Author Topic: Sizing the dump load (Read 19536 times)

Speo · « **on:** October 16, 2007, 09:28:10 PM »

Hello,

I was trying to figure out the size of the dump load for my little system (12V, maximum 100W).

Because I have around a 4.4 Ohm resistor, I started with this one, trying to see if it fits:

dumping at 14.4 V means:

I = 14.4/4.4 = 3.27 Amps

P = 14.4 * 3.27 = 47.1 Watts

Does that mean that my resistor can handle 47 watts ? Is this big enough?
What is going to happen if the gennie is putting out 100W and I dump only 47?
If I paralel another one, identica, will the calculation be similar, but the resistance will be 2.2 Ohms, so I can dump (14.4/2.2)* 14.4 = 94 W?
Can I parallel with mine one with a different resistivity, 1 Ohm for example, resulting in (1/4/4 + 1) = 1.22 Ohm and dump 170W ?
Below is my resistor, but beside the 4.4 Ohm, I don't understand the other marks. Is anybody aware of what's the meaning of ROS-150? Also there is another mark "Y.EL".

Thaks,

Speo

ghurd · « **Reply #1 on:** October 16, 2007, 03:57:03 PM »

It will only dump 47W. It is probably rated for 150W. 47W is safe for a 150W resistor.

However if the windmill is making 100W and the resistor is dumping 47W, then 53W will still be going into the battery.

It would be better to have 2 of those resistors. In parallel. Dumping 94W of 100W.

It will be easier to think of each parallel resistor as separate. Do not exceed the rated watts of an individual resistor.

Do not make the bolt and nut in the center too tight.

G-

Speo · « **Reply #2 on:** October 16, 2007, 04:10:01 PM »

Thanks for the reply, ghurd.

You were saying: "However if the windmill is making 100W and the resistor is dumping 47W, then 53W will still be going into the battery."

Why are the 53 watts going into the battery if the C40 switched to "dump" the power? I was expecting eventally for the gennie to speed up because not all the power is used, but not the 53 watts to make it to the battery.

Speo

ghurd · « **Reply #3 on:** October 16, 2007, 04:35:55 PM »

The windmill will still be charging the battery. Possibly at up to 100W.

The resistor will be draining the battery, but only draining the battery at 47W.

The other 53W will be overcharging the battery.

G-

Speo · « **Reply #4 on:** October 16, 2007, 04:39:49 PM »

Yes, it makes sense now, after I looked at how the c40 is connected to the mill and battery as dump load.

Thanks,

Speo

SamoaPower · « **Reply #5 on:** October 17, 2007, 12:02:47 AM »

Speo,

You, along with Ghurd, seem to have a bit of a problem understanding the concept of a wind turbine's power output.

The rating of a given power at a certain windspeed is simply an upper limit imposed by the capability of the machine and the prudence of the user. It does not mean that, at the rated windspeed, it will always be delivering that rated power.

The power delivered under those conditions will primarily be determined by the load, whether batteries, resistors or a combination determined by a controller such as a C40.

The C40, in Diversion Mode, shunts a certain amount (via PWM) of the WT available current to the dump load resistor to maintain the battery terminal voltage for the charge algorithm phase it happens to be in at the time. This is true whether in absorption or float phase. Thus, the actual power delivered by the WT depends on these loads.

You may notice that I tend to talk in terms of current rather than power. It clarifies the issue.

Ghurd: "However if the windmill is making 100W and the resistor is dumping 47W, then 53W will still be going into the battery."

This is incorrect. The WT will only deliver as much power as the loads demand. If the battery is far from full, all of the current available will continue to charge the battery with nothing going to the dump load, or if it's close to full, it will divert some of the current to the dump load while continuing to charge the battery according to its needs, or if the battery is full, it will divert all of the available current (less a little) to the dump load.

Your assumption that with that resistor, the turbine will speed up and will not be protected against higher windspeeds, is correct. Not recommended.

