ALU Organization Michael Vong Louis Young Rongli Zhu Dan
Overall ALU Organization The output lines (Y3 … Y0) run all the way through
ALU Organization: One Function Per Column Control signals will enable all transmission gates in a column
ALU Organization: One Bit Per Row Only one transmission gate in a row will be turned on. Only one function will drive Y.
Adder Logic Design
BK Cell States Our adder uses BK Cells. For each column of addition, there are three possible states (0 + 1) or (1 + 0) is carry propagate = P (1 + 1) is carry generate = G (0 + 0) is carry kill = K
BK Cell Truth Table More Significant InputLess Significant InputOutput KKK KPK KGK PKK PPP PGG GKG GPG GGG Each BK cell looks at the carry status of two networks and generate a single carry status.
BK Cell Boolean Equation Y1 = BD + AD + AB Y0 = BC + AC + AB Note: The encoding used: G = 11, K = 00, and P = 10 or 01 Y1 and Y0 are the same Boolean function. Just do the layout for Y1 and replicate it twice to get a BK cell This is the same function as the ripple adder’s carry out
Using BK Cells to make an Adder There is only one rule to using BK cells: To compute the carry of C i, you must have enough BK cells to reach all preceding bits, from bit (i-1) to bit 0. You can have just enough BK cells to compute the final carry, or you can have lots of BK cells to compute all carries.
BK Cell Example (part 1 of 2) If you just want the carry out of an 8 bit addition operation, then you will need 7 BK cells.
BK Cell Example (part 2 of 2) Note that the first input into the first BK cell on the right (the C and D of the red box), must be either G (11) or K (00). Let say the number we are adding are called A and B, this input is C = D = A 0 B in + A 0 C in + B 0 C in. The final output, the Y1 and Y0 of the yellow box, is also either G(11) or K(00).
Our Adder’s BK Cells This adder is around the same speed as a ripple adder. The entry into the red cell has the same delay as a BK cell. Red cell’s C = D = A 0 B 0 + A 0 C in + B 0 C in. So from input to C 3 there are really 3 BK stages. Each stage is the same as a carry out of a ripple.
Other BK Cell Examples Our adder does not benefit from the BK cells because it’s only 4 bits wide. Larger adders do benefit. Screen shots are taken from: Sklanski's adder: Problem: high fan-out for the lowest C 8 BK cell.
Another BK Cell Example Kogge & Stone adders: This one has more BK cells but less fan-out.
Our Adder --- After the BK Tree After the carries are generated, add them to the xor sums. If we are add A and B, and let the answer be SUM: SUM 0 = (A 0 xor B 0 ) xor C 0 SUM 1 = (A 1 xor B 1 ) xor C 1 SUM 2 = (A 2 xor B 2 ) xor C 2 SUM 3 = (A 3 xor B 3 ) xor C 3 This operation of two xor gates is called the “summer” in our adder.
Summer Schematic The idea is to have A and B preset a path so that when C is correctly set, it will show up at Y really fast. It didn’t work out that well. Y = (A XOR B) XOR C
Adder Logic Summary A tree of BK cells are used to compute all of the carries. The final sum for the i-th bit is A i xor B i xor C i, where A and B are the numbers that we are adding, and C is the carry computed by the BK cells.
