4 active low inputs to 7 active high outputs decoder

Thread Starter

0Wafi0

Joined May 15, 2015
25
hello there!
This is m first post here. I'm Wafi and I'm a second year aerospace avionics major. I'm posting my first question about active low and active high inputs and I've struggling to understand this for a while now!
Any help or push in the right direction would be greatly appreciated.
 

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MrCarlos

Joined Jan 2, 2010
400
Hello 0Wafi0

The first thing you should do is fill in the truth table in columns a to g.
For example, in the first row, from A to D contain 0’s. It means that no input is selected.

The next line says: 0001, means that the switch has enabled the input D. So that segments that will be active high Are: b c d e g.
And so on.

Does not look complete truth table but should be considered, when more than one entry is Active Low. In this case should light segments to form the letter "F".
 

Thread Starter

0Wafi0

Joined May 15, 2015
25
awesome thanks...I knew about the truth table but I was confused about this active low inputs with active high outputs...It's easy just work through that...but I wanted to know how the active low input affects this whole thing with the active high output...
 

MrCarlos

Joined Jan 2, 2010
400
Hello 0Wafi0

Ok. Normally in positive logic:
"Active Low" is a voltage level close to zero volts,
while:
"Active High" is a voltage close to the bias voltage (VCC).

So:
Active Low = 0
Active High = 1

Note that the truth table "A", "B", "C" and "D" column are represented in the other direction, it means that:
1 Yes, yes it is active.
0 No, it is not active.
 

WBahn

Joined Mar 31, 2012
24,700
The normal convention is that 0 means LO (low voltage) and 1 means HI (high voltage). Active-LO means LO is TRUE (hence, "active-LO) so that 0 means TRUE and 1 means FALSE. In this case, inputs 0 0 0 0 means that ALL inputs are being asserted, while if the switch is open-circuited it means that the inputs are 1 1 1 1 (as a result of the pullup resistors).
 

Thread Starter

0Wafi0

Joined May 15, 2015
25
awesome guys!!! thanks for the help!!! so basically in the truth table I would have to flip everythin for ABCD? correct?
 

MrCarlos

Joined Jan 2, 2010
400
Hello 0Wafi0

Yes, of course

Note now that the inputs A, B, C and D are as mentioned in the phrase that we see in the image that you enclose in your post #1.

Now 0 represents the letter is active.

In that image there is a box to the right of the 7-Segment Display.
Under the "Sw Input" column are the letters when they are "Active Low" light segments named in the "Segs" column.

For example: for A, the segments should light are abcefg.
They will be "active high".

Remember:
When no input (ABCD) is activated, the display must be turned off.
When more than one input are active the display should show a F.
 
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