My question is how would I connect D0 and D1 on the first AND gate with D2 and D3 on the second AND gate. ( further questions would be ask)
Blue mark = 74LS32
Red mark = 74LS08
Yellow mark = 74LS04

 

Welcome to AAC!

Assign component designators to each gate, label each pin, and start wiring.

U3-2 to U1-1
U3-4 to U1-2
U1-3 to U1-9
U1-8 to U2-1
. . .

 

Hello,

As @dl324 said, by assigning numbers to the gates and pins, you can follow the schematic and make the connections.

Bertus

 

Hello,

As @dl324 said, by assigning numbers to the gates and pins, you can follow the schematic and make the connections.

Bertus

Hi thank you so much for replies both of you! And thank you for the welcome. I only just started working on things like this and new knowledge just seems to fascinate me. Anyway for the 7404 do I just connect a jumper wire to 1A(pin1) to 1Y(pin2) to create D3 and the inverted D3(D3#)?

 

Welcome to AAC!

Assign component designators to each gate, label each pin, and start wiring.
View attachment 135678
U3-2 to U1-1
U3-4 to U1-2
U1-3 to U1-9
U1-8 to U2-1
. . .

Thank you for the wonderful explanation dl324! Im starting to understand little by little. Now I just have a few questions on how the starting inputs would be connected(D3, D2, D1, D0)

 

have a few questions on how the starting inputs would be connected(D3, D2, D1, D0)

Using the designators in my example, D0 would be connected to U3-9, U1-5, and any other gates using that signal. D1 would be connected to U3-5, etc.

 

Thank you so much for the fast response! Ive grasped the situation firmly now. Hope to see you again on future questions!

 

Using the designators in my example, D0 would be connected to U3-9, U1-5, and any other gates using that signal. D1 would be connected to U3-5, etc.

Since im using a common anode D3, D2, D1, D0 must be connected to negative?

 

Since im using a common anode D3, D2, D1, D0 must be connected to negative?

If I understand what you're saying (inverting the inputs), that won't work. It would just invert the decoding so that 1110 would decode as 0001.

I'd approach it this way. If I had a logic HIGH signal, but my device/gate required the opposite polarity; what could I do to the signal to change it to a LOW?

 

If I understand what you're saying (inverting the inputs), that won't work. It would just invert the decoding so that 1110 would decode as 0001.

I'd approach it this way. If I had a logic HIGH signal, but my device/gate required the opposite polarity; what could I do to the signal to change it to a LOW?

Okay I finally had the time to try the problem out and got this as a result. On the 7 segment display (anode) everything is all lit up. Now my question is how would I connect a DIP switch to shuffle around the 7 segment like for example in the dip switch. I switched 0101 then the display would result in a 5. 1010 = A. Would I connect it in the IC 04?

 

Would I connect it in the IC 04?

If that's the 'LS04, yes.

For ease in setting the numbers, connect like this:

EDIT: And rotate the switch 180 degrees...

BTW, the 100 ohm pull up resistors can be replaced with 1k to reduce power dissipation.

 

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