Keithley 197 Repair

[staze]

Not unless something very weird happened. =P

That said, I haven't tried checking to make sure the TLC271CP I replaced (U107) isn't bad. *shrugs*

[staze]

Anything connected to the -9V / -6.4V rail may act dodgy as long as that rail is out of spec. You can either cut pins / tracks to isolate sections, or try to figure out where the current is going. To pull up the negative rail, a fair amount of current needs to be flowing from a more positive rail. Maybe you can find this point? The more positive the negative rail, the closer you are to the culprit. Of course if the negative rail is positive, ESD protection diodes may start conducting and make life interesting. I would consider powering the -V or -9V rail from a lab power supply again (make sure you disconnect the zener). This will allow more current to flow. You may then be able to use voltage drop (the more positive the closer to the culprit) to find the culprit. If only you had a high-resolution DMM . It might be something stupid like a short somewhere on the PCB.

If the -9 V rail is that unstable, then either something's drawing a serious amount of current (does any part feel hot?), or the -9V rail has a high impedance. Does the same happen to the -6.4V rail? Labeling the circuit around U107/Q128 (where these rails are generated) with voltages and studying that may provide some clues. Resistors are convenient gauges to indicate current direction and quantity. Also check with a scope: is anything oscillating? What's the cause and what's just an effect? Any way to verify the operation of the -9V / -6.4V circuit without the load connected?

Okay, in a position to work on this a bit more, and a couple questions.

1. When you say power the -V or -9V rail via a PS, and to disconnect the zener, I would assume you mean disconnect the cathode, then hook the PS up to the circuit referenced to gnd (which is easy on the 197 since they're all a shared ground). So for the -V rail, try disconnect VR105, or with the -9V, disconnect VR101? What should I limit the current to?
2. Measuring the voltage relative, as in, where it's the most positive, I would measure that relative to the cathode of the zener for that rail, yes?

So, for example, let's say I disconnect the -9V zener, hook the negative rail of my PS to that spot, positive size to ground. That should leave the -9V and -6.4V running (though I'm still confused as to what U107 and Q128 are going to do in that case, since they're getting power from -V). Then, clip the positive probe to the negative rail of the PS, and go and probe all the -9V rail, and see where things get higher than at the PS lead. Highest spot should be my problem.

Or if I wanted the -6.4V rail, just disconnect VR102, and do the same.

My biggest question is, what current do I limit the PSU to (though, I suppose if I DON'T limit it, I could do the whole "smoke it out" method, though my PSU only does 2A, which might not be enough to smoke anything)?

One question I have is the bridge rectifier is measuring +15V on one side, and -19V on the other. Is that "normal"? I would think the zeners for the rails would pull both down to ±15V.

Okay, think that's it for now. I'm going to try doing some probing tonight, but it all depends on my son's sleep patterns. =)

[alm]

1. When you say power the -V or -9V rail via a PS, and to disconnect the zener, I would assume you mean disconnect the cathode, then hook the PS up to the circuit referenced to gnd (which is easy on the 197 since they're all a shared ground). So for the -V rail, try disconnect VR105, or with the -9V, disconnect VR101? What should I limit the current to?

For V- I would disconnect VR105 and R132, for -9V disconnect VR101 and Q128. Incrementally increase the current, monitor the voltage and check if anything gets hot. If anything gets hot, then you've likely found a/the culprit. A thermal imager would also be very helpful for this.

2. Measuring the voltage relative, as in, where it's the most positive, I would measure that relative to the cathode of the zener for that rail, yes?

I would measure relative to ground, although since you're only looking for relative differences, it doesn't really matter as long both are stable. Note that I would try measuring voltages before going through the trouble of disconnecting rails. Maybe the voltage drops are already large enough with the current current (no pun intended). Especially if you could use some sort of 5.5 digit DMM .

