Digital Logic Problem

muckleluck

May 24, 2005
15
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May 24, 2005
Messages
15
I'm messing around with some logic gates trying to get a basic understanding or digital logic but for some reason I can't seem to get them to work.  The specific chip that I am using is a 7408 TTL AND gate. For some reason when I connect the power to the chip the voltage drops from 5 to 2.3. I can't figure out whether I have a power supply problem or I'm doing something wrong with the set up. If anyone has an idea of whats going on your help would be appreciated.

 

MP1

Dec 7, 2003
3,399
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3,399
Not much current draw on this chip. How did you connect the supply? Pin 7 for GND and 14 for VCC? If you think you have a supply problem, you can try battery power. Be careful to not exceed the recommendations.

MP

 

muckleluck

May 24, 2005
15
Joined
May 24, 2005
Messages
15
Yeah it seems to be a power supply problem.  I have another question about these chips. I want the inputs to be low normally and high when I press a button. How can I achieve this?

Muckleluck

 
A

Alun

Jan 1, 1970
0
Connect them to +V with a pull up resistor (about 10K) then connect the switch from the input to 0V.

 

muckleluck

May 24, 2005
15
Joined
May 24, 2005
Messages
15
With this arrangement the chip is acting like a NOR gate. If I reverse the connections so that the resistors are connected to the ground and the switches to the V+ the led will never turn off because the pins float high. Thanks for bearing with me on this line of basic questions - this is the result of a the elec education they give mech eng students.

Muckleluck

 

MP1

Dec 7, 2003
3,399
Joined
Dec 7, 2003
Messages
3,399
Yes, he gave you backwards instructions. To have the pin low normally, then high when you press the button, you will want to connect the pin to ground via a resistor. Then with a switch connected from the input to VCC, you will have high only when pressing the switch. If the pin is floating, you need to change the value of the resistor. The state of the pin should be 0 when the switch is open. This should take place because the pin of the chip finds an easy path to ground through the resistor. You can also use a debounce circuit on the pin to keep the signal more stable.

MP

 
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