DDC+ Downgrade to DDC - W/ Pics!

[fr0zt]

Hi, I'm attempting to downgrade my DDC+ (MCP355) to a DDC (MCP350). I've read this article (http://forum.effizienzgurus.de/f23/howto-laing-pro-zur-ultra-umloeten-t1106.html#post19296), but my pump has an orange impellor and the contact points are different. Can anyone link to somewhere or describe the process of downgrading to a lower wattage on this revision pump? Thanks a lot. Attached are pics of my pump.

[fr0zt]

Just to clarify, this is reducing the wattage from 18W to 10W. I guess this shouldn't be hard, with the right technical knowhow. 1W = 1J/S right? Would a resister limit the wattage or does a resistor just limit the voltage? Are the voltage and amps the same for the 18w and 10w versions? Thanks

[Circaflex]

put it on a fan controller?

[fr0zt]

Hmm.. I think I've got it...

P (W) = I (A) x V(v)
V (v) = I (A) x R (ohms)

So if you double the resistance, you get half the voltage. With half the voltage, you get half the power (wattage). So according to this, I'd have to put a 2 ohm resistor rated for 10 or 18W on the power line. Is there an easier way to do this as I don't have a 2 ohm resistor and one rated for 10 or 18W would be huge wouldn't it?

As for the fan controller, I'm assuming it works with a potentiometer.... I guess that would work but is it rated for such high power?

[tdunks]

Hmm.. I think I've got it...

P (W) = I (A) x V(v)
V (v) = I (A) x R (ohms)

So if you double the resistance, you get half the voltage. With half the voltage, you get half the power (wattage). So according to this, I'd have to put a 2 ohm resistor rated for 10 or 18W on the power line. Is there an easier way to do this as I don't have a 2 ohm resistor and one rated for 10 or 18W would be huge wouldn't it?

As for the fan controller, I'm assuming it works with a potentiometer.... I guess that would work but is it rated for such high power?

resistors dont have poer ratings i dont think...but 10-18w is NOTHING in power they make resistors for huge watages and voltages.

[Graystar]

resistors dont have poer ratings i dont think...but 10-18w is NOTHING in power they make resistors for huge watages and voltages.
Resistors always have power ratings, and a 10 watt resistor is BIG! (relatively speaking.)

It will also get very hot and probably should have a heat sink on it.

I tried to reduce the voltage on a DDC-2 from 12 volts to 9.2 with 4 diodes. It worked but the diodes got so hot that the solder joints melted. I ended up buying a 10 watt DDC.

[Pete]

Hello can non of you fools read!!!!!!

That link is for the newest Laing DDC pump, the latest 10w which has the blue impler but the real important bit is the pcb being a 3.3 revision.

Yours is way older than that, not possible at all, not worth doing.

I know the review is both in english and german so how the hell you got confused beats the hell out of me.

I strongly advise you to leave it well alone. Sell it and buy a 10W one.

How hard it is to read the page where it say use only the latest rev to mod and it also say 10W too 18W not 18w too 10W..why anyone would want to again beats me

[Fossil]

You would need a proper pump controller to reduce the speed of the pump, otherwise it will not be able to start reliably, and that would be bad.

Controller boards that do both fans and pumps are not hard to find.

Resistors most certainly do come with wattage limits - but they are typically for the resistor running uncooled in air, unless it is sold with a build in cooling product. If you look at the data sheet for the product line it will all be there.

[ign0]

find a pump controller if you want to put it onto a controller.

don't just put it on a normal fan controller (which applys LM317 chip or its like), since it may only handle 10w or less. it'll be less reliable, or just burn.

[Pete]

It'll die and burn no matter what on a controler..they dont take too kindly to be fanny farted about with!

So sell it or buy a new one but why you dont what an 18w version beats me

[Graystar]

Hello can non of you fools read!!!!!!

That link is for the newest Laing DDC pump, the latest 10w which has the blue impler but the real important bit is the pcb being a 3.3 revision.

Yours is way older than that, not possible at all, not worth doing.
Could you please explain what the "Laing V6.0" in the lower left is referring to, and how to do tell what the actual revision is?

[SiGfever]

Hi, I'm attempting to downgrade my DDC+ (MCP355) to a DDC (MCP350). I've read this article (http://forum.effizienzgurus.de/f23/howto-laing-pro-zur-ultra-umloeten-t1106.html#post19296), but my pump has an orange impellor and the contact points are different. Can anyone link to somewhere or describe the process of downgrading to a lower wattage on this revision pump? Thanks a lot. Attached are pics of my pump.
Here is a quote from the article...

