how to create an electronic circuit

Thread Starter

Darkwingduck

Joined Jul 28, 2012
4
Hello all,

I am trying to learn about electronics and I am attempting to create an electronic circuit that fallows the Q= A AND NOT B logic show by the image below.



In order to do this I figure I should have a AND gate with inputs A, and B. With an inverter on the B input before it reaches the AND gate.

With the risk of looking foolish I am wondering how I can go about doing this. I have found the Inverter and AND gate componets I would like to buy (listed at the end of post) but I am wondering how to attach them to a circuit board for proper use. I know I will have to solder the components to the board but what kind of board should I buy and were can I find one..

Thanks for any help it is greatly appreciated!!

:D



Inverter I want to use = http://http://www.jameco.com/webapp/wcs/stores/servlet/Product_10001_10001_873830_-1

AND gate I want to use= http://www.jameco.com/webapp/wcs/stores/servlet/Product_10001_10001_873725_-1
 

Wendy

Joined Mar 24, 2008
23,415
Welcome to AAC!

Since the parts you have selected are surface mount, you will either have to make a PCB, or find an adapter (they are available).

Just curious, why are you stuck on these parts? They are hard to use, and one you solder them in will not be very reusable.

A through hole set of components and a protoboard are 100% reusable. The book links shown above have some good experiments on the subject.

DIGITAL INTEGRATED CIRCUITS
 

crutschow

Joined Mar 14, 2008
34,285
As Bill noted you want to use DIP type through-hole packages, such as can be found in CD4000 series CMOS digital circuits (Here's a list of common CD4000 circuits). And a protoboard (solder-less breadboard) is any easy way to connect them together.

One Note: Make sure you ground all unused inputs on any CMOS circuits. Otherwise you may get erratic results.
 

MrChips

Joined Oct 2, 2009
30,714
If you wish to reduce your chip count you can do this with a single QUAD 2-INPUT NAND Gate,
one NAND gate and two gates as inverters for B' and Q'. You will have one spare NAND gate.
 

Thread Starter

Darkwingduck

Joined Jul 28, 2012
4
Thanks for the hearty welcome Bill_Marsden!

I am not particularly set on using the parts I listed so I will deffinantly do more research on the protoboard setup and see what that is all about, thanks for the help!!
 

Thread Starter

Darkwingduck

Joined Jul 28, 2012
4
As Bill noted you want to use DIP type through-hole packages, such as can be found in CD4000 series CMOS digital circuits (Here's a list of common CD4000 circuits). And a protoboard (solder-less breadboard) is any easy way to connect them together.

One Note: Make sure you ground all unused inputs on any CMOS circuits. Otherwise you may get erratic results.
Ok thanks for the info, after doing some searching on the internet for the protoboards they seem to be exactly for the type of thing iam looking for. Also I appreciate your advice to ground all unused inputs for a CMOS and will make sure to do so!
 

Thread Starter

Darkwingduck

Joined Jul 28, 2012
4
If you wish to reduce your chip count you can do this with a single QUAD 2-INPUT NAND Gate,
one NAND gate and two gates as inverters for B' and Q'. You will have one spare NAND gate.
Yes reducing the chip count is a great idea, I think I see what your saying.. I worked it out on a piece of paper and if I connect the B input to both inputs on one NAND gate and then connect that output to another NAND gate with the other input being from A and then connect that output to both inputs of another NAND gate it will fallow the same truth table.. thanks a lot for the help!
 

WBahn

Joined Mar 31, 2012
29,979
Yes reducing the chip count is a great idea, I think I see what your saying.. I worked it out on a piece of paper and if I connect the B input to both inputs on one NAND gate and then connect that output to another NAND gate with the other input being from A and then connect that output to both inputs of another NAND gate it will fallow the same truth table.. thanks a lot for the help!
And what you have discovered, more or less, is that any logic circuit can be implemented using only two-input NAND gates (or, it turns out, only two-input NOR gates).

Here's a little exercise to help you explore this. Imagine taking the circuit you started with and changing it so that it was an OR instead of an AND. In other words, what if you had wanted to implement Q= A OR NOT B? Can you figure out how to do that using one a single quad NAND gate package? Hint: You could actually implement two of these circuits that are completely independent. In other words, you only need two two-input NAND gates which makes trying a brute force search of all possible configurations pretty easy.

If you can figure out how to do that, then you have just figured out how to apply what is known as DeMorgan's Theorem. So you might want to read about it in the near future.

While expressing things as

Q = A AND NOT B

is okay in some circumstances, you will want to start getting used to using the more common Boolean algebra notation. In this, AND is represented the same way as normal multiplication, so AB, (A)(B), A(B) and A*B would all be seen as (A AND B). The OR is represented the same way as normal addition, so A+B would be see as (A OR B). The NOT is represented a few ways. The one I like the best is using a trailing apostrophe. So A' is the same as (NOT A). Other methods are to use the asterisk, so A* is NOT A. For this reason, I tend to not use the asterisk at all, because some people look at it as indicating an AND and others as a NOT, so if YOU use it, it is a good practice to include a note indicating what you mean by it. The other common way to show the NOT is with a leading slash, either forward or backward. Hence, A', A*, /A, and \A are all common representations of (NOT A).

One thing to watch out for is that the inversion does not distribute, meaning that AB' is the same as (A AND (NOT B)). If you meant to say NOT (A AND B), then that would be expressed as (AB)'. And, again, the NOT does not distribute, so (AB)' is not the same as A'B'.

Don't worry about getting all of this down right now. Just keep in the back of your mind that when working with logic expressions, you need to be careful and explicit. You'll learn the specific rules as you go forward.
 

MrChips

Joined Oct 2, 2009
30,714
Yes reducing the chip count is a great idea, I think I see what your saying.. I worked it out on a piece of paper and if I connect the B input to both inputs on one NAND gate and then connect that output to another NAND gate with the other input being from A and then connect that output to both inputs of another NAND gate it will fallow the same truth table.. thanks a lot for the help!
btw, there are two ways to create a NOT gate from a NAND gate. You have shown one way. Can you show the other to indicate that you understand?

Then do the same with a 2-input NOR gate.
 

MrChips

Joined Oct 2, 2009
30,714
What WBanh is leading up to is something called De Morgan's Theorem (or De Morgan's Laws) or rather, De Morgan's Identities.

Check this out on the web and let us know if you need help.
 
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