Ghurd: "The windmill will still be charging the battery. Possibly at up to 100W.

The resistor will be draining the battery, but only draining the battery at 47W.

The other 53W will be overcharging the battery."

This is also incorrect. It wouldn't be much of a controller if it allowed this to happen. The dump load resistor does not "drain the battery". The controller only diverts enough current from the WT to maintain a given battery voltage according to its algorithm and does not allow battery overcharging.

As to your basic question, the Xantrex C40 manual says that a good rule of thumb is to size the dump load to 80% of the controller rating, i.e., 32 amps. At 14.4V, that means 0.45 ohms. This would also allow for a larger mill expansion than you now have. The power rating should be at least twice your current WT power. Personally, I would chose 1000W to allow for future expansion and a 100% safety factor.

By the way, your formula for parallel resistors is incorrect. It's R total = the recipical of the sum of the recipicals, i.e., Rt = 1/((1/R1)+(1/R2)+(1/R3)...+(1/Rn)). For two resistors it simplifies to Rt = (R1*R2)/(R1+R2).

ghurd · « **Reply #6 on:** October 17, 2007, 12:13:11 PM »

If you are using a C40, it is a 'shunt' instead of a 'dump'.

If you are going to use it with your 100W Ametek windmill, a 125W load should be enough.

G-

Speo · « **Reply #7 on:** October 17, 2007, 10:06:23 PM »

SamoaPower,

Thank you for the input.

Below is the C40 connection in Diversion Mode:

From what I understand from this image, the generator is hooked directly to the battery and the C40 is diverting, as you said, as much as the battery cannot handle, but, up to 47W.

You did not agree with ghurd's affirmation: "However if the windmill is making 100W and the resistor is dumping 47W, then 53W will still be going into the battery."

As per the image above, what keeps the 53 watts away from the battery when the battery is full and the other 47 are diverted, in full wind? There is nothing "between" the generator and the battery. In this case, I think ghurd was right, it will boil the battery and overspeed the mill. Right?

My formula for paralel resitors was typed as "(1/4/4 + 1) = 1.22 Ohm" and actually there is a typo, I accidentally typed / instead of . and it should be (1/4.4 +1) = 1.22 Ohm which is equivalent with your formula, but I used 1 instead of 1/1

Thanks again for the input,

Speo

Speo · « **Reply #8 on:** October 17, 2007, 10:33:00 PM »

ghurd,

what is the difference between shunt and dump? Is it just called differently by certain brands?

SamoaPower · « **Reply #9 on:** October 18, 2007, 12:02:52 AM »

Speo,

Thanks for the response and the C40 hookup diagram which I'm quite familiar with, having two of them, one of which is in diversion mode.

Apparently I didn't do a very good job of explaining what your problem is in understanding WT power output.

"...what keeps the 53 watts away from the battery when the battery is full and the other 47 are diverted, in full wind?"

There is no "extra" 53W to worry about - it doesn't exist. Under these conditions, the WT will NOT deliver 100W, but only 47W since that is what the load requires. The WT output is determined by the load, not by it's maximum capability. It will only deliver 100W if that is what the load demands and the wind is blowing enough. To deliver 100W at 14.4V would require a resistor of about 2.1 ohms. R=V^2/P.

Your parallel resistor formula is still incorrect. For resistors of 4.4 ohms and 1 ohm, it would be Rt=1/((1/4.4)+1)= 0.81 ohm. Or, for two resistors only, Rt=(4.4x1)/(4.4+1)=4.4/5.4=0.81 ohm.

Did I do any better this time?

DamonHD · « **Reply #10 on:** October 18, 2007, 03:17:22 AM »

The only difference that I can imagine that the shunt happens upstream of the blocking diode, stopping the excess power ever getting to the battery, whereas a dump would be from the battery.

