#### pinkoryx

Joined Dec 11, 2017
47
I am studying a lesson on decoders, when I came across this exercise. I do have a brief understanding on decoders and what they do, but I don't understand this example in particular:

We have available 74138 decoders and wish to select one of 32 devices.
(What is meant by this?)

We would need four such decoders.

(On what basis did they assume we needed four decoders?)

Typically, one of these would select one of the first eight addressed devices; another would select one of the next eight, and so forth.

(what are the first eight addressed devices?)

Thus, if the address were given by bits a, b, c, d, e, then c, d, e would be the inputs (to C, B, A in order) for each of the four decoders, and a, b would be used to enable the appropriate one.

(On what basis did we decide we needed this many bits?)

Thus, the first decoder would be enabled when a b 0, the second when a = 0 and b = 1, the third when a = 1 and b = 0, and the fourth when a = b = 1.

Since we have two active low enable inputs and one active high enable, only the fourth decoder would require a NOT gate for the enable input assuming a and b are not available.
(The answer is attached to this post)
Any help is greatly appreciated!

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#### absf

Joined Dec 29, 2010
1,968
[post deleted] didnt notice this is homework help.

#### LesJones

Joined Jan 8, 2017
3,655
You are very close but not quite right. The A to E inputs are not in the correct sequence. Also your statement
"Thus, the first decoder would be enabled when a b 0, the second when a = 0 and b = 1, the third when a = 1 and b = 0, and the fourth when a = b = 1." is not quite right. Tip. Write down the table for the sates inputs A to E for all 32 output states. Are you sure that you are allowed to use the extar inverter ? (This is the most efficient way to do it.) If you are not then you would have to use another 74138.

Les.

#### pinkoryx

Joined Dec 11, 2017
47
You are very close but not quite right. The A to E inputs are not in the correct sequence. Also your statement
"Thus, the first decoder would be enabled when a b 0, the second when a = 0 and b = 1, the third when a = 1 and b = 0, and the fourth when a = b = 1." is not quite right. Tip. Write down the table for the sates inputs A to E for all 32 output states. Are you sure that you are allowed to use the extar inverter ? (This is the most efficient way to do it.) If you are not then you would have to use another 74138.

Les.

#### LesJones

Joined Jan 8, 2017
3,655
You have used bit E as the least significant bit. It is normal to use A as the least significant bit. This is the way the inputs on the 74138 are labeled. I had not noticed the that you had you had reversed the sequence in your schematic. I think using your convention with E as the least significant bit it will work correctly.

Les.

#### pinkoryx

Joined Dec 11, 2017
47
On what basis do the output pins display 0-31 if the inputs (C B A) are always 000-111 (0-7)?
(I understand why it displays 0-7 on the first decoder but I don't understand the remaining 3)
(Please refer to the attached screenshots. Both were given to us during class - neither is my work)

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#### pinkoryx

Joined Dec 11, 2017
47
You have used bit E as the least significant bit. It is normal to use A as the least significant bit. This is the way the inputs on the 74138 are labeled. I had not noticed the that you had you had reversed the sequence in your schematic. I think using your convention with E as the least significant bit it will work correctly.

Les.
Both A and e are least significant bits. Attached below is the truth table for the cascade circuit based on my understanding so far

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#### LesJones

Joined Jan 8, 2017
3,655
If you have been tought to use "a" as the most significant bit then it think it is correct for that convention. (I may have confused you by using upper case letters for your inputs where you used lower case letters.)
To answer your question in post #6 When a decoder is not enabled by the enable inputs none of the outputs will be selected so they will be at a high logic level. (As they are active low signals.) You are correctly using the 2 most significant bits (a and b in the convention that you are using.)
Les.

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