 04, 2012, 06:55:47 PM »
so if R1 and R2 are your existing resistors then add R3 and R4 as in the diagram.

cdog

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Re: xantrex c35 question
« Reply #10 on: March 04, 2012, 07:08:42 PM »
Awesome!
Thanks for taking the time to illustrate that....greatly appreciated!

ghurd

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Re: xantrex c35 question
« Reply #11 on: March 04, 2012, 10:45:17 PM »
Now comes the Money part of it, if that is OK with you?

Two 2 ohm resistors in series = 4 ohms.
28V / 4ohms = 7A
7A x 7A x 4ohms= 196W in those 2 resistors.

It seems silly to use two 2 Ohm 300W resistors in series, where a single 4 Ohm 225W would do the same thing but cheaper!

"In the past i tried hooking another one up but got the orange flashing light.
Is this because the ohms are too low?"
Yes.
Three 2 Ohm resistors in parallel = 44A, 14A per resistor, which makes the orange light flash in a 35A max controller.

G-
(math adjusted for my thought processes)
www.ghurd.info<<<-----Information on my Controller

cdog

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Re: xantrex c35 question
« Reply #12 on: March 04, 2012, 10:48:44 PM »
A few hundred for a proper diversion load controller is money well spent....please spit out your ideal scenario....i will happily take your advice....

ghurd

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Re: xantrex c35 question
« Reply #13 on: March 04, 2012, 11:33:47 PM »
For a 13'
Depending on the situation, a name brand MPPT suitable for wind power would be dandy.

Next would be a MorningStar PWM dump load controller.

Then my little kit, with a whole bunch of extra parts, which is not exactally ideal for a 10' or 13', but it is better than a relay!
G-
www.ghurd.info<<<-----Information on my Controller

cdog

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Re: xantrex c35 question
« Reply #14 on: March 05, 2012, 10:53:55 AM »
I checked out your kit and i would guess this must itself switch something larger to handle the current?

zvizdic

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Re: xantrex c35 question
« Reply #15 on: March 05, 2012, 07:22:30 PM »
My setup is so simple and it works on 10' 48V mill.

I use old dryer heating element cut to form a 4.5 ohm resistor .
GHurd 48V controller switching 4 IRFP150N in parallel 56 ohm resistor on each gate of Mosfets.
GHurd controller works so good  and reliable not even considering using anything else.
If you buy your battery bank from same batch equalizing is no issue not for me anyway .

I am saying you do not need big money only some skills.
GHurd tried to talk me out of using his controller for 10' mill , he is so modest or he is not realizing how good his controller is and what it's limits are.
Like to hear if any one is using Ghurd controller on a bigger mill.

cdog

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Re: xantrex c35 question
« Reply #16 on: March 05, 2012, 07:30:50 PM »
This sounds very much like the direction i want to go....
Unfortunately i would have to have someone assemble the cintroller and spoon feed me exactly what you did....

ghurd

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Re: xantrex c35 question
« Reply #17 on: March 17, 2012, 08:33:00 PM »
Like to hear if any one is using Ghurd controller on a bigger mill.

It is being used on quite a few 10', a few 13', and a 17'.
Not sure how many 48V units are up and running now.  Maybe 10 or 15?

cdog,
The 24V version is a lot simpler than the 48V version.
The kit switches the fets.  Use a lot of fets, each with their own individual loads, and it makes it simple to control a large total amount of power.

Here is a 24V 1650W assembly,
http://s701.photobucket.com/albums/ww20/ghurd1/Large%20Dump%20Load%2024V%201650W/

G-
www.ghurd.info<<<-----Information on my Controller