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2.4.1 Pushing Bubbles

The AOI and OAI logic cells can be built using a single stage in CMOS using series–parallel networks of transistors called stacks. Figure 2.13 illustrates the procedure to build the n -channel and p -channel stacks, using the AOI221 cell as an example.

FIGURE 2.13  Constructing a CMOS logic cell—an AOI221. (a) First build the dual icon by using de Morgan’s theorem to “push” inversion bubbles to the inputs. (b) Next build the n -channel and p -channel stacks from series and parallel combinations of transistors. (c) Adjust transistor sizes so that the n- channel and p -channel stacks have equal strengths.

Here are the steps to construct any single-stage combinational CMOS logic cell:

1. Draw a schematic icon with an inversion (bubble) on the last cell (the bubble-out schematic). Use de Morgan’s theorems —“A NAND is an OR with inverted inputs and a NOR is an AND with inverted inputs”—to push the output bubble back to the inputs (this the dual icon or bubble-in schematic).
2. Form the n -channel stack working from the inputs on the bubble-out schematic: OR translates to a parallel connection, AND translates to a series connection. If you have a bubble at an input, you need an inverter.
3. Form the p -channel stack using the bubble-in schematic (ignore the inversions at the inputs—the bubbles on the gate terminals of the p -channel transistors take care of these). If you do not have a bubble at the input gate terminals, you need an inverter (these will be the same input gate terminals that had bubbles in the bubble-out schematic).

The two stacks are network duals (they can be derived from each other by swapping series connections for parallel, and parallel for series connections). The n -channel stack implements the strong '0's of the function and the p -channel stack provides the strong '1's. The final step is to adjust the drive strength of the logic cell by sizing the transistors.

2.4.2 Drive Strength

Normally we ratio the sizes of the n -channel and p -channel transistors in an inverter so that both types of transistors have the same resistance, or drive strength . That is, we make b _n = b _p . At low dopant concentrations and low electric fields m _n is about twice m _p . To compensate we make the shape factor, W/L, of the p -channel transistor in an inverter about twice that of the n -channel transistor (we say the logic has a ratio of 2). Since the transistor lengths are normally equal to the minimum poly width for both types of transistors, the ratio of the transistor widths is also equal to 2. With the high dopant concentrations and high electric fields in submicron transistors the difference in mobilities is less—typically between 1 and 1.5.

Logic cells in a library have a range of drive strengths. We normally call the minimum-size inverter a 1X inverter. The drive strength of a logic cell is often used as a suffix; thus a 1X inverter has a cell name such as INVX1 or INVD1. An inverter with transistors that are twice the size will be an INVX2. Drive strengths are normally scaled in a geometric ratio, so we have 1X, 2X, 4X, and (sometimes) 8X or even higher, drive-strength cells. We can size a logic cell using these basic rules:

• Any string of transistors connected between a power supply and the output in a cell with 1X drive should have the same resistance as the n -channel transistor in a 1X inverter.
• A transistor with shape factor W ₁ /L ₁ has a resistance proportional to L ₁ /W ₁ (so the larger W ₁ is, the smaller the resistance).
• Two transistors in parallel with shape factors W ₁ /L ₁ and W ₂ /L ₂ are equivalent to a single transistor (W ₁ /L ₁ + W ₂ /L ₂ )/1. For example, a 2/1 in parallel with a 3/1 is a 5/1.
• Two transistors, with shape factors W ₁ /L ₂ and W ₂ /L ₂ , in series are equivalent to a single 1/(L ₁ /W ₁ + L ₂ /W ₂ ) transistor.

For example, a transistor with shape factor 3/1 (we shall call this “a 3/1”) in series with another 3/1 is equivalent to a 1/((1/3) + (1/3)) or a 3/2. We can use the following method to calculate equivalent transistor sizes:

• To add transistors in parallel, make all the lengths 1 and add the widths.
• To add transistors in series, make all the widths 1 and add the lengths.

We have to be careful to keep W and L reasonable. For example, a 3/1 in series with a 2/1 is equivalent to a 1/((1/3) + (1/2)) or 1/0.83. Since we cannot make a device 2 l wide and 1.66 l long, a 1/0.83 is more naturally written as 3/2.5. We like to keep both W and L as integer multiples of 0.5 (equivalent to making W and L integer multiples of l ), but W and L must be greater than 1.