Rgds

Damon

Flux · « **Reply #11 on:** October 18, 2007, 03:35:00 AM »

This is all a bit confusing.

The dump load should be chosen to divert the maximum mill current and still maintain the volt drop across it to no more than the controller set voltage. Best to choose a slightly lower resistor to be safe.

Assume that the controller is set to 14v, then the power produced by the mill will be 14 x the maximum amps it can produce. It doesn't know whether it is charging the battery or cooking a resistor.

With the correctly sized dump resistor, a low battery will accept all the charge current. A completely charged battery will accept virtually nothing if the controller is set at the correct voltage and at this point all the input power will be diverted to the resistor. Somewhere in between there will be a state where low currents are accepted by the battery but the higher currents will be largely diverted to the resistor.

If the load resistor value is too high you will have occasions where the battery voltage will be dragged above the set point. Things now change a bit as the dump load resistor current will now rise as the voltage across it rises so that it diverts more than its calculated power, but the increase in battery voltage will result in the battery gassing.

The balance comes when the battery current (electrolytic and heating) plus the new resistor current at the higher voltage equal the incoming current from the machine.

Ghurd is essentially right that the excess power goes to the battery ( less the increase in load power due to rise of volts). It is not charging current as the battery is fully charged.

For flooded cells and a diversion resistor that is only just too high in value the effect will not be serious, it will happen infrequently and will just be a bit of unintended equalising. Sealed batteries will object violently. If you take it to extreme flooded cells will object as well.

You should choose a resistor low enough not to let this happen. Make sure it is within the resistor rating and make sure the resistor current at set point is less than the full load rating of the controller.

I don't follow Samoa's argument that the wind turbine will not produce its rated power. I do agree that any load will not see more than the power that it can receive from the nominal 14v source whatever the wind does, but with a diversion control the turbine will still deliver what power the wind can provide as long as the load is clamped at a nearly constant voltage.

Flux

SamoaPower · « **Reply #12 on:** October 18, 2007, 09:04:12 AM »

Flux,

You're right in that the issues are a bit confusing and in my attempt to clarify, perhaps I've only muddied the waters. Apologies to all concerned.

From the above discussion, it appeared that the assumption was being made that the WT output was independent of the presented load. Ignoring a battery for the moment and looking at resistive loads, the two limiting cases of a load equal to zero ohms and infinite ohms obviously result in zero power delivered.

Also obviously, somewhere in between, we face our (your's and mine) favorite topic of load matching. The point I was trying to make, in my bumbling way, is that the load presented to the WT DOES make a difference to delivered power and does have an impact on sizing a dump load resistor value.

In the cited case of a 12V system, a 4.4 ohm dump resistor, a C40 controller in diversion mode and a full battery, I don't believe the WT will deliver 100W, even though capable, unless the wind speed is significantly greater than the rated (matched) condition.

Now, what this leads to, of course, is that the dump load is simply too large in value, in this case, to absorb the potential output, as we both pointed out.

Unless the dumped power is to be utilized, a lower value resistor sized to the controller capability (0.45 ohm) is indicated. Otherwise, it should be matched to the WT for maximum dump energy harvest.

Thanks for stimulating the gray matter.

Speo · « **Reply #13 on:** October 18, 2007, 01:00:48 PM »

SamoaPower,

You are right, my formula was wrong. I forgot to divide 1 to those values.

"There is no "extra" 53W to worry about - it doesn't exist."

Is it possible that "some" (15%) of this extra power available in wind will raise the battery voltage and boil the batteries and the rest of the extra power available in wind (85%) will just speed up the generator instead of being generated?

Thanks for the input,

Speo

Flux · « **Reply #14 on:** October 18, 2007, 01:32:13 PM »

Speo

This was intended for SamoaPower and he may wish to come back, but as I see it, the only real conclusion we can draw is that it is inadvisable to use a dump resistor capable of dumping less than the machines output.

