SN74AS04N - TI Inverter

Hello, I have a question about my 7404 Inverter and was wondering if any of you could give me some help. So here is whats going on...
on my solderless breadboard I have a 9 volt battery connected sending power and ground up the rails as is standard, off of that I am feeding the 9 volts into my 7805 5 volt regulator which is also grounded. The output pin from the regulator goes to pin 14 on my 7404 while pin 7 of the 7404 is also grounded to the rails. Lastly, I have a magnetic door switch made by GE which goes from the +9V rails to one side of the switch then the other side of the switch goes back to the breadboard and to a 1MΩ resistor (I have tried many different values of resistor). The resistor I have connected to many pins but when I apply my voltmeter, even without any connections besides the power and ground to my 7404 I notice that Pin 1 reads about 1.62V then Pin 2 reads about .15V Pin 3 is at 1.62V Pin 4 is about .15 etc... this pattern continues all the way around the chip except obviously the 5V in and 0V for my ground. Can someone please help me figure out what is going on here? Did I buy 20+ broken 7404's? =/

If you let the inputs "Float", you will get very unpredictable outputs from your logic chips. THe inputs need to be either at logic "0" or logic "1", (+5V or GND). It's also not a good idea to try to pull the logic gate input to a logic "1" state through a resistor to a higher voltage, unless you have a way to limit the voltage (not the current, as you are doing here). Try setting your unused login inputs to gnd, then pull the one input to +5V through a small value resistor such a 100 ohm, the GATE output should follow correctly in an inverted fashion (Logic "0")...

Happy Switching!
Mad Soundman

The 7400 series gates will float to a logic "1" when left floating. It's traditional to pull it "down" to introduce a signal.

Connect one side of the switch to GND and the other side to the 7404 input (e.g. pin 1). It is a good idea to pull this pin up to 5V to make sure it sees "1" when the switch is open. You should see 0V on pin 1 when the switch is closed and 5V when it is open. (Without the pull up you would see the 1.62V you saw before. It would usually work without any pull up, but you want it to always work.)

100 ohms is too low a value, since it would use 1/4 watt with 5V across it, and you probably only have 1/4 watt resistors unless you specially asked for larger ones. (And your 9V battery won't last very long ;-) 1 Meg ohm is too high, since it only gives a few microamps of extra current and isn't enough to guarantee any logic level. Anything between 470 ohms and 10k makes a nice pull up.

On the output of the 7404 (e.g. pin 2) you will see about 0.15V for a logic low and 2.5V or more for logic high. It will not reach 5V.

You might need some capacitors on the input and output pins of the 7805 depending on how long the wires are and who made the part. See the data sheet (Fairchild, pg 22, figure 7)

Thank you very much I really appreciate your help. Just to clarify, I should use a resistor between 470ohm and 10k on all of the unused pins on my inverter and set those all to ground? Again thank you both very much!!

The idle state of TTL inputs is "1". Resistor from input pins to 5V is best.

You only need pull ups on inputs, not outputs

Unused inputs can be tied together and share one resistor.

alright got ya. Thanks again!