Logic ICs suffering from Leak Current

[Zander]

I've brought 25 three input NOR gate ICs (74HCT27), and it doesn't function the way it's supposed to. It behaves more like a NAND gate, anything to high switches it off.

My allowable supply voltage is 4.5 to 5.5, and mine (through a voltage divider) is approx. 4.6 .

For example, if I leave the out put wires out ( L L L) the LED doesn't turn on. It should though since it's a nor gate.


When I insert a wire and just lightly touch it (on the insulation), the capacitance just on the surface of the insulation turns it on then it fades back off.

I have a feeling however, it might have to do with my input voltage. Which is the same as supply.

Data Sheet:
http://www.nxp.com/documents/data_sheet/74HC_HCT27.pdf

I'll post the test diagram later.

 

[LTX71CM]

You have pull-up (or pull-down) resistors on your inputs? Right? No input should be left floating anywhere on the chip. If you're only using one gate tie all other inputs high or low. Your input voltage is fine according to the datasheet. Being fed from a voltage divider sounds a bit dicey but I'll hold further comment until you post your schematic.

You seem to make contradictory statements regarding gate behavior so I'll just remind you the output will only ever be high if all three inputs are low. If you have your LED wired from your 4.6 divider to the gate output that would explain it's apparent inversion.

 

[(*steve*)]

Two input NOR gate will (assuming positive logic) have a low output if either input is high. That defines NOR

A B | Q
----+--
0 0 | 1
1 0 | 0
0 1 | 0
1 1 | 0

With +ve logic, 0 corresponds to 0V and 1 corresponds to Vcc.

With -ve logic it's a NAND gate, but thinking in negative logic can really mess with your mind, so don't do it unless you have to

 

[Zander]

I think I just realized what I've been doing wrong.

I'm using only one gate, that means that 8 pins are left ungrounded.

I'm thinking about just using a voltage regulator Mosfet, and abandoning the whole voltage divider thing

Hear is the test Diagram.

Two input NOR gate will (assuming positive logic) have a low output if either input is high. That defines NOR

A B | Q
----+--
0 0 | 1
1 0 | 0
0 1 | 0
1 1 | 0

Yeah I totally forgot about the whole truth table.

I'll figure this next question out later on in the day;

I have 40 inverters (4 per IC) and I have 75 three Input Nor gates ( 3 per IC)

How many full-adders can I make from them?

 

[Resqueline]

Yes, that power supply scheme is rubbish, it couldn't possibly work well from that - even driving a LED too. You'll need a proper voltage regulator. Unused inputs = high.

 

[Zander]

alright, I have a single nor gate working nor, all I had to do was ground the unused pins.

I also have a 7805 supplying a constant 5volts.