RE: CPLD Controlled Door Lock - Input Circuit

michaelkellett — Sat, 25 Apr 2015 17:28:03 GMT

I think I would do it like this:

[View:/resized-image/__size/620x514/__key/commentfiles/f7d226abd59f475c9d224a79e3f0ec07-4d91809f-fb87-4911-a7f5-417e829ce5bd/4073.contentimage_5F00_178888.jpg:620:514]

Much less power wasted in the LED resistor and the input to the CPLD goes from low impedance 0V when button pressed to pull up to its own 3.3V supply. It will be less susceptible to noise and variations in the resistor values.

If you will have long wires to the button something else might be needed.

If you want to detect a 24V (or other voltage) signal then replace the button with an NPN transistor with a resistor from base to ground and an input resistor to the base, collector to R1 and R2, emitter to ground.

By setting the input resistor correctly you can have almost any input voltage and the CPLD input is quite well protected.

If the electrical environment is awful then add a capacitor from the transistor base to ground.

You'll need to debounce the signals but you can do that in logic in the CPLD.

RE: CPLD Controlled Door Lock - Input Circuit

oldmanraskers — Sat, 25 Apr 2015 14:14:36 GMT

I know there is no junction at IN01 but I have since fixed it.

RE: CPLD Controlled Door Lock - Input Circuit

clem57 — Sat, 25 Apr 2015 14:10:10 GMT

Good now

RE: CPLD Controlled Door Lock - Input Circuit

michaelkellett — Sat, 25 Apr 2015 11:47:27 GMT

I can't see anything when I click on your link - can you embed the picture using the little camera icon ?