I have a 4.5V source. I connected + terminal to a 15ohms resistance the resistance is connected to the + end of a green LED, the - end of green LED is connected back to the battery to make the circuit complete . I observe the green LED lights up. The drop across the resistance is 1.3V and the led is 3.2V.

Now I connected the Vcc of NTE74LS08 ("and gate") to the +-terminal of battery. The ground of NTE74LS08 is connected to the negative terminal of the battery.

pin 1 and 2 are 1A and 1B pin 3 is 1Y of NTE74LS08.

I connect 1Y with another + end of a red LED, the - end of the red LED is connected to 15 ohms resistance. The end of the resistance is connected to the negative terminal of the battery.

I observe even without the 1A and 1B connection (that is both the A and B input being zero) the red LED is lighting up.

My question is what is the mistake of my design ? Why is the any gate not working ?

  • 3
    \$\begingroup\$ Edit the question, click on the schematic button, and insert a schematic. \$\endgroup\$ – Ignacio Vazquez-Abrams Mar 5 '14 at 0:49

The inputs of TTL logic gates are emitters of NPN transistors. They are current sources so when not connected they act as High level or logic "1", so your AND gate sees two high inputs, and naturally outputs a High. It is good practice to connect inputs you want to be high to the positive supply, either directly or through a resistor.

For the inputs to act as Lows, they must be connected to the negative side of the supply (normally considered "ground").

The above applies to 74xx, 74LS, and 74ALS logic families.

The CMOS logic families 74AC, 74C, 74HC (and anything else with a "C" in the middle) have very high impedance inputs which, if left unconnected, will assume random logic levels. With these parts, inputs must be connected to either supply or ground, either directly or through a resistor, and not left unconnected.


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