Need value of resistors connected to f16-100hip bios chip

[McKay420]

Well long story short, while attempting to replace the bios chip on my nv53a laptop, I accidentally ripped off 2 pads. They seemed easy enough to just drag a line of solder from the pin directly to the nearby resistor, since I could tell they were connected. My main problem is that I got the resistors too hot and they must have stuck to the soldering iron and fell off somewhere. It’s chip footprint U27 and I need the values of the resistors connected to pins 5,6, and 7. Again, the chip is f16-100hip. The component connected to pin 5 MAY have been a capacitor but I’m not very sure. Thanks

 

[Larry Sabo]

What's the make and model of the computer?

 

[McKay420]

It’s a Gateway NV53A. There were a couple other letters or numbers after the NV53A, but I won’t be able to look until tomorrow. The resistors should be the same either way.

 

[Moltuae]

F16-100HIP appears to be a package marking for an EN25F16 serial flash memory IC.
100HIP is just the spec/variant info (100HIP = 100Mhz, 8pin, Industrial, RoHS compliant).

Data sheet here: http://www.dzsc.com/uploadfile/company/123460/201226164345523.pdf

So, on the 8 pin package, pins 5, 6 and 7 are Data In, Clock and Hold respectively -- effectively they're all data-type inputs. In which case, any resistors connected to those pins will almost certainly be weak pull-up/down resistors. To confirm, you could check that the other side of the resistors went to VCC or VSS. It's a pretty safe bet they're just pull-up/down resistors though because series resistors would almost never be used between data connections, except in a few rare exceptions to act as voltage dividers where it has been necessary to address a voltage mismatch (very unlikely on a motherboard I think).

There's also a small possibility that the 'resistors' were in fact bypass capacitors (very unlikely). But if they were, the value would've been low (picofarads or low nanofarads), meaning that they were merely precautionary and that in most cases the circuit will work fine without them. Ether way, replacing them with weak pull-up/down resistors won't cause any damage or have any detrimental effect on the circuit.

Assuming they are just weak pull-up/down resistors, typical values will be in the range of 4K7 to 10K. An exact value won't be important, especially not at the relatively low data rates involved here. However, you might be able to find out what value was originally fitted by following the other inputs (CS# and WP#) which may (or may not, depending on this particular operation/configuration) have weak pull-up/down resistors fitted too.

 

[Larry Sabo]

Here you go...

 

[McKay420]

Ok thank you both for the replies. So from the data I gathered earlier and what you’re telling me, 4.7kohm should work for all three pins. I will report back tonight and let you know how it goes. Thanks again

 

[Moltuae]

4K7 will be fine for a pull-up resistor (such as R480/482 in the schematic Larry posted). Is that part of the NV53A motherboard schematic @Larry Sabo?

However, since the schematic above suggests R481 is in series with pin 6, I would check first to see if the other side of the resistors go to VCC/VSS. If the schematic is correct, R481 appears to be a zero ohm link in series with the SPI clock signal (ie, it does not connect to a power rail).

 

[Larry Sabo]

Is that part of the NV53A motherboard schematic @Larry Sabo?

Yes, @Moltuae, found here.

 

[McKay420]

The chip in Larry’s diagram appears to be different from the one on my board. The pin layout is different and the chip markings. The U27 is the same though.

If larry’s diagram is correct and that is a zero ohm resistor. Is it acceptable to just jumper the pin to the far resistor pad?

 

[McKay420]

Ahh I see now that the pins are are scrambled, so I’m guessing that’s just how schematics are laid out. Ok so pin 7 can have a 4.7kohm, pin 6 is zero ohm/jumper, but I am still not clear on what pin 5 would be. It looks like there should be no component there, but there definitely was either a resistor or a capacitor. I found one component on my table that almost looks tan like a capacitor, but I remember it as a resistor when it was on the board.

 

[Moltuae]

If the schematic is correct, pins 3, 6 and 7 are connected to resistors (rather than 5, 6 and 7). Are you sure about pin 5?

(pin 8 is also connected to resistor R479, although that's a slightly larger 0603 package, according to the schematic).

It's normal/common for pin layouts to differ on schematics for convenience so you can ignore that. It's only the pin numbers that matter, not the order.

And yes, you could replace the zero ohm resistor with a link/jumper. Personally I would replace it with the correct size (0402) zero ohm resistor because it would be a neater repair, plus they're very inexpensive.

Have you tried checking the resistance of the component you found?

 

[McKay420]

If the schematic is correct, pins 3, 6 and 7 are connected to resistors (rather than 5, 6 and 7). Are you sure about pin 5?

(pin 8 is also connected to resistor R479, although that's a slightly larger 0603 package, according to the schematic).

It's normal/common for pin layouts to differ on schematics for convenience so you can ignore that. It's only the pin numbers that matter, not the order.

