cmosdigital-logiclayout
Can someone please clarify me what the rightmost strip of this layout is?
I know the red ones are poly, the squares are contacts, in green the n-well, in brown the p-diffusion and in blue with diagonal lines is metal 1. But what is that rightmost long strip connected to the output of the cell (output of the last inverter). Is it metal too? Why is the texture different?
Extra comments:
They are asynchronous PRESET and CLEAR (active low). In this setup, there is the constraint that PRE and CLR cannot both be active (low) at the same time (or else both Q and Q' are 1). You could give one priority if you wanted by modifying the topology a bit.
Additionally, if either PRE or CLR are active (low) at the rising clock edge, the output states will not necessarily be inverted (as you pointed out). But, because the edge detector pulse is narrow, the PRE or CLR will quickly propagate through Q and Q' after the edge detector pulse ends assuming they are held through the length of the pulse.
In essence, whichever 'holds' longer will win: either the data will pass through if the edge detector pulse stays active longer than the PRE/CLR signals stay active (low), or the PRE/CLR signals will stay active longer than the edge detector pulse and over-write whatever D put in there.
In practice, these constraints would be represented the library characterization files. There would be a setup and hold arc defining the timing between the clock, d, PRE, and CLR to prevent any unwanted states.
Or, if the circuit was used in a more custom way, its designers would need to make sure they understood the operation of the pulse latch (not really a flip flop, imo) and how to properly enable or reset it.
You can tell the PRE and CLR are asynchronous easily by looking at the signal flow. PRESET and CLR pass to the output without any gating by the CLK signal (which would make them synchronous).
To prove this to yourself, assume the clock is not toggling, and both PRESET and CLEAR are '1' (inactive). Also, this circuit cannot have both PRE and CLR low at the same time.
Also assume the initial states for Q and Q':
PRE = 1, CLEAR = 1
Q = 1, Q' = 0
As long as you don't touch anything, everything will stay as it is (latched).
Now, pull CLR down to '0' without toggling the clock or data.
As shown in the image above, this clears Q from '1' to '0'. And, clock has not toggled. This means the circuit is asynchronous. From here the CLR signal can be inactivated (returned high) and the circuit will still hold its state.
An easy well to tell is the gating of the clock relative to the clear/enable/reset signals.
Best Answer
The rightmost strip is likely a higher metal layer than the metal strip you see in blue with diagonal lines. In this case, it is likely metal 2 (M2). By going up M2, they are able to cross over the M1 horizontal strip at the top without forming a connection. Another reason we can assume this is likely metal 2 is that it is a vertical strip. The convention for many PDKs is to use even metal layers for vertical routing and odd layers for horizontal routing.
Image Source: https://my.eng.utah.edu/~cs6710/handouts/floorplanning_3LM.pdf