One wouldn't expect that; it changes the transmission-line impedance, so your flat frequency response depends on the source and/or termination impedance. For anything critical, it's worthwhile to buy a few hundred (or thousand...) feet of your wire on a spool, measure with a few different source/terminate values, and label that spool with the REAL characteristic impedance. A good noise generator and FFT oscilloscope should handle the job easily...
With high Z like your source, it's usually most effective to shield both the differential-pair wires,and DRIVE EACH SHIELD from a follower or low-gain amplifier + attenuator. Minimizing/zeroing the effect of stray capacitance is better than minimizing the capacitance (which will never be zero).
========================= ==
========================= ==
In your drawing, C1 and C3 are in series and across C2. To the signal itsel f, if C1 and C3 are equal, it should not matter if the shield is grounded o r not as it is nulled out.
When CAT cable was the cat's ass, well that was alright. But now the LVDS c ables like in monitors and TVs as well as the HDMI video cables generally r un like that, but have reduced the voltage.
The fact that they are extremely well balanced is responsibe for the noise reduction, the lower voltage allows for higher speeds/frequencies because e ven though there is capacitance there, you simply do not have to charge and discharge it as much.
If the pair is twisted right, noise should be hitting both polarities at th e same level and in the same phase, effectively nulling out. The only reaso n to use shielding is if you have reason to believe that the common mode no ise will swamp the amp by exceeding its input limits or CMRR. Ergo, if you deem shielding necessary, it should be grounded. It should not make a diffe rence in the capacitance between the two of the twisted pair. The capacitan ce seen by the driving side should be the same.
Should that is.
Bottom line - yes. And it does not matter if it is grounded or not. It is strictly a function of the cable you use.
So if you deem shieding necessary you might as well ground it.
On 13.06.2015 17:12, George Herold wrote:
Ok, now I think that I'm starting to understand what's happening.
Anyway, what follows is just my semi-educated wild guess, so take it with a grain of salt and feel free to correct me where needed.
You see mains static on your signal lines. Your signal line is supposed to be differential in nature and it's located in a grounded metal tube. This means that something doesn't add up. Mains static normally comes in through the air via capacitive coupling, thus having an extremely high "source" impedance. In a fully differential amplifier system that's very unlikely to be an issue, unless there's a gotcha with the signal source or the amplifier construction somewhere. So I'll try to list some possible "gotchas" that may be lurking, hoping that something from that list may be useful.
Basically, signals can propagate in a twisted pair cable in common mode and in differential mode. Normally they always propagate in both modes, even when one tries to make a fully differential setup, real-world parasitics will still make sure that at least some small rest of common mode signal will still remain. That rest is what needs to be minimized as far as practical, and that rest comes from a variety of sources:
- limited common mode suppression of amplifiers
Normally modern op-amps are quite good when it comes to CMRR, but CMRR will always be finite. However, if both the signal source and the inputs to the amplifier have a very high common mode (to ground) impedance (say, a floating source connected to an amplifier with megaohm inputs), then a situation can (and usually will) arise, where there exists no low impedance common mode driver for the transmission line at all (that is the source does not strongly force the common mode voltage of the line to any definite value and neither do the inputs of the amplifier do so). This leads to the transmission line being essentially "undefined" common-mode-wise. That is, the source may be driving a good and valid differential signal, but in addition to that, the transmission line can also easily pick up a common mode portion from stray fields, because there is no low-impedance common-mode driving source that could "overpower" those floating strays and force the transmission line to some fixed and definite common mode voltage. This situation normally results in the common mode strays becoming far higher than the wanted differential signal, and even (easily!) hitting the amplifier's rails. Remember that an opamp's CMRR practically drops to zero as soon as the inputs hit the power rails. So an amplifier with, say, 90 dB CMRR, will suddenly find itself having 0 dB CMRR when a high impedance differential signal "floats away", out of the amplifier's permitted common mode range simply because no low impedance common mode driving source has prevented it from floating so far.
Now you ground one of the signal lines - and voila - the strays disappear - which is easily explained since a ground connection is indeed a (very) strong low impedance common mode driving source that does not permit your transmission line to float freely.
