# Number of CMOS transistors required for...

Discussion in 'Homework Help' started by jegues, Sep 29, 2010.

1. ### jegues Thread Starter Well-Known Member

Sep 13, 2010
732
45
See first figure attached for problem statement.

Second figure for my "attempt".

I can't figure out how to calculate the correct number of transistors for a given gate.

I figured well a NOT will require 1 CMOS transistor, and a NAND will require 2 CMOS transistors.

So if I do a NAND followed by a NOT I'll get an AND.

So a 2 input AND would be 3 transistors by this logic.

Then a 4 input AND would be 6 transistors, but the answer says it's 10 transistors.

Similarily, a NOR will require 2 CMOS transistors and if I follow this NOR by a NOT I'll get an OR.

So a 2 input OR would be 3 transistors.

Then a 3 input OR would be ? transistors.

I need a better method for finding out how many CMOS transistors are required for each logic operation.

Any ideas/suggestions/tips/hints?

Thanks again!

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2. ### jegues Thread Starter Well-Known Member

Sep 13, 2010
732
45
Bump, still can't figure this one out. Is there a systematic way of computing this?

3. ### EngIntoHW Member

Apr 24, 2010
128
0
How about drawing Karnaugh map for each function you wanna implmenet, find a minimized expression, of 2-input gates, and implement it that way.

4. ### jegues Thread Starter Well-Known Member

Sep 13, 2010
732
45

The function is already simplified. It's asking how many transistors are required to build that function.

I don't know how many transistors are required for each logical operation.

5. ### Ghar Active Member

Mar 8, 2010
655
73
I would say there's a mix up here in terminology... "A CMOS transistor" is super confusing since there's no such thing.
CMOS is made up of NMOS and PMOS transistors.

A NOT gate requires 2 transistors, 1 NMOS and 1 PMOS.
A NAND gate requires 4, a 2 input AND requires 6.

A 4 input NAND gate requires 8 transistors, add an inverter and you have 10 transistors.

6. ### jegues Thread Starter Well-Known Member

Sep 13, 2010
732
45
Okay so if we have 3, 3-input OR gates to make:

A NOR gate requires 4 transistors. If I follows a NOR by a NOT I will get an OR. A NOT contains 2 transistors.

Therefore a NOR followed by a NOT will give me a OR.

4 + 2 = 6 transistors.

But I have a 3-input OR not a 2-input.

So I should have 4 + 2 + (4/2) = 8 transistors.

8 times 3 will give me 24 transistors.

Now, if we have 1, 2-input OR gate:

1 2-input NOR requires 4 transistors. If I follow this by a NOT I will get a 2-input OR so,

4 + 2 = 6 transistors for 2-input OR.

Lastly, if we have 1, 4-input AND gate:

A 2-input NAND gate requires 4 transistors. This followed by a NOT will give me,

4 + 2 = 6 transistors.

But I have a 4-input AND gate so,

4 + 2 + 4 = 10 transistors.

So that total number of transistors for this function will be,

24 + 6 + 10 = 30 transistors.

How does this look?

Morris Kal B likes this.