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[HELP] - Basic AND Gate problem

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Carl James Jumawan

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I'm supposed to make an AND GATE for this circuit. My problem is that the DIP Switches seem to play no role in the circuit. Removing it still keeps the LED on. My goal is that the switches 1, 2 and 3 should be switched on before the LED will light up.

BTW, I'm only allowed to use the LS7408 IC, 4 bit Dip switch, the battery, an LED and jumpers.
No further resistors and whatnot.

**broken link removed**
 

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The very old 74LS08 is a TLL AND gate. Its output is high when both inputs are 2V or higher. But its maximum allowed output current is only 0.8mA when high and is recommended to be only 0.4mA which will barely light an LED. It needs a resistor in series with the LED to reduce the output high current so that the gate is not destroyed.
 

Follow Audioguru's advice about the LED series resistor, try 470 Ohms, it should be enough.

The reason why the logic isn't working is you are assuming the an input not connected to VCC through the switch is logic zero. Opening the switch does not make it drop to zero, in fact on most TTL devices the input assumes a high level if it isn't connected to anything. You need to add 'pull down' resistors to the gate inputs so they actually go low when the switch disconnects them from VCC.

Try adding a 1K resistor from each input from the switch to 0V (pin 7) and see if that fixes it.

Brian.
 

I do not know why your teacher did not read the datasheet and why you did not read the datasheet of the old 74LS08 TTL gate. Your teacher and you know nothing about its details. It shows its maximum input low current and voltage which requires a 1000 ohm resistor to ground for you to make one input at a logic low.

EDIT: Also your teacher and you should follow the recommended output high current of only 0.8mA which requires using a 1.6k ohms or higher resistor in series with the yellow LED to limit its current. It will make the LED appear very dim.
 
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