# Connecting AND Gate Chip to an Integrated Circuit

I am trying to teach myself how to make integrated circuits and I'm having trouble integrating an AND gate into the circuit though.

I took a picture of it. The voltage running across the power rails is 4.7 V (the chip is TTL logic, I figured it would be enough). The gate is an AND gate 2 input 1 output (7408).

My question is why, with both DIP switches turned off, is that LED shinning?? It seems that the current doesn't run through the 'AND' circuitry but through the VCC and out through the supposed to be 'output of the inputs A and B'. If the connections are wrong what's the proper way to integrate the gate into IC?

Original

Yellow- Path current is supposed to follow

Red- Path current apparently takes -.-

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Great illustrations :) –  abdullah kahraman Jun 17 '12 at 17:20
I think you have a fundamental misunderstanding about how the circuit works. The current that flows through the LED is not also flowing through the switches. The switches only provide control signals that direct circuits inside the AND gate to either provide current to the LED or not. Therefore, we can connect the switches to ground and still get the control signals that will ultimately cause current from Vdd to be directed to the LED. –  Joe Hass Feb 15 at 12:05

You connected the DIP switches between Vcc and the AND gate's input, and that's wrong. A floating TTL input (DIP switches off) is seen as logical 1, and when you close the switch you just enforce that 1. So inputs are always seen a 1, and output will be 1, and the LED will light.

Two things:

1. connect the DIP switches between the inputs and ground
2. connect the LED between output and Vcc. The logic will be inverted, but the output can sink more current than it can source, and your LED will light more visibly. Check the LED's polarity: the anode goes to Vcc. You also have to add a 150 $\Omega$ resistor in series with the LED to limit it's current.

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hmmm... I agree with the resistor bit, but I'm having a hard time understanding what you mean by connecting the DIP switches between the AND gate's inputs and the ground? Would that be after the AND gate's output and the ground? so wouldn't that defeat the purpose of the DIP switches' ability to control whether the input would be 1 or 0 (supposedly) –  Valentine Bondar Jun 18 '12 at 4:52
You want to use the DIP switches to switch between 1 and 0. But like I explained, at the moment inputs will always be 1, also when the switches are off. So connect them between the inputs and ground, so that closing a switch will make that input low. The output has nothing to do with it. –  stevenvh Jun 18 '12 at 5:04
Thanks for writing back so quickly! OK so atm my input flows in like this: V+ --> DIP --> AND Gate Inputs. How do you get ground (V-) involved? or by ground do you mean something other than V-? –  Valentine Bondar Jun 18 '12 at 5:29
@abdullahkahraman: but look how prettier they are now, with the dots :) and without the black bar –  clabacchio Jun 18 '12 at 8:40
@valentine - the dashed rectangle in clabacchio's schematic is your 7408, so the pull-up resistors are inside the IC, you don't have to add them yourself. –  stevenvh Jun 18 '12 at 12:04