So, for example, let's say I disconnect the -9V zener, hook the negative rail of my PS to that spot, positive size to ground. That should leave the -9V and -6.4V running (though I'm still confused as to what U107 and Q128 are going to do in that case, since they're getting power from -V). Then, clip the positive probe to the negative rail of the PS, and go and probe all the -9V rail, and see where things get higher than at the PS lead. Highest spot should be my problem.

Yes, you should also disconnect Q128. Don't really care about U107 at that point.

Or if I wanted the -6.4V rail, just disconnect VR102, and do the same.

I would also disconnect R125J. I would probably start with the -6.4 V rail and work my way up. It may very well be that the -9 V rail recovers if you power the -6.4 V rail seperately, then you've localized the problem to the -6.4 V rail.

My biggest question is, what current do I limit the PSU to (though, I suppose if I DON'T limit it, I could do the whole "smoke it out" method, though my PSU only does 2A, which might not be enough to smoke anything)?

The only thing I would be worried about killing are PCB traces. You're limiting the voltage to that nominal for that rail, so no component should damaged. Anything drawing enough current to be damaged is likely already dead anyhow. But based on the schematic I would expect the current draw to be very low: all current for the negative rails has to flow through R132, a 330 Ohm resistor, so they are unlikely to be designed to draw more than maybe 20 mA. You can measure the voltage across R132 to calculate the current it's drawing now. I certainly wouldn't go up to 2A, I expect things to get hot well below that. Something like 100 mA would be unlikely to damage any PCB traces, but would produce a ridiculous voltage drop of 33 V across R132.

One question I have is the bridge rectifier is measuring +15V on one side, and -19V on the other. Is that "normal"? I would think the zeners for the rails would pull both down to ±15V.

VR104 will pull the voltage on its side of R131 to +15V, and will drop the rest of the voltage across R131. So I would expect the voltage at the rectifier output to be a little higher that +15V. I would expect at least 100 mV or so drop across R131, possibly with an AC component.

[staze]

1. When you say power the -V or -9V rail via a PS, and to disconnect the zener, I would assume you mean disconnect the cathode, then hook the PS up to the circuit referenced to gnd (which is easy on the 197 since they're all a shared ground). So for the -V rail, try disconnect VR105, or with the -9V, disconnect VR101? What should I limit the current to?

For V- I would disconnect VR105 and R132, for -9V disconnect VR101 and Q128. Incrementally increase the current, monitor the voltage and check if anything gets hot. If anything gets hot, then you've likely found a/the culprit. A thermal imager would also be very helpful for this.

A thermal imager would be very helpful for many things. =P

2. Measuring the voltage relative, as in, where it's the most positive, I would measure that relative to the cathode of the zener for that rail, yes?

I would measure relative to ground, although since you're only looking for relative differences, it doesn't really matter as long both are stable. Note that I would try measuring voltages before going through the trouble of disconnecting rails. Maybe the voltage drops are already large enough with the current current (no pun intended). Especially if you could use some sort of 5.5 digit DMM .

Yeah, figured I'd start that way. =)

So, for example, let's say I disconnect the -9V zener, hook the negative rail of my PS to that spot, positive size to ground. That should leave the -9V and -6.4V running (though I'm still confused as to what U107 and Q128 are going to do in that case, since they're getting power from -V). Then, clip the positive probe to the negative rail of the PS, and go and probe all the -9V rail, and see where things get higher than at the PS lead. Highest spot should be my problem.

Yes, you should also disconnect Q128. Don't really care about U107 at that point.

Good point.

Or if I wanted the -6.4V rail, just disconnect VR102, and do the same.

I would also disconnect R125J. I would probably start with the -6.4 V rail and work my way up. It may very well be that the -9 V rail recovers if you power the -6.4 V rail seperately, then you've localized the problem to the -6.4 V rail.

That becomes a bitch, if I have to pull that resistor network.

My biggest question is, what current do I limit the PSU to (though, I suppose if I DON'T limit it, I could do the whole "smoke it out" method, though my PSU only does 2A, which might not be enough to smoke anything)?

