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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?

Alen
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1 Answers1

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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

Trevor_G
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  • 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 Jan 14 '18 at 15:53
  • Also why don't you consider that EN_A is coming from Y3? That should also be considered... – Alen Jan 14 '18 at 15:54
  • @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 Jan 14 '18 at 16:01
  • @Ayya see my edit if it helps.. – Trevor_G Jan 14 '18 at 16:25
  • Thank you so much. It really helped me. Can you post your table for EN_C so I can compare with mine? :) – Alen Jan 14 '18 at 16:28
  • @Ayya since this smells like homework I am reluctant to do that, but if you show me yours, and it is right, I will add the final table. – Trevor_G Jan 14 '18 at 16:40
  • No, it's not, I'm learning for one exam, well I would put that A15 =1, A14=1, A13=0, A12=0 AND A11=1, and A10 A9 A8 for Y0 - 000, Y1= 001 and so on... So I will have for start adress (Y1) 1100100100000000 and (Y3)=1100101100000000 and than OR from that. Is this good? – Alen Jan 14 '18 at 16:56
  • @Ayya well done. You passed. My test at least :) Though, you really should use the x's (Means don't care) – Trevor_G Jan 14 '18 at 16:58
  • @Ayya don't forget to accept this answer if you are happy with it. – Trevor_G Jan 14 '18 at 17:34
  • I'll accept it. Thank you. I could put X to put this was for the start adress. – Alen Jan 14 '18 at 17:36
  • @Ayya yup I realized that, but if you put (Y3)=1100101100000000 as an answer in an exam you would get a "WRONG" mark for it. It should be (Y3)=11001011xxxxxxxx or (Y3) = CBxx – Trevor_G Jan 14 '18 at 17:38
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    Yes, I know that :) – Alen Jan 14 '18 at 17:58
  • I have another question. In EN_B is it correct to say that 5 adress ranges activate EN_B? – Alen Jan 15 '18 at 11:12
  • @Alena NO.. I have no idea where you are getting that idea from. – Trevor_G Jan 15 '18 at 14:57