What happens if you do is rather speculation. What do you mean by the machines rated power. If you take the figure for power at rated wind speed you get one answer. If you take a nominal figure and don't qualify it it may be the thermal limit on some machines, on others they may become reactance limited and just can't produce any more.

It also matters a great deal how badly you make the discrepancy. Within reason you will not force a battery voltage drastically above its equalising voltage. You will typically have trouble forcing a 12v battery to 18v and at this level most machines will still be able to produce their nominal power even if it is a bit fast and at slightly higher wind speed. If you go crazy then you may force the battery to 30v and at this point the machine may not be able to reach its nominal power out.

Samoa is correct that it has to do with matching of the load. I don't think you can give definite figures for the bits that go into gassing and which bits reduce the matching to the point that power out is restricted.

I think that the confusion in the first place came from thinking of the extremes of what could happen. Just avoid the situation and parallel enough resistors to avoid the situation.

Flux

SparWeb · « **Reply #15 on:** October 20, 2007, 11:19:43 PM »

Here's a more straightforward suggestion:

I think your resistor has too much resistance for your purpose. The C40 will not dump much load through that.

Try getting a resistor like one of these: http://www.avtron.com/edgewound_resistors.htm

You could also look at Ohmite, Post-Glover, or Powerohm.

Like SP's preceding calculation showed, you need about 1/2 Ohm of resistance. If it's being dumped by a C40, then a resistor rated for 40 amps is adequate. All the de-rating business is about the generator side of the machine. Don't put 40 amps into it (your Ametek can't anyway). But still size the resistor according to what the C40 is designed for.

As for the resistor you have, I spent a bit of time trying to identify it, but the top google hits don't look like the one you got. I wondered if the "ROS" means RoHS, but that didn't seem to come up, either. Don't worry. You may find a use for that one some other day. If you reverse the calculations, (speculating here):

squareroot(150W/4.4Ohm) = 5.8 amps

150W / 5.8A = 26 Volts

So maybe that resistor is actually more suitable for a 24V system.

Speo · « **Reply #16 on:** October 21, 2007, 08:13:29 PM »

ghurd, DamonHD, SamoaPower, Flux, SparWeb,

Thank you for all your helpful answers. I got the ideea.

To be on the safe side, the controller should easily handle the maximum power generated(and dumped) and the resitor should be choosen to be able to handle all the power dumped.

Thanks again,
Speo

defed · « **Reply #17 on:** October 09, 2010, 05:41:50 PM »

ancient thread, i know, but it pertains to what i am looking for. and after reading all of it, i'm STILL not sure what i need.

currently i have 700w solar on an xantrex mppt, 4' turbine, 24v system. i need a dump load, mainly for the turbine, not that it makes much, but just in case. how the heck do i size it? there are so many resistors and the more i read, the more confused i get!

for example, i found a resistor described as: 10ohm, 100w, 24v.

if i've got it right, if i'm dumping at 28.5v, it would look like this: 10ohm 28.5v 2.85a 81.23w

these figures vary depending on the voltage. am i correct im presuming that the resistors aren't voltage specific, just their values change as per voltage? i mean, ohms of the resistor is constant, so its wattage and amperage would vary depending in the volts.

i found a page that describes how to size the dump load, and it says this: To figure out the resistance you ultimately need with the combination of dump loads, use the formula Resistance Needed = (DC Voltage of your system)^2 / Peak Power of Your Turbine

according to this, ohms = 576/300 ( "rated" output is 200w, have seen close to 500w once, usually caps in the mid 200's for brief moments).

ohms = 1.92 (2ohms)

2ohms, dumping at 28.5v is 14.25a, 406w...

so, does this mean i need 400w of dump, which would be 5 of the resistors previously reference? or have i got it all wrong?

as always, thanks!

