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 Post subject: DIY voltage drop
PostPosted: Tue Nov 01, 2016 5:41 pm 
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I am very inexperienced with regards to electronics and solderings. It seems most if not all voltage reduction cables just have a single resistor on the wire for the yellow (+V) or red (+V), depending on which you have; the +V cable has a resistor to it.

If you're talking about 2 resistors Ohm's law applies and the voltage drop is the ratio between the resistors. So a resistor of 1 Ω and one of 2 Ω will see 1/3 of the voltage being consumed by the first, and 2/3 by the second.

This also means that the current going through the circuit is going to be e.g. 12V / 3 = 4 amps? That means that this circuit would draw 4 x 12 = 48W ??.

But Ohms law does not apply to dynamic materials. A fan rated at e.g. 0.33 A on 12V, would, according to Ohms law, have a resistance of 36 Ω. Then, if you wanted to drop the voltage by 1/3, you would ensure that the preceding resistor has a resistance of 18 Ω and then it would consume 4V, leaving 8V for the Fan. If the Fan is still drawing 0.33 A, the resistor would consume 4V * 0.33 = 1.33 W.

When I look online at selections of resistors they are all rated for 0.25W, they are those tiny things. But you can also easily find .e.g 3W online. Looking at some of the loudest fans they draw at most 4W at 92x92mm. Yes, this is for a graphics card and I haven't checked but its fan might go up to 3600 RPM. It's 80 degrees value was 2250 RPM I believe, and its 100 degrees value was 3600.

Other fans that can do at most 2500 RPM draw at most 2.1W. So for a resistor I assume for a 1/3 voltage drop you will be looking for some 1-2W maximum for whatever fan you hook up. A 2W resistor is simply going to be smaller.

However I have no clue what would happen with PWM. If resistance of the 2nd component (the load) goes up and down, then the voltage drop will differ from time to time. If the thing uses a lot of current the resistance will be low and the voltage drop of the resistor will be higher, which means that when it draws more current, it will get less voltage, and that might cancel it out in some way.

Then, to get a fixed voltage drop, I would have to employ this thing. To create a voltage divider I must employ two resistors in parallel where... I mean. They have to be in sequence but the gap between them must feed to the fan. So now I need two resistors where one goes back to the black wire (ground) and the other one is in sequence to the fan, that also feeds back on the black wire.

However the second resistor (that feeds back to ground) is in effect part of a parallelel circuit that also contains the load:

Code:
PSU ---> resistor 1 -------> resistor 2 -------> ground
                       \---> load ----------/


Since the voltage drop for resistor 2 and load is going to be identical its size (magnitude) is going to depend on the resistence of the parallel block with respect to resistor 1. If the maximum power draw (current draw) for the load is say 0.30 A with accompanying resisance of perhaps 40 Ω, then at maximum draw if you put a 40 Ω resistor at the 2nd spot and a 20 Ω resistor at the first spot, you will have 8V for the fan.

But if the amperage goes down and resistance goes up then the weight of resistor 1 becomes heavier and ... the voltage drop will be more pronounced. But I am already mistaken. Two identical ohms will see the resistance of the second block go down to 20 Ω in our example. Thus, the first resistance must be 10 Ω. The resulting resistance of the 2nd block (the parallel block) will always be less than its components. But if the resistance goes up.... let's say it goes up to 80 Ω for the fan.

1/40 + 1/80 = 3/80 --> 80/ 3 = 26.667 Ω for the second block with the first resistor at 10 Ω gives a 10/36 * 12 = 3.3V voltage drop, leaving 8.7V for the fan, so I am mistaken, the voltage drop is less pronounced. The weight of resistor 1 does not become heavier obviously, it becomes lighter. My mistake. With infinite resistance the voltage drop approaches 1/2, or 6V.

So we'd end up with a voltage of 8V (when the fan is spinning fastest) down to 6V (when the fan is spinning very slowly). That means the speed reduction will be more pronounced in the lower speed range. Building on this, you could reduce the voltage reduction by placing e.g. some 5 Ω resistor at first spot at which point the maximum voltage drop will only be 1/5.

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 Post subject: Re: DIY voltage drop
PostPosted: Tue Nov 01, 2016 5:55 pm 
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I just keep messing this up but I can't delete the post.

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 Post subject: Re: DIY voltage drop
PostPosted: Tue Nov 01, 2016 7:13 pm 
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I'm gonna have to stop thinking because I'm not doing very well at it.

The maximum resistance minimum voltage drop is 1/4 (for these figures) at 3V up to a fan voltage of 9V.

The speed reduction is then less pronounced at lower speeds, but more pronounced at higher speeds.

3V of 12V is still 1/4 lower. For a more pronounced voltage drop the first resistor must be higher; 20 Ω would give 6V reduction at full speed.