The only thing I would be worried about killing are PCB traces. You're limiting the voltage to that nominal for that rail, so no component should damaged. Anything drawing enough current to be damaged is likely already dead anyhow. But based on the schematic I would expect the current draw to be very low: all current for the negative rails has to flow through R132, a 330 Ohm resistor, so they are unlikely to be designed to draw more than maybe 20 mA. You can measure the voltage across R132 to calculate the current it's drawing now. I certainly wouldn't go up to 2A, I expect things to get hot well below that. Something like 100 mA would be unlikely to damage any PCB traces, but would produce a ridiculous voltage drop of 33 V across R132.

okay, I think.

One question I have is the bridge rectifier is measuring +15V on one side, and -19V on the other. Is that "normal"? I would think the zeners for the rails would pull both down to ±15V.

VR104 will pull the voltage on its side of R131 to +15V, and will drop the rest of the voltage across R131. So I would expect the voltage at the rectifier output to be a little higher that +15V. I would expect at least 100 mV or so drop across R131, possibly with an AC component.

Okay, but what about the negative side of the rectifier producing that -19V. Is that just because the -V rail is running at -12V vs -15V?


okay, so, I'll start with just looking at voltages on the -6.4V with everything still connected. -6.4V isn't connected to much (U101, R117C). Though, like you said, since that rail is completely reversed at this point, things are going to be conducting weird. Pretty sure U101 is going to be where that voltage is higher (unless I'm missing somehting else that aril is attached to).

-9V should be easier. Though, now that I think about it, the closer to the issue I look, shouldn't the -9V be MORE negative rather than positive? So if at the zener it's -4.8V or whatever, at the source of the issue, it might be down at -5V (or whatever)? There are quite a few things on the -9V rail.

I'm really hoping it's U103, U112, or U109, etc. Some jellybean part. I was looking at options for replacing something like U110, and having a hell of a time finding something that would work (main issue being slewrate). *sigh*

I'll poke at things tonight. Try to keep good notes.

[alm]

Okay, but what about the negative side of the rectifier producing that -19V. Is that just because the -V rail is running at -12V vs -15V?

The transformer is producing a higher voltage than the regulated voltage, and the excess is dropped across R132, that's how a zener regulator works. As long as the voltage is high enough so enough current can flow through R132 to keep VR105 conducting and the rest of the circuit powered, you don't care. The magnitude of the voltage is likely higher because the load is much lower (compare R131 to R132).

okay, so, I'll start with just looking at voltages on the -6.4V with everything still connected. -6.4V isn't connected to much (U101, R117C). Though, like you said, since that rail is completely reversed at this point, things are going to be conducting weird. Pretty sure U101 is going to be where that voltage is higher (unless I'm missing somehting else that aril is attached to).

-9V should be easier. Though, now that I think about it, the closer to the issue I look, shouldn't the -9V be MORE negative rather than positive? So if at the zener it's -4.8V or whatever, at the source of the issue, it might be down at -5V (or whatever)? There are quite a few things on the -9V rail.

-6.4 V is the reference for the -9 V rail, so if the -6.4 V rail is at +6.2 V, then Q128 is not conducting, and both the -6.4 V and -9 V rails will be off. Something is still pulling current from the -9 V rail, so VR101 is conducting and is dropping about 8 V, resulting in a -9 V rail that's 8 V above the V- rail (which is also overloaded through VR101). Of course with -4.8 V across the series combination of R125J and VR102 will not make VR102 conduct, so that rail is essentially floating. Most likely some current is flowing from the devices that are connected to this rail through R125J, dropping -4.8 - 6.2 V across R125J (about 1 mA). So it seems that something is loading down the -9 V rail by feeding current into the -9 V rail from a more positive rail (ground or one of the positive rails). The closer you get this something, the closer you get to the positive rail, and the more positive the voltage will be. After this analysis, I think you can ignore the -6.4 V rail. It's likely one of the many devices on the -9 V rail that's the culprit.

The idea behind using a sensitive multimeter and forcing more current through it is that PCB traces will drop more voltage the more current flows through them. So you are using the PCB traces as current shunts.