Tritium · « **Reply #18 on:** October 09, 2010, 06:07:38 PM »

5 of your 10 ohm 100 watt resistors in parallel should work fine (you will only dissipate 406 watts at 2 ohms and 28.5v )

Thurmond

SparWeb · « **Reply #19 on:** October 09, 2010, 10:55:03 PM »

Tritium's answer is directly what you are looking for, but allow me to add some detail...

You do have the right facts in front of you, and now it's a matter of sorting out what you need, in the proper order. The real starting point is the turbine itself, then the charge controller, and doing your calculations on that basis you can then narrow down what you need by way of resistors.

Turbine: Let's say it can put out 300 watts in a peak or a little more if the furling got tripped up for a moment. Your machine may be different.

Controller: There is a limit on what the Xantrex can send to the diversion load. The new Schneider website is a total mess so I don't know what you have. Lets assume it 60 Amps for the sake of getting on with the explanation.

Battery: As you say, 24V. The Xantrex regulates that somewhere around 27 to 28 volts, depending on temperature and battery type. If we were to assume flooded lead acids, and a cold day, then it can go over 30V.

Put the maximum system voltage (30V) and the maximum controller current together (60A) and you have 1800 Watts that can be diverted to the resistance load. It is much higher than the power that the turbine can supply, however from one controller it is possible to mix up settings and accidentally dump the solar to the diversion, too. Been there done that. Combining 700w solar and 300w wind means that your system won't even dump more than 1kW, so finding a 1800 watt dump load sounds like overkill. On the other hand, it will be safe from errors, allows expansion of the solar or wind equpment in the future.

Back to the math: 1800W / 30V / 30V = 2 Ohms

You could do it the other way: 60 amps / 30 V = 2 Ohms

The result is the same, you need a resistance dumpload of 2 Ohms and however you assemble the pieces they must dissipate a total of 1800 watts or so.

If you use the restors you were looking at before, for example:

5 resistors in parallel, 10 ohms each, you get 2 Ohms. The power dissipated by each when exposed to 30V will be 90 Watts; all five will dump 450 watts together. This isn't enough to meet the requirement of the charge controller, it also prevents much expansion of the wind power system, and in the case of mis-configuring the Xantrex the resistors could overheat. The charge controller will keep trying to deliver the current, even as they melt/sag/break/burn. Excessive temperatures in any material will increase the resistance, meaning that over-heated resistors (despite being made of materials that have a low thermal coefficient) will suffer the same fate, too.

So at this point you can choose to take a chance with resistors as you've found them - but have no margin for error (it isn't a big risk, just making you aware of it).

Or you can keep looking (Digikey / Mouser / Post-Glover / Avtron) because what you really need is out there!

If you are keen to build it yourself, by the way, you can always take the element out of a dryer (~2400W) and cut the series wires into several parallel elements that will work at 24-30V.

dave ames · « **Reply #20 on:** October 10, 2010, 02:25:25 AM »

sparweb,

always a big fan of your work and appreciate your tireless efforts explaining some of this stuff or pointing the way to further information. and agree with some of your post here...but

can we have another look at what we have here?

Quote from: SparWeb on October 09, 2010, 10:55:03 PM

Back to the math: 1800W / 30V / 30V = 2 Ohms

You could do it the other way: 60 amps / 30 V = 2 Ohms

The result is the same, you need a resistance dumpload of 2 Ohms and however you assemble the pieces they must dissipate a total of 1800 watts or so.

is it possible we should work those numbers like this?

