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  #1  
Old 11-24-2008, 04:58 AM
Meloncoly Meloncoly is offline
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Default Designing a 3-8 decoder with multiple 2-4 decoders

Hello,

The problem asks me to make a 3-8 decoder (no enable input required), with inputs x,y,z and 8 active high outputs labelled 0-7. The only building block I can use is a 2-4 decoder with active high enable.

I didn't listen much in class, regrettably so I don't know much of the terms nor how I'm really supposed to approach this, so please help me along the way.

Right now I have two 2-4 decoders, one representing 0-3, and another representing 4-7. Input x is connected to both the decoder's active high enable. Input Y and input Z are connected in the regular inputs of the respecting decoders.

I think this is correct, but what am I missing. I was thinking, that if I didn't place a inverter before the enable on the top or bottom decoder, then the decoders won't be labelled 0-7 but rather two seperate entities. Am I correct in thinking that? Or is there a way where I can "assign" one decoder to be 0-3 without the use of an inverter?

Also, how is active high and active low represented. If a bubble is on the enable of the decoder, does that mean active high or active low? How about when the bubble is on the output of the decoder? Is that representing active high or active low?

Thank you for the assistance.
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  #2  
Old 11-24-2008, 12:34 PM
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beenthere beenthere is offline
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Our chapter on logic will answer you question on the bubbles - http://www.allaboutcircuits.com/vol_4/chpt_3/1.html.
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Old 11-11-2010, 02:06 AM
johncurran94 johncurran94 is offline
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Quote:
Originally Posted by Meloncoly View Post
Hello,

The problem asks me to make a 3-8 decoder (no enable input required), with inputs x,y,z and 8 active high outputs labelled 0-7. The only building block I can use is a 2-4 decoder with active high enable.

I didn't listen much in class, regrettably so I don't know much of the terms nor how I'm really supposed to approach this, so please help me along the way.

Right now I have two 2-4 decoders, one representing 0-3, and another representing 4-7. Input x is connected to both the decoder's active high enable. Input Y and input Z are connected in the regular inputs of the respecting decoders.

I think this is correct, but what am I missing. I was thinking, that if I didn't place a inverter before the enable on the top or bottom decoder, then the decoders won't be labelled 0-7 but rather two seperate entities. Am I correct in thinking that? Or is there a way where I can "assign" one decoder to be 0-3 without the use of an inverter?

Also, how is active high and active low represented. If a bubble is on the enable of the decoder, does that mean active high or active low? How about when the bubble is on the output of the decoder? Is that representing active high or active low?

Thank you for the assistance.
If you checkout sn74138 and a and sn74139 and compare the circuits and the truth table it will give you some help
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Old 11-11-2010, 02:15 AM
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shteii01 shteii01 is offline
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Scroll down to Demultiplexers section, it shows the 3 to 8 using two 2 to 4: http://www.doc.ic.ac.uk/%7Edfg/hardw...eLecture12.doc
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Old 11-11-2010, 03:09 AM
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?? 2 of the same exact threads, with different ops???

Must be close to final exam time.

@shteii01 That is a very informative and easy to grasp lesson. Good find.
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Old 11-11-2010, 04:18 AM
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shteii01 shteii01 is offline
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Quote:
Originally Posted by retched View Post
?? 2 of the same exact threads, with different ops???

Must be close to final exam time.

@shteii01 That is a very informative and easy to grasp lesson. Good find.
Thank you.
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