Quote:
"The circuit boards of both water pumps are equipped identically, so higher amperage after the mod is no problem. By the way, this modification is easy to reverse, just remove the solder bridge. You can also downgrade your DDC-1plusT to a DDC-1T water pump if you want."

A de-soldering bulb will do the trick.

http://img142.imageshack.us/img142/8342/modlaing5mp5.jpg

http://img265.imageshack.us/img265/8117/modlaing4lm9.jpg

[Graystar]

Here is a quote from the article...
The problem, though, is that I believe the mod was performed on an old board design, despite what the article says.(but that just my personal opinion.) What I believe to be the newer boards, which fr0tz has, don't have the pad required for the solder bridge. (the board in the article just has an older, less sophisticated look to it, so I think it's older.)

Here is a picture of a DDC-2 board from a pump with a manufacture date of 10/06. It's a V6.0
http://home.nyc.rr.com/graystar/watercooling/IMG_0030.JPG


Here's a pic of a DDC-1 board from a pump with a manufacture date of 9/06. It's a V5.4.
http://graystar.homeip.net/watercooling/IMG_1788.JPG

The blue, yellow, and black wires are all in the same place on both pumps. There are no obvious differences in soldering to indicate what you'd change, but it's also clear that the boards are different.

[Pete]

The mod is only possible on a 3.3 pcb, it's totaly differnt all round!

The atrical is about modding to an 18w and back all with the 3,3 pcb version 10w pumps nowt mopre nowt less.

Can i not make it any more clear on sell the bloody pump buy a new one or just use the dam bloody thing.

If you want slow flow then make the loop restrictive

[Graystar]

Someone figured out how to mod one version...why do you think it's impossible for someone else to figure out how to mod the versions we have? Why are you so negative?

[Pete]

Why buy an 18w pump when you want it 10W duh

Like buying a DB9 nd only driving it at the speed of the whitehaired goon inthe nursing home (20mph)

[Graystar]

Why mod a 10w when you could have bought an 18w to begin with...they're usually the same price or within 5-10 bucks. See...that argument can go both ways. Regardless, it's not yours or anyone else's place to judge what someone wants to do.

[Pete]

Only if it was tha way that there is small change in price!! £35+ for a 18w ontop of a 10w in the UK and were closer to the sodding Germans that any one in the states!

If it can be done it'd be shown by now and hasn;t and how long has everyone had these sodding things for?? Ages thats what!

Pay uber lower price for a 10w and less than 20 seconds with soldering iron have a 18w and thats it. One under ya belt, many pros too it!

Same as all this case modding lark, car modding, bike modding and redocrating ya house...want me to go on cos i bloody well will.

Ask the Germans, they know these better than anyone. Fact sod it, i will ask Robin directly of Lain UK/EU what he has to say then maybe you febale gray matter might wake up and go ah okay.

My point still yet to be answer is why the sodding hell buy an 18w and want to run it at 10w. Makes no blooody sense at all

[fr0zt]

It seems like everyone's asking the question "why?" The DDC+ is much louder compared to the DDC, I want to reduce the wattage to reduce the noise level, as everything else in this case is pretty much completely silent (17db fans). Also, the wires are not in the same location on all boards, you can see that the colours are reversed on the following 2 pictures.

http://img265.imageshack.us/img265/8117/modlaing4lm9.jpghttp://home.nyc.rr.com/graystar/watercooling/IMG_0030.JPG

The v6 being my pump. So you guys are telling me an 18/10 ohm resistor rated for 20 watts will not be sufficient as >10W is needed for startup? After startup the wattage is lowered back to 10W? I guess one option would be to have a switch between 18W and 10W, I could start it at 18W and once the pump is on, lower it to 10W... might be a bit dangerous as I'd have to be careful to never reboot without switching back to 18W... Can anyone link to a device that can switch automatically?

You'll probably be asking this so I'll answer it just to clarify. Why did I buy the DDC+ instead of just buying the DDC? I got the DDC+ very cheap locally (equivilant to $63 US brand new from a retail store [no shipping]) and would have had to wait 2 weeks to backorder the DDC from the same store or pay quite a bit extra for the DDC from Petras shop. This pump is non-returnable so please don't link to other places to purchase it. Thanks!

[Graystar]

The problem with the resistor is the voltage drop. You need at least 9 volts at the pump to start it. A resistor with a high enough resistance to lower the current draw to 10 watts will very likely drop the voltage across the pump to 8, even 7 volts. So the pump won't start.

