IC 74HC08 can the output drive the input without destroying it? [closed]

When your 1Y output drives HIGH, the 74HC08 5 V supply must be dropped across three things in its series circuit:

• The 74HC08 1Y internal output stage (see its datasheet)
• The LED (see its datasheet for voltage drop at a certain current)
• The current limiting resistor

The resistor value is selected to set the LED current to something (a) the 74HC08 1Y output stage is capable of and (b) that gives the wanted LED brilliance if possible.

As an example, let's assume the LED current is 1 mA, the LED drops 2.1 V and the 1Y output stage drops 0.4 V. That leaves 2.5 V across the series resistor. That's well above the 1B threshold for a logic HIGH so the LED on will always drive 1B HIGH. When the SW11 is pressed, 1B will also be driven HIGH. Only when SW11 is released and 1Y drives LOW will 1B be pulled LOW by its resistor.

Your present 20K pull-down is far too high to allow 1 mA of LED current, so it won't glow much. The resistor can be changed to 2K2, allowing just over 1 mA with the above example characteristics.

Neither the current circuit nor this revised one would damage the 74HC08.

But your current question doesn't state your application, so it's not possible to say if it can be done better or if this gives the function you need.

There's something else for you to consider. Unlike normal diodes, a lot of LEDs can only withstand a low reverse voltage, often less than 2.5 V. Look up VRRM in your LED's datasheet. When SW11 is pressed and 1Y is driving LOW, LED D5 will have a reverse of about 4.7 V across it and may be damaged.