- common-to-differential and differential-to-common mode conversion
Now, grounding one end of a differential transmission line is indeed a practicable way to introduce a common mode driving source where one is needed and no "naturally" existing one is already available. However that way is usually not the best way to do it. The reason is, that as soon as you ground one end, the transmission line essentially stops being differential and becomes single-ended. In order to operate a differential transmission line properly, one has to take care to avoid unwanted mode conversions. A conversion between differential and common mode of signal propagation occurs at all asymmetrical impedance discontinuities, so in order to avoid it, one has to insure that the entire differential system (including the source, the line, and the amplifier) have equal impedances (including stray capacitive ones) from each of the signal wires to ground (and to all other non-differential signals). Grounding one wire sets its impedance to ground to zero, while the other wire's impedance to ground remains in the megaohms, thus creating an asymmetry and introducing differential to common mode conversion. Unfortunately the conversion works both ways: while it converts the signal from differential to common mode, it also converts various induced strays from common mode to differential, thus allowing the amplifier to pick them up. That way, while grounding one wire will prevent a floating line from railing the amplifier's inputs and thus degrading the amplifier's CMRR, it will also introduce an undesired common to differential mode conversion of stray signals and thus put these strays into the amplifier, where you don't need them.
The classic solution to both problems is to: a) have a common-mode driver to prevent a floating line and b) have this driver be symmetrical to avoid mode conversion.
Unfortunately for high impedance signal sources implementing this solution properly and reliably is not easy.
There's a simple approach, which has some practical limits, and many more elaborate approaches, usually application-specific.
A simple approach is to introduce a weak connection between the signal wires and ground, for example a 10 megaohm resistor instead of the direct short. This approach can even be made fully symmetric (use 2 resistors with 20 megaohm each, one from each line to ground). This greatly reduces the differential to common mode conversion. The symmetric version even almost eliminates it, but has the unintended consequence of extra loading of the signal source via the 40 megaohm (line to line equivalent). Still, if your source can stand driving an extra 40 megaohm in parallel, that may be a viable alternative.
A significant downside of this is, that the common-mode driver has a high impedance, not a low impedance (10 megaohm effective in both examples above). If your setup is shielded well and the stray fields are limited then that's all you need. If you have horrible field strengths from nearby high voltages that you can't shield, then these fields can overpower the weak common-mode driver and still "float" your signal "away", out of the amplifier's permitted common mode range. That's obviously bad news for signal integrity.
There are other, more complex solutions, but they tend to be more specific to particular signal types and source topologies, so less universal.
One way is to use a transimpedance amplifier. These can be single-ended (with a normal opamp) or differential (with a fully differential one). Their input impedance is essentially zero (on both signal lines if the architecture is differential), and the common-mode impedance is ery low too. Thus a differential transimpedance amp makes a very good common mode low impedance driver that can "protect" the line against strong common mode strays, while at the same time it amplifies the signal. That would be a classic example of the "naturally existing common mode driving source" mentioned above. Unfortunately, being TIA-based, it only works with current source type signal sources, and it also can't provide a DC bias voltage if the signal source happens to need one.
You said that your sensor is something "photoconductive", so I'll hazard a guess that it can provide a current (like a photodiode) and that it can make use of a DC bias voltage (like a photodiode too).
A current can be amplified by TIAs - they have nice low impedances both in common and (if differential) in differential mode, and they are quite difficult to "rail" from stray fields. So that may be a viable option for you. Their only downside is that they don't provide a DC bias by default.
Besides, running your signals into a TIA "virtual ground" should greatly increase the available bandwidth, possibly by factors of magnitude - since the voltages that charge cable capacitance are essentially zero.
You didn't say if your signal source needs or uses a DC bias, but if it does, it may be possible to provide that bias by putting the sensor between 2 TIAs that "sit" on different voltage rails. Then on has to subtract the constant difference voltage between the TIA outputs later.
As an alternative, you can consider using 2 TIAs and a discrete current mirror. That way you can run both TIAs at the same DC bias and you won't have to subtract the bias voltage from between their outputs. The current mirror can be fed from a supply rail located considerably "higher" in order to provide any DC bias for the sensor.
