# logic gate's open drain output pulled to voltage higher than supply voltage

I am using 74AHC1G09, it has an open drain output. It says the maximum Vo is 7V, but also says "The input and output voltage ratings may be exceeded if the input and output current ratings are observed." My question is, what's the output current ratings means? Is it refer to "output current" or "output clamping current"? I think it is the output clamping current, because if Vo is pulled to a voltage higher than the IC's supply voltage, the output clamp diode will conduct, the datasheet want the user to protect this diode from overcurrent.

However, I see similar post "link", NXP's employee says it is the "output current" which should be observed. That doesn't make sense, because if that is true, the datasheet can only list the maximum ratings of Io, but it listed both the Io and Vo.

So can I pulled up to for example 12V through 10k resistor on the output of the gate, while only powers it from 3.3V? When doing this, will the 1mA current sourcing from 12V to 3.3V cause problem for the 3.3V supply, a linear regulator?

Thank you

You are correct, that the overvoltage figure basically does not matter as long as the current is bounded to safe limits. All the absolute maximum ratings I have seen still mention the voltage limit. Guess it cannot hurt to avoid at least a few dumb user mistakes.

Pulling up the output to 12 V via 10k sounds tolerable. 1 mA is far below the maximum rating. However, it is probably not conducive for long term reliability to constantly conduct a significant current through the clamp diodes, even if staying below the limiting values. In fact I would consider at least 100k as the pullup resistor in this application.

I would not exceed the datasheet values. When the output is off, then the full voltage from the pullup will be present on the pin. If this exceeds the $$\Vds\$$ voltage of the output transistor, the transistor will fail.
The chip manufacturer might use a higher voltage process on the die in some fabs and a lower voltage process in other fabs. Rather than make two datasheets they are going to spec the lower voltage one, which works for 99% of the users out there.

Also, there might be ESD protection diodes on the pin, which would be forced into conduction if the voltage on the pin exceeded Vcc.

If you need higher voltage, you can have the chip drive an external FET that is rated for the voltage that you need.

When doing this, will the 1mA current sourcing from 12V to 3.3V cause problem for the 3.3V supply, a linear regulator?

No it won't hurt the regulator. Assuming that the output transistor didn't fail, the voltage would be dropped across the 10k resistor. It would pose no difference to the regulator than if you pulled the pin up to 3.3v with a 3.3k resistor (1mA).

Other manufacturers (TI, Diodes) for the same part say that output max voltage is VCC+0.5, which implies there is protection diodes on output and the part is really meant to drive loads that are connected to same VCC or lower voltage. Usually it would be clearly indicated if the part can tolerate higher voltages on output than it's VCC supply, even if VCC would be zero.

• Both TI and Diodes unconditionally allow up to 5.5 V in the recommended operating conditions, so there must be an error in those datasheets.
– CL.
Nov 15, 2019 at 12:38