![]() Hopefully, that gets the idea across and you can manage to extend it to 5 bits. Then you will need \$N-1\$ adder tiers, arranged something like the following 4-bit example (I used 4 bits because 5 bits would have taken too much room on the page.) It's just repetitive.ĮDIT: For an unsigned \$N\times N\$ multiplier, where \$N\ge 1\$, you will need \$N\times N\$ AND gates to create the \$N\$ partial products for the first stage. Since half adders and full adders are made from simple gates, too, you should be able to generate the entire thing in very simple gates. The carry will then ripple over to the next full-adder needed, as you add the remaining 3 bits of \$PP_0\$ to the associated ones of the lane-shifted \$PP_1\$. The algorithm used here is a simple one that uses repeated addition. This will result in \$R_1\$ as the next lowest order result bit. Implementation of 4 bit array multiplier using Verilog HDL and its testing on the Spartan 2 FPGA The aim here is to take you through the design and implementation steps of FPGA implementation for 4-bit binary multiplier. At this point, you need to use a half-adder to add the next lowest order bit of \$PP_0\$ (now the lowest bit of the shown 4-bit bus) to the lowest order bit of \$PP_1\$. However, I decided to show that the lowest order bit of \$PP_0\$ is actually the lowest order bit of the result. But I show it as just 5 bits since there's no reason wasting a "0" when all you are doing is a lane-shift. ![]() \$PP_1\$ is shifted left by 1 bit, so you might think of it as a 6-bit binary value. Note that each bit of the 5-bit value of \$B\$ is used to either gate, or not gate, the 5-bit value of \$A\$ to its associated partial product term. Simulate this circuit – Schematic created using CircuitLab ![]() To build on Eugene's comment, look at the following schematic: ![]()
0 Comments
Leave a Reply. |