1800W/(30x30) =0.5 Ohms

30V / 60 amps = 0.5 Ohms

will not a 2 ohm resistor @ 30v always dump about 450 watts no matter how much current is available? and the real issue with our charge controller trying to dump more has to do with the duty cycle of the pwm dump load reaching 100% and alarming out the controller? is there some way to force more than about 15 amps into a 2 Ohm load at 30v? i can't see the "The charge controller will keep trying to deliver the current, even as they melt/sag/break/burn" scenario.

please help point the way to further understanding

kind regards, dave

defed · « **Reply #21 on:** October 10, 2010, 07:18:00 AM »

the controller is 60amps. there is an aux output, that when set up (voltage and time delay for 'on', voltage and time delay for 'off') will power a separate relay. i have a SSR that the controller will trigger to carry the full amount of power necessary from the batteries. that brings up another question....what exactly will be the power running thru the relay to the resistor? if it the dump turns on at 28.5v and off at 28v (for example)....i guess it's just like any other load and the resistor draw what it is rated for.

one interesting thing about the controller when used w/ the turbine...i was in float, which is about 2-3 amps...when the turbine kicked up and made 3 amps, the solar controller would cut back the panel output to 0amps. if the turbine made 1 amp, the controller would have the panels output 2amps. i believe that this works in all charge modes, so even w/o a dump load, the controller still regulates the total amperage to the batteries by regulating the the solar portion. this is why i'm not too worried about having a dump load for the 4' turbine, as it can barely make over float amps on a constant basis. eventually want a 10' turbine, which would definitely need dump protection.

so if i want 1800w of dump, i can basically use any resistor (ohm/wattage rating), and just do the calcs to find how many i need? when they list a wattage on the resistor....that is more of a factor of the ohms and voltage input than an actual 'what it will dump' figure?

thanks for all of the good info.

dave ames · « **Reply #22 on:** October 11, 2010, 02:06:40 PM »

defend,

i'd be tempted to "save" that aux trigger output for some other function (something it's better at) and get one of those two for $25 kits from the hurd corporation. set it up to dump ~300 to ~400 watts a bit above the mppt absorb voltage.

Quote from: Tritium on October 09, 2010, 06:07:38 PM

5 of your 10 ohm 100 watt resistors in parallel should work fine (you will only dissipate 406 watts at 2 ohms and 28.5v )

can always go with something bigger (a store bought dump controller) or add on to this one when you get the 10' machine on line, and these kits are nice to have in the tool box for lots of other uses.. seems to be a new flavor for these every month or so!

like you said .."not too worried" with the limited output of the four footer but still nice to have a "safety valve" in place for peace of mind and not having to run around turning stuff on when an extended blow comes thru

it's all good fun!

ghurd · « **Reply #23 on:** October 11, 2010, 02:28:28 PM »

Defend,
Simplified:
You need to dump a little more amps than the 4' turbine can make.

It would be a lot easier to explain so you can understand it if we knew more about the specific turbine.

I agree with Dave. I would save the aux output for something better.
Plus I am not exactly sure the relay output will do what you want to do.
G-

defed · « **Reply #24 on:** October 11, 2010, 04:08:36 PM »

Quote from: ghurd on October 11, 2010, 02:28:28 PM

Defend,
Simplified:
You need to dump a little more amps than the 4' turbine can make.

It would be a lot easier to explain so you can understand it if we knew more about the specific turbine.

I agree with Dave. I would save the aux output for something better.
Plus I am not exactly sure the relay output will do what you want to do.
G-

i have a 4' Piggott at 24v. it has made 16 amps once (was overnight, so i presume VERY briefly) and i have also seen 9amps, but again, very short burst. in a nice wind (15-20mph), i see 1 to 6 amps typically. this does fluctuate quite a bit as the wind gusts and settles, gusts and settles.

the only reason i want to use the aux output is because it is a temporary setup, the easiest way to get it done quickly, and i don't have anything else to use it for at this time (tho it does have alot of other uses).

one function is a dump load. the parameters are voltage to turn on, and time to be at that voltage before it comes on (another set of parameters for voltage off). when the parameters are met, the aux output puts out a voltage (5-13v @ up to 200ma) which can power something direct (a small, low wattage, dc fan) or it can power a relay. i have some SSR's that i would use. the aux parameters are met, send out 12v to SSR control (normally open), this closes the relay and passes my 24v dc from the battery to the load.

when i get a 10' dan's machine, i would probably do it differently...but for now, i think this is the quickest and easiest way.

ghurd · « **Reply #25 on:** October 11, 2010, 04:59:29 PM »

I expect that is intended for more of an oportunstic load, and not really a dump load used to control the battery voltage. It is not going to be a very great assembly.

