09/18/2006
A student writes:
If A=1 and B and C both = 0 then what is the output of all three at
the intersection. (in the squiggly oval)
(The student included a .jpg file with the above question)
Thank you,
<<name withheld to protect one who takes accurate notes in class>>
Good! You have captured exactly the three-input AND gate that I put
on the board last week. Unfortunately, the email program I am using
does not allow me to simply attach the .jpg file you included, so
I will have to draw its essence here:
_______ ________ _______
| | |
_| _| _|
|| || ||
A --o|| B --o|| C --o||
||_ ||_ ||_
| | |
| | |
---------------------------------------- the point in question.
|
_|
||
A --||
||_
|
|
_|
||
B --||
||_
|
|
_|
||
C --||
||_
|
|
---
\ /
Let's apply the 0s and 1s you suggest to the gates of all transistors
and see what we get.
With A=1, the corresponding P-type is an open circuit, and the
corresponding N-type is a closed circuit.
With B=0 (same for C), the P-type is closed and N-type is open.
If I now redraw the circuit, replacing the "opens" with broken connections
and the "closeds" with pieces of wire, I get:
_______ ________ _______
| | |
| | |
| |
1 -- 0 -- | 0 -- |
| |
| | |
| | |
---------------------------------------- the point in question.
|
|
|
1 -- |
|
|
|
|
0 --
|
|
|
0 --
|
|
---
\ /
Finally, we analyze what is going on. From "the point in question," there
are two direct connections (wires) to the power supply (2.9 volts). Also,
from "the point in question," the path to "ground" is broken so there is no
path to ground. Therefore "the point in question" is at 2.9 volts.
Therefore: the value 1.
Got it?
Yale Patt