;-).

No longer full speed then.

And 1/3 at lowest speeds is the 4V we were aiming for. The 3600 rpm fan will then only spin at 1800 max which is a bit slow but yeah. But the 1500 base will then go down to 1000 :p. More likely is...

30 Ω at lowest speeds for the entire block. So the lowest resistance is the 20 Ω we started out with, the highest is 30 Ω, not 40. The fan block + 40 Ω resistor then hovers between 30 Ω and 20 Ω. To get 1/3 at 30 Ω we have to use a first resistor of 15 Ω. The drop at 30 Ω is then going to be 1/3 and the drop at 20 Ω is going to be 15/35 = 43%. So our fast spinning fan then spins at 2052 rpm instead of 3600 :p.

Provided that anything I've said here is even moderately correct.

This is assuming that the resistance at 43% fan speed is going to be 1/0.43 times as high for the fan, than it is at 100% fan speed.

1 / 40 Ω + 1 / (2.33 x 40 Ω) = (2.33 + 1) / (2.33 x 40 Ω) = 3.33 / (2.33 x 40Ω) = 3.33/2.33 / 40.

Let's assume x = 0.43 is the minimum fan speed ratio. The above becomes:

1 / 40 Ω + 1 / (1/x * 40 Ω) = (1 + 1/x) / (40 / x Ω) = (1 + 1/x) / (1/x) / 40 = (1 + 1/x) * x / 40 = (x + 1) / 40 at which point the new resistance becomes 40 / (x + 1) = 40 / 1.43 in this case = 28 Ω and not the 30 Ω I wrote earlier.

Assuming the base fan has a resistance of 40 Ω at full speed. At 28 Ω you utilize a 14 Ω resistor to get the same values as above.

So low speed resistance = high speed resistance / (x + 1) and the first resistor becomes high speed resistance / (x + 1) / 2. This gives a 1/3 voltage drop at low speeds. That voltage drop at high speeds becomes hi-speed / (x + 1) / 2 / (hi-speed / (x + 1) / 2 + hi-speed / 2) yields in the end 1 / (x+2).

So my fan spinning at 43% now sees a 41% voltage drop if we use the 43% voltage drop to be 1/3. Which results in 2124 RPM at full speed for this 3600 RPM fan. That is still reasonable I think.

So: assuming the fan has 40 Ω resistance at full speed. Assuming its current draw doesn't go down as voltage drops, and assuming we create a voltage divider circuit with one 40 Ω resistor in parallel with the fan, and one 14 Ω resistor ahead of it, both consuming....

7V x 0.3A for the fan is 2.1W, 7V / 40 = 0.18 A for the resistor, at full speed, x 7 V = 1.23W for the resistor (2nd).

And now my calculations and my understanding defeats itself because the current draw among both components (first resistor, and second block) must be the same for this "serial" circuit. As the second block draws 7V the current draw ought to be 0.3 + 0.18 = 0.48 A, but because my hypothetical fan does not honour Ohms law as voltage drops, it is more than the 0.35A that the first resistor is supposed to draw (or see pass through). If the amperage of the fan also goes down then both fan and resistor (2nd) draw 0.175 A and the entire thing is right again at 0.35 A for the first resistor.

Otherwise it would draw more than the resistor (1st) is supposed to be able to do. For the 5V it has at this point.

Anyway some 0.35A x 5V = 1.75W for the first resistor. So at least one thing should be true, and that is that a fan rated at at most 0.3 A for 12V should be easy to tackle with two 2W resistors.

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 Post subject: Re: DIY voltage drop
PostPosted: Tue Nov 01, 2016 7:53 pm 
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So, at the 10,000ft level instead of sea level, what fans are you working with what are they doing today and what are your goals (voltage/rpm/etc)?

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 Post subject: Re: DIY voltage drop
PostPosted: Wed Nov 02, 2016 3:55 am 
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Lol I think I have already said it. It's a graphics card fan that won't honour the Maxwell architecture BIOS values (that I modded, and which it recognises, to some extent) so it is stuck at a base minimum of 1500 RPM and has a fan that can do reportedly do 3600 RPM max.

It uses JST PH connectors that you are familiar with (here is an image of the PWM connector you would see on any graphics card mostly) -- just a random image of a PHR-4 contact -- that I found after hard searching ;-).

And it is not hard, in principle, to take 2 wires out of the housing, give them its own housing, put a header to it, and now you have 2 (or 4) pins that you can use to solder something onto. The easiest thing is to just solder a wire (or 2, actually) to some regular 3-pin fan plug so you can attach it to an ordinary voltage controller. Then you'll ordinarily simply have a resistance on the wire in between that, with a fixed fan voltage, would just lower the fan's voltage by a fixed amount, what I mean is that the power (current) draw of the fan is ordinarily going to be fixed and so the speed reduction will also be fixed although it will vary by fan (the voltage drop will actually vary by fan, I believe).

