Is it okay to use a transistor to connect a terminating resistor between the A and B lines of a differential signal? I'm not sure because I know that at the source/emitter of an NPN transistor must be at ground potential for the base/gate to turn the device on. My instinct says no but please help. If this won't work is there of something other than a relay that would work to do this with a GPIO pin of a processor?
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\$\begingroup\$ What are the voltage range of your differential signals (regarding gnd)? Each signal is 0 to 5V for instance, or AC signal around 0V? \$\endgroup\$– Blup1980Mar 21, 2014 at 9:30
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\$\begingroup\$ Your A and B suggests RS 485? \$\endgroup\$– gwidemanMar 21, 2014 at 9:31
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\$\begingroup\$ The reversal of polarity of the differential signal may not be your only concern. Differential sometimes implies floating at an average voltage which you can't predict before hand. In that case the "switch" part of your intervention may need to be isolated from your controlling device (GPIO pin)... a relay is looking pretty good. What do you have against it? \$\endgroup\$– gwidemanMar 21, 2014 at 9:33
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\$\begingroup\$ Yes, RS485 or CAN \$\endgroup\$– PugzMar 21, 2014 at 9:43
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1\$\begingroup\$ There's also this: "R@485 Transceivers with Integrated Switchable Termination" cds.linear.com/docs/en/lt-journal/… \$\endgroup\$– gwidemanMar 21, 2014 at 10:01
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5 Answers
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I'm not going to explain why is this a bad practice, but rather mention why a single transistor won't do the job.
Say you have a setup like the following:
simulate this circuit – Schematic created using CircuitLab
Obviously, you need two FETs simply to tackle the fact, that reverse-biasing the Source-Drain junction triggers the device's intrinsic diode. Second, you'd have to make sure, that at all times, your enable would be higher/lower than your maximum/minimum single-ended voltage + \$V_{th}\$ of the FETs.
Say if your signals swing from 1-2 Volts single-ended, then your Enable signal would have to go from 1V to at least \$2V+V_{th}\$ (in respect to ground).
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Analog switches seem to be suitable for this task. There are also over the rail products from various manufacturers.
answered Mar 21, 2014 at 11:14
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Many differential cables are terminated in two resistors; one on each leg to ground\$^1\$. There is a good reason that they do this - if the cable is shielded or screened and the signal is not truly differential there can be reflections on each wire to ground that upset things if only terminated with a single differential resistor.
If the differential resistor is 120 ohms then the more practical solution is to use 2 x 60 ohms to signal common. This: -
- Terminates the differential pair at 120 ohms and
- Partially terminates each wire individually to ground to reduce the problem mentioned above.
How does this help? If the signals on the differential pair are logic levels (or biased above ground) then using two transistors simultaneously to connect each wire to ground should work.
It should work but to be sure I'd need to see the voltage waveforms to 0V.
\$^1\$ For ground also read screen, shield, 0V, earth, common, return etc..
answered Mar 21, 2014 at 12:42
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Assuming you need to preserve the typical common-mode range of RS485 signals: -7V to +12V and have only a single supply available, this is not all that easy. It's also connected directly to the outside world, so ESD immunity has to be superb, and the resistance of the switch should be perhaps 10 ohms or less or its variability will affect the accuracy of the termination.
@gwideman's suggestion LTC2854/55 looks like a good way, if you don't mind using a "boutique" component.
There are well proven small telecom-style latching or non-latching relays, but they're fairly large.
Finally, there are analog switches with built-in charge pumps, such as the MAX14759, but availability may be an issue.
Of course there are lower tech non-software-programmable solutions such as jumpers and switches, and those ones will withstand ESD that will kill the receiver dead.
answered Mar 21, 2014 at 13:24
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Short answer is: no you can't use a single bipolar (NPN or PNP) transistor in this role. The characteristics of a transistor are wrong for this role for a bunch of reasons. Just one reason is that the transistor only functions a particular way with current in a particular direction from collector to emitter, whereas a differential signal reverses polarity.
As to a reasonable way to do this... head scratching in progress.
