Digital Logic Tutorial

2 bit to 4 line decoder with active low output

One thing that is really common in computers are sequences. Things happen, one after the other, in an ordered, regular, pattern.

Sequences are formed in computers when a binary number is decoded onto a set of lines. Each of the output lines is activated, in turn, when the binary number on the inputs lines counts up.

Try it in the simulator to see what we mean. Press the up arrow over the right hand digit in the double digit input on the left of the screen and watch the lights go off in sequence.

We call this an "active low" output because... well actually it's because we were too lazy to invert the output of each NAND gate with an NAND-as-NOT.

Advanced

Can you modify this circuit to have "active high" outputs? Warning: You won't be able to save your work here; it's just to learn how to do it.

• First click on the word "simulate" to turn off the simulation.
• Next, use the right mouse button to click on each light and select "Remove Gate"
• On the far left side, click on "NAND" then place the NAND gate 4 dots to the right of the first NAND gate.
  Do that again for each of the 4 NAND gates. You should have 2 columns of them now, with 2 gates on each row.
• Click on the output of the first original NAND gate, then click about 2 dots to the right, then 2 dots up, then click on the top input of the new gate.
  If you don't have all the wires straight, just drag the corners to where you want them.
• Click on the output of the first orginal NAND gate again, then click 2 dots to the right again, but now click 2 dots down, the click on the bottom input of the new gate.
• Repeat that process for each of the rows.
• Now click on "LED" in the far left list and place a few dots to the right of the output of your topmost new NAND gate.
• Click on the output of your new NAND gate and click again on the input to the LED.
  Repeat that for each row
• Press the "simulate" button again and try out your new circuit.

See also:

Questions:

• ice_chokSPAM at KILLspamhotmail.com asks: " hi..can u teach me how to design 4-to-16(active low output) using 2-to-4(active low output and enable)" James Newton of Massmind replies: It's really pretty easy when you think about it: Use 5 2-to-4 decoders, the first one, called "E", is connected to the upper 2 bits of your 4 bit input and it then selects which of the other 4 chips will be enabled. The remaining 4 2-to-4 decoders, called "A", "B","C", and "D", each connect to the lower 2 bits of your 4 bit input and are enabled only when the 4 bit input is on their "bank" of 4 possible outputs. The first of the low bit decoders decodes 0,1,2 and 3, the second decodes 4, 5, 6, and 7, and so on.

  input		enable low bit chip #
  decimal	binary
  0	0000	A
  1	0001	A
  2	0010	A
  3	0011	A
  4	0100	B
  5	0101	B
  6	0110	B
  7	0111	B
  8	1000	C
  9	1001	C
  10	1010	C
  11	1011	C
  12	1100	D
  13	1101	D
  14	1110	D
  15	1111	D

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