For temporary, and with the small turbine, maybe dump 10A.
Dumping will be at about 29V.
29V / 10A = 2.9 ohms.
10A x 29V = 290W.

Reality-
Power resistors are expensive.
290W is pretty close (too close for me) to 300W, which is a standard watt value.
A 3.2 or 3.3 ohm 300W resistor will dump about 9A or 8.8A, which is about 260W, which is OK in a 300W resistor IMHO.
That should keep it under control for most conditions.

Next reality is SSRs have a pretty substantial forward voltage drop in low voltage systems. Loosing 2~3V in a 48V system is a lower percentage than in a 12V or 24V system.
Meaning the resistor will not see full voltage, and it will dump less amps.
But if you size the power resistor for partial voltage, then it can not be used later with a decent controller.
Also meaning the voltage drop makes the SSR get hot, and needs a big expensive heat sink.
I dislike relays. A lot.
But a regular old relay may be the best option for you, unless you want to use a fet or 2. Fets are about as easy to connect as SSRs, and easier than working with big heat sinks for me!
G-

TomW · « **Reply #26 on:** October 11, 2010, 07:34:23 PM »

Quote from: ghurd on October 11, 2010, 04:59:29 PM

I expect that is intended for more of an oportunstic load, and not really a dump load used to control the battery voltage. It is not going to be a very great assembly.

For temporary, and with the small turbine, maybe dump 10A.
Dumping will be at about 29V.
29V / 10A = 2.9 ohms.
10A x 29V = 290W.

Reality-
Power resistors are expensive.
290W is pretty close (too close for me) to 300W, which is a standard watt value.
A 3.2 or 3.3 ohm 300W resistor will dump about 9A or 8.8A, which is about 260W, which is OK in a 300W resistor IMHO.
That should keep it under control for most conditions.

Next reality is SSRs have a pretty substantial forward voltage drop in low voltage systems. Loosing 2~3V in a 48V system is a lower percentage than in a 12V or 24V system.
Meaning the resistor will not see full voltage, and it will dump less amps.
But if you size the power resistor for partial voltage, then it can not be used later with a decent controller.
Also meaning the voltage drop makes the SSR get hot, and needs a big expensive heat sink.
I dislike relays. A lot.
But a regular old relay may be the best option for you, unless you want to use a fet or 2. Fets are about as easy to connect as SSRs, and easier than working with big heat sinks for me!
G-

G;

Isn't some guy using your kit to dump a couple kilowatts or something?

Seems I heard or read that someplace?

Tom

defed · « **Reply #27 on:** October 11, 2010, 07:57:26 PM »

Quote from: ghurd on October 11, 2010, 04:59:29 PM

I expect that is intended for more of an oportunstic load, and not really a dump load used to control the battery voltage. It is not going to be a very great assembly.

i would have to disagree with its intended use based on this from the manual:

----

Hi Battery Voltage - Activates the aux output when the battery voltages rises above the trigger setting for the trigger delay time.

- Installations that have another external charging sources such as a wind generator or hydro generator connected directly to the batteries. The XW SCC aux output can control a relay to disconnect the external charging source from the battery when the battery is in danger of being overcharged, or control a relay to turn on a diversion load.

