7
\$\begingroup\$

Many times when we design a PCB we might get stuck with PCB board real estate shortages. Of course we can overcome it many times by increasing the number of layers, subsequently making it more complex and expensive.

But, a few times the size of the components and the board size availability is just not balanced. For example, I have a board of 10x10 (centimeters).

I calculated the area occupied by all the components (including passives) and I come up with a number. The available area is 100 cm2. What should be an ideal total component size (total area of all components)? Is there a rule of thumb as such to get away with?

Like if the total component size area is 'x' then a minimal of 15-20% extra board space is required more or so. This argument would be helpful in tackling customers who keep asking for smaller PCBs to fit their enclosures.

I understand that this could be very vague, but I would appreciate any points you can kindly share please.

\$\endgroup\$
  • 2
    \$\begingroup\$ Hmm some requirements have a big impact on how you can use the board space, sometimes you need a 10 mm gap around a specific part of the circuit (isolation) which results in way more board space usage than in other cases. Maybe some clearance requirements can be used to calculate a minimal distance between components which would give a lower boundary for the extra needed space, but then the number of components plays a role as it is different placing 100 small ones compared to 1 big one. \$\endgroup\$ – Arsenal Oct 23 '15 at 9:19
  • 1
    \$\begingroup\$ I assume the area for each part includes the footprint and spacing between components. That would give an optimistic lower-bound. A crude 'guesstimate' for the track area might be count the number of pins, and assume a track connects two pins. Assume all tracks are on the same side as the components. Assume tracks on average are 1/2 the shortest dimension of the board, each is track-width + 1 space-width wide. This give an area for tracks estimate. You might be able to do better by routing tracks under parts, or using vias to reach the other side, but vias consume space, and it's a SWAG. \$\endgroup\$ – gbulmer Oct 23 '15 at 9:29
  • \$\begingroup\$ I voted to close because I don't think there's a general rule. Any real answer would be a thick book describing how best to layout a board including info on high voltage, high current, temperature, heat removal, etc. \$\endgroup\$ – JRE Oct 23 '15 at 10:02
  • \$\begingroup\$ @JRE. Thank u for your opinion. You say "You dont think there is a general rule". This post has been around for like 1 hour or so. Let others put in some of their thoughts and if indeed my post is useless close it. I did mention that the post could be vague. \$\endgroup\$ – Board-Man Oct 23 '15 at 10:03
3
\$\begingroup\$

I think connection density (number of connections divided by PCB area) is used as proxy metric. There is of course no general rule for what density is the most appropriate, but you can occasionally find statistics for certain types of electronics, and compare yours with that (if appropriate, e.g. don't compare an ATX power supply with a mobile phone). An example from Economics of Electronic Design, Manufacture and Test (1994):

enter image description here

\$\endgroup\$
  • 1
    \$\begingroup\$ As addendum, for stuff where component density really matters a lot, like satellites, people do crazy shit like run genetic algorithms on thousands of layouts to select the one that minimizes area (among other objectives); e.g. books.google.com/books?id=U2LLBQAAQBAJ&pg=PA383 \$\endgroup\$ – Fizz Oct 23 '15 at 12:14
4
\$\begingroup\$

enter image description here

Obviously if you have no traces between components, parts can be spaced extremely tight. The top figure a) assumes a clearance of 0.15 mm, or 6 mils, which is the default for Eagle. The middle figure b) includes traces and vias again with spacing and trace size assumed to be 6 mil..

It's possible to make a reasonably good estimate for a board, but it will take a fair amount of work if traces have to be accounted for. If assumptions can be made about the number of traces (because it's a multi-layer board), then things become much easier.

For one component, the space required (assuming no traces or vias) is:

$$A = (l+c)\times(w+c)$$

where A is the area, l = length, w = width, and c is clearance.

For parts with at least one trace on one or both sides, it is:

$$A = (l + c + t2c + x2c)\times(w + c + y2c + z2c)$$

where A is the area, l = length, w = width, and c is clearance, t and x are the number of traces or vias across one or both ends, and y and z are the number of traces or vias along one or both of the sides. A via feeding into a resistor counts the same as a trace.

This could be done either in mm or mils, as long as the same system is used for everything. I'm assuming the clearance between traces and the trace width are the same.

If this is going to be a multi-layer board, then one can assume almost all of the traces will be on the bottom or middle layers and the top will be just vias, like figure c). In that case, as a first approximation just assume a via for each pin.

For large parts, like a µC which has a lot of pins spaced closely together, allow for two rows of vias. I'm not allowing for the fact that parts like that can have several vias underneath them and/or there are no pads visible (like a QFN package).

So now you have a lot less work -- just the area of the components, plus the room needed for the vias. The formulas reduce to these four simplified ones (no estimating the number of or likely placement of traces):

For rectangular parts with pads at each end, like resistors or caps:

$$A = (l+5c)\times(w+c)$$

For rectangular parts, with pins along the side (like SOIC packages etc.):

$$A = (l+c)\times(w+5c)$$

For rectangular parts, with pins along the sides and at the ends:

$$A = (l+5c)\times(w+5c)$$

For large square components, with pins all around:

$$A = (l+9c)^{2}$$

These assume there are hardly any traces on the component side(s); i.e. at least a four layer board with two signal layers. So these areas are fixed for each type of component; they don't vary by the number of traces.

If this was fed into a spreadsheet, with formulas set up for each component size (like a 0805 resistor or cap), it won't be too bad. You only need one value for each type of component. Count the number of each different size and add them up (e.g. both resistors and capacitors may come in 0603 packages -- you don't need separate entries for each).

Remember to allow space for the pads at either end of a component -- for example the resistor length only includes the body. A 0603 (1608 metric) resistor is 1.6 mm long by 0.8 mm wide, but including the pads, it is 2.8 mm long. These footprints will be defined in the component library for your PCB layout program.

So for a 0603 resistor, the area needed (including the pads and vias at each end) would be:

$$A = (l+5c)\times(w+c)$$

$$A = (2.1 + 5\times 0.15)\times(0.9 + .15) = 2.85\times 1.05 = 2.99\space mm^{2}$$

The area of just the body of the resistor (1.6 mm x 0.8 mm) is a little less than half that -- 1.28 mm². The footprint (including the pads) is 2.1 mm x 0.8 mm or 1.68 mm². So using either of those plus a fudge factor of even 20% is not going to work. The minimum clearances needed for just one via or trace constituent almost half of the area.

With smaller parts, the clearances dominate even more. For a 0201 resistor (0603 metric, i.e. 0.6 mm by 0.3 mm), the required area is:

$$A = (0.75 + 5\times 0.15)\times(0.3 + .15) = 1.5\times 0.45 = 0.68\space mm^{2}$$

compared to a component size of 0.18 mm² and a footprint of 0.22 mm². In this case, the footprint (including recommended pad area) only accounts for 1/3 of the needed area.

With a spreadsheet you could vary the numbers around (making some worst and best case estimates) and at least get a rough idea. Perhaps change all of the resistors from 0603's to 0201's and see what difference that makes -- or go from a 100-pin µC to a 64-pin one -- or change the clearance size.

\$\endgroup\$
  • \$\begingroup\$ Small correction : 8 mil equals 203.2 µm. 0.15 mm are a bit smaller than 6 mil (152.4 µm) \$\endgroup\$ – Grebu Oct 23 '15 at 14:41
1
\$\begingroup\$

Im doing a generic PCB design, and working on specifying an enclosure to match a schematic, and I do use a general rule.

As background - our designs do not require isolation or large traces on the board. We are working with 3-5VDC circuits that are self contained and do basic logic. These are generally mounted into an enclosure and have to match the enclosure size. Managing the two processes in parallel is advantageous as it can save design iterations and accelerate the time to market.

The approach is that to add up all the the area that will be consumed by non passive components and double it.

So for example, if I need 5,000mm2 for components, we estimate 10,000mm2 for the board size. Take the square root of that and there is an estimate of the board size if it is square.

This allows for doing rough comparisons to enclosures. This at least gets us in the ballpark, and we work forward from there. It allows the process to go in parallel between the schematic and other areas of the design.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.