CAN Filter and Mask Calculator

What a CAN acceptance filter does

A CAN controller sees every frame on the bus. Acceptance filtering is the hardware deciding, before any interrupt fires, whether a frame is worth handing to the software. The test is one comparison:

(received_id XOR filter) AND mask == 0

Wherever the mask says to compare, the identifier must equal the filter. Wherever it says to ignore, the identifier can be anything. That is the whole mechanism, and everything awkward about filters follows from it being this simple.

The mask convention is not universal

Two families of parts disagree about what a mask bit of 1 means, and nothing in a hex value tells you which one you are holding:

• 1 compares — STM32 bxCAN, MCP2515, and most modern controllers. A mask of all ones matches exactly one identifier.
• 1 ignores — SJA1000 in PeliCAN mode and parts that follow it. A mask of all ones accepts everything.

The failure this causes is unhelpfully quiet. A mask taken from the wrong reference does not warn, does not error, and usually does not partially work: it either blocks everything or accepts everything, and both look like a wiring or bitrate problem long before anyone suspects the filter. Setting the convention is the first control on this page for that reason.

Why one filter cannot match an arbitrary list

A filter compares bit positions, not values. It can only compare a bit when every identifier you want agrees on it, and it must ignore any bit where they differ. That makes the tightest filter for a set unique — there is nothing to choose — and it makes a certain amount of waste unavoidable.

Take 0x100 and 0x200. They differ in two bit positions, so both must be ignored, and ignoring two bits admits four identifiers: 0x000, 0x100, 0x200 and 0x300. There is no filter and mask pair that accepts the two you asked for and rejects the two you did not. The only options are to accept the extra frames and discard them in software, or to use a second filter bank.

Contiguous aligned blocks are the happy case. 0x100 through 0x103 differ only in the low two bits, so a filter of 0x100 with the low two bits ignored admits exactly those four. This is why identifier schemes that allocate a node a block of consecutive IDs are easier to filter than schemes that scatter them, and it is worth knowing while the identifiers are still being assigned rather than afterwards.

Reading the result

The number that matters is how many identifiers the filter admits against how many you wanted. A filter admitting four to deliver two is usually fine — two spurious frames per cycle cost almost nothing. A filter admitting 512 to deliver five is a filter that is not doing its job, and on a loaded bus it will show up as interrupt load rather than as a functional bug.

When the waste is too high, the set has to be split. Group the identifiers that share bits, give each group its own filter bank, and check each one separately. If the controller offers list mode, a small scattered set fits there exactly, with no false accepts at all.

Filtering is load management, not validation

It is worth being clear about what a filter is for. It reduces the frames the CPU has to look at, which on a busy bus is the difference between a node that keeps up and one that drops messages. It is not a safety or security boundary: anything that passes the filter still has to be checked, and a hostile or faulty node can transmit any identifier it likes.

Nor does a filter change what is on the wire. Every node still receives every frame electrically, still participates in arbitration, and still signals errors. Filtering happens after all of that.

A note on counts

An open filter on extended identifiers admits 536,870,912 of them, and this page says so rather than rounding it off to "everything". The count is the honest measure of how much a filter is actually constraining, and on a 29-bit identifier the numbers get large quickly: ignoring ten bits already admits over a thousand identifiers. Listing them all is not useful, so the first thirty-two are shown and the total is stated exactly.