We are doing a project about a multiplier. Two of my teammates are assigned to make the transistor level designs of the adders, subtractors and few barrel shifters. My prof has told me to make a structural level Verilog modelling of the same multiplier. He then instructed " When your teammates reduce the number of transistors for the adders and subtractors, the boolean equation changes. Put those boolean equations in the Verilog Code and we'll have the optimized multiplier". Our team is slightly confused, since the Boolean equation has nothing to do with the transistors but just the logic to be implemented.

So TLDR ; If the number of transistors in a Static CMOS logic design are reduced, will the Boolean equation change significantly?

  • \$\begingroup\$ Ask your instructor. We can only make assumptions about what the instructor wants you to do. Asking here might get you some good advice but a poor grade because you didn't do what the instructor asked. \$\endgroup\$ Apr 2, 2021 at 11:51

1 Answer 1


Sure it does. An unsimplified boolean expression has redundant operations hidden inside it. If you implement that unsimplified boolean expression, then you will have extra logic gates/transistors implementing those redundant operations.

If you did things correctly, the end result doesn't change, but the path to get there does.

the Boolean equation has nothing to do with the transistors but just the logic to be implemented.

The boolean equation is used to pick the logic gates, and the logic gates are made of transistors.

  • \$\begingroup\$ For example, If there is a full adder whose sum and carry outputs are already in their simplified forms, will removing the transistors still simplify the already-existing boolean equation ? I'm sorry if i'm not able to express the doubt much clearly, This doubt is being difficult to express as is. Thanks in advance \$\endgroup\$ Apr 2, 2021 at 11:42
  • \$\begingroup\$ @KuchiYashwanth The only case I can think of there is that if the transistors you are removing are of the same type, in parallel and have the same input; Then that should not change the boolean expression because the transistor is serving double-duty for more than one gate. It would be a type of simplification beyond that which appears in the boolean expression, being at the transistor level which is lower level than both gates and boolean math. \$\endgroup\$
    – DKNguyen
    Apr 2, 2021 at 19:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.