I have a dead HDD I am trying to fix for a friend. I have a replacement PCB to see if that sorts it but it involves desoldering the ROM chip from the original board and soldering it on to the new board.
When I was taking the ROM chip off the new board I inadvertently took off a nearby component.
I have uploaded a couple of pictures here
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One photo shows the new board with the original ROM soldered in place and a red arrow showing where the missing component is. The other photo shows the original board with the ROM removed but you can see the component that is missing on the new one - it looks like a black bar.
I have been in discussion with the PCB supplier who specialises in these boards as well as fixing them. He said that this component is not critical and I should try the PCB without it.
Couple of questions
Any idea what the component is? A resistor?
On the new board you can see that the pads the component sits on are now not connected as it would be through the missing component. If the component is not critical, my guess would have been that I at least need to connect the pads together?
Another option is to take the one off the original board and use that. Is that viable with a regular soldering iron? Also, depending on what it is I guess it could have a particular orientation which adds more complexity
Also, any view from the experts on what best to do?
Thanks in advance
Lee.
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alan_m
The photo is of such poor resolution that it it is impossible to see the component, the tracks to it or any of the legend in the IC to which it is connected.
The component may be a power supply decoupling capacitor so on no account short out the pads. If it is a capacitor then i) it probably will not make any difference to the operation of the circuit ii) shorting the pads may short one of the power supplies and destroy other components on the board.
T
T i m
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As has been mentioned I don't think there is sufficient detail to be sure but if it's only two legs it should only be a cap, resistor or diode. ;-)
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Probably definitely not. ;-)
Why don't you (try to) measure the value of the one you still have. If you are lucky it will be a resistor and read a preferred value, like
1k, 4.7k etc? If you have a DMM, do it on the std resistance range, not the diode test. If you measure both ways round, get the same reading of a sensible value, the chances are it was a resistor. If you just get a pulse as you go either way round, it's more likely a cap and if the values differ each way, it might be a diode (but could also be the circuitry around it).
You sometimes have to be careful measuring stuff in circuit, less you damage anything (sensitive circuitry) around it.
One with a reasonable fine tip yes. If your eyes are good enough, you can sometimes put some flux on the area and (quickly) flood the component with solder (so it bridges it) and then you can (quickly) lift it off with some fine tweezers.
Yeah, not generally an issue re which way round if it's a resistor but might be if it's a cap and deffo a diode. ;-)
We will have to wait and see. ;-)
Cheers, T i m
T
Theo
I had to delete the -- from the URL, but by the looks of it that's the firmware flash chip for the processor which has the thermal pad on it (which obscures its part number).
Assuming the flash is the standard SPI NOR flash pinout, that would be pin 7 which is HOLD# - see eg:
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The other end disappears in a via so hard to know where it goes, but that file suggests a 10K pullup resistor, which would fit your board if the the resistor's via goes to VCC (usually an inner board plane so no tracks visible on the other side).
Theo
D
Dave W
Usually on such boards resistors have values marked but caps don't. Your component is certainly not a link as there is no track underneath that requires a link to bridge it. It looks like the same kind of cap as is across the narrow pair of tracks under your arrow that go into the corner of the square IC. Probably you could use that to replace the missing one.
F
Fredxx
The pinout seems consistent with a SPI Serial Flash memory where pin 3 is a nWP or write protect pin.
There might be a internal block protection bits that is used in conjuction with this pin.
If the device is pre-programmed then the level shouldn't matter, otherwise the pin has to be held high and I would expect the resistor is passively holding the pin low to prevent inadvertent programming.
Without more details of the device the above is an educated guess.
P
Paul
It looks to me like this. Your second picture shows 3 is joined to 7, so /WP is joined to /HOLD, and it's like the functions are not being used. They would be going to a pullup to put them in the negated (not used) state.
This is just to show the basic idea.
+-----+----- 10K ---- 3.3V | | 3 7 /WP /HOLD I/O I/O <=== outputs in quad SPI mode, so only want vanilla SPI operation while wired this way.
I can't find an exact match for the part number on Winbond, so this is close to it but not the precise item.
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What I seem to be seeing in there, is no definition of any built-in pullup on those pins. Which is passing strange. Normally, chips have a weak function for pins, such that no resistors are needed externally. It's unusual for any modern chip to be completely devoid of a function like that. Tying the pin to the deasserted state, would be a fairly common practice.
Typically, people seem to remove some ROM from controller boards, and transfer it from the old board. There's some kind of check that the controller and disk are matched somehow. Forcing people who do controller swaps, to bring over the old chip. Now, if the chip was dead, I don't know what they're supposed to do with regard to recovery. Each SPI flash has a custom (Winbond factory) written serial number, so it is possible for a check to be done.
The code in the flash, isn't the whole code load. The service area has code too. The code in the flash is sufficient to move the arm over to the service area and read the rest of the code content. It's something like that.
If you copied code from one flash chip to another, the serial number of the second chip would still be wrong. It wasn't always necessary to swap those chips, but it's become a "thing" on the more modern drives.
This is information I've been able to glean from a couple data recovery forums on the web. They don't come right out and explain stuff, so you have to collect enough tidbits from what they say, to spot a pattern.