Did you already make a thermal image with the Mk. I finger to see if anything gets hot?

[staze]

okay, so, I'll start with just looking at voltages on the -6.4V with everything still connected. -6.4V isn't connected to much (U101, R117C). Though, like you said, since that rail is completely reversed at this point, things are going to be conducting weird. Pretty sure U101 is going to be where that voltage is higher (unless I'm missing somehting else that aril is attached to).

-9V should be easier. Though, now that I think about it, the closer to the issue I look, shouldn't the -9V be MORE negative rather than positive? So if at the zener it's -4.8V or whatever, at the source of the issue, it might be down at -5V (or whatever)? There are quite a few things on the -9V rail.

-6.4 V is the reference for the -9 V rail, so if the -6.4 V rail is at +6.2 V, then Q128 is not conducting, and both the -6.4 V and -9 V rails will be off. Something is still pulling current from the -9 V rail, so VR101 is conducting and is dropping about 8 V, resulting in a -9 V rail that's 8 V above the V- rail (which is also overloaded through VR101). Of course with -4.8 V across the series combination of R125J and VR102 will not make VR102 conduct, so that rail is essentially floating. Most likely some current is flowing from the devices that are connected to this rail through R125J, dropping -4.8 - 6.2 V across R125J (about 1 mA). So it seems that something is loading down the -9 V rail by feeding current into the -9 V rail from a more positive rail (ground or one of the positive rails). The closer you get this something, the closer you get to the positive rail, and the more positive the voltage will be. After this analysis, I think you can ignore the -6.4 V rail. It's likely one of the many devices on the -9 V rail that's the culprit.

The idea behind using a sensitive multimeter and forcing more current through it is that PCB traces will drop more voltage the more current flows through them. So you are using the PCB traces as current shunts.

Did you already make a thermal image with the Mk. I finger to see if anything gets hot?

Okay, that makes way more sense. I thought it would be more negative since things go back to ground, but I guess since the offset is coming from the positive rail... =/

I measured around with my multimeter temp sensor, and the only things that were hot that I found were listed earlier (though, I was mainly focused on ICs and diodes).

So, should I just skip trying things as they are, disconnect VR101, and Q128, and feed the negative -9V rail with my PSU with like 50mA or so, and slowly ramp that up (to 100mA or so)? Or start how we are and just see where I get looking at voltages? I just printed out the schematic on 11x17 (A3) so I can easily mark voltages.

Yay having a 5.5 digit multimeter.

[alm]

The ground is also positive relative to the negative rail .

I would probably start measuring voltages. Start at the cathode of VR101, and see if you can measure a voltage drops across any of the traces originating from there. If they're all essentially zero, try disconnecting the zener/transistor and increasing the current. Current is now approximately 20 mA, so 50 mA sounds like a good starting point. Any part that gets hotter as you increase the current is suspect.

[staze]

The ground is also positive relative to the negative rail .

I would probably start measuring voltages. Start at the cathode of VR101, and see if you can measure a voltage drops across any of the traces originating from there. If they're all essentially zero, try disconnecting the zener/transistor and increasing the current. Current is now approximately 20 mA, so 50 mA sounds like a good starting point. Any part that gets hotter as you increase the current is suspect.

"part" being IC's, I would assume. I would imagine resistors are going to get hotter. =)

Does it make sense to measure at VR101 cathode, then hit all the IC's connected to that -9V rail? I would assume one should be higher. Then start looking around the trace between that and VR101 to make sure it's the highest point? Or am I just giving myself more trouble since I'm skipping resistors, and the like? Can I assume that the point of "leakage" is going to be substantially (1V or more) higher than the -4.8V that I'm seeing at the cathode of VR101, or are we thinking it's going to be rather minimal and the balance point just happens to be around that -4.8V?

[alm]

"part" being IC's, I would assume. I would imagine resistors are going to get hotter. =)

Technically yes, but I don't expect any of the resistors to get hot, since I don't expect any low value resistors across the -9 V rail.

