Snap 705N/4100 lives and thrives!

[jontz]

Originally Posted by blue68f100

OK, Here is the rest of the nickle.

Power up the drives 2 at a time, then the controller board.

That sounds like it would work, but I have to agree with Pheonix on this one. Seems like upgrading the power supply would be the best way to go all around.

And correct me if I am wrong, but isn't the power switch on these units soft? If so, don't you need the controller board to have power so that you can turn the power on? (wow...chicken and the egg...)

[Phoenix32]

Originally Posted by blue68f100

OK, Here is the rest of the nickle.

Power up the drives 2 at a time, then the controller board.

A couple of 555 TImer circuits maybe?

[Phoenix32]

Originally Posted by rpmurray

I'd love to see someone try the Hitachi drives as a solution, since they seem to draw less power on startup. At the moment I can't afford to get four of them to try it out.

Ditto...

[Phoenix32]

Originally Posted by jontz

Probably, but what fun is that? And since when are we interested in doing things the EASY way around here?

Good point...

[Phoenix32]

Originally Posted by jontz

That sounds like it would work, but I have to agree with Pheonix on this one. Seems like upgrading the power supply would be the best way to go all around.

Now wait one second here buckaroo! You just said, and I agreed, about that not being fun or interesting... LOL

Originally Posted by jontz

And correct me if I am wrong, but isn't the power switch on these units soft? If so, don't you need the controller board to have power so that you can turn the power on? (wow...chicken and the egg...)

I think it is a soft on/off also. But that would not be too hard to get around I think. Once the power switch is pressed, the controller still waits for the power to get turned on. We just have to intercept that power before it gets to the controller and delay it a bit.

For example:

1- Break all the power to the controller board EXCEPT THE STANDBY VOLTAGE (used for the soft on) and put in a relay or SCR circuit.

2- Use a couple of 555 timer circuits to power up a pair of drives, one pair at a time, also using either a relay or SCR circuit.

3- Use a 555 timer circuit to kick in the power relay or SCR for the controller board after the drive pairs are spun up.


There are most likely a hundred ways to do this, but this was one example. 555 timer circuits are easy to design and/or can be found all over the internet.

Now don't start complaining either, you said you wanted to make it fun and interesting....

[jontz]

Originally Posted by Phoenix32

Now wait one second here buckaroo! You just said, and I agreed, about that not being fun or interesting... LOL

RATS! Caught again

Originally Posted by Phoenix32

Now don't start complaining either, you said you wanted to make it fun and interesting....

Funny, because I WAS going to say that your solution sounded a bit complex, but when you put it that way... LOL

[Phoenix32]

Well, for a Snap 4000, it seems to me from what we have here in this thread is;

1 - Use Hard Disks using less than or equal to approx 2.25 A startup power

2 - Change the power supply

3 - Install a delay circuit to starts Hard Disks in sequence


Also, since we do not want to exceed say 75% full load on the power supply during normal operation, we want to use Hard Disks that use no more than 1.125 A during normal post spin up operation for options 1 and 3 above.

To summarize, since most users will not change the power supply or install a delay circuit, this means a stock Snap 4000 needs Hard Disks that use no more than 2.25 A on the 12 volt rail during spin up and no more than 1.125 A on the 12 volt rail during normal post spin up operation. So the question is, what large IDE Hard Disks meets these requirements, using REAL specs, not just default whole series specs listed by most manufacturers?

[blue68f100]

Most users have been able to use 250-300 gigs without any problems. These drives proably only have 2 maybe 3 platters. The next step up (4 platters, 400 gig) puts them over the edge. The new perpinduclar technology may help solve this.

You know the 2.5" notebook drives may be an option. My Hatichi 7K100 pulls 1.1 amps. But the capacity is well below that of 3.5" drives.

[Phoenix32]

Originally Posted by blue68f100

You know the 2.5" notebook drives may be an option. My Hatichi 7K100 pulls 1.1 amps. But the capacity is well below that of 3.5" drives.

Interesting idea... Didn't think of that...

[jontz]

I had given it some thought, but now you are talking about 100-120GB a drive, so you might was well stick with 3.5's as they are typically faster than laptop drives.

[blue68f100]

You are correct on the speed but they are getting faster with the perpendicular technology. I think Samsung just anounced 200gig.

I wonder it we could use a current limiting resistor on the 12v to kill the inrush????

[jontz]

Originally Posted by blue68f100

I wonder it we could use a current limiting resistor on the 12v to kill the inrush????

Well, you could....but you would actually amplify your problem. Putting a resistor on the 12V side would decrease the voltage, which would in turn actually increase the current draw by the drive. Putting a capacitor on the 12V rail, as I have stated above, would be the better solution.

[Phoenix32]

Originally Posted by jontz

Well, you could....but you would actually amplify your problem. Putting a resistor on the 12V side would decrease the voltage, which would in turn actually increase the current draw by the drive. Putting a capacitor on the 12V rail, as I have stated above, would be the better solution.

