OPA818: Whether it's high or low to active of the PD pin?

Hello Rain,

Great question!  I can certainly understand your uncertainty present here in the Pin Functions table.  

First, please take a look at this excerpt from the Layout Guidelines section, 11.2 Layout Example:

As you can see, the PD (bar) pin must be connected to VS+ to enable the amplifier.  The PD (bar) pin is active low, in the case of the OPA818.  The values present under the Power Down table under Electrical Specifications match the 'active low' nature of the PD (bar) pin: a voltage no less than 1V lower than the VS+ supply is required to power on the amplifier; a voltage no greater than 3V lower than the VS+ supply is needed to turn off the amplifier.  

Put another way:

To enable the amplifier: PD(bar) > (VS+ - 1V); which would be PD(bar) > 5V for a 6V single supply (VS+ = 6V).

To disable the amplifier, using the active low PD(bar) pin: PD(bar) < (VS+ - 3V); which would be PD(bar) < 3V for a 6V single supply (VS+ = 6V).

These stated values align with the datasheet recommendation in section 11.2 of setting PD(bar) = VS+ supply.

Now for the explanation:

The Pin Functions table states: "Power down (low = enable, high = disable); internal 1-MΩ pull-up allows floating this pin".

The terms 'low = enable' & 'high = disable' are referring to the PD(Bar) function, not the power state of the amplifier.  The PD(bar) pin (& function) is enabled when at a low voltage, which would disable the amplifier.  The PD(bar) pin is disabled at a high voltage, such as VS+, which would enable the amplifier.

If you are ever unsure about the meaning of a specification or device function, using the E2E forum is an excellent way to find an answer.  There is a good chance a similar question has already been asked; if your question is new/unique, we will work on finding a solution for you.

Please continue with your design and evaluation of the OPA818 with this knowledge.  If you have further questions, please come back to the forum.

Best,

Alec