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VDD A M5 B A B C M6 C M4 M3 M2 M7 D M8 A R5 B A R3 B R2 C D M1 R1 D C3 VDD R6 C R4 R7 D F CL R8 in Microsoft Office Make ECC200 in Microsoft Office VDD A M5 B A B C M6 C M4 M3 M2 M7 D M8 A R5 B A R3 B R2 C D M1 R1 D C3 VDD R6 C R4 R7 D F CL R8




How to generate, print barcode using .NET, Java sdk library control with example project source code free download:
VDD A M5 B A B C M6 C M4 M3 M2 M7 D M8 A R5 B A R3 B R2 C D M1 R1 D C3 VDD R6 C R4 R7 D F CL R8 using barcode generating for microsoft office control to generate, create data matrix ecc200 image in microsoft office applications. code 39 Figure 6.11 Four input NAND gate and its RC model. Section 6.2 Static CMOS Design The propagation del ay can be computed using the Elmore delay model and is approximated as:. t pHL = 0.69 ( R 1 Data Matrix 2d barcode for None C 1 + ( R 1 + R 2 ) C 2 + ( R 1 + R 2 + R 3 ) C 3 + ( R 1 + R 2 + R 3 + R 4 ) C L ). (6.3). Notice that the res Data Matrix 2d barcode for None istance of M1 appears in all the terms, which makes this device especially important when attempting to minimize delay. Assuming that all NMOS devices have an equal size, Eq. (6.

3) simplifies to. t pHL = 0.69R N ( C + 2 C + 3 C + 4 C L ) 1 2 3. (6.4). Example 6.4 A Four- Input Complementary CMOS NAND Gate In this example, the intrinsic propagation delay of the 4 input NAND gate (without any loading) is evaluated using hand analysis and simulation. Assume that all NMOS devices have a W/L of 0.

5 m/0.25 m, and all PMOS devices have a device size of 0.375 m/0.

25 m. The layout of a four-input NAND gate is shown in Figure 6.12.

The devices are sized such that the worst case rise and fall time are approximately equal (to first order ignoring the internal node capacitances). Using techniques similar to those employed for the CMOS inverter in 3, the capacitances values can be computed from the layout. Notice that in the pull-up path, the PMOS devices share the drain terminal in order to reduce the overall parasitic contribution to the output.

Using our standard design rules, the area and perimeter for various devices can be easily computed as shown in Table 6.1 In this example, we will focus on the pull-down delay, and the capacitances will be computed for the high-to-low transition at the output. While the output makes a transition from VDD to 0, the internal nodes only transition from VDD-VTn to GND.

We would need to linearize the internal junction capacitances for this voltage transition, but, to simplify the analysis, we will use the same Keff for the internal nodes as for the output node.. GND A B C D Figure 6.12 Layout a four-input NAND gate in complementary CMOS. DESIGNING COMBINATIONAL LOGIC GATES IN CMOS Table 6.1 Area and perimeter of transistors in 4 input NAND gate. Transistor 1 2 3 4 5 6 7 8 W ( m) 0.

5 0.5 0.5 0.

5 0.375 0.375 0.

375 0.375 AS ( m2) 0.3125 0.

0625 0.0625 0.0625 0.

296875 0.171875 0.171875 0.

296875 AD ( m2) 0.0625 0.0625 0.

0625 0.3125 0.171875 0.

171875 0.171875 0.171875 PS ( m) 1.

75 0.25 0.25 0.

25 1.875 0.875 0.

875 1.875. 6 . PD( m) 0.25 0.25 0.

25 1.75 0.875 0.

875 0.875 0.875.

It is assumed that DataMatrix for None the output connects to a single, minimum-size inverter. The effect of intra-cell routing, which is small, is ignored. The various contributions are summarized in Table 6.

2. For the NMOS and PMOS junctions, we use Keq = 0.57, Keqsw = 0.

61, and Keq = 0.79, Keqsw = 0.86, respectively.

Notice that the gate-to-drain capacitance is multiplied by a factor of two for all internal nodes and the output node to account for the Miller effect (this ignores the fact that the internal nodes have a slightly smaller swing due to the threshold drop).. Table 6.2 Computati on of capacitances for high-to-low transition at the output. The table shows the intrinsic delay of the gate without extra loading.

Any fan-out capacitance would simply be added to the CL term. Capacitor C1 Contributions (H L) Cd1 + Cs2 + 2 * Cgd1 + 2 * Cgs2 Value (fF) (H L) (0.57 * 0.

0625 * 2+ 0.61 * 0.25 * 0.

28) + (0.57 * 0.0625 * 2+ 0.

61 * 0.25* 0.28) + 2 * (0.

31 * 0.5) + 2 * (0.31 * 0.

5) = 0.85fF (0.57 * 0.

0625 * 2+ 0.61 * 0.25 * 0.

28) + (0.57 * 0.0625 * 2+ 0.

61 * 0.25* 0.28) + 2 * (0.

31 * 0.5) + 2 * (0.31 * 0.

5) = 0.85fF (0.57 * 0.

0625 * 2+ 0.61 * 0.25 * 0.

28) + (0.57 * 0.0625 * 2+ 0.

61 * 0.25* 0.28) + 2 * (0.

31 * 0.5) + 2 * (0.31 * 0.

5) = 0.85fF.
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