It would not increase the current draw, but it would lower the voltage to the Drive, even after spin up, and make the drive not work well I would say.

The CAP would increase the initial current draw and thus increase the original problem.

BUT, you could use a current limiter (not a resistor). The drives would take a bit longer to spin up (not much I would bet), but might work as long as they are not being restricted too much (slowing their spin up too much and making the drive look like it failed). But this solution is too easy, and thus not fun...

[jontz]

Originally Posted by Phoenix32

It would not increase the current draw, but it would lower the voltage to the Drive, even after spin up, and make the drive not work well I would say.

The CAP would increase the initial current draw and thus increase the original problem.

BUT, you could use a current limiter (not a resistor). The drives would take a bit longer to spin up (not much I would bet), but might work as long as they are not being restricted too much (slowing their spin up too much and making the drive look like it failed). But this solution is too easy, and thus not fun...

I'm going to have to disagree with you about the current situation. If you drop the voltage to a device, it will draw more current to compensate. That's why when you run your A/C on a 300' long 16ga extension cord, the compressor will burn up. The voltage drop imposed by the long, small gauge cord (which can be thought of as a resistor) will cause the compressor to draw more amperage, which in turn will burn up the compressor motor because it starts to draw more amperage than it was designed to. That's also why brown outs are so dangerous.

A cap would cause an initial current rush, IF it was not charged ahead of time. The cap would have to be installed and then charged with a resistor in line (to keep the current rush down so the cap will not blow the power supply). Once the cap is charged, and resistor removed, then the drives could be hooked up and used as normal. The cap levels out the initial current inrush, and charges once the voltage/amperage levels return to "normal" after the drives spin up.

Yeah, the current limiter WOULD be too easy We're looking for something that will keep us down in the basement tinkering for longer than 15 mins...

[Phoenix32]

Originally Posted by jontz

I'm going to have to disagree with you about the current situation. If you drop the voltage to a device, it will draw more current to compensate.

Sorry, but that is not how it works. Ohms Law on this is

I = E / R (Current equals Voltage divided by Resistance)

The "R" part of the equation is increasing, thus causing the "I" part of the equation to decrease in return.

Example:

In our Snap Server 4000, we are looking at the 12 volt rail and it's current. While a Hard Disk is not a resistor, it is a Load, and this Load can be shown as a resistance in Ohms Law equations. In this case, we know the Voltage (E) as 12 Volts and we know we are talking about a max Current (I) of 2.25 Amps. Using Ohms Law, R = E / I, we can determine the Load Resistance as 12 / 2.25 (12 divided by 2.25). This means at spin up, the Load resistance is 5.34 Ohms.

5.34 = 12 / 2.25

Now this Load resistance is not going to change, the drive motor is what it is, period. So we want to know what will happen if we add a current limiting resistior in the circuit. Let's say just for this example, a 10 Ohm Resistor. This time we will use the equation I = E / R to find this new Current drain. The voltage is still 12 volts, but now our resistance is 15.34 Ohms. This is from adding the 10 Ohm resistor with the 5.34 Ohm Load Resistance of the drive motor at spin up (adding when in series, dividing when parallel). So we can determine this new current load as 12 / 15.34 (12 volts divided by the 15.34 total resistance). This means our new current with the current limiting resistor is .78 A, a far cry from the 2.25 we started with before. But the point is, it went down, not up. As the resistance increases, the current will decrease with the same fixed voltage source (in this case, the 4000 power supply is a fixed voltage source).

Also, you said, "If you drop the voltage to a device, it will draw more current to compensate". This is also wrong, sorry. Given a Load Resistance (R), as the Voltage (E) goes down, so will the Current (I). This is shown in the same equation I = E / R of Ohms Law.

Example:

12 volts and 10 Ohms. This is 12 / 10 = 1.2 which means 1.2 A current.
10 volts and 10 Ohms. This is 10 / 10 = 1 which means 1 A current.

As voltage went down, so did current.

See what happens when you argue electronics with an electronic engineering tech? (all in fun, honest)



NOTE: For those who may not know, and actually care, Ohms Law is a set of 12 equations used in electronics to help determine unknown values in a circuit using known values. The four values concerned are;

P = Power in Watts
E = Voltage in Volts
I = Current in Amps
R = Resistance in Ohms

The primary equations are;

P = I * E (Power equals Current times Voltage)
E = I * R (Voltage equals Current times Resistance)

The other 10 equations can be determined with basic algebra by these two base equations. The idea here is to be able to determine any one unknown value given two known values.

For example, in the message above, I used the equation E = I * R to determine the equations required for the situation. I used E = I * R to derive the two equations I = E / R and R = E / I.

Another example would be to determine Power (P) with the known values of Current (I) and Resistance (R). We know P = I * E and we know E = I * R. We can substitute the E in the first equation with the equal values to E of the second equation, leaving P = I * (I * R), or P = I * I * R or P = I Squared * R (sorry, I have no idea how to show a squared on this keyboard).

[jontz]

Originally Posted by Phoenix32

See what happens when you argue electronics with an electronic engineering tech? (all in fun, honest)

Must've been that bad crack I took before I wrote that last post...