And yes, you could replace the zero ohm resistor with a link/jumper. Personally I would replace it with the correct size (0402) zero ohm resistor because it would be a neater repair, plus they're very inexpensive.

Have you tried checking the resistance of the component you found?

I have checked the resistance and it was in the megaohm range so I’m assuming that was the zero ohm. Having a neater repair is pretty much impossible at this point but I did find some zero ohm resistors. The resistors I have are one size bigger, but I would assume that a higher wattage is ok as long as the resistance is the same. I’ve included the only pic I have of the chip before starting the repair. I’ve labeled the three pins I’m talking about along with ohm ratings for 6 and 7. Let me know if this looks right. I already have the 4.7kohm soldered.

 

[McKay420]

Here is a pic of where I’m at now. Pretty embarrassing to post this, but it may help clarify. The resistor is tipped on its side because it was easier to hold with tweezers and will not short out as easily with neighboring resistors. I have checked my soldering connections with a multimeter and it checks out.

 

[Moltuae]

A larger, higher wattage resistor is ok, as long as you can make it fit but a resistor of the correct size (0402, according to the schematic) would make the job easier and look a lot more professional. It'll cost pennies to buy the right size resistors so it's not worth struggling.

If the component you measured had a very high (>10Mohm) resistance, it's unlikely it was a zero ohm link. More likely it's a capacitor or a faulty/broken resistor.

It's difficult to make out what's connected to what looking at those photos but it looks like at least two of the missing components connect directly to pins 5 and 6. Did you check what the other side connects to and if it compares with the schematic? If the other side connects to VSS or VCC I would NOT connect a zero ohm resistor or link. The safest thing to try in that case would be 4K7 resistors.

 

[Larry Sabo]

I am still not clear on what pin 5 would be. It looks like there should be no component there, but there definitely was either a resistor or a capacitor.

If you go to page 28 of the schematic, it shows an option resistor mod inline with pin 5, IIRC.

 

[McKay420]

I’m not seeing this mod you are talking about. I’m not very good with schematics yet, but on page 28, it seems that pin 5 has nothing going on. I think you may be referring to pin 6. Is that a 10 ohm resistor that it shows?

 

[McKay420]

It seems as though the other side of the resistors do not go to VCC or VSS, but I’m not very confident in my conclusion. I figured I should try the 4.7kohm first. What will the computer behave like if something is wrong? Is it possible that the programming could be wiped? I don’t know how to program. Had to order on ebay from Sweden.

 

[Larry Sabo]

I’m not seeing this mod you are talking about. I’m not very good with schematics yet, but on page 28, it seems that pin 5 has nothing going on. I think you may be referring to pin 6. Is that a 10 ohm resistor that it shows?

Yes, my error. R4919 is the one I was referring to but I'm not sure of it's value, nor the value of R481 for that matter.

 

[McKay420]

10_0402_5% seems to be a 10ohm resistor as far as google tells me. I’m not aware of any 10ohm resistors that small though. Is it likely that they just forgot to add a k after the 10?

Edit: I see now that 10 ohm resistors in that size do exist. Isn’t that practically a zero ohm resistor? I read that zero ohms can technically be up to 50ohms. I probably do have the proper ones though. Just curious. Do I dare try a 10ohm on pin 6 on my first attempt?

 

[Moltuae]

10 ohms (and much lower) are common in all sizes. In the circuit above the resistor forms part of a simple series RC (resistor/capacitor) circuit, which is effectively acting as a low-pass filter (ie it passes low frequencies and blocks high frequencies).

At DC (zero Hz) and low frequencies, the capacitor has little effect. The resistor, being a relatively low value (relative to the low-current/high-impedance data circuit that is) is able to deliver more current than the circuit demands so (in accordance with Ohm's law) there is negligible change in voltage. You can think of the capacitor a bit like a variable resistor, the value of which decreases with frequency. So, as the frequency increases the capacitor load becomes greater. At some point, as the frequency increases (you'd have to do the maths to work out precisely when), the effective resistance (impedance) of the capacitor becomes great enough to have a significant effect, providing a low resistance path to ground and therefore reducing the voltage of the signal.

It's likely that the low-pass filter circuit above is largely precautionary, to ensure that any high frequency interference (from external noise or much faster neighbouring data) is removed. As such, replacing the 10 ohm resistor with a link may have no noticeable affect but it's impossible to predict that noise won't get through in some circumstances, which could result in weird operation and instability if/when it does.

The flash memory uses a Serial Peripheral Interface (SPI), which is a relatively old and slow interface (by modern standards), commonly used for IC-to-IC communication. Most of the resistors you are missing appear to be connected to the SPI pins. If a low value resistor is required and you incorrectly substitute with a high value resistor, the most likely outcome will simply be no SPI communication. However, if a high value resistor is required (such as a VCC/VSS pull up/down resistor) and you substitute it with a low value resistor or link, an IC may be internally damaged when it attempts to drive the data line in the opposite direction.