Does it make sense to measure at VR101 cathode, then hit all the IC's connected to that -9V rail?

You can also probe the ICs. But keep the PCB traces in mind, since these are going to drop the voltage. Power pins on multiple ICs all connected to the same plane are going at the same potential. I haven't checked the PCB layout. Long skinny PCB traces make this much easier than fat traces and polygon fills.

I would assume one should be higher. Then start looking around the trace between that and VR101 to make sure it's the highest point? Or am I just giving myself more trouble since I'm skipping resistors, and the like? Can I assume that the point of "leakage" is going to be substantially (1V or more) higher than the -4.8V that I'm seeing at the cathode of VR101, or are we thinking it's going to be rather minimal and the balance point just happens to be around that -4.8V?

Even if that 20 mA is all flowing through the same part, then 1 V drop would require 50 ohm of resistance. No PCB trace is going to be 50 ohm. So you're most likely looking at millivolts, which is why I'm suggesting that you may need to increase the current.

[staze]

"part" being IC's, I would assume. I would imagine resistors are going to get hotter. =)

Technically yes, but I don't expect any of the resistors to get hot, since I don't expect any low value resistors across the -9 V rail.

Does it make sense to measure at VR101 cathode, then hit all the IC's connected to that -9V rail?

You can also probe the ICs. But keep the PCB traces in mind, since these are going to drop the voltage. Power pins on multiple ICs all connected to the same plane are going at the same potential. I haven't checked the PCB layout. Long skinny PCB traces make this much easier than fat traces and polygon fills.

I would assume one should be higher. Then start looking around the trace between that and VR101 to make sure it's the highest point? Or am I just giving myself more trouble since I'm skipping resistors, and the like? Can I assume that the point of "leakage" is going to be substantially (1V or more) higher than the -4.8V that I'm seeing at the cathode of VR101, or are we thinking it's going to be rather minimal and the balance point just happens to be around that -4.8V?

Even if that 20 mA is all flowing through the same part, then 1 V drop would require 50 ohm of resistance. No PCB trace is going to be 50 ohm. So you're most likely looking at millivolts, which is why I'm suggesting that you may need to increase the current.

Okay, that makes sense. It's a doubled sided pcb, so there's no power "plane". But it's not a star topology really, so I'm not going to see one leg then another. I guess I'm in for some tracing. Most of the traces are long and skinny, and snake from one part to the next, basically. I wish the component layout image was a bit higher detail... looks like of like a scan of a copy of a copy. The traces are pretty much illegible.

So you're saying U104 pin 4 is going to read the same as U103/112 pin 12, or U110 pin 4? I don't doubt you, I just don't understand how that could be the case. I would think the problem component would have it's -9V connection be up at say -4.75V, vs others being down from that. It should be the "high" spot. Or at least direct me to look at the traces in that area (since it could be a short, or something... though I would think a true short would do more than just bring it up to -4.8V).

Sorry, I know I'm asking all these questions without the board/meter in hand... I'm just trying to understand.

[alm]

There is no -9 V point, there is -4.8 at the cathode of VR101, and it only goes up from there, assuming my analysis is correct and Q128 is not conducting. If two pins are connected by a trace, than the voltage between those two pins follows from Ohm's law. So if it's 20 mA through a short, fat trace with less than 1 mohm resistance, then the voltage drop between those points is going to be less than 20 uV. A long skinny trace with a resistance of 0.1 ohm will create a much larger voltage drop, however.

From the PCB layout it looks like these parts are not that close together, so I think you may be able to measure a drop between those. Though maybe not without increasing the current.

[staze]

oh, one other question...

how long should I let the unit warm up before doing the measurements? just so things, hopefully, stabilize? I've been worried about leaving it on TOO long in it's current state thinking something hard to replace might get toasted...

[alm]

I don't think you need much of a warm up. You're mostly looking at voltage drops across traces, not absolute voltages. And parts that overheat will get hot within seconds.