Confirming the Logic with Verilog module bk(Y1, Y0, A, B, C, D); input A, B, C, D; output Y1, Y0; assign Y1 = (B&D) | (A&D) | (A&B); assign Y0 = (B&C) | (A&C) | (A&B); endmodule module summer(Y, A, B, C); input A, B, C; output Y; assign Y = (~A & ~B & C) | (~A & B & ~C) | (A & ~B & ~C) | (A & B & C); endmodule
adder.v (page 1) module adder(SUM, COUT, A, B, CIN); input [3:0] A, B; input CIN; output [3:0] SUM; output COUT; wire c1, c2, c3, c4; assign c1 = (A[0] & B[0]) | (A[0] & CIN) | (B[0] & CIN); wire bk1_0, bk1_1, bk2_0, bk2_1; wire bk3_0, bk3_1, bk4_0, bk_1; bk bk1(bk1_1, bk1_0, A[1], B[1], c1, c1); bk bk2(bk2_1, bk2_0, A[2], B[2], bk1_1, bk1_0); bk bk3(bk3_1, bk3_0, A[3], B[3], A[2], B[2]); bk bk4(bk4_1, bk4_0, bk3_1, bk3_0, bk1_1, bk1_0);
adder.v (page 2) assign c4 = bk4_1; assign c3 = bk2_1; assign c2 = bk1_1; assign COUT = c4; summer s0(SUM[0], A[0], B[0], CIN); summer s1(SUM[1], A[1], B[1], c1); summer s2(SUM[2], A[2], B[2], c2); summer s3(SUM[3], A[3], B[3], c3); endmodule
test.v module testbench; wire [3:0] SUM; wire COUT; reg [3:0] A, B; reg CIN; adder adder1(SUM, COUT, A, B, CIN); reg [4:0] i, j, k; initial begin CIN = 4'd1; for(i = 0; i < 16; i = i + 1) begin for(j = 0; j < 16; j = j + 1) begin A[3:0] = i[3:0]; B[3:0] = j[3:0]; #20; k = i + j + 1;
test.v (page 2) if(SUM[3:0] != k[3:0]) begin $display("At time %t, A = %d, B = %d, CIN = %d, SUM = %d, COUT = %d \n", $time, A, B, CIN, SUM, COUT); end else begin $display("A = %d, B = %d, CIN = %d tested \n", A, B, CIN); end $display("end of test \n"); end endmodule
Simulation with ModelSim
Adder Circuitry
Layout Guidelines PMOS: L = 0.6 um, W = 5.4 um NMOS: L = 0.6 um, W = 3 um Transistor Sizes (most of the time): Cell height: Total Height: 27 um VDD and GND path width: 1.5 um
Cell Hierarchy zproj_adder4b –zproj_bk zproj_bk_y1 –zproj_summer Zproj_mux2b
The bk_y1 Cell AOI Schematic Recall that the BK cell has a Y1 and Y0
The bk_y1 layout Note that metal 2 can route vertically through almost all of the cell.
The Complete BK Cell Schematic
The bk Cell Layout View 1
The bk Cell Layout View 2 A and B is the same all the way across while C and D swap rows Y1 is in the middle while Y0 is at the far right
Multiplexer Schematic The multiplexer is the basic cell for the summer
Multiplexer Schematic
Multiplexer Layout Note that vertical routing of metal 2 is possible in less than half of the cell.
Multiplexer Test Setup Note how ideal sources are fed directly into the mux
Multiplexer Power Usage
Multiplexer Test Result The load capacitance is 30 fF SONEZEROYTime (ns) 55-> ?? 50-> ?? 00-> ?? 05-> ?? 0-> >5055-> > >0050->50.133
Multiplexer Test Results (Page 2) Power = W/cm 2 The first group of results highlighted in red cells turned out to be inaccurate. The ONE and ZERO lines are not gate terminals. When the path way is set (S held steady), the rate at which the output changes is actually proportional to the change in input. The output is changing rapidly in the test because the input is an ideal voltage source with a rise time of 200 ps. A more realistic switching time can be obtained by passing the ideal input through two inverters before sending it to the “ONE” or “ZERO” line of the multiplexer.
Summer Schematic
Summer Layout View 1 The (NOT C) is labeled as C’
Summer Layout View 2 The right most Y is the sum
Putting the Support Cells Together to Form a 4 Bit Adder
Adder Schematic Page 1 This is the BK tree part of the adder
Adder Schematic Page 2 Output from the BK tree, and the original A and B bits are passed into the summer cells.
Adder Layout Note the long distance metal2 vertical routing
Adder Layout View 2
Adder Layout Input View
Adder Layout Output View
Adder Layout Area
Adder Test Setup I used VDC for 1 and VPULSE for Each output pin is loaded with 30 fF capacitors.
Adder Testing Results ABCINCOUT > 2.5v SUM3 < 2.5V SUM3 < 1V ns3.264 ns4.192 ns ns3.111 ns4.039 ns ns2.33 ns3.150 ns ns1.906 ns2.726 ns ns2.22 ns3.069 ns
Adder Worst Case A = 1111 B = 0001 CIN = 0 Note the sagging SUM3 output.
SUM 3 output is sagging because it is 4 Transistor Away from GND
Speeding up SUM3’s Rate of Change with a Multiplexer The capacitor now charges and discharges faster because it is closer to VDD and GND. However, the multiplexer will be an extra delay Effect of using an extra multiplexer at the output: -Y_fast will arrive at 2.5V 0.35 ns later. -Y_fast will arrive at 1V ns earlier.
Timing without a Multiplexer Buffer SUM3 is changes slowly if its output is used to charge 30fF of capacitance directly. Note the time scale for this test goes up to 16 ns.
Timing with a Multiplexer Buffer Note the time scale for this test goes up to just 12 ns. Passing SUM3’s output to a multiplexer buffer delays the wave but increase the rate of change.
ALU Schematic
DRC of ALU
Extracted View
LVS