# BCD Counter

#### eXodus

Joined Mar 26, 2007
1
I've got most of the formulas worked out, but the Counter still isn't working properly. The ones that I think are wrong are:

K' = B'C'D'

I came up with

K = BCD

But I am sure that is incorrect. I just don't know what to do with it.

K' = A'D' + A'C'

I came up with

K = (A+D)(A+C)

Which I figured out after I posted this is correct! Yay for me!

Everything else I did looks good to me, so if it still doesn't work after some help, I will post more information.

Thanks in advance to anyone who helps out!

#### thingmaker3

Joined May 16, 2005
5,084
K' = B'C'D'
The output is low when all three of the inputs are low. What kind of gate does that?

K' = A'D' + A'C'

I came up with

K = (A+D)(A+C)

Which I figured out after I posted this is correct! Yay for me!
Yay indeed! It is correct. But is it simplified? Can you generate K with two gates instead of three by applying one of the Boolean Algebra Laws?

#### Dave

Joined Nov 17, 2003
6,970
K' = B'C'D'

I came up with

K = BCD
NOT both sides:

K'' = (B'C'D')'

Think of DeMorgan's Theorum: (A.B)' = (A' + B')

(B'C'D')' = (B'' + C'' + D'')

B'' = B
C'' = C
D'' = D

And

K'' = K

Therefore:

K' = B'C'D' simplifies to K = B + C + D

Dave