Understanding Address decoder

Question

Given is 3-to-8 decoder. There is 3 selection lines C, B, A (C=MSB and A=LSB) and one enable input. In total here are 16 adresses. enter image description here

So first we're going to discuss only EN_A.

1. How many address ranges are there in which the above address decoder activates the line EN_A? Can someone explain me how can I see that?

2. If EN_A active is, what is beginning & ending address? How big is the address range?

Address lines - A15 A14 A13 A12 A11 A10 A9 A8 A7 A6 A5 A4 A3 A2 A1 A0

 Y3 

EN_A - Start Address - 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 -> 6000h

EN_A - End Address - 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 -> 7FFFh

7FFFh-6000h -> 2^13 = 4kiByte

Is this good?

Also if we have EN_B is this adress going to be calculated the same (if my solution is good) because I'm little bit confused with the enable input of the second decoder?

Answer 1

You are getting confused by Address 0. It helps if you use address state tables.

En_A goes high under the following condition.

enter image description here

That is it goes high from address C000 - FFFF for an address range of 16K.

Therefore En_B has this truth table. Note you copy EN_A state from above first then add the new address bit states.

enter image description here

I am sure you can take it from there.

Update:

The above is short form, you might understand better if you chart the complete truth tables, as so.

enter image description here

• Alen: I don't understand why do you use in EN_B that A13=0, A12=0, A11=1 is? How many address ranges are there in which the above address decoder activates the line EN_A? Or how many address ranges activate EN_A?
• Alen: Also why don't you consider that EN_A is coming from Y3? That should also be considered...
• Trevor_G: @Ayya because Y1 of the second decoder is only enabled when A13=0, A12=0 AND A11=1. Note the first table Y3 IS en_A and is only enabled when A14 AND A15 are high.
• Trevor_G: @Ayya see my edit if it helps..
• Alen: Yes, I know that :)