Problem set is in the attached picture.
My working is as such:
This is a CMOS OR AND INVERT (OAI) gate. Traditionally, this has 4 transistors for 2 input. From what I read from text, you would need 2^n transistors (PMOS+NMOS), where n is number of inputs.
The logic expression is as follows: Z=[(B+C)A]'
So for 2 input NOR and AND logic gate, you would need 4+4=8.
Inverter/Nand require 2 transistors (PMOS+NMOS). Which would make the total 10 transistors.
How do you get 6 as the answer? Does the question mean 6 set of transistors? Or does it involve open-drain/collector NMOS where the PMOS is removed and pull up resistor attached to it?
I have trouble understanding the concept. The answer is 6 minimum transistors
My working is as such:
This is a CMOS OR AND INVERT (OAI) gate. Traditionally, this has 4 transistors for 2 input. From what I read from text, you would need 2^n transistors (PMOS+NMOS), where n is number of inputs.
The logic expression is as follows: Z=[(B+C)A]'
So for 2 input NOR and AND logic gate, you would need 4+4=8.
Inverter/Nand require 2 transistors (PMOS+NMOS). Which would make the total 10 transistors.
How do you get 6 as the answer? Does the question mean 6 set of transistors? Or does it involve open-drain/collector NMOS where the PMOS is removed and pull up resistor attached to it?
I have trouble understanding the concept. The answer is 6 minimum transistors
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