'LS373 is a 'transparent' latch.
This means that while your Enable is active (and in your circuit it is always active - pin 11 high) then the data presented to an input will always immediately get reflected to the output. As long as output is enabled (which it is - pin 1 is Low)
For this latch to hold data, you must do the following :
1. Enable gates (pin 11 high)
2. Place data on input (pin 3) - i.e. sw1 is closed (hi) or open (low)
3. remove enable (pin 11 low)
4. release data on input pin 3 - now it won't matter what you do on pin 3, output will remain as it was just before step 3
Have a look at the truth table. You have LE fixed High, hence output equals input....
Need to either set LE low to latch the input or choose different latch.
If you have CD4011, the remaining 2 gates can be used to combine the 2 switches in one. If you like I will draw the schematic for you.
Note: Since the output current of the CMOS series 4000 is relatively low, a transistor should be added as a buffer to drive the LED.
Out of curiosity, why you don't like using simple 7 1P1T switches only?
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