Here's a (simplified) example:
A matched transistor pair current mirror provides DC bias to the sensor on the other end of the shielded line. Two transimpedance amplifiers amplify the low current signal from the sensor, one directly and one mirrored. The differential useful signal appears as the sum of the TIA outputs and the common mode stray field "signals" appear as the difference between the TIA outputs. To get the signal and reject the noise you only need to sum these outputs together via 2 equal resistors. Because the transistor current mirror is non-ideal, one of the feedback or summing resistors may need a slight trimming (I've put a "*" next to a feedback resistor in the schematic, but it would actually be more practical to trim one of the summing resistors instead).
If your sensor works like a photodiode (thus relatively voltage-independent), that may be OK. If it's more like a photoresistor and it needs a better stabilized and "stiffer" bias voltage, then you can add a little stabilization to the current mirror feeding the DC bias to the sensor. This can (for example) be done like this:
Here, a low leakage opamp is fed some stable reference voltage from a low noise supply and actively keeps the sensor's DC bias at precisely this level (obviously it also needs a low-noise power supply with the appropriate voltage in order to operate, I skipped it on the drawing for simplicity).
Well, that was quite long. I hope, there's at least one idea of practical use to you somewhere :)
Regards Dimitrij
Sorry, missed the line wraps. Try again.
Ok, now I think that I'm starting to understand what's happening.
Anyway, what follows is just my semi-educated wild guess, so take it with a grain of salt and feel free to correct me where needed.
You see mains static on your signal lines. Your signal line is supposed to be differential in nature and it's located in a grounded metal tube. This means that something doesn't add up. Mains static normally comes in through the air via capacitive coupling, thus having an extremely high "source" impedance. In a fully differential amplifier system that's very unlikely to be an issue, unless there's a gotcha with the signal source or the amplifier construction somewhere. So I'll try to list some possible "gotchas" that may be lurking, hoping that something from that list may be useful.
Basically, signals can propagate in a twisted pair cable in common mode and in differential mode. Normally they always propagate in both modes, even when one tries to make a fully differential setup, real-world parasitics will still make sure that at least some small rest of common mode signal will still remain. That rest is what needs to be minimized as far as practical, and that rest comes from a variety of sources:
- limited common mode suppression of amplifiers
Normally modern op-amps are quite good when it comes to CMRR, but CMRR will always be finite. However, if both the signal source and the inputs to the amplifier have a very high common mode (to ground) impedance (say, a floating source connected to an amplifier with megaohm inputs), then a situation can (and usually will) arise, where there exists no low impedance common mode driver for the transmission line at all (that is the source does not strongly force the common mode voltage of the line to any definite value and neither do the inputs of the amplifier do so). This leads to the transmission line being essentially "undefined" common-mode-wise. That is, the source may be driving a good and valid differential signal, but in addition to that, the transmission line can also easily pick up a common mode portion from stray fields, because there is no low-impedance common-mode driving source that could "overpower" those floating strays and force the transmission line to some fixed and definite common mode voltage. This situation normally results in the common mode strays becoming far higher than the wanted differential signal, and even (easily!) hitting the amplifier's rails. Remember that an opamp's CMRR practically drops to zero as soon as the inputs hit the power rails. So an amplifier with, say, 90 dB CMRR, will suddenly find itself having 0 dB CMRR when a high impedance differential signal "floats away", out of the amplifier's permitted common mode range simply because no low impedance common mode driving source has prevented it from floating so far.
Now you ground one of the signal lines - and voila - the strays disappear - which is easily explained since a ground connection is indeed a (very) strong low impedance common mode driving source that does not permit your transmission line to float freely.
