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Op-amp clarification in table

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Full Member level 6
May 10, 2020
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I'm using this Op-amp with Vcc of 5V and GND, voltage follower configuration.

Input to the non-inverting terminal is 0-2V (from a DAC). Output pin is connected to the inverting terminal.

Since, it's a voltage follower, I expect the output voltage to follow the input voltage.

But in this table below,

enter image description here

Can someone explain me, how Vout=V+/2 (Which is 2.5V)? Like, since its a voltage follower, shouldn't the output follow the input? If the input is, say 2V, then how can vout be 2.5V?

Like, is OUTPUT HIGH=2.5V? If so, why is Vout mentioned as 4.9 or 4.85?

I'm genuinely confused with the Vout and OUTPUT HIGH?

Also, how much is the maximum output current that I can drop from this op-amp for an output swing between 0-2V? how to get this value from the datasheet?

As for Vout usage, the table header tells "unless otherwise indicated". Think again.

For your application, the most relevant information is output low restriction. The actually achievable voltage follower output with 0V input depends however on load conditions. A pulldown resistor and no current driven from the load usually allows voltage starting below 1 mV.


Read about op amp test set-ups used in datasheets. It should help. I seem to remember that Bruce Trump (TI engineer and OA expert) has a blog post that has a paragraph or two about that somewhere in his very, very helpful series on op amps - very readable and clearly understandable explanations.

Just looked, can't find the exact one mentioned above, but this one is also interesting and links to the series:

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