In Figure 2.13(c) the transistors in the AOI221 cell are sized so that any string through the p -channel stack has a drive strength equivalent to a 2/1 p -channel transistor (we choose the worst case, if more than one transistor in parallel is conducting then the drive strength will be higher). The n -channel stack is sized so that it has a drive strength of a 1/1 n -channel transistor. The ratio in this library is thus 2.

If we were to use four drive strengths for each of the AOI family of cells shown in Table 2.10, we would have a total of 224 combinational library cells—just for the AOI family. The synthesis tools can handle this number of cells, but we may not be able to design this many cells in a reasonable amount of time. Section 3.3, “Logical Effort,” will help us choose the most logically efficient cells.

2.4.3 Transmission Gates

Figure 2.14(a) and (b) shows a CMOS transmission gate ( TG , TX gate, pass gate, coupler). We connect a p -channel transistor (to transmit a strong '1') in parallel with an n -channel transistor (to transmit a strong '0').

FIGURE 2.14 CMOS transmission gate (TG). (a) An n- channel and p -channel transistor in parallel form a TG. (b) A common symbol for a TG. (c) The charge-sharing problem.

We can express the function of a TG as  

Z = TG(A, S) ,

(2.27)

but this is ambiguous—if we write TG(X, Y), how do we know if X is connected to the gates or sources/drains of the TG? We shall always define TG(X, Y) when we use it. It is tempting to write TG(A, S) = A · S, but what is the value of Z when S ='0' in Figure 2.14(a), since Z is then left floating? A TG is a switch, not an AND logic cell.

There is a potential problem if we use a TG as a switch connecting a node Z that has a large capacitance, C _BIG , to an input node A that has only a small capacitance C _SMALL (see Figure 2.14c). If the initial voltage at A is V _SMALL and the initial voltage at Z is V _BIG , when we close the TG (by setting S = '1') the final voltage on both nodes A and Z is

		C BIG V BIG + C SMALL V SMALL
V F	=	–––––––––––––––––––––––––	.	(2.28)
		C BIG + C SMALL

Imagine we want to drive a '0' onto node Z from node A. Suppose C _BIG = 0.2 pF (about 10 standard loads in a 0.5 m m process) and C _SMALL = 0.02 pF, V _BIG = 0 V and V _SMALL = 5 V; then

		(0.2 ¥ 10 –12 ) (0) + (0.02 ¥ 10 –12 ) (5)
V F	=	––––––––––––––––––––––––––––	=	0.45 V .	(2.29)
		(0.2 ¥ 10 –12 ) + (0.02 ¥ 10 –12 )

This is not what we want at all, the “big” capacitor has forced node A to a voltage close to a '0'. This type of problem is known as charge sharing . We should make sure that either (1) node A is strong enough to overcome the big capacitor, or (2) insulate node A from node Z by including a buffer (an inverter, for example) between node A and node Z. We must not use charge to drive another logic cell—only a logic cell can drive a logic cell.

If we omit one of the transistors in a TG (usually the p -channel transistor) we have a pass transistor . There is a branch of full-custom VLSI design that uses pass-transistor logic. Much of this is based on relay-based logic, since a single transistor switch looks like a relay contact. There are many problems associated with pass-transistor logic related to charge sharing, reduced noise margins, and the difficulty of predicting delays. Though pass transistors may appear in an ASIC cell inside a library, they are not used by ASIC designers.

FIGURE 2.15 The CMOS multiplexer (MUX). (a) A noninverting 2:1 MUX using transmission gates without buffering. (b) A symbol for a MUX (note how the inputs are labeled). (c) An IEEE standard symbol for a MUX. (d) A nonstandard, but very common, IEEE symbol for a MUX. (e) An inverting MUX with output buffer. (f) A noninverting buffered MUX.

We can use two TGs to form a multiplexer (or multiplexor—people use both orthographies) as shown in Figure 2.15(a). We often shorten multiplexer to MUX . The MUX function for two data inputs, A and B, with a select signal S, is  

Z = TG(A, S') + TG(B, S) .

