Tips for test?

1. Mar 11, 2004

david90

My intro to digital system teacher is hard. He gives question that ask u design circuit such as make a counter that count 3 5 9 0 3 using mod-6 counter + encoder + decoder. Do question like this have a systematic way of solving or is only solvable if u have mad experience? He seems to give question that test ur EXPERIENCE rather than steps. What should I do? I don't think I can do every problem cuz heisquestions have a big range.

2. Mar 11, 2004

Guybrush Threepwood

ok, some ideeas
a mod-6 counter is easy to made by resetting the counter when the output reaches 6 = 110b. So it will count 0, 1, 2, 3, 4, 5 for ever ... btw are you sure you didn't forget a number in your series?

the output of the counter goes into a decoder and the outputs of the decoder are scrambled to the inputs of the encoder according to the series you want to get.

PS: i would draw a picture but I'm a bit busy so I'll try later if you didn't understand.....

3. Mar 11, 2004

david90

the example that I gave you was just a sample. here is the real problem.

Using a MOD-8 counter, 3 to 8 decoder and a 8 to 3 priority encoder, make a circuit that will count 0 > 5 > 2 > 7 > 4 > 1 > 6 > 3 > back to zero

4. Mar 12, 2004

Guybrush Threepwood

same things apply just use a 3 bit counter (that would be a usual mod-8 counter since it counts from 0 to 7)

5. Mar 13, 2004

david90

elaborate or give an example of what u mean by scamble

6. Mar 15, 2004

Guybrush Threepwood

OK, so your 8-bit counter counts from 0 to 7
When you put n (n beeing the output of the counter) on the selection inputs of the 3to8 decoder you get an level of 0 on decoder output n.
So in our example you will get the outputs of the decoder to have 0 logic in the order 0 to 7.
Then you connect output 0 of the decoder to input 0 on the priority encoder. Output 1 of the decoder will be connected to input 5 on the priority encoder. Output 2 of the decoder will be connected to input 2 of the priority encoder. And so on ... get the picture?
In this way when your counter is working at the output of the priority encoder you'll find your series. If you want another series just connect the decoder and priority encoder in a different way....

7. Mar 15, 2004

david90

"OK, so your 8-bit counter counts from 0 to 7
When you put n (n beeing the output of the counter) on the selection inputs of the 3to8 decoder you get an level of 0 on decoder output n.
So in our example you will get the outputs of the decoder to have 0 logic in the order 0 to 7.
Then you connect output 0 of the decoder to input 0 on the priority encoder. Output 1 of the decoder will be connected to input 5 on the priority encoder. Output 2 of the decoder will be connected to input 2 of the priority encoder. And so on ... get the picture?
In this way when your counter is working at the output of the priority encoder you'll find your series. If you want another series just connect the decoder and priority encoder in a different way...."

Thank you Guybrush Threepwood for saving my ass. I totally get what ur saying but there is a problem.

If you connect output 0 of decoder to input 0 of the PE, what will you do when the counter outputs a 4 and into the decoder and decoder output is 5? Not sure if i'm right or not but I think the output 0 of the decoder shouldn't be connected to anything to make the PE output a zero. countter > 0, input decoder >0, output of decoder is 0 and if u connect that to 0 input of the PE, wouldn't that make the PE output a 1 instead of zero?

Last edited: Mar 15, 2004
8. Mar 15, 2004

Guybrush Threepwood

not sure I completely understand you but once the counter starts there is only one output active at one time for the decoder. So there's only one input for the priority encoder to interpret. Maybe you should tell how your decoder works (there are a lot of models of decoders out there....)

PS: may work computer is broken so expect some delay in my response .....