[staze]

Okay, so, measured all the -9V egress points, and they all are basically -4.2V (some a bit higher, some lower). Hard to tell, becasue the bitch is, that rail is oscillating like crazy, so it's hard to get a good reading. So... going to have to power directly, I think. Guessing it's oscillating since the -V rail is as well. Doh!

The only thing definite I found is VR101, is pretty warm. After a couple minutes running, it was over 45C (everything else I measured except Q123, and R132 were down around 27C (a bit higher than the room temp (~25C)). That seem normal?

Once I get a chance, I'll pull VR101, and Q128, and try powering the -9V rail directly from my PSU at like 20mA or so, at first. See where that gets me. Hopefully I can do that tomorrow...

Thanks!

[staze]

Alrighty, progress, I think.

So, seeing things oscillate, I pulled Q128, and VR101 (anode side), and hooked up my bench PSU to the former location of Q128's emitter.

I set the PSU for 20mA, and fired everything up. PSU goes into current limit mode, and is only pumping about -1.8V. I have to crank it up to 40mA to get a solid -9V out of it. Hmmm. Though, PSU says only 30mA, but I'm guessing that's because it doesn't have that 1mA resolution. Hooked the 199 in series, and yeah, it's drawing 27.5mA (and change). Guessing my PSU is not 100% accurate on it's amp reading.

So, checked a few of the -9V rail points, and they're all at -9V, and not oscillating. Cool. The -V rail is back at -15V too.

But, meter still shows OL, until I switch to the 200V range, in which case it reads 63-70V oscillating all over. Awesome. (note, after about 5mins running, it's settled around 58V, with a couple volts oscillating... and yes, the inputs are shorted).

Check the -6.4V rail. +6.2V. That's... still not right. Which, might make sense since U101A/B look like they're involved in the mux switching. I also note that R125J is relatively warm. It also only reads 1.02K rather than 1.1K. Obviously that's only 80R, but I'm not sure how accurate these resistor networks should be.

So... thoughts? Am I looking at U101 being bad (I hope not, since I can't seem to find what it actually is). Schematic seems to indicate a dual op-amp, but the parts list just says it's an integrated circuit. It LOOKS like it's labeled SN103152P (TI, obviously). I don't see anything else on that rail, but I could be blind.

So, then I went back through the voltage checks. U109 pins 2 and 3 look good, +6.4V on both pins 2 and 3. The -6.4V shows up on pin 2 of U107, but I see +6.2V on pin 3. Which makes sense since it's coming from that zener. Could that zener be bad? Also interesting, pin 12 of S101 shows -19V... is that just because I'm largely bypassing the -V rail at this point?

I did see that -6.4V at +6.2V with both U107 in, and out of circuit. So that doesn't seem to be it. Either U101 is bad and pushing it up, or that zener is bad. =/ I don't have a second bench PSU to feed that rail with though... so, maybe hook VR101 and Q128 back up, and disconnect VR102 and inject -6.4V?

Thoughts?

[staze]

So I just realized... why the hell are pins 2 and 3 on U107 different? I'm no opamp expert, but my understanding is they should pretty much always be the same (assuming they're within the rails, etc).

I'm guessing this must be because U101 (or some other device I'm not seeing) is pushing back hard enough that it's swamping out the ability for pin 3 to come down to that voltage (since pin 2 is set at the -6.4V by a voltage divider off the negative rail)?

If that's the case, I'm guessing that's what's causing the -9V rail to oscillate (if those two inputs are that far off, and U107 is basically set up like an integrator, it's pushing Q128 to be on/off ramped like that. So that would explain why the -V rail (and -9V rail) look like integration, no?

[alm]

I don't have the schematic in front of me, but I don't think U107 has feedback with Q128 disconnected (its output is floating), so pins 2 and 3 may be different (and the output will be close to either its positive or negative rail).

