Joe,

Who says it has to be a "3mm thickness" (I think you may actually mean that
the trace has to be 3 mm wide)?

Is this requirement imposed on you by an Engineer, or is it some requirement
found in a datasheet for some specific circuit or device?

A "current sense resistor" has to be "in series" ("in line" like a fuse
would be) with the "load" (the circuit or voltage that is being sensed) so
that it can determine how much current is being used. In most cases a
current sense resistor is used so that there can be some "feedback" to the
circuit that is controlling the voltage output of some type of voltage
regulator (or voltage distribution), however, in few cases, it may be used
so that the current can be measured for some other reason such as to take a
measurement for a remote "monitor" of some kind.

In most, if not all cases, there needs to be a direct and unobstructed path
back to the "feedback" input of the controlling circuit (which in your case
is the amplifier input) which has no additional "loss" (such as a "resistive
drop" in the trace) which could otherwise affect the measurement. Typically,
this requirement may be met by  specifying a wide trace (such as your 3 mm
requirement) for the "feedback" signal, but more often, a very short and
direct trace can be used, which can  usually be accomplished by having a
good component placement so that the current sense resistor is not too far
away from the circuit involved (however, you must still watch out for any
"direct" "feedback" to the amplifier input from the other end of the current
sense resistor, which is usually the "output" of the "controlling" circuit
(in other words, the location of the sense resistor can be very critical)).

As stated above, the current sense resistor is "in series" with the "load",
and this means that it is only there so that you can measure the current
going thru it (by forcing a very small and controlled amount of current
"limiting"), as opposed to being in the circuit for some other kind of
current "limiting", which is the normal function of a resistor. This means
that the "load" side of the sense resistor (where you are also taking the
trace back to the amplifier input (for a feed back measurement)) is also in
most cases the "supply" voltage for whatever circuit is connected to this
"load" side of the current sense resistor. This would mean that it should be
a wider than normal trace so that it can handle the "power distribution" to
that circuit, just as if it were the normal "VCC" (or other power supply)
trace in your circuit if you were not using any internal planes for power or
ground distribution.

With that said, the real question that needs to be addressed here is just
how much current is going thru the sense resistor to the "load".

It may just be that the 3 mm requirement in your case is for the "load"
itself, as opposed to the "feedback" line from the "load".

On the other hand, if there is a very large "load" (on the "output side of
the current sense resistor) that is being "measured" by the "feedback"
trace, then 3 mm might be a very appropriate width for the trace.

What is critical here, is that you do not want any "loss" or any other
"variables" in this "feedback" trace (which would affect the "feedback"
measurement) that may vary from board to board due to such things as
manufacturing processes (etching or plating differences or minute
differences it the copper (trace) thickness), or which may vary in the same
board from such things as the resistivity of the trace varying due to
changes in ambient temperature during operation.

All of this boils down to having a very good and direct path back to the
"current sensing" input of the amplifier in your application which will not
have any resistive "loss" or "drop".

Thus the 3 mm width requirement.

In this case, I would additionally say that you do not want to have any
feedthrus or vias in this "feedback" trace, nor do you want to have anything
else in the trace that might in someway affect the resistivity of the trace.

A parallel response to this post, which deals with your wanting to use two
"2 different nets" (the subject of this thread), suggests that you might be
able to use two different nets as you request in your original post, and
then using the "Lomax Virtual Short" method to join your two 2 different
nets together. This would be a very good solution to your problem in just
about any other application but this one, since the "Virtual Short" plays
some tricks on Protel by having a very very small gap between the traces to
overcome DRC errors and objections, but which itself is reliant on the two
traces actually "bridging" or "shorting" the small gap (or more accurately,
not etching it thru completely, or bridging it with solder) during the
manufacturing of the board, which unquestionably will have a major impact on
the actual "resistivity" of the "feedback" trace involved. For this reason,
I would strongly advise you to not use the "Lomax Virtual Short" scheme, or
in any other way attempt to isolate the "feedback" trace from the "load"
side of the current sense resistor, in your present design.