- common-to-differential and differential-to-common mode conversion
Now, grounding one end of a differential transmission line is indeed a practicable way to introduce a common mode driving source where one is needed and no "naturally" existing one is already available. However that way is usually not the best way to do it. The reason is, that as soon as you ground one end, the transmission line essentially stops being differential and becomes single-ended. In order to operate a differential transmission line properly, one has to take care to avoid unwanted mode conversions. A conversion between differential and common mode of signal propagation occurs at all asymmetrical impedance discontinuities, so in order to avoid it, one has to insure that the entire differential system (including the source, the line, and the amplifier) have equal impedances (including stray capacitive ones) from each of the signal wires to ground (and to all other non-differential signals). Grounding one wire sets its impedance to ground to zero, while the other wire's impedance to ground remains in the megaohms, thus creating an asymmetry and introducing differential to common mode conversion. Unfortunately the conversion works both ways: while it converts the signal from differential to common mode, it also converts various induced strays from common mode to differential, thus allowing the amplifier to pick them up. That way, while grounding one wire will prevent a floating line from railing the amplifier's inputs and thus degrading the amplifier's CMRR, it will also introduce an undesired common to differential mode conversion of stray signals and thus put these strays into the amplifier, where you don't need them.
The classic solution to both problems is to: a) have a common-mode driver to prevent a floating line and b) have this driver be symmetrical to avoid mode conversion.
Unfortunately for high impedance signal sources implementing this solution properly and reliably is not easy.
There's a simple approach, which has some practical limits, and many more elaborate approaches, usually application-specific.
A simple approach is to introduce a weak connection between the signal wires and ground, for example a 10 megaohm resistor instead of the direct short. This approach can even be made fully symmetric (use 2 resistors with 20 megaohm each, one from each line to ground). This greatly reduces the differential to common mode conversion. The symmetric version even almost eliminates it, but has the unintended consequence of extra loading of the signal source via the 40 megaohm (line to line equivalent). Still, if your source can stand driving an extra 40 megaohm in parallel, that may be a viable alternative.
A significant downside of this is, that the common-mode driver has a high impedance, not a low impedance (10 megaohm effective in both examples above). If your setup is shielded well and the stray fields are limited then that's all you need. If you have horrible field strengths from nearby high voltages that you can't shield, then these fields can overpower the weak common-mode driver and still "float" your signal "away", out of the amplifier's permitted common mode range. That's obviously bad news for signal integrity.
There are other, more complex solutions, but they tend to be more specific to particular signal types and source topologies, so less universal.
One way is to use a transimpedance amplifier. These can be single-ended (with a normal opamp) or differential (with a fully differential one). Their input impedance is essentially zero (on both signal lines if the architecture is differential), and the common-mode impedance is ery low too. Thus a differential transimpedance amp makes a very good common mode low impedance driver that can "protect" the line against strong common mode strays, while at the same time it amplifies the signal. That would be a classic example of the "naturally existing common mode driving source" mentioned above. Unfortunately, being TIA-based, it only works with current source type signal sources, and it also can't provide a DC bias voltage if the signal source happens to need one.
You said that your sensor is something "photoconductive", so I'll hazard a guess that it can provide a current (like a photodiode) and that it can make use of a DC bias voltage (like a photodiode too).
A current can be amplified by TIAs - they have nice low impedances both in common and (if differential) in differential mode, and they are quite difficult to "rail" from stray fields. So that may be a viable option for you. Their only downside is that they don't provide a DC bias by default.
Besides, running your signals into a TIA "virtual ground" should greatly increase the available bandwidth, possibly by factors of magnitude - since the voltages that charge cable capacitance are essentially zero.
You didn't say if your signal source needs or uses a DC bias, but if it does, it may be possible to provide that bias by putting the sensor between 2 TIAs that "sit" on different voltage rails. Then on has to subtract the constant difference voltage between the TIA outputs later.
As an alternative, you can consider using 2 TIAs and a discrete current mirror. That way you can run both TIAs at the same DC bias and you won't have to subtract the bias voltage from between their outputs. The current mirror can be fed from a supply rail located considerably "higher" in order to provide any DC bias for the sensor.
Here's a (simplified) example:
A matched transistor pair current mirror provides DC bias to the sensor on the other end of the shielded line. Two transimpedance amplifiers amplify the low current signal from the sensor, one directly and one mirrored. The differential useful signal appears as the sum of the TIA outputs and the common mode stray field "signals" appear as the difference between the TIA outputs. To get the signal and reject the noise you only need to sum these outputs together via 2 equal resistors. Because the transistor current mirror is non-ideal, one of the feedback or summing resistors may need a slight trimming (I've put a "*" next to a feedback resistor in the schematic, but it would actually be more practical to trim one of the summing resistors instead).
