Equivalent logic function

[dhruvabedre]

hi guys,

please can you help me with the equivalent logic functon of the given circuit which is in attached file.

I am confused about it.

Thanks in advance

 

[andre_luis]

Seems as that:

F = (A.B)'.(A.B)

 

[albbg]

I named each transistor as well as the "control" node "X".



F will be low if X is high or TR9 and TR10 are both ON, that means A and B are high
F will be high if TR7 is conducting (X is low) and at least one between TR5 and TR6 is ON (that means at least one between A and B low)
That is:

X A B F
0 0 0 1
0 0 1 1
0 1 0 1
0 1 1 0
1 0 0 0
1 0 1 0
1 1 0 0
1 1 1 0

We have now to see the behaviour of X

It will be low if both A and B are high and will be high if at least one between TR1 and TR2 is ON (that means at least one between A and B is low), Then:

A B X
0 0 1
0 1 1
1 0 1
1 1 0

I'll rewrite it with X at the beginning to compare with the first table.

X A B
1 0 0
1 0 1
1 1 0
0 1 1


This means in the first table the rows:

X A B
0 0 0
0 0 1
0 1 0

will never be generated, then F=0 regardless to the status of A and B

Other analysis:

case A=0, B=0
TR1, TR2 are ON, TR3 and TR4 are OFF, then X is high and TR8 is ON while TR7 is OFF
TR5 and TR6 are ON (but are disconnected due to the status of TR7)
TR9 and TR10 are OFF
then F=0 (TR8 is conducting)

case A=0, B=1
TR1, TR4 are ON, TR2, TR3 are OFF, then X is high and TR8 is ON while TR7 is OFF
TR5 in ON and TR6 is OFF (but are disconnected due to the status of TR7)
TR9 is OFF, TR10 is ON
then F=0 (TR8 is conducting)

case A=1, B=0
TR2, TR3 are ON, TR1, TR4 is OFF, then X is high and TR8 is ON while TR7 is OFF
TR6 in ON and TR5 is OFF (but are disconnected due to the status of TR7)
TR10 is OFF, TR9 is ON
then F=0 (TR8 is conducting)

case A=1, B=1
TR1 and TR2 are OFF, TR3 and TR4 are ON, then X is low and TR7 is ON while TR8 is OFF
TR5, TR6 are OFF
TR9, TR10 are ON
then F=0 (TR9, TR10 are conducting)

F=0 regardless to the status of A and B

NOTE: is the same result find by andre_teprom, since X.X'=0 I just detailed more.

 

[dhruvabedre]

Thanks andre_teprom:smile: