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 > > * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * To post a message: mailto:[EMAIL PROTECTED] * * To leave this list visit: * http://www.techservinc.com/protelusers/leave.html * * Contact the list manager: * mailto:[EMAIL PROTECTED] * * Forum Guidelines Rules: * http://www.techservinc.com/protelusers/forumrules.html * * Browse or Search previous postings: * http://www.mail-archive.com/[EMAIL PROTECTED] * * * * * * * * * * * * * * * * * * * * * * * * * * * * * *