[Sparky]

The new pump with the blue impeller can be modded to and from DCC+, the older one with the orange impeller cannot be. Putting a resistor in would be a pain because of the resistor wattage requirements - it would be pretty big - and the pump wouldn't start. It needs close to 12V to start up IIRC.

Here is an idea for you:
Wire up a switch that will switch the pump between 12V and 7V. You need a double pole switch with 6 terminals. In one position it connects the center terminals with the lower terminals, in the other it connects center terminals with the upper terminals. Both sides are separate from each other.

Wire up the switch like this:
http://i58.photobucket.com/albums/g270/SparkyJJO/pump12to7.jpg

Now, you have to have the pump on 12V in order to turn the PC on. Once it is on though, you can flip the switch to run it on 7V and thus will be quieter.

[Graystar]

Here is an idea for you:
Wire up a switch that will switch the pump between 12V and 7V.
I'd say try it, but at first thought I don't think it will work. The problem is the pumps soft-start circuitry. In the instant that the relay switches over, I'm guessing that the circuitry will reset and attempt a soft-start when the power comes back on.

Whether the pump will reset or not, and whether it succeeds in starting up even if it does reset, is something that we'll only learn by trying it out.

I actually had a similar idea but by using the power resistor. Maybe I'll draw up a diagram.

[fr0zt]

Thanks, that looks like a good idea. Still not immune to turning on w/ 7v as a soft reboot or power spike might cause it to turn off and on. I think I may end up doing this, just gotta find a switch like that. What would happen if something like a soft reboot or power spike happened and the pump was only given 7v on startup? Would it cause permanent damage?

And I'm just trying to understand your drawing, let me know if the following logic is right:


yellow wire = 12v
red wire = 5v
black wire = ground

In the down position, the pump draws 12v from the yellow line and grounds through the black, giving a total of 12v.
In the up position the pump draws 12v from the yellow line, but grounds off the 5v line, a difference in pottential of 12v-5v = 7v.

[Sparky]

Thanks, that looks like a good idea. Still not immune to turning on w/ 7v as a soft reboot or power spike might cause it to turn off and on. I think I may end up doing this, just gotta find a switch like that. What would happen if something like a soft reboot or power spike happened and the pump was only given 7v on startup? Would it cause permanent damage?
Soft reboot - from restarting windows for example - won't cause it to reset the pump because the PSU is still running. A power spike might be a different matter, but that is what a battery backup is for ;)
Those switches are usually pretty easy to find. Check out radioshack. You don't want a double pole double throw switch, because it goes through an off in the middle. You want a double pole 3 way switch, not sure what they are called exactly but I have a few of them here. They are slider switches instead of toggle but same concept.


And I'm just trying to understand your drawing, let me know if the following logic is right:


yellow wire = 12v
red wire = 5v
black wire = ground

In the down position, the pump draws 12v from the yellow line and grounds through the black, giving a total of 12v.
In the up position the pump draws 12v from the yellow line, but grounds off the 5v line, a difference in pottential of 12v-5v = 7v.

Correct.

[fr0zt]

Alright.. ya don't have a battery backup, not too worried about a power spike though because usually the power goes off for more than an instant, when the power comes back on I don't think the computer turns on automatically.

What do you think of what Graystar said? I guess that's all up to the pumps circuitry. Graystar, I think I know what you mean but with using the resistor, instead of grounding on the red wire, ground on the black wire with the resistor? That would be susceptible to the same problem of the pumps circuitry though...

Thanks again for everyone's help, it's greatly appreciated.

[serialk11r]

Uh doesn't the DDC not start up without enough voltage?

[Graystar]

Here's a very rough circuit I threw together. I made it as visual as possible.

This is a pretty simple circuit. It uses a single-pole/double-throw relay to do two things. First, when the computer is powered off the relay will bypass the resistor. When you push the button, the relay disconnects the bypass.

That changes the pump voltage without actually ever disconnecting the pump. That should keep the pump running.

The second thing the relay does is power the coil. So when you push the button the relay actually will stay on until you turn the computer off.

Jameco.com has a 2 ohm, 25 watt power resistor that will do the job. They also have the relay with 12v coils.