If your sensor works like a photodiode (thus relatively voltage-independent), that may be OK. If it's more like a photoresistor and it needs a better stabilized and "stiffer" bias voltage, then you can add a little stabilization to the current mirror feeding the DC bias to the sensor. This can (for example) be done like this:
Here, a low leakage opamp is fed some stable reference voltage from a low noise supply and actively keeps the sensor's DC bias at precisely this level (obviously it also needs a low-noise power supply with the appropriate voltage in order to operate, I skipped it on the drawing for simplicity).
Well, that was quite long. I hope, there's at least one idea of practical use to you somewhere
Regards Dimitrij
But one problem people often overlook is that LVDS is not really differential like i.e. Ethernet. There you have transformers that force the currents through both wires to be equal. For LVDS you have two independent drivers. They are just in one chip and hopefully closely matched. But any mismatch will create current flowing in the ground wire. What I have often found that bad ground wires (one for the whole cable or using the shield as ground connection) can create "nice" diagrams during EMC testing. Good for the lab, they earn more money. :-(
-- Reinhardt
The LVDS internal circuits that I've see have a Vcc-side current source and a ground-side sink of some sort, with esentially an h-bridge between to steer the currents into the output pair. So it's not two independent drivers.
There's a loop off to the side that tweaks the dueling current sources to set the output common-mode voltage.
See fig 1.
Fig 5 shows the common-mode loop.
-- John Larkin Highland Technology, Inc picosecond timing laser drivers and controllers jlarkin att highlandtechnology dott com http://www.highlandtechnology.com
Just like many standards for commercial equipment. USB ist another one. just connect a USB hub to a laptop and you have a nice 120MHz radiation source. I had this several times with commercial and self designed hubs.
Which is why all the talk and layout guidelines about differential impedance is more or less nonsense. Treat the signals as two independent signals and take care not to make them too different regarding propagation delay to avoid skew.
I had the situation several times during EMC testing. The radiated emissions came from internal LVDS cables going to a display. My rule now is the pairs have to be individually shielded, the shield is used not as shield but for return ground. This way the two signals and the return "wire" are closely together. The sum of the currents is really zero then. Shielding the cable is nearly impossible and does not help that much, because the cables get too clumsy and there usually is no provision at the display for shield connection. This increases the price for cable by about 10$ but saves at least
3-5000$ for the re-testing in the lab, not even counting my time. Since most devices are low volume avionics stuff in the rage of n*100 this easily computes.-- Reinhardt
Not quite; they simply filter out the common mode. Indeed, there's a 1n cap to ground (after some common mode chokes), there mainly for ESD absorption.
What people rarely appreciate is that the inputs (and even the outputs) have a common mode range, and if you exceed that, piff, nothing, it goes to zero. Differential only ever works if it's fully true at all times. Any lapse in that and you get problems. Since you don't get much overhead on LVDS, it's a rather poor standard for signaling over extended lengths that might be influenced by EMC.
As I recall, the equivalent common mode resistance is around 500-1500 ohms. Kind of weird. Ideally, you'd like it to be about 100 ohms each end, for best common mode termination...
Which is something people also don't understand. Common mode is a signal just like any other. You want to terminate it like any other. Which really means, almost always, you want to simply treat each signal independently, then take the electrical difference once all is said and done.
That should give ~0 ohms common mode, but that does depend on whether TI and others actually use the circuit or not.
Tim
-- Seven Transistor Labs, LLC Electrical Engineering Consultation and Contract Design Website: http://seventransistorlabs.com
...
+42-- Reinhardt
LVDS receivers seem to be r-r input comparators. No piff.
Differential only ever works if it's fully true at all times.
The signal is differential and the termination is differential. There's no need for common-mode termination.
Which isn't done in LVDS. There is no reason to do it. It's just some wimpy DC voltage.