But I'm not all too keen on how it works so I was just asking.

Also with PWM if it works by changing the effective resistance on the fan then the voltage drop is going to differ according to the resistance of the fan which is why I wrote that.

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 Post subject: Re: DIY voltage drop
PostPosted: Wed Nov 02, 2016 4:27 am 
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Someone wrote in 2014:

Quote:
A resistor divider circuit is for creating a voltage that's a fraction of an input
voltage, and expects to be driving a very high impedance load (otherwise the
divide ratio is influenced by the load).


which is basically what I described in my post.

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 Post subject: Re: DIY voltage drop
PostPosted: Wed Nov 02, 2016 8:20 am 
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Sorry - didn't have time to read the wall of text in the earlier posts for the hidden overview :)

Pulse Width Modulation (PWM) is pretty much what it says. At the most basic, the fan controller PWM signal is an on/off switch for the fan that varies in width. When 12V is applied and if the PWM control is on 100% of the time, the fan runs at max rpm. The fan control is operating at some clock rate (let's call it 100Hz for grins). So, for 50%, the fan control is on for half of the time and off for half of the time for each of these 10ms periods. The fan averages out these on/off pulses, so instead of zero->max rpm and repeating at 100Hz, it runs at roughly 50%.

That's for 12V. Now, what if you apply less than 12V to the fan? You've set a new lower maximum rpm for 100% PWM signal. So, a 50% PWM will run at roughly 50% of the new maximum. Note that fans aren't terribly linear and they aren't a purely resistive load. So, using a resistive splitter can be a trial by error process.

I've wanted to mess around with using diodes to drop the 12V rather than resistors. Don't have to scramble the brain trying to figure out resistor values. :) Haven't had the time. A basic 1N914 diode has a forward drop of 0.6 to 0.7V. So, you could stack these in series between the 12V line and your fan until you get the new minimum. Just a thought. Would need to know the fan's amperage rating and get the appropriate diodes.

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 Post subject: Re: DIY voltage drop
PostPosted: Wed Nov 02, 2016 10:24 am 
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Ah. I regret not just going ahead with my voltage divider plan: I think it would have worked :).

Today I seem to be going for a voltage regulator but the annoying thing is that you have to get a fixed value (or get a variable one with a potentiometer to go with it, which then has to be a form of turny button).

Basically if you get one of these:
Image

As a 7808 package you get this:
Code:
                        ___
fan |-----red -------- |---|*
    |---- black ------ |---|*--------- black ------ | PSU
                       |___|*--------- red   ------ | PSU


Which, at a cost of some 0,70 €, will dyamically reduce 12V to 8V at a fixed output value with the same amperage, and the 4V is just dissipated at heat x the same amperage. No matter the load.

Quite boring, isn't it.

Too much trouble to look into the variable ones now. I don't even know for how much the fan is rated. I will assume it is rated at 40 Ω nominal at the high end but I've since learned that resistance goes UP when the fan spins FASTER. If I assume a startup amperage greater than the rated amperate (of say 0.16A as I saw one 2500 rpm fan to be) the resistance at start is going to be lower than that 40 Ω, suppose it takes 0.5A at start = 12 / 0.5 = 24 Ω at the low end.

This fan will then have a low resistance of 24 Ω and a high resistance probably at its rated amperage, which would be 75 Ω for a 0.16A fan.

Since this is a 2500 rpm fan at 0.16A, I will assume my 3600 rpm fan will draw upto 3.8W as I've seen the Papst 3412N/2GHHP being mentioned somewhere. That one is set at 3250 RPM. So 3.8W -4W should be a safe power draw to assume... That's that maximum 0.32A you could assume. Now 0.3A gave that 40 Ω which it should have while spinning really fast at the maximum PWM. You'd assume this goes up with lower PWM. So it could be as high as 93 Ω at that 43% minimum PWM that my fan actually does.

Assuming normal operation between 40Ω and 93Ω is what I already had as an assumption it's just that at 93Ω the voltage drop will be very modest ...err, wrong again, it will be 28 Ω at 93 Ω for 1/3 if the first resistor of the voltage divide is 14Ω.

Code:
Nominal A: 0.3A, startup: 0.5A, lowest: as low as it can get.

Resistance nominal: 40Ω, resistance 43% PWM: 93Ω, resistance startup: 10Ω (?)

Circuit: 14Ω first resistor, 40Ω parallel resistor, divide at:
- 40Ω: 14/34 = 41% voltage drop
- 93Ω: 14/42 = 33% voltage drop
- 10Ω: 14/24 = 58% voltage drop (will it start?)


