I'm trying to understand how the 74LS241 works. From what I read, if two input are high on a buffer, then the output will be high. For some reason, this isn't working on my circuit.
So here is what I did:
I have put a led on the 3 output pin, in order to test the output. Then i wired the 19 and 17 pin with 3.5v, but my led isn't showing any high input.
(red where the input are at 3.5v, green where the expected output should be high)
And when I connect the Vcc pin, without 19 and 17, my led is on, where I expected this to be off.
I have the feeling that I have understood the schematic, but the fact is this is not working. So I clearly don't understand at all.
My question is: did i break the chip not protecting the input ? (I didn't put any resistors on the input). But I thought gate like those didn't need to be protected.
update
Here is my actual scheme:
Best Answer
A tristrate output can have 3 different states: low, high, or disconnected (also sometimes called 'floating', 'high-z'). In the first two instances the output is actively pulled low or high using a transistor; in the disconnected state, neither transistor is activated.
Pins 1 and 19 of the LS241 control this state for the buffers inside the chip. These 'G' inputs must be activated in order to have the buffers connected to them pull their outputs low or high, depeding on their input. Without it, the outputs float. Note the invertor symbol (tiny o ring) at the control gate connected to pin 1; you must pull pin 19 high to activate, and pin 1 low to activate.
This circuit will help:
simulate this circuit – Schematic created using CircuitLab
If you pull pin 19 down, the output will float and both LEDs will glow dimly because current is passing through both of them. Pull pin 19 high, and the output will follow the input (17) and either the green or the red LED will glow brightly; the other one will be off.