So my initial analysis was wrong, and it's not just that VR101 did not have enough voltage across it. So there's a huge voltage drop across R125J? This means that something is drawing too much current from the -6.4 V rail. I wouldn't worry about the accuracy of R125J, it just has to provide an approximate current for the zener and whatever else is connected to the -6.4 V rail. If that's just U101 (I haven't checked, but I expect you did a thorough job), then that is the likely culprit unless there's a short between the -6.4 V trace and some other trace. I'm guessing no low resistance from the -6.4 V rail to any of the positive rails? Low as in much less than 1 kohm.

~27 mA on the -9 V rail is quite high, though, given that it all has to flow through R132, which would drop about 10 V across this resistor.

You keep mentioning oscillation, is one of the opamps actually oscillating (eg. U101)? Did you check with the scope? I forgot if you had access to a scope. Something oscillating could easily explain the high current draw.

[staze]

I don't have the schematic in front of me, but I don't think U107 has feedback with Q128 disconnected (its output is floating), so pins 2 and 3 may be different (and the output will be close to either its positive or negative rail).

So my initial analysis was wrong, and it's not just that VR101 did not have enough voltage across it. So there's a huge voltage drop across R125J? This means that something is drawing too much current from the -6.4 V rail. I wouldn't worry about the accuracy of R125J, it just has to provide an approximate current for the zener and whatever else is connected to the -6.4 V rail. If that's just U101 (I haven't checked, but I expect you did a thorough job), then that is the likely culprit unless there's a short between the -6.4 V trace and some other trace. I'm guessing no low resistance from the -6.4 V rail to any of the positive rails? Low as in much less than 1 kohm.

~27 mA on the -9 V rail is quite high, though, given that it all has to flow through R132, which would drop about 10 V across this resistor.

You keep mentioning oscillation, is one of the opamps actually oscillating (eg. U101)? Did you check with the scope? I forgot if you had access to a scope. Something oscillating could easily explain the high current draw.

I mention oscillation because the -V rail is/was oscillating. See picture attached to https://www.eevblog.com/forum/testgear/keithley-197-repair/msg278450/#msg278450.

Attached I've included the schematics. To me, it looks like U107 DOES get feedback through C139, which makes it kind of look like an integrator to me, which is what I thought the -V rail looked like in the picture referenced. So, oscillation, yes?

I don't see any low resistance between the -6.4V rail and any other power rail (closest is between it and the -9V, which is the value of R125J).

I also see, now, the -6.4V rail is hooked up to R117C (second page, top, where the input comes in from the mux), pin 10 of S101. I'm not quite sure why...

So... next step? re-install Q128 and VR101, and try injecting -6.4V at... ?

[staze]

Though, thinking about it, I'm not sure how I can disconnect the -6.4V rail... other than cut the trace, or cut the pins on U101, and R117C.

Otherwise, I'd say hook the -9V back up, and disconnect the -6.4V and see what happens. But looking at the schematic more, the -6.4V rail is kind of tied to a lot through that second page. It's not a rail for anything, but it seems to be involved in offsetting the input somehow. It's connected to U105, U117 (both relatively easy to replace), and Q121, which I can't quite read the label other than P842. Schematic makes it look like a dual fet. It's a 6 pin can package. ?

Update: It's an MP842 (just looked at the parts list).

[staze]

So, figuring the issue might be with the input on the other side of the buffer, it looks like R117C is supposed to be ~28K, but it's reading 2.2K. So I'm guessing there's some other pathway that I'm not seeing. Would this be low enough for the rail to be offset by a more positive rail?

On a related note, what if I hooked the -9V rail back up (Q128, and VR101), then used my bench PSU to basically push the -6.4V rail harder? Is that feasible, or likely to break something? I would think it might help "smoke" whatever has failed.

[staze]

I just don't understand this schematic sometimes... I think I'm going to try cutting pin 4 on U101 soon, since I found that the TLC27L2CP should be a good replacement (as noted in the pictures here: http://kuzyatech.com/keithley-197-microvolt-dmm-in-pictures/olympus-digital-camera-188

But I'm not sure that's the issue. It SEEMS like the issue is on page 2, where the -6.4V rail is attached to the main signal path, and R117C showing as 2.2KR. I also note that R117 pin 4, to ground is only 30R, which seems to be because Q119 is partially conducting when off, which seems weird since as an N channel FET, it should only conduct when it's base is positive. Guess I could ground the base and check again.