Not all the controller boards are recovered from dead drives. Some of the controller boards are available as brand new items. But that still does not remove the need to swap SPI, if that's what the design era calls for. The 4GB WDC I have here from the year 2000, would not need that. A straight up controller swap would work there. The controller board is likely around 6 years old (just a very rough guess).
Paul
A
Andy Burns
It's not unusual for zero ohm resistors to be used as links, but the photo isn't good enough to make out any markings, a camera with a macro lens and some strong lighting might help ...
B
Brian Gaff (Sofa
No don't short it, it might be a capacitor. I hate surface mount. Brian
C
Chris Green
I've always found "zero ohm resistors" a rather wierd concept, surely that's either a link or just a piece of wire. Why call it a 'resistor' when it's not?
P
Paul
It's a surface-mountable wire, which fits into the whole manufacturing process. It can be placed at high speed, and doesn't slow down the pick-and-place when used. The SMT item can be affixed with a glue dot like any other resistor, and won't fall off going through the line. What's not to like ? You put a reel of 3500 of those on the machine, press the button and off it goes.
Generally you try to use those, in places where it won't affect circuit performance. It can be used on a Bill Of Materials (BOM) to control the addition or subtraction of optional circuit blocks.
And just because it's zero ohms, you don't go around running 20 amperes through it on purpose :-/ Although some of my colleagues have probably thought about doing that. Because they are zero ohms and its magical. They're like tiny superconductors, right ? I can think of at least one engineer who would fall for that. And no, I've never ohmed one out to see what its real resistance is. I'm sure there's an estimate around somewhere.
As another example of a function, I could put a zero ohm today, and if I get in the lab and discover a logic signal needs series damping with a 33 ohm, change the BOM to read
33 ohms, and then when manufacturing starts, I look like a genius. You'll find all sorts of lands on PCBs that are optional, and are a form of "hedging" against unexpected surprises. You'll find people putting extra lands on boards, even without thinking about it (like, they've copied a chunk of design from some place else, complete with options they don't understand).
You can even do zero ohms like this, but manufacturing is going to be phoning you if they spot stuff like this. The rules set for PCB design, is several feet thick. I'm sure that breaks some rule or rules. You can do stuff like this if the land is square, and rotating a component is footprint compatible.
If in the BOM, I set R2 to zero ohms, then my signal is going off to destination #2. And then I depop R1 so that leg is open circuit. Voila, circuit change after schematic capture is finished. I can put R1 or R2 in the BOM, but not both.
The opportunities for abuse are endless. Especially if the engineer doesn't document what all that crap is for.
On things like computer motherboards, you'd remove stuff like that for the third and final design cycle, before high volume manufacturing starts. Because a zero ohm resistor costs money and "it has to go". But on "material cost is no object" circuit boards, where you're only making 300 boards a year, you'll see stuff like that. Like, I make a logic board with $10K worth of logic chips, and I charge the customer $100K for it. The cost of a zero ohm really doesn't matter then.
Paul
T
T i m
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This is often a problem with 'technical people' in general.
1) They are often of a mindset that has led them to be in that field.
2) Because they are often immersed in that field they forget how 'most other people' wouldn't have a clue when stuff is seen in abstract.
3) Because they are in that field and of that mindset they aren't good at writing user manuals for others not like them. ;-)
All this was reinforced when I moved from a lifelong role in 'IT Support' to 7 years as an IT Trainer, seeing how many people hadn't really used a screwdriver or a mouse, let alone understood static electricity or binary.
Where this appears at it's worst (the assumptions re baseline understanding) is websites offering fairly technical solutions where unless you knew what the software did, you wouldn't be able to work it out from the site.
It tells you that V3.5 is now available and that it does XYZ better than the old version but with no mention of what it does, unless you knew what all the acronyms meant. DAW, OBD etc.
This is why I can rarely make any practical progress with Linux related things.
With Windows / OSX you generally download a 'Setup' file of some sort, run it and away you go. They might have included some installation instructions because there are likely to be millions of users.
For Linux the chances are there will be fewer users and (and 'what Linux' exactly) so you are often left installing / building / finding / installing dependencies yourself and the only chance of me being able to do that with any hope of success is pretty low. Unless I can find a walkthrough, written by someone similar to me who actually covers *all* the steps in a form that someone not into the details can use, even if they don't understand.
For many the point is the end, not the means, as really should be the case for an ordinary user and an OS.
Cheers, T i m
F
Fredxx
I'm now a fan of surface mount. Far easier to remove a component off without taking the lands with it. Multi-pin components come off with a hot air gun.
What I do hate is BGAs.
L
leen...
Hi all,
I have taken a few more photos with our camera (had to find it and charge it first as haven't used in a couple of years). They are in the same location as the others
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Trying to get a photo of what is written on the chip is next to impossible. So best I can make out the following is written on the top of the 2 chips Chip 1 -
25FS406 06 2PHON
Chip 2 - Winbond
25040BW607 1244
In terms of the circuit board. As has been mentioned above, it looks like it goes pin 7 -> missing component -> pin 3 of same chip pin 7 -> hole in circuit board but nothing the other side.
