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EDIT: Clarified the output I’m seeing at the oscilloscope.

EDIT 2: I've changed the input signal to be a different square wave with V_high = 2V and V_low = 0.5V. It seems that I may have a hidden connection between all the ports as I am seeing the same square wave from all the ports, even the ground port. At the Vcc+ port where I supply a 5V_DC, I'm seeing the square wave superimposing on the supply voltage. Thanks for suggesting ways to test and highlighting the issue regarding input voltage being beyond the limit.

Moving forward, how can I improve the design so that I don't end up getting this hidden connection? (It's on a PCB)

  1. Make sure traces don't criss-cross and separate them as far as possible?

  2. Separate the power and signal ports as far as possible?

Side note, have I correctly implemented the pull up resistor?

I'm testing out a comparator (TS880ICT) where I connect the following pins as follows:

  • Non-inverting pin via the 'signal out port' - square wave 4Vpp centered at 0V

  • Inverting pin via the 'threshold' port - ground

  • Vcc+ - 5V

  • Vcc- -ground

  • Output via the 'OUT' port - passed a pull up resistor of 1k ohm. Pulled up to Vcc+

I'm expecting to see one half of the square wave but I am seeing a slightly attenuated square wave instead. I suspect that I may have pulled up the output incorrectly. The graph below is the output after the pull up resistor, at the OUT port.

Can someone advise if this is so?

enter image description here

This is the circuit I'm working with.

enter image description here

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    \$\begingroup\$ Looks like your oscilloscope input is set to AC (coupled) meaning the average voltage will be zero. Can you confirm? \$\endgroup\$
    – Spehro 'speff' Pefhany
    Commented Aug 23, 2021 at 15:08
  • \$\begingroup\$ It’s set to DC, 1M ohm. The red square wave is the output at the OUT port. Sorry I didn’t make it clear. I’ve edited the original post \$\endgroup\$
    – Ryan Yong
    Commented Aug 23, 2021 at 15:11
  • \$\begingroup\$ Ok, can you make sure the probe ground reference is actually connected to ground and try channel B to ensure there isn't some kind of math going on. Your screen capture is quite uninformative- does "show details" provide more info? \$\endgroup\$
    – Spehro 'speff' Pefhany
    Commented Aug 23, 2021 at 15:18
  • \$\begingroup\$ "Non-inverting pin via the 'signal out port' - square wave 4Vpp centered at 0V" So from -2V to +2V? -2V is way outside the absolute maximum ratings for this device. \$\endgroup\$
    – Finbarr
    Commented Aug 23, 2021 at 15:32
  • \$\begingroup\$ If you are pulling the input negative, you are outside the operating area for this device. Look at the "Input Voltage Range" on your datasheet \$\endgroup\$
    – evildemonic
    Commented Aug 23, 2021 at 15:32

1 Answer 1

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To understand your confusion, view each node and check if it conforms to the specification and understand if there are any hidden ESD protection diodes.

If not , then linear superposition applies.

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