I'm not entirely sure if I'm doing this right and would like someone to check my work. I have no answer for this homework, so here it goes

I have a circuit using 3 not gates, 4 and gates, and 1 or gate.

From what I understand from another thread is if its implemented into a CMOS I would need 1 PMOS transistor and 1 NMOS transistor so 2 transistors per 1 gate.

So it would be a total of 8 gates times 2 for the amount of transistors, this would be 16 total transistors correct?

the question specifically asks: "If each gate in the circuit is implemented as a CMOS gate, how many transistors are needed?"

 

And one more thing, how many transistors are in multiplexors in CMOS?

 

And one more thing, how many transistors are in multiplexors in CMOS?

A CD4051, CD4052 and CD4053 are Cmos multiplexers. Their datasheets do not show their schematics.

 

Heeey... Another homework assignment, some more time on AAC.

http://www.allaboutcircuits.com/vol_4/chpt_9/7.html

I think I'm reading this wrong, but it looks like it's two AND gates and an OR. Which I believe would require a hell of lot more than six. Thinking about what a multiplexer does, I come up with something completely different. I can see how an OR gate would be implemented, it act as the switch, and then the two inputs...

Is that how it's six? Two transistors used for the inputs, one for the switch, and then the compliments for each of them?

 

For the discussing and have a clear conception, you have to use the right name to describe the parts.

Transistor - Called Transistor or BJT or Bipolar junction transistor.
http://en.wikipedia.org/wiki/Transistor

MOSFET - Called MOSFET or MOS-FET or The Metal–Oxide–Semiconductor Field-Effect Transistor.
http://en.wikipedia.org/wiki/MOSFET.
although MOSFET is one kind of transistor, but we almost call it MOSFET not Transistor.

I think In your question that you said transistor should be a MOSFET not a transistor.
The following info could be useful for you.

Analog device MT-088 Tutorial.
Analog Switches and Multiplexers Basics.
http://www.analog.com/static/imported-files/tutorials/MT-088.pdf

CD4028 datasheet, logical structure.
http://www.datasheetcatalog.org/datasheets/120/108942_DS.pdf

 

You are studying CMOS logic, and you have to discover basic gate structure by yourself?

 

I'm not entirely sure if I'm doing this right and would like someone to check my work. I have no answer for this homework, so here it goes

I have a circuit using 3 not gates, 4 and gates, and 1 or gate.

From what I understand from another thread is if its implemented into a CMOS I would need 1 PMOS transistor and 1 NMOS transistor so 2 transistors per 1 gate.

So it would be a total of 8 gates times 2 for the amount of transistors, this would be 16 total transistors correct?

the question specifically asks: "If each gate in the circuit is implemented as a CMOS gate, how many transistors are needed?"

Ask yourself if it makes sense that you can implement an AND gate with the same number of transistors as a NOT gate?

In CMOS, basic inverted-logic gates require two transistors (1 NFET and 1 PFET) per input. So a NOT takes 2 transistors and while (two-input) NANDs and NORs each take 4 transistors. The noninverted-logic gates take an additional two transistors, since they are constructed using an inverted-logic gate followed by an inverter.

For a general logic function implemented in straight CMOS (not built up from standard gates) you generally need four transistors per input - two to get the complement of each input and then two in the actual complementary logic chains.

Now, as to how many transistors a given circuit that is presently implemented with 3 NOT gates, 4 AND gates, and 1 OR gate would require, that depends on whether you have to use those gates, or whether you can use NAND/NOR gates to both reduce the transistor count and speed the circuit up.

What constraints are you under?

 

You are studying CMOS logic, and you have to discover basic gate structure by yourself?

not so much that its more that our teacher assigns homework without explaining stuff sometimes so we have to discover it. like the amount of transistors in gates and what mux's do in a circuit/how to get a function from a circuit that has mux's in it

Ask yourself if it makes sense that you can implement an AND gate with the same number of transistors as a NOT gate?

In CMOS, basic inverted-logic gates require two transistors (1 NFET and 1 PFET) per input. So a NOT takes 2 transistors and while (two-input) NANDs and NORs each take 4 transistors. The noninverted-logic gates take an additional two transistors, since they are constructed using an inverted-logic gate followed by an inverter.

For a general logic function implemented in straight CMOS (not built up from standard gates) you generally need four transistors per input - two to get the complement of each input and then two in the actual complementary logic chains.

Now, as to how many transistors a given circuit that is presently implemented with 3 NOT gates, 4 AND gates, and 1 OR gate would require, that depends on whether you have to use those gates, or whether you can use NAND/NOR gates to both reduce the transistor count and speed the circuit up.

What constraints are you under?

i understand what your saying here thank you so much in general not gates have 2 resistors, and gates have 4 per input?

constaints are our teacher left us to discover this on our own and thats about it.

 

RCA invented CD4xxx Cmos logic ICs. About 40 years ago they published a book that told all about it. The 1983 copy of the book is here:
http://archive.org/details/RcaCmosI.c.Databook1983

 

not so much that its more that our teacher assigns homework without explaining stuff sometimes so we have to discover it. like the amount of transistors in gates and what mux's do in a circuit/how to get a function from a circuit that has mux's in it




i understand what your saying here thank you so much in general not gates have 2 resistors, and gates have 4 per input?

constaints are our teacher left us to discover this on our own and thats about it.

No, simple NAND and NOR gates require 2 transistors per input. To convert these to AND and OR, an inverter (NOT gate) must be added, which requires 2 more transistors.
So, inverting gates require 2N transistors, where N is the number of inputs. Noninverting gates require 2N+2 transistors.
Here is a good tutorial:
http://www.ittc.ku.edu/~jstiles/312/handouts/Examples%20of%20CMOS%20Logic%20Gates.pdf

There are others. Google is your friend.

 

i understand what your saying here thank you so much in general not gates have 2 resistors, and gates have 4 per input?

No, not quite. inverted logic functions take a total of two transistors per input. So a NOT gate has 2 transistors, a 2-input NAND gate has 4, and a 5-input NOR gate has 10. If you want a non-inverted gate, you need to add two more. So a 2-input AND (or OR) gate has 6 while a 5-input AND (or OR) has 12.

 

Oh, I see Ron H beat me to the punch. Oh well, consider it "reinforcement learning".