SN74AXC8T245-Q1: Current consumption during VCCB is 0V

On the powered side, the leakage current is I_OZ; on the unpowered side, I_off. (But they have the same values.) These specifications are per pin.

Hi Clemens,

Do "per pin" mean we need to doble current consumption at VCCB due to there is 2 VCCB pin?

I guess that VCCB is 0V so leakage current shouldn't happen.
So I think talk current consumption during VCCA = 3.3V VCCB =0 V is as below.

12uA * 1pin(VCCA) + 0uA * 2pin(VCCB) + 1uA * 8ch(A/B port) + 1uA *3pin(OE/DIE1/2) = 23uA?

Also, how about following point?
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Also, in above condition, A port may apply HIGH signal during VCCB is 0V.
Is there any risk of causing leakage current by this situation?

We want to make sure HIGH input during OUTPUT is Hi-z won't cause additional current consumption.
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Best regards,
Kazuki Kuramochi

This computation is correct.

When VCCB is 0 V, then the I/Os on the A side are disabled. So you get 8 × I_OZ for those. (The leakage current flows out of GND for a high I/O voltage, or into VCC for a low I/O voltage. But the current is already considered for the I/O pins, so you do not need to count it again for VCC/GND.)

Hi Clemens-san,

Thank you for your description.
I found following app note which is explaining parameter's definition.
https://www.ti.com/lit/an/szza036c/szza036c.pdf

Based on your description and app note I recalculate worst case as below.

12uA(Ioz)*8port + 25uA(ICCA during VCCA=3.6V/VCCB=0V) = 121uA.

Best regards
Kazuki Kuramochi