What this means from a practical standpoint in Protel 99 SE is that you are
going to probably have to route this trace manually, and manually give it a
different "width" than the other part of the "load" supply distribution.
This also means that you will probably have to live with a few DRC "Errors",
concerning the "width" of this portion of "net". There is nothing new or
sacred about this, since Protel 99 SE is really pretty "dumb" when it comes
to handling "special circumstances" such as this that actually happen in the
"real world", and you often have to override or ignore what Protel wants you
to do in some these special situations.  Just learn to accept these minor
little shortcomings in Protel 99 SE and learn to live with them.

You might just want to put a note in a text file in the "database" or
directory for this particular design which explains the "problem" and
reasons for the DRC "Errors" so that it will be clear to yourself in a few
years when you or someone else has to come back and modify the design for
some reason.

Lastly, since it is not totally out of the question that your design may
have the "load" side of the current sense resistor going into some sort of
an internal "power plane" for the distribution of that "power" to the
"load", you need to remember that you will probably need more that just one
feedthru or via to to get that "power" into the plane, so that there will be
no "loss" in getting the "power" into the plane. It is always a very good
idea to use multiple feedthrus or vias to get "power" or "ground" into an
internal "power" or "ground" plane to avoid loss in that single feedthru or
via or possible fusing. By multiple, I mean many, not just 2 or 3, but
depending on the current that you need to get into the plane, you may want
to have a very large number of feedthrus or vias. Also remember that you
additionally may not want to have any "thermal reliefs" on any feedthrus or
vias that are used in a situation such as this on the "input" (for "power")
or "output (for "ground") to the internal planes (which is another whole can
of worms in Protel, since you usually do want them (thermal reliefs) on most
of the other connections to that plane on the same net), where you are not
soldering a component (which needs the thermal isolation from the plane),
but rather trying to get "power" and "ground" to or from the internal planes
without any losses (or fuses). I usually will build up a special little
group of "pads" and "fat traces" in a special little block or grouping
outside of the perimeter of the board where I can work on it and select it
and copy it and as a "group" so that I can change its net name and copy it
anywhere I want it for just this purpose of getting to the internal planes,
and yet since it is not a "component", it will not give me problems with the
schematic or netlist. You have to be careful when using these little "izmos"
since Protel will try and eliminate the "extra" feedthrus or vias that are
in it if you do any manual routing on that net after placing one of these
(which you do by copying it) if you have the "remove loops" function set
(which only means that you may have to delete the little "gizmo" that lost
whatever got lost and copy in a new one in its place).

Anyway, enough for one post (I know, I know, too much already), and
hopefully it is not too complicated or hard to understand.

I hope this information will help you understand the requirement for a 3 mm
trace width, and just what is really going on with the current sense
resistor and the "feedback" from the "load" side to the input of the
amplifier in your design.

JaMi

----- Original Message -----
From: "Joe McCauley" <[EMAIL PROTECTED]>
To: "Protel EDA Forum" <[EMAIL PROTECTED]>
Sent: Wednesday, September 03, 2003 4:09 AM
Subject: [PEDA] Joining 2 different nets keeping seperate identifiers?


> I need to have a current return line with 3mm thickness. This line has a
net
> identifier of 'Iret'.
> This line connects to a current sense resistor. I need to take a line from
> this resistor to an amplifier input.
> There is no need for this amplifier input line to be 3mm thick, in fact
from
> the point of view of routing it would be better if it were not! Is there a
> way of joining 2 different nets in the schematic while keeping seperate
> identifiers? If there were then I could setup the design rules in PCB to
> always have the 'Iret' net 3mm thick, while the other one which connects
to
> it could be (say) 0.35mm. Am I over complicating things by trying to do it
> this way?
>
> Thanks for any pointers,
>
> Joe
>
>



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