At 10:43 PM 9/3/2003, JaMi Smith wrote:
[in the original inquiry it was stated:]

> >I need to have a current return line with 3mm thickness.

"return" generally refers, in my experience, to the wiring which returns to ground or other more negative voltage. I'm sure someone can come up with a more precise definition. But it might be used, in this case, to refer to the return line coming from the load, not to ground, but to a sense resistor and from there to ground or common. The sense resistor is of a low value, and the voltage drop created by the current through it will be small compared to the voltages involved in driving the load.


[...]
> >This line connects to a current sense resistor. I need to take a line
from
> >this resistor to an amplifier input.

The writer did not state which side of the resistor is connected to the amplifier input. The imprecision of the question has led to a bit of confusion in the answers, but it does not really affect the most important part of the answer, only the dicta.


Since the writer only refers to one sense line, I'm assuming that the voltage drop in the current path from the resistor to ground may be neglected; otherwise one would need to use *two* sense lines and a differential amplifier, though it might also be possible to calibrate the circuit and get away with only a single sense line. In other words, the true sense resistor would be the assembled part plus the resistance of the remainder of the ground circuit.

Quite clearly, however, the inquirer was concerned with a voltage sense line and its different width requirements from what is otherwise the same net, Iret. Perhaps the "amplifier" is located on the other side of the circuit board, or even on another circuit board.

[I wrote:]
> Presumably there will be negligible current in the sense line (the
> "amplifier input line.") At least whatever measures that voltage should be
> designed to minimize the current. So the line can be narrow, really it
only
> needs to be wide enough to be reliably fabricated.
>

I would disagree here, in that I believe that the trace should be large
enough to not contribute any "losses" of its own by being so narrow that
differences in manufacturing runs may produce traces which may have
differences in their own resistivity, which will in fact affect the circuit.

Mr. Smith must be having a bad day. (Or I've really lost it myself, certainly a possibility....)


The resistivity of this line in a current measurement application will have no effect on the voltage at the amplifier, since, in a properly designed circuit, there will be no current in the trace. If there is no current, there is no voltage drop. True, if the current is rapidly varying, resistance could create some problem, but it would be unlikely in a normal power application for this to be an issue.

There needs to be a good direct path from the "load" side of the current
sense resistor back to the input of the amplifier, and it needs to have no
problems of its own such as losses or crosstalk from other circuits.

Crosstalk might be an issue, I suppose, but, again, in power circuitry, it would be unusual for the problem to be such that a bit of capacitance at the amplifier input would not eliminate it. We might be talking DC, here. If crosstalk is an issue, I'd think the thickness of the trace might be irrelevant.... If we are talking RF, all bets are off.


Assuming that "Iret" is in fact the "return net" from the "load",

Yes, I'm sure that's what he meant. He's free to chime in with a correction, of course!


 and is a
large trace connected to one end (the "load" end) of the current sense
resistor, with the other end of the current sense resistor being connected
to ground (the negative supply),

Yes. It's a large trace for heat reasons, less likely losses might be an issue. But losses in the Iret trace will not affect the current measurement if the sensing point is right at the high end of the sense resistor.


Drop in the ground end net *would* be an issue, though it might be possible to ignore it, as was implied by the question.

 I would say that there should be another
trace going from the same "Iret" end of the current sense resistor to the
amplifier input.

"Another" trace? That's the trace he was asking about!


Perhaps Mr. Smith intended to refer to a trace from the ground end (not the "Iret" end) of the sense resistor, which would then be used in a differential voltage measurement, i.e., a differential amplifier would be used. But this was not the question we were asked. However, the trace size and naming and DRC issues would be the same, only it would be with two traces instead of one.

 This trace is the "feedback" portion of the "Iret" trace,
or what I would call the "feedback" trace, but there is absolutely no reason
in the world that this trace should have a different "net" name, or have any
"virtual short" involved with it.

I must say that Mr. Smith has lost me here. "feedback"? Feedback would be involved in the amplifier circuitry, but the feedback loop would not go back to the sense resistor. Besides asking for trouble with noise, there is simply no reason for it.


All you need is a good clean direct trace of moderate dimension (to avoid
any losses) that goes back to the amplifier input and avoids any crosstalk
from other traces. It is really that simple. If you do anything else, you
are shooting yourself in the foot.