(2.30)

We can write this as Z = A · S' + B · S, since node Z is always connected to one or other of the inputs (and we assume both are driven). This is a two-input MUX (2-to-1 MUX or 2:1 MUX). Unfortunately, we can also write the MUX function as Z = A · S + B · S', so it is difficult to write the MUX function unambiguously as Z = MUX(X, Y, Z). For example, is the select input X, Y, or Z? We shall define the function MUX(X, Y, Z) each time we use it. We must also be careful to label a MUX if we use the symbol shown in Figure 2.15(b). Symbols for a MUX are shown in Figure 2.15(b–d). In the IEEE notation 'G' specifies an AND dependency. Thus, in Figure 2.15(c), G = '1' selects the input labeled '1'. Figure 2.15(d) uses the common control block symbol (the notched rectangle). Here, G1 = '1' selects the input '1', and G1 = '0' selects the input ' 1 '. Strictly this form of IEEE symbol should be used only for elements with more than one section controlled by common signals, but the symbol of Figure 2.15(d) is used often for a 2:1 MUX.

The MUX shown in Figure 2.15(a) works, but there is a potential charge-sharing problem if we cascade MUXes (connect them in series). Instead most ASIC libraries use MUX cells built with a more conservative approach. We could buffer the output using an inverter (Figure 2.15e), but then the MUX becomes inverting. To build a safe, noninverting MUX we can buffer the inputs and output (Figure 2.15f)—requiring 12 transistors, or 3 gate equivalents (only the gate equivalent counts are shown from now on).

Figure 2.16 shows how to use an OAI22 logic cell (and an inverter) to implement an inverting MUX. The implementation in equation form (2.5 gates) is  

ZN	=	A' · S' + B' · S
	=	[(A' · S')' · (B' · S)']'
	=	[ (A + S) · (B + S')]'
	=	OAI22[A, S, B, NOT(S)] .	(2.31)

(both A' and NOT(A) represent an inverter, depending on which representation is most convenient—they are equivalent). I often use an equation to describe a cell implementation.

FIGURE 2.16 An inverting 2:1 MUX based on an OAI22 cell.

The following factors will determine which MUX implementation is best:

1. Do we want to minimize the delay between the select input and the output or between the data inputs and the output?
2. Do we want an inverting or noninverting MUX?
3. Do we object to having any logic cell inputs tied directly to the source/drain diffusions of a transmission gate? (Some companies forbid such transmission-gate inputs —since some simulation tools cannot handle them.)
4. Do we object to any logic cell outputs being tied to the source/drain of a transmission gate? (Some companies will not allow this because of the dangers of charge sharing.)
5. What drive strength do we require (and is size or speed more important)?

A minimum-size TG is a little slower than a minimum-size inverter, so there is not much difference between the implementations shown in Figure 2.15 and Figure 2.16, but the difference can become important for 4:1 and larger MUXes.

2.4.4 Exclusive-OR Cell

The two-input exclusive-OR ( XOR , EXOR, not-equivalence, ring-OR) function is  

A1 ⊕ A2 = XOR(A1, A2) = A1 · A2' + A1' · A2 .

(2.32)

We are now using multiletter symbols, but there should be no doubt that A1' means anything other than NOT(A1). We can implement a two-input XOR using a MUX and an inverter as follows (2 gates):  

XOR(A1, A2) = MUX[NOT(A1), A1, A2] ,

(2.33)

where  

MUX(A, B, S) = A · S + B · S ' .

(2.34)

This implementation only buffers one input and does not buffer the MUX output. We can use inverter buffers (3.5 gates total) or an inverting MUX so that the XOR cell does not have any external connections to source/drain diffusions as follows (3 gates total):  

XOR(A1, A2) = NOT[MUX(NOT[NOT(A1)], NOT(A1), A2)] .

(2.35)

We can also implement a two-input XOR using an AOI21 (and a NOR cell), since  

XOR(A1, A2)	=	A1 · A2' + A1' · A2
	=	[ (A1 ·A2) + (A1 + A2)' ]'
	=	AOI21[A1, A2, NOR(A1, A2)],	(2.36)

(2.5 gates). Similarly we can implement an exclusive-NOR (XNOR, equivalence) logic cell using an inverting MUX (and two inverters, total 3.5 gates) or an OAI21 logic cell (and a NAND cell, total 2.5 gates) as follows (using the MUX function of Eq. 2.34):  

XNOR(A1, A2)	=	A1 · A2 + NOT(A1) · NOT(A2
	=	NOT[NOT[MUX(A1, NOT (A1), A2]]
	=	OAI21[A1, A2, NAND(A1, A2)] .	(2.37)