If you don't normally do this kind of thing...it's a project. But it's very straight forward. The diagram has all you need and I can show you the parts from Jameco.

http://graystar.homeip.net/watercooling/pumprelay.jpg

http://www.jameco.com/Jameco/catalogs/c271/P81.pdf

http://www.jameco.com/Jameco/catalogs/c271/P200.pdf

http://www.jameco.com/Jameco/catalogs/c271/P176.pdf

[fr0zt]

Thanks a lot for drawing that all out. I don't completely understand the schematic, I know that the top part is the resistor, not sure what the 6 - and the 1+ mean. Not sure about everything else though. I'd probably buy the parts from www.digikey.com as I've bought from them before and it's in Canada, I think I found an appropriate resistor, 35W 2 ohm:

http://dkc3.digikey.com/PDF/T071/P1513.pdf

Not sure what to search for with the relays though. Can you explain the circuit step by step? Thank-you for helping me with this.

[Graystar]

I added some text to the drawing.

The relay has 5 connections. Two are for the coil, which is what turns it on and off. Those are the connections on top of the relay schematic.

The three connections on the bottom is the two-position switch. When the relay is off, COM (for common) is connected to NC (which means normally closed). When the relay is on, COM is connected to NO (normally open.)

Here’s how it works. When the computer is powered down and the relay is off, the COM / NC connection provides a route around the resistor. So the resistor is bypassed and the pump gets the full 12 volts from the power supply.

So you turn on your computer and the pump has enough voltage to turn on.

Once the computer and pump are running, you push your push button. That energizes the coil and turns on the relay. The COM / NC connection will now open, disabling the bypass circuit. Now the electricity has no choice but to go through the resistor. The pump slows down.

If you now release your push button, the relay would turn off and the pump would speed up again. So we need a way to keep the relay turned on. That’s the Self-Power on Loop in the diagram. You’re just connecting two pins on the relay to provide the coil with power.

The specs on the relay are: SPDT (single pole, double throw) and the coil voltage must be 12 volts.

The type of push-button you want is MOMENTARY, NORMALLY OFF...like the power button on your PC.

And one other thing. I would recommend using a 5 Ohm resistor to get the voltage across the pump at 7 volts. A 2 Ohm resistor isn't going to make that much of a difference because the pump will be running at nearly 10 volts.

[fr0zt]

K, thanks! I understand the resistor and switch now, I understand what the relay does, but not really how it works. For the switch I'll probably use my reset button, as I rarely use it.. (just use the power button or ctrl-alt-delete if in bios).

As for the magnitude of the resistor, I'm trying to go from 18W to 8.3W (taken from the Swiftech site here - MCP350 (http://www.swiftnets.com/products/mcp350.asp) and MCP355 (http://www.swiftnets.com/products/mcp355.asp)).
P = I V
V = I R
P = I² R
Now to go from 18W to 8.3W
18 = I² R
8.3 = I² (X)

X = 2.17 ohms, if we give R a value, 2.17x greater than it's current value, the power will go from 18W to 8.3W.

Shouldn't that mean 2.17 ohms as opposed to 5? Also, how bout just finding a way to ground it off the 5v wire if the switch is engaged, that would bring it to 7v.

Looking at the Swiftech site again, it says the operating voltage of the MCP350 is 9-13.2v and the MCP355 is 8-13.2. If I reduce it to 7v, will the pump die/break. Even 8v may not be sufficient due to voltage fluctuations.

I've tried to draw up my understanding of the schematic, but I'm not sure where the relay goes, I know it's wrong in this picture, as the electricity would still have to flow through the resistor.

Thanks again for all your help!

[Graystar]

For some reason I thought the minimum voltage was 7 volts. So I’ll recalculate.

V=IR
I=V/R
R=V/I
P=VI

DDC-2
V = 12 volts
I = 1.46 amps
R= 12 / 1.46 = 8.2 ohms

Minimum operation level
V = 8 volts < the minimum necessary to operate the pump
R = 8.2 ohms
I = 8 / 8.2 = 0.98 Amps
P = 8 * 0.98 = 7.84 Watts

Operation level at 8.3W
V = 8.249 volts
R = 8.2 ohms
I = 8.249 / 8.2 = 1.006 Amps
P = 8.249 * 1.006 = 8.3 Watts

What’s the size of the resistor for 8.3 W operation?
Resistor voltage = 12 – 8.249 = 3.751 volts
Resistor current = 1.006 Amps
Resistor resistance = 3.751 / 1.006 = 3.73 Ohms

What’s the size of the resistor for 7.84 W operation?
Resistor voltage = 12 – 8 = 4 Volts
Resistor current = 0.98 Amps
Resistor resistance = 4 / .98 = 4.08 Ohms

I think 2 of those 2 ohm resistors in series will give you what you want.