I think that Mr. Smith has missed the basic issue. Totally missed it. This becomes clearer:


With no offense intended, I would say that this is one place that the "Lomax
Virtual Short" should not even be considered, let alone discussed, since
even by  your own discussion here you must admit that you cannot really
control what is happening at manufacturing level with the gerbers etc., and
you cannot explicitly rule out that there could be some etching into the
trace at the point of the "gap" during manufacturering, which will in fact
unquestionabaly affect the resistivity of the trace, and therefore the
operation of the circuit.

As I said, either I've lost it totally or Mr. Smith has completely missed the point. And now he shows that he doesn't understand how a "virtual short" works. Typically a "virtual short" is nothing more than two pads placed close enough to each other that the gap cannot be fabricated unless a fabricator decides, without consulting you, to increase it. Normally he would not even have the occasion to do this unless you have neglected the way in which you generate the gerbers. Properly done, the pads touch precisely on the film, no gap at all, so they are just solid copper when fabricated.


A virtual short is not a trace. It is, rather, a shorted jumper. You'd have to work hard to make any variation in its fabrication detectable as a variation in resistance -- again, unless a fabricator opens it up, in which case it will not merely "affect the resistivity," it will increase it to, well, practically speaking, infinity. The circuit won't work *at all*, it will be quickly discovered and someone will take a soldering iron to it (typically the pads are exposed unless one has deliberately tented them, which I wouldn't recommend) and put it out of its misery. I.e., short it.

The trace we are discussing is the sense trace. In this application it would be connected to the virtual short on the other side from the virtual short's connection to the sense resistor load end. Again, it should not be carrying any significant current in a properly designed circuit. As long as it connects at all, it will be sufficient. Really. I'd be amazed if several hundred ohms had any effect. And you won't get anything like that with a fabricatable copper trace on a circuit board. I suppose if you made it as thin as you could and wound it around the board a few hundred times....

With that said, I think that almost everything else said here is not
relevant to the real problem at hand, which is really boiled down to two
very simple issues, the width of the trace from the "load" end of the
current sense resistor to the "load" itself, and the width of the trace from
that same "load" end of the current sense resistor back to the input of the
amplifier that is sensing the voltage drop across the resistor.

Yes. However, the width of the high-current trace (the first of the two mentioned by Mr. Smith) was not the question. That width was fixed in the question at 3 mm. Without knowing the currents involved and some other aspects of the application, I could say nothing more about it, except that 3 mm is a pretty normal power trace.


Everything else here is totally irrelevant and confusing the issue.

The inquirer wanted to be able to short two nets together, and really he asked about how to do this. It seems Mr. Smith thinks that this is unnecessary. But, to have critical aspects of the design checked by DRC, it *is* necessary to either do this or manipulate the to-from width rules. For best accuracy, one would want the sense trace to be picked off of the Iret trace right at the load end of the sense resistor. Otherwise the voltage drop in the Iret trace itself will show up in the sense voltage. Virtual shorts, among other things, control the exact place at which two nets are shorted.


Why are two nets needed? Well, first of all, one wants to control the pick-off point. If there is only one net, as far as DRC is concerned the voltage sense line could come from anywhere on Iret. It's the same problem as with the original virtual short application, with, say, subnets we might call GND and AGND. Typically one wants the two subnets to be shorted at a very precise point, usually the ground leg of the main power bypass capacitor.

Using two nets with a virtual short and placing the virtual short at that location will accomplish this. In this case, not only is the connection location controlled, but the width of the sense trace can be independently controlled. It does not need to be fat unless it is carrying current....

If you don't care about DRC, then you would simply draw the sense trace at whatever width you want (within the range allowed by your design rules). But DRC is a very valuable tool, as we have learned the hard way, by neglecting it in the name of getting the job out quickly. Especially when the next revision comes around....

I've stated that the current in the sense trace would be very low. Mr. Smith has not addressed this issue at all; rather, he seems to simply assume that change in resistance of the sense trace will affect the circuit. I'm not an electronics engineer, or I might describe a typical low-input-current amplifier that would be used in a circuit like this. If I was stuck on a desert island and I had to build one of these, I suppose I'd use an op amp with very high imput impedance, the sense line would connect with the non-inverting input, and the output would feed back directly to the inverting input. So the output voltage would be equal to the input voltage while no current is drawn from the sense line, except for leakage into the amp. Then I'd amplify the signal with another stage, since current could now be drawn from the buffered signal. I think there might be simpler ways. If there were any risk of noise, I'd put a modest amount of capacitance on the sense line at the op amp input.




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