...to be honest, I should have known better. It has been waaayyyy too long since I have had to apply ohms law to anything. If I would have thought about it for 5 seconds before posting we could have avoided this whole mess

Hey, never any offense taken. We are all here to learn and share. Thanks for the info!

[Parity Error]

Originally Posted by Phoenix32

Sorry, but that is not how it works. Ohms Law on this is

I = E / R (Current equals Voltage divided by Resistance)

The "R" part of the equation is increasing, thus causing the "I" part of the equation to decrease in return.
.

You fell into a logical trap that Prof K. in EE413 would have rapped our knuckles over....

The fixed factor is the MECHANICAL power needed to spin the drive.
THAT begets the electrical power needed, in watts.

When you lower the available voltage, and require the same output power, the current needs go UP, not down.

Reduced-volated starting is always harder on the motor than across-the-line, for that reason. [It's easier on the grid, and sometimes on the shaft load...]

Where lowering the Ein works is with a dependent load, say a lamp. THERE, reducing Ein will reduce the Wout, and thus Iin as well..

[jontz]

Exactly, and I thought about it after Phoenix took me behind the woodshed

Now that we have that all cleared up, what was this thread about anyway???

[Phoenix32]

Originally Posted by Parity Error

You fell into a logical trap that Prof K. in EE413 would have rapped our knuckles over....

The fixed factor is the MECHANICAL power needed to spin the drive.
THAT begets the electrical power needed, in watts.

When you lower the available voltage, and require the same output power, the current needs go UP, not down.

Reduced-volated starting is always harder on the motor than across-the-line, for that reason. [It's easier on the grid, and sometimes on the shaft load...]

Where lowering the Ein works is with a dependent load, say a lamp. THERE, reducing Ein will reduce the Wout, and thus Iin as well..

Um, no, it doesn't work that way (as you described it), but it is not worth arguing over. Go ahead and try it, see what happens.

I will say this, you are correct in that the power requirment to spin up the motor according to spec has not changed, but if that power (i.e. voltage to drive the current needed to accimplish this power requirment) is not available, this does not mean it will magicly happen anywise. The motor in this case will just spin up slower than in spec, using a lower available power. Power requirements do not drive Current or Voltage. If the voltage is just not available, or if the current is just not available, then the power will not be there. It really is that simple. And for the record, there is a big difference between what happens in DC motor circuits and AC motor circuits (hint hint). But again, this IS NOT worth arguing over. If you think it will work, go ahead and try it, and please be sure to post your results. (even when you find out I was right, LOL, heh heh)

[Phoenix32]

Originally Posted by jontz

Now that we have that all cleared up, what was this thread about anyway???

I think it was about this;

Originally Posted by soundidea

Question for the group..... Until some clever person is able to extend the drive limit of 137GB on the 4100/705N, what is the best Snap choice to hold 4 BIG drives?

And then developed into other units (like the 4000), where I asked basicly the same thing about the 4000 (and other SNAPS). What larger drives work, or work best in the various SNAPS, including my 4000? Would be good info to have before I, or anyoen else, goes out and buys 4 drives only to find out they won't work for this or that reason.

Then it made a short detour into the land of, "how do we get around current requirements of the largest drives"?

Which leads to discussions on electronic theory.

Hey! You asked...

Oh and by the way, yes, I am a smart ass.

LOL, gotta have fun in life, right?



(my apologies to the admin for the use of that one bad word, I could not figure a way around it)

[jontz]

Originally Posted by Phoenix32

Oh and by the way, yes, I am a smart ass.

LOL, gotta have fun in life, right?

Hey, if you weren't a smart ass I probably wouldn't like you

[re3dyb0y]

Originally Posted by Phoenix32

(my apologies to the admin for the use of that one bad word, I could not figure a way around it)

Joe doesn't mind

He'd filter it if he did

[jontz]

So what we have ended up with in this thread is....upgrade the power supply or get yourself some low power drives. We have overlooked one valuable avenue to solve the problem:

Why not wire up one of those little 12V bicycle generators, attatched to a stationary bike, to the power supply? Just peddle your butt off until the drives spin up! Problem solved, and it also helps with the sometimes sedentary lifestyle of the techie/geek.

[blue68f100]

This thread is gone down hill, but it has been fun.

There nothing says you can't use a external power power supply. As long as the grounds are tied together for reference it should work. Put all of the drives off the external.

Buy a 8 bay drive unit w/ps then you expand the unit with a 2 drive ribbon connector. Expanding the raid 5 array up to 7-8 drives ( 1 spare). 7 x 300 = 2.1 T - overhead may be 2 tera bytes.

How sweet would this be .........

[Parity Error]

Originally Posted by jontz

So what we have ended up with in this thread is....upgrade the power supply or get yourself some low power drives.

Yes, this is the right approach. Attempts to limit the starting surge will not solve the bigger issue that you really do NOT want to screw around with marginal power supplies, period. Murphy will bite you on the *ss for doing so.