breadboard - 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 -.-

Answer

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.