Also, of note, if I get all the buttons to be in their "out" position except the power, I do get some voltage fluctuation on the -6.4V rail (very minor). So something is feeding back into it slightly, rather than it being completely floating.

I'm just not really sure where to go at this point.

[staze]

BAM! So, knowing that I could replace U101 with a TLC27L2CP, I threw caution to the wind and cut pin 4 on U101. Guess what? The -6.4V rail came back, and the meter reads 0.0000V!

So, ordering a replacement as we speak.

Guess the question is, do we think U107 blew out U101, the other way around, or do we just think shit happens?

Thanks alm for the help! Once I get the parts I'll put everything back together and post back.

Thanks again!

[alm]

I was planning to spend some time to actually study all your results as opposed to just some quick remarks, but you already found the problem, darn .

If you're careful cutting pins can be kind of reversible. I received an instrument like that once that had all positive power pins cut in an attempt to track down a shorted opamp. Only the person doing the repair didn't solder them back together properly, so it still didn't work after replacing the damaged part . Worked fine after resoldering the connections.

Was U101 actually shorted or was it just oscillating/driving a heavy load? The 7 mA current it was drawing doesn't seem totally insane to me for a healthy opamp. Can you supply 15V from a bench supply to the power pins and short its inputs and see what current it draws, or is it too far gone for that? U101 is driving the guard, so I can also see a short on the PCB causing high current draw or oscillation. Is the resistance between the guard traces around the voltage and resistance traces and the traces around it as high as you would expect based on the schematic?

[staze]

I was planning to spend some time to actually study all your results as opposed to just some quick remarks, but you already found the problem, darn .

If you're careful cutting pins can be kind of reversible. I received an instrument like that once that had all positive power pins cut in an attempt to track down a shorted opamp. Only the person doing the repair didn't solder them back together properly, so it still didn't work after replacing the damaged part . Worked fine after resoldering the connections.

Was U101 actually shorted or was it just oscillating/driving a heavy load? The 7 mA current it was drawing doesn't seem totally insane to me for a healthy opamp. Can you supply 15V from a bench supply to the power pins and short its inputs and see what current it draws, or is it too far gone for that? U101 is driving the guard, so I can also see a short on the PCB causing high current draw or oscillation. Is the resistance between the guard traces around the voltage and resistance traces and the traces around it as high as you would expect based on the schematic?

I'll try hooking it up on a breadboard tonight. I only cut that one pin with an x-acto (needed two blades), the rest I unsoldered pretty cleanly. I don't have a split bench supply (though I have a rail splitter I bought a few weeks ago that I've been meaning to try).

I'm not sure what you mean by it's driving the guard... but if I recall, the resistance between it's inputs (on the board) seemed reasonable, as well as the feedback loop being intact.

Just checked, the inputs, on the PCB, for U101A (pins 5,6) show 1MR between them, and for 101B (pins 2,3) show 3.72MR. Haven't checked voltages as I'm getting ready for work, but can do that tonight. I was actually kind of curious what they all should be since it seems like U101 is involved in the system for switching FETs on and off (tied to the comparators), but not exactly sure what they were doing.

Will post back tonight with breadboard results, and voltages.

[alm]

A quick glance at the schematic suggested to me that they were driving the guard (see Fig 23 a here): driving the guard trace with a buffered version of the input signal. Could have been wrong, though.

Why do you think that you need split supplies to drive an op-amp? An op-amp does not have a ground pin. It can't tell whether you connected it to +15 V and ground, ground and -15 V or +7.5 V and -7.5 V. You can just short the inputs or configure it as a buffer and feed it a voltage from a divider between the power rails (eg. pot). If it's really broken, then I expect it to draw several mA regardless of the input and output connections.

It's mainly the outputs that are interesting, since any current flowing from the output will directly affect current draw.