There is no writing on the missing component that I can see so I assume it is a resistor? Checking the resistance with a DMM I get 1880 in both directions.
Based on this and from what I understand from the various replies, it sounds like this is to ensure the chip doesn't go into reprogram mode and therefore safe enough for me to test the board without the component. Is this correct?
Thanks again for all your help
Lee.
T
T i m
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Or a cap and so you are measuring 'resistance' of whatever else appears across those two connections. The only true way to tell would be to disconnect that component and measure it out of circuit. Or, all other things being equal, measure across where the component was on the other board and see what that reads (may not be possible if you have removed associated components).
That may confirm it's not a diode at least, and that it's not a link. ;-)
Cheers, T i m
L
Lee Nowell
Aha yes good point. With the chip removed, those pads read 1.7M is one direction and 2.4M is the other. Not that helps us though?
F
Fredxx
Which is a Sanyo Serial Flash memory as suggested:
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On Semi bought Sanyo and not sure how to access the old Sanyo parts directly.
Which I haven't looked up but assume it is a volatile memory.
Which does suggest a 1.8k resistor. Possibly a bit lower in value than I would have used and higher tolerance. 1% resistors are the norm. It costs more to put them on a PCB than the actual cost of the component.
What it does do is ensure the default level is write-protect, so that if there is a corruption in the micro the memory won't be overwritten.
Writing to a Flash memory is non-trivial, with specific codes being sent to enable writing, even when the nWP pin is high.
I would say it is normally safe, but would be safer with the resistor fitted.
T
T i m
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Well, a bit, in that 'nothing else' (other than the component or chip) is impacting your resistance reading to the 2k level. Although that may have been know by those who have checked out the chip and looked at the tracks and concluded it goes nowhere else (other than the chip) that has a 2K resistance between those points.
Another way could be to cut a track to isolate that component in a way that it would be easy to reinstate afterwards (easier than removing and replacing the component, other than you then have one to attach to the other board ;-)
Cheers, T i m
P
Paul
One side of the *resistor* should go to the VCC rail (as long as the logic type is still CMOS/TTL like). That's how you can get some idea of the function, is to see what the <unknown component> is joined to. Placing a cap on something whose state is steered by leakage, isn't going to be much of an improvement. Whereas a resistor is a cinch for the job (of defining state in a safe way).
+-----+----- 10K ---- 3.3V | | 3 7 /WP /HOLD I/O I/O <=== outputs could go quad SPI mode, so only want vanilla SPI operation while wired this way. Otherwise, I/O would fight with one another during quad SPI read mode.
Some really high speed diff interconnect are AC coupled, so as to not upset the DC bias on the receiver. And the coding of the signals can be DC balanced dynamically (the link encoder tracks DC balance so that an "ideal" signal is seen at the other end).
But around a chip like that one, there's no reason for super-high I/O speeds.
25FS406
25040BW607
*******
25FS40 is a 512KB flash-able code chip. Someone was taken advantage of by a data recovery firm here (chip swap into another board by the looks of it). You can see a resistor next to the 8 pin DIP, for pin 3 and 7.
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The 25Q40BW datasheet is here (the zero above is a Q). Note the extremely low operating VCC voltage. Which means you'd have to check the datasheet carefully for levels allowed on the outside.
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When ohming the resistor, if present, you'd want to use a meter with "low power ohms" (I'm saying this in the general-practices sense - this has nothing to do with VCC value). This means, a multimeter which when placed on the ohms scale, applies no more than
1V max to the circuit. This is intended to avoid causing semiconductors to start conducting. My old analog multimeter for example, applies up to 9V (from its 9V battery) to the circuit. You can either consult the multimeter manual for details/specs of the ohms function. Or, you can connect two harbor freight multimeters red to red, black to black, place one on ohms, one on volts, and discover the open circuit voltage specs that way.
Now that a 1.65V chip has entered the arena, I haven't a clue what they'd tie that signal pair to. Certainly you could ohm from one end of the resistor, to the VCC pin on the flash, and identify that the hot end of the resistor is tied to the most positive supply on the flash chip. I think that one end of the resistor goes to pin 8.
The chip leakage is a couple microamps (leakage goes worst case at high temp), and the pullup used needs to be "strong enough" to override that. The resistance value is immaterial as long as its low enough - only the leakage current will be flowing in the resistor. Resistance times leakage-current-value gives voltage drop across it, and the voltage cannot be so large as to not allow the input signal to be recognized as a logic level.
2ua * 10K = 2e-6 * 1e4 = 2e-2 = 20mV, and is only a fraction of 1650mV.
You might be able to use a 100K resistor, but it would be a pointless application of such, when 10K swamps it nicely.
Paul
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Lee Nowell
Thanks very much Paul the first link is talking about exactly the same HDD ?? To be honest you have gone a bit more technical than I can follow sorry - not very experienced with this sort of thing . My goal for the drive is just to get it up and running long enough to get the data off it If this link is about switching the chip into program mode given it has already been programmed, am I safe enough to give it a try to see if the drive is detected / works or am I likely to cause further issue to the PCB / chip?
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