As for your understanding of the relay...don’t say, “when switch is on.” Instead, say, “when coil is energized.” That’s the distinction between powered and not powered.

Remember that the switch has two poles. Call them poles 1 and 2. When pole 1 is on, pole 2 is off…when pole 2 is on, pole 1 is off.

So, knowing all this we would say...when the coil is de-energized (off) the switch is in the COM / NC position (NC being our pole #1.) When the coil is energized the switch is in the COM / NO position (NO being our pole # 2.) That’s the basic operation of a relay.

In your drawing, change “switch is off” to “de-energized” and change “switch is on” to “energized.” That’s how you should think about a relay.

Also remove the word Relay. I believe you’re referring to the coil there, but the coil is in parallel with the entire circuit because the coil needs 12 Volts to operate. So you would need to include the black wire to properly show the coil.

Otherwise you got the jist of the right. When the NC switch is closed (when the relay is off) electricity will go around the resistor (electricity always follows the path of least resistance, right?) When the relay turns on, the NC switch opens (and the NO closes) and now electricity has to flow through the resistor.

[Sparky]

Uh doesn't the DDC not start up without enough voltage?
Right, it won't start. But that is why we are coming up with a way that it will start on 12V but then afterwards be able to reduce the voltage the DDC gets. Once it it running you can slow it down some.

I like that relay idea Graystar, prevents any trouble from switchovers or brownouts. One question though, why a resistor? Is there a way you could use the 7V connection, that way removing a component and not having to deal with the power lost to heat from the resistor?

[fr0zt]

K, so what are the components in this, is there the coil, the relay, the resistor, and the switch, or is the coil a part of the relay? Or the coil is the relay? Sorry for being so slow with this, I think I'm also missing the self power on loop in my drawing. Thanks!

Sparky: Even if that would be possible and it could be grounded on the 5v line to make 7v, the pump can't operate at 7v.

[Graystar]

K, so what are the components in this, is there the coil, the relay, the resistor, and the switch, or is the coil a part of the relay? Or the coil is the relay? Sorry for being so slow with this, I think I'm also missing the self power on loop in my drawing. Thanks!

The coil is inside the relay. The coil is a little magnet. When it's energized it will flip the magnetic switch in the relay.

Yes, you're missing the self-power connection. Fortunately it's very simple. Just connect NO to the + on the coil.

The push-button connection and the self-power connection are actually in parallel. They both start at the 12 V yellow wire, and end at the positive end of the coil. They both do exactly the same thing...connect the 12 V wire to the plus side of the coil. One does it through a manual push button and the other does it through the switch in the relay.

The resistors you linked to look just fine for this.

[fr0zt]

Alright, I've revised it, still seems wrong though. When the button is NC, it'll get 12v, when it is NO it'll go through the resistor, but once it is released (because it is a push button) it'll go back to 12v.

[Graystar]

The drawing is wrong. The relay has 5 contact points that you need to connect. Those 5 points are not represented properly in your drawing.

The drawing I posted shows exactly what needs to be connected and where. What don’t you understand about it?

Here's another. Maybe this is more clear.

http://graystar.homeip.net/watercooling/relaydiagram.jpg

Notice how the COM / NO connection does the same thing as the push button connection. That's how the relay stays on after you push the button. Kinda like the relay is pushing its own start button.

[fr0zt]

Ohhhhh, I understand now, thank-you. I just didn't understand it with the other drawing, this one makes a lot more sense to me. I didn't realize the relay had 5 terminals, but I guess it has to if it's a switch and needs power. Thanks a lot, I think I'll order the parts from digikey now.

[Graystar]

No prob! :thumbsup:

When I shop for relays I like to find the one with the highest coil resistance. That way you use as little power as possible. The coil resistance will be listed in the specs.

[Quazi]

Wouldn't just connecting it to a Rhebus and taking it from 12w down to 7w after it cranks up do the same thing?

[Graystar]

This relay resets itself when you turn the power off so you always have 12v to start the pump. With the Rhebus if you forget to set it back to 12v after you turn off your computer, the pump may not start the next time you boot up.

[Sparky]

K, so what are the components in this, is there the coil, the relay, the resistor, and the switch, or is the coil a part of the relay? Or the coil is the relay? Sorry for being so slow with this, I think I'm also missing the self power on loop in my drawing. Thanks!

Sparky: Even if that would be possible and it could be grounded on the 5v line to make 7v, the pump can't operate at 7v.

Ohh ok I wasn't aware of that.