The funny thing is that the more amps the fan tries to draw, the lower its voltage will be :p. It seems like a power delimitor to me, you can probably calculate the maximum Wattage this whole circuit will be able to draw ;-). At 0Ω the first resistor will take 12V * (12/14) = 10W which is quite hefty and would burn the resistor. At 10Ω it would take 14/22 = 7.63V square / 14 = 4.165W which would still burn the transistor... ehm, resistor? I am not sure how a transistor would come into play but a MOSFET is one and got some use somewhere... but I lost my links at this point (closed everything).

Basically with 12V and a 14Ω resistor you'd be in trouble. If you increase the resistors to match the 93Ω what you get is:

Code:
Circuit: 23Ω first resistor, 93Ω parallel resistor, divide at:
- 40Ω: 23/51 = 41% voltage drop
- 93Ω: 23/69 = 33% voltage drop
- 10Ω: 23/32 = 72% voltage drop which is even MORE pronounced lol. But this time it takes 3,23W on the resistor. But the fan is left with 3,36V and will never start with resistance this low.


So lol, this gets harder and harder :p. But actually a resistor should be able to withstand a bit higher wattage for a second.

I don't know how low the starting resistance of the fan goes. I could create one circuit with 14Ω and 40Ω and one with 23Ω and 93Ω, I don't know.

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 Post subject: Re: DIY voltage drop
PostPosted: Wed Nov 02, 2016 7:04 pm 
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Let me summarize that again:

* A voltage regulator is a cheap solution that looks ugly on a wire.
* The earlier ohmages I chose should already do the trick provided the fan still starts.
* A start amperage of high value might burn the first resistor if it takes too long
* Raising the resistor values reduces this threat but also lowers the start voltage beyond reasonable amount.

Well I ordered a bunch of stuff. Enough to make 3 different 'voltage regulators' ;-).

My god, the nonsense I do. Can make a few more if anyone likes.

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 Post subject: Re: DIY voltage drop
PostPosted: Thu Nov 10, 2016 3:32 pm 
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I got some of the items I ordered.

I am really good at screwing things up, and this time I used a screw for it.

Actually a drill which is pretty much the same thing for this respect.

I got these little boxes where you could put electronics into. They are only 35mmx35mm x 15mm high.

Drilling in plastic always has the risk of your drill getting a life of its own if you don't hold the item tight enough and it is slung along the track of the drill and this can basically create a hole too large or just gaps to the side.

My first drill was perfect. 1. Small wood drill to make the initial hole, 2. 5mm metal drill to complete it.

Second hole ruined it. (It was a 6mm metal drill). I used a 6mm wood drill and not only had I miscalculated the epicentre of the hole, due to instability it also took too large a gap. The plug was drifting in the hole. I fixed it with glue and then I got impatient and put the thing on top of a light bulb and the thing melted --- slightly.

Enough to ruin it for me.

Second attempt.

ImageImage

First hole was perfect. Second attempt the drill got a life of its own and "upped" the epicentre of the hole by 1mm and that is why you see it sticking out so much.

Actually the other way around. First was wrong, second was perfect. It's really not that hard to get it right. But if you're me ;-).

This is just really tiny. Maybe I should have put some coin next to it but this is really the size of two fingers in width (and length).

The cable plugs are what they call "glands" and are just the cheapest here:

http://www.conrad.com/ce/en/product/1095600/Cable-gland-Polyester-Black-KSS-SR-611-1-pcs

It's not actually that cheap, these little boxes are more than 3 euro each I believe. Oh it's not too bad, 2,22 at present. So for about € 2,50 you have one of these little crackers, but this one will have to do for now.

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 Post subject: Re: DIY voltage drop
PostPosted: Thu Nov 10, 2016 8:10 pm 
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I hadn't seen the "build your own fan controller" thread.

My GPU is SO QUIET NOW it is unsettling.

The fan now starts off at about 860 RPM (instead of 1580) and you just can't hear it, but then again my 7200 rpm 2.5" HDD is the loudest component now by far.

I'm running in open case but with less airflow (one fan turned off) and the fan goes up to ... well the temp goes up to 66 degrees (C.) under a fan load of 1260 RPM (35%) and this is a 100% Furmark stress test.

So I don't know what the guys at MSI were thinking (they were probably guys you know) but this:

MSI GTX 950 2GD5 OC (white version, one fan) that you know as:

Image

This card.

And is VERY LOUD at 1500.

I've been running Furmark for some 10 minutes now at it doesn't go above 66° C.

What's UP with this.

Why do manufacturors create lousy loud components that do not need to be???

I just asked MSI about it. I EXPECT ANSWERS :p.

Well, this was another nice waste of time as death is looming and destruction is nigh.

Took a bit longer than I thought because I broke off the black wire of the fan by accident and it took me at least an hour to fix a new contact onto it :(.

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