Which
The common-mode setting loop in the transmitters is weak. It only needs to set some reasonable cm voltage for the high-impedance receivers.
All the LVDS receivers that I've tested behave like super fast high input impedance rail-to-rail-input comparators who actually don't care much what the common-mode voltage is. They are actually usable as cheap, fast comparators in a lot of situations.
-- John Larkin Highland Technology, Inc picosecond timing laser drivers and controllers jlarkin att highlandtechnology dott com http://www.highlandtechnology.com
You can run the diff pair close on a pc board, in which case it behaves like a real differential pair, with trace-trace coupling. Or you can run the two far apart and treat them as independent. I think the close case has electrical and EMI advantages, electric and magnatic coupling that improves symmetry, in addition to the obvious geometric advantages.
There should be no common-mode signal. That's the advantage of the current source/h-bridge circuit; it's current sources looking up and looking down.
-- John Larkin Highland Technology, Inc picosecond timing laser drivers and controllers jlarkin att highlandtechnology dott com http://www.highlandtechnology.com
You cannot JUST measure the inter-lead capacitance as the capacitance from each lead to the shield is always there (and "interfering").
Sure I can. Do a proper 3-lead capacitance measurement. My Boonton 72 can do that, measure the lead-lead capacitance C2 and ignore both capacitances to ground.
Ground the shield. Apply an AC voltage to one lead. Connect an AC ammeter between the other lead and ground. Measure the current and compute the resulting capacitance. That's my C2. The measurement is unaffected by the values of C1 or C3.
That's what the Boonton does. It can accurately measure a fraction of a pF at the end of a couple of coax cables that each have many 10s of pF to ground.
-- John Larkin Highland Technology, Inc picosecond timing laser drivers and controllers jlarkin att highlandtechnology dott com http://www.highlandtechnology.com
I've seen two values stated for the capacitance between Earth and the moon. I think one is the 2-terminal capacitance and the other is the
3-terminal value.-- John Larkin Highland Technology, Inc picosecond timing laser drivers and controllers jlarkin att highlandtechnology dott com http://www.highlandtechnology.com
If you have common mode voltage range problems, use galvanic isolation.
In industrial environment for RS-422/485/CanBus (which have a larger CCM range than LVDS) I recommend using galvanic isolation if the line is longer than 10-20 m or if the line goes to a different room with different electric feed system.
No objection against the advantage of better immunity against EMI and EMC.
That's what I believed, too. Reality proved me wrong.
-- Reinhardt
Third terminal is the sun? :-)
-- Reinhardt
PCB pairs are even *worse* than anything you can get in a cable, save for full-on coax pair. The common mode impedance is effectively due to the sum trace widths, which as you can imagine, is quite wide for a pair of traces side by side, comparable to the laminate thickness. So the common mode impedance is quite low, and the "coupling" between traces is, at best, modest.
A diff pair over bare PCB would be quite comparable to Ethernet (which, if wired as independent pairs through space rather than packed in a cable, would have a common mode impedance well over 100 ohms, or if considering the cable as a whole). The traces have to be rather wide to maintain 100 ohms differential, since edgewise coupling is poor (which is my whole point here..).
It's rarely talked about, but both USB and Ethernet (the PHY side) specify common mode impedance. USB is fairly modest (the cable and traces should be something like 70 ohms, so a little more differential character than not). Ethernet would actually be best satisfied with no differential routing at all, since each pin driver is expecting 50 ohms, and the pins are individually terminated into 50 ohms (against whatever VREF is, usually a 3.3V filtered rail).
From the driver, perhaps. (In reality, the CM tracking won't be perfectly fast, and there will be some slop in the transitions of each pin. But they'll also be pretty well matched (~100ps?), by design.)
I'm talking more about interference. Even if you can guarantee your signal sources and loads are absolutely ideal (no common mode), you can't escape the inevitability that, somewhere down the cable, someone is going to f*ck up everything. :) And that's where you need to take account of common mode impedance, termination, filtering and shielding.
Tim
-- Seven Transistor Labs, LLC Electrical Engineering Consultation and Contract Design Website: http://seventransistorlabs.com
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