I have two questions under same heading: First why the led connected to terminal 3 and ground is always on when terminal 14 is connected to 5V and 7 to ground when there is no any input at 1 and 2?
Second As we give inputs to 1 and 2 and have output at 3 why we need to connect 14 to power supply and 7 to ground?
Thanks!
The 7400 series (including the 7432) are TTL (Transistor-Transistor Logic) devices.
The inputs are multiple emitter transistors which assume a logical HIGH if not connected. So by not connecting anything to the inputs you have both inputs HIGH and so the output will be HIGH.
In order for the transistor circuit to work (the gate) we have to supply external power. This is done through pin 14 (+5V) and pin 7 (0V). The chip does not take power from the inputs.
1st question:
First of all, all unconnected pins are like little antennas, and maybe can have a floating voltage on it, possible from every radiation caught around it from wifi, tv, radio and so on. So to assure the LED will go off you have to short pins 1 and 2 to the GROUND. Only connecting they to zero voltage will guarantee the led to turn off.
2nd question: The OR gate actually look like this. You can see this circuit has 4 terminals besides the output: two digital inputs A and B (1 and 2 inputs at 7432) and two feeding pins (Vcc and Gnd). The OR gate does not copy the input as you were taught. Actualy it sends the same voltage expected by the OR table but from the Vcc(from the pin 14)
Old-style TTL gates interpret an open input as a logic '1' so the output of an OR or AND gate will be '1' with the inputs open. The inputs are always high but somewhat susceptible to noise so a 1K pull-up resistor is recommended, or they may be connected directly to GND if a logic '0' is desired. The 1K is recommended for TTL rather than a direct connection to Vcc for TTL only.
As you can see from the above schematic, the 4K resistors source current out the inputs, so about 1.1mA flows when an input is grounded.