----

upon reading further, it looks like out of the 6 function that the aux port can take on, only 2 seem like they have much use....the high battery voltage and there is a low battery voltage that would disconnect loads from the battery to prevent over-discharge. both of which are functions that your controller can handle, and maybe handle better. i'll mess w/ it and see. eventually, i will probably want one or 2 of your controllers to play with as well.

defed · « **Reply #28 on:** October 11, 2010, 08:07:10 PM »

Quote from: ghurd on October 11, 2010, 04:59:29 PM

Next reality is SSRs have a pretty substantial forward voltage drop in low voltage systems. Loosing 2~3V in a 48V system is a lower percentage than in a 12V or 24V system.
Meaning the resistor will not see full voltage, and it will dump less amps.

i'm not sure where you measure for this voltage drop. i am currently using an SSR in my 'headlamp heating system'...an electronic thermostat switches on to energize the SSR. the 24v is on the output side that goes to the headlamp load. when i measure voltage at the SSR output terminals, it is 24v. not sure what it is AT the lamps because i have the connections inside the headlight housing. but i think they draw the same amps as another pair i have hooked directly to the batteries. the pair of lamps is 70w, i have the SSR on a heatsink (tho it claims to not need one under 10a), and it never gets warm to the touch when left on all day. granted, it is only ~2.7a @ 26v.

ghurd · « **Reply #29 on:** October 12, 2010, 09:28:02 AM »

Aux output- Just because it can do it does not mean it will do it well.
But like you said, it is temporary.

SSR forward voltage can be measured between the output terminals.
Meter leads on the screw going to the battery pos and the screw going to the load pos. Possibly marked output, or #1 and #2.
What comes out of the SSR is what is important concerning the SSR. If the bulbs are seeing much less than that, then it is a wire issue instead of SSR issue.

The 1st data sheet I came acrossed this morning showed Vf between 1.6V and 5.7V.
Most I play with are about 2 to 3.5Vf.
I am on the road this week, or I would have some better numbers at hand.
http://www.crydom.com/en/Products/Catalog/1_dc.pdf

That could mean... dumping at 29V, but the resistors only see 23.3V.
That really screws up the resistor math.

SSR Vf will change with the amps.

Yes, there certainly are guys dumping a couple kilowatts using my kit.
It holds the voltage rock soild as long as the dump load amps are greater than the charging amps.

A relay on a timer dumping power is not going to keep the battery voltage very stable.
G-

TomW · « **Reply #30 on:** October 12, 2010, 09:47:56 AM »

Quote from: ghurd on October 12, 2010, 09:28:02 AM

Yes, there certainly are guys dumping a couple kilowatts using my kit.
It holds the voltage rock soild as long as the dump load amps are greater than the charging amps.

A relay on a timer dumping power is not going to keep the battery voltage very stable.
G-

Yeah, on my system the Hurd Solar (tm) controller driving a massive IGBT brick holds the voltage rock solid at the set point. Even with 100+ amps blowing in at 28.8 volts long term. I completely forgot the actual size of my resistor pack now but it is north of a KW I "think". Glen probably remembers as he designed it! For the cost it is the best deal going. Before and after the sale service is at a level unheard of from most sellers / manufacturers.

Just my experience.

Tom

ghurd · « **Reply #31 on:** October 12, 2010, 10:23:58 AM »

Tom,
And the guy that designs them is damn good lookin' too!

All the specific details were lost in my recent crash.
I probably have a hard copy (notebook scribbles) at home.

I watched the Bengals in stereo Sunday, on TV and out the window.
G-

SparWeb · « **Reply #32 on:** October 12, 2010, 03:36:04 PM »

Quote from: dave ames on October 10, 2010, 02:25:25 AM

is it possible we should work those numbers like this?
1800W/(30x30) =0.5 Ohms
30V / 60 amps = 0.5 Ohms
kind regards, dave

Yes Dave you're right! I was so involved in the explanation I didn't let little things like mathematics stand in my way!

It is always wise to consult the formula and exchange the variables across the "=" sign in proper order.
Ohm's law is simple, so a mistake like that is even more embarassing.
Everybody seems to have carried on despite my blunder - thankfully I haven't tripped you all up.

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Author Topic: Sizing the dump load (Read 19536 times)