I'm a newbie engineer. I have been assigned a design to modify for a different use. A have a question about a circuit that has nothing to do with the changes I must make. The original designer is gone.
I have an active low 2.5V open drain output with VOL of 0.4Vmax @ 8mA and VOH of 2.1Vmin @ -8mA that drives a LVTTL input with a VIL of .8Vmax and VIH 2.0Vmin (10uA input leakage current). The signal is pulled up to 3.3V with a 4.75K resistor.
Is the open drain output signal pulled up to 3.3V because the input is LVTTL? Wouldn't pulling it up to 2.5V be OK, since VOH is 2.1Vmin and VIH is 2.0Vmin? I'm wondering why the origianl engineer did this because he had to get 3.3V to the pullup, when 2.5V was already there.
I'm confused about the voltages. Why is it OK to pull the open drain output up to 3.3V even though it comes from a 2.5V output. Is this because it's an output? What if I pulled it up to 5V? Is that OK? Why?
Thanks, John