# Demultiplexer

Discussion in 'Homework Help' started by Dorumon, Oct 8, 2009.

1. ### Dorumon Thread Starter New Member

Oct 8, 2009
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Updated. See post #6 for questions. Thanks.

Last edited: Oct 10, 2009
2. ### beenthere Retired Moderator

Apr 20, 2004
15,815
282
A circuit diagram is worth several Kwords.

3. ### beenthere Retired Moderator

Apr 20, 2004
15,815
282
Obtaining a data sheet for each IC will explain how and why each one works. That should cover the next four questions.

One hint, though - A LED is a current operated device. Each will need a limiting resistor in series if you wish to keep them and the 'LS138 operating. About 300 ohms should be good.

1. Vcc is the positive voltage supply that operates the TTL IC's. A logic "1" will approach Vcc, but the data sheets will show you that LSTTL can't output the full Vcc level.

2. Ground is defined in lots of ways. It is a measurement reference for logic levels. The ground bus sources electrons to run the IC's. That is covered in basic electricity texts.

3. The square wave is known as a "clock source". The counter can't count without one.

4. A timing diagram shows the progression of outputs as clock pulses are received. A truth table gives all possible output states. A timing diagram on paper has no effect on or by an actual logic output from an IC. Both inputs and outputs can be active high or low. That is why understanding a data sheet is really important.

5. Again, look at the data sheet. You are using the Q1 - Q3 outputs. The count rolls over to 000 from the maximum of 111. It's not an up/down counter.

Last edited: Oct 8, 2009
4. ### beenthere Retired Moderator

Apr 20, 2004
15,815
282
The 'LS138 is a shorthand notation. All logic IC's work as the data sheet says they do.

5. ### JDT Well-Known Member

Feb 12, 2009
658
85
You are missing something important on this circuit.

Where the select inputs on the mux (A0 A1 A2)? When you find them, what are they fed from? The binary counter?

Need to know.

6. ### Dorumon Thread Starter New Member

Oct 8, 2009
24
0

The Diagram is updated correctly. And I think I figured it out AFTER 3 FREAKING DAYS, to make things simple, I drew a Truth Table, please reassure my assumption whether its correct or not.

2 Tables, top for Enabled, below for Disabled.

1) When the DEMUX is enabled and a Binary Counter is continously pulsing through it, all the LEDs will be Light Up and the Output will be the LED that is Not Light Up and running through all 8 Output LEDs. This is because the DEMUX has active low outputs where Logic 1 does not light up the LED.

2) When the toggled is switch on, the DEMUX is disabled. All 8 LEDs are Light Up. However, as the Binary Counter is continously pulsing through it, the Enable/Disable state switches instantly. So the running LED continues to run, but skips when it (For example we use Switch I0) reaches I0 and LED I0 never switches off because when the DEMUX is enabled, it is Switch On and the running LED never reaches it because when it reaches its turn and when the DEMUX is disabled, it also remain Switch On.

Am I right? I just need reassurance?

And also 1 more thing I want to confirm is for active low activation, a 1 is still required for activation right? Its just that a 0 is needed to be pulse in for it to be activated. But the 0 is than conveted into a 1 and activates.

For example for an active low enabled, E = 0, so E* = 1 so it activated
E = 1, so E* = 0, so it does not activate right?

Last edited: Oct 10, 2009
7. ### Dorumon Thread Starter New Member

Oct 8, 2009
24
0
Its abit longwinded but can any1 confirm if I am right?