WELL.......NOW guess what THIS does....Warning 56K / photos...

Quote:

---

Originally Posted by ZMarre

Hey ehm, is their a chance we will be able to buy a do-it-all-by-yourself-packet? :)

To reduce the costs for those who can solder, and want to adept the frontplate to their own demands :)

So just PCB and components (not nessesary the leddisplays, you find them everywhere, I even have them 24/7 at home ;) )

If(do-iet-all-bai-jourselve-pakkett = true)
{
teller++
printf("big kisses from belgium! ,% :banana: ")
}

:toast:

---

Any comment plz Hippro? :stick:

Quote:

---

Originally Posted by ZMarre

Any comment plz Hippro? :stick:

---

I don't think so coz it includes a lot of SMD parts too....... ;)

Any ETA yet?

Quote:

---

Originally Posted by hipro5

I don't think so coz it includes a lot of SMD parts too....... ;)

---

OK, you got point DOH :toast:

wahh

this device looks awesome :D i'm really amazed what ppl do with there knowledge ;)


how long do we have to wait till retail :D ? i just want one i will perfectly fit into my lianli and to my rig :D

:banana:

and back to the do it yourself version i also know how to solder smd :D (just solderd a 0.5mil tqfp ic :D )

GL Hipro5

I've always liked you bud, always willing to share the knowledge you have garnered without question.

m

Hipro, can I be your sole us. distributor :D j/k. Can't wait to try it....congrats

So... when do I give you my money? I want on with blue LEDs as well... :slobber:

What I don't understand is how things like this thing and the DDR booster from OCZ work. Doesn't the motherboard already give a certain voltage? You can't just give it "more", because you'd have a short (copper is pretty conductive ;)), right?

its not a short, youre just feeding it twice
and the Maximer/booster just gives a bit more juice than the mobo.
But that isn't a problem :)

If the mobo is still giving +2.8 (or whatever) compared to ground, and you're putting 3.2 on the same rail, you've got a (Delta)V of 0.4, so why doesn't it short?
The only thing I can think of is that the memory slots are connected serially, but that'd seem stupid to me, because if one slot is borked the rest is too...


"More juice" isn't a technical explanation IMHO. The only thing you could do to feed it "more juice" is to put the exact same voltage (same source) through a thicker wire so that you can pull a bit more amps through it.

[dutch]
Of mis ik delen van m'n fysica? Zou voor een aspirant regent toch echt niet mogen ;) Je gaat op een of andere manier de stroom die het moederbord zelf levert moeten tegenhouden... Wat wel kan is dat de slots in serie staan en dat je dus het doorlussen tegenhoudt.

Quote:

---

Originally Posted by racoontje

If the mobo is still giving +2.8 (or whatever) compared to ground, and you're putting 3.2 on the same rail, you've got a (Delta)V of 0.4, so why doesn't it short?

---

a short only happens when a current is directly connected to the ground. That doesnt happen when you feed the mem twice.

Quote:

---

"More juice" isn't a technical explanation IMHO. The only thing you could do to feed it "more juice" is to put the exact same voltage (same source) through a thicker wire so that you can pull a bit more amps through it.

[dutch]
Of mis ik delen van m'n fysica? Zou voor een aspirant regent toch echt niet mogen ;) Je gaat op een of andere manier de stroom die het moederbord zelf levert moeten tegenhouden... Wat wel kan is dat de slots in serie staan en dat je dus het doorlussen tegenhoudt.

---

Quote:

---

Originally Posted by Jack

a short only happens when a current is directly connected to the ground. That doesnt happen when you feed the mem twice.

---

I'm pretty sure that a short circuit is when two electrical conductors are connected by another conductor and when R = ~0. For example, you can turn a motherboard on by shorting the power pins. You can short two poles, with neither being ground (think power outlet).

Also, the problem isn't that the memory is fed twice (it's highly credible that the current LM

It is because the FETs cut off when the voltage on the reduced side is higher than the desired one, to make the votage go down..
so it is not fed twice..
that is how i understood it at least.

that makes sense

Yeah, I guess that could be right, the only thing that made sense would be that the IC senses Vdimm (and if Vout>Vdimm, disconnect it)

For jack though
You can't feed a part "more" voltage by feeding it 2.8 AND 3.2... You WILL create a short, and due to the very low resistance of copper you WILL cause a lot of current to flow and you WILL most likely blow a copper rail off your board.

Quote:

---

Originally Posted by racoontje

Yeah, I guess that could be right, the only thing that made sense would be that the IC senses Vdimm (and if Vout>Vdimm, disconnect it)

For jack though
You can't feed a part "more" voltage by feeding it 2.8 AND 3.2... You WILL create a short, and due to the very low resistance of copper you WILL cause a lot of current to flow and you WILL most likely blow a copper rail off your board.

---

okay :D
i'm no electronic engineer and my explanation seemed logical to me.
But FET cutting off voltage when voltage is higher on the reduced side, makes more sense indeed :)

Isn't it just like putting another batterey in a simple circuit, only in a different place? For that matter, why should it matter where in the circuit the battery is in the first place?

Quote:

---

Originally Posted by racoontje

Yeah, I guess that could be right, the only thing that made sense would be that the IC senses Vdimm (and if Vout>Vdimm, disconnect it)

For jack though
You can't feed a part "more" voltage by feeding it 2.8 AND 3.2... You WILL create a short, and due to the very low resistance of copper you WILL cause a lot of current to flow and you WILL most likely blow a copper rail off your board.

---

That is correct because its 1 circuit with a comon ground. try connecting your 5v rail to your 12v rail to proove it :eek: dont do that!!

the reason why this works is because the mosfets stop switching when the voltage on the drain pin is above what it is set to.
:confused: hmm or was it becasue the mosfet then acts as a diode and blocks the current from flowing backwards? i think that was it. kinda forgot this stuff aswell.
it could also be that once the voltage feedback chip detects a higher voltage than what it is set to produce it will make the mosfets produce less and less volts until the mosfets produce 0v and stop switching.

1 of those 3 atleast. :D

The important bit is that the MOSFET stops working when it senses voltage (coming from the booster) :)

any updates?? The updates ran dry for a while. WHen can we expect these things on the shelf? I have some BH-5 hiding in my drawer.

:confused:

Hi there and HAPPY THE NEW YEAR to all of us...... :D :toast:


I'm waiting for the "CE" certificate so as to be ready for selling.......I think any time soon........You know , you have to have a LOT of money so as to "push" a bit those things......IF you want them to take a "good and FAST care" of you , you have to "feed with money" some employs of the gov. to do so.......or you hust waiting for your turn to come...... ;)

Cool I didn't know it worked like that, interesting......

awesome hipro, so does that mean production is already underway?