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?


1 Answer 1


The key words are "unless otherwise stated". They are written on the 2nd line of your picture. In other words, for most of the parameters, it is assumed that Vout is mid-rail (2.5 volts in your case) but, for output amplitude tests, it makes no sense to assume the default words of Vout = V+/2.

Regarding how much the output voltage might be when under load, take a look at figure 2-6: -

enter image description here

This tells me that if the output were at 0 volts and a sink load of 100 mA were applied, the output could be dragged up to typically 1 volt. It also tells me that if the output were 5 volts, then the output could be dragged down by 1 volt when sourcing 100 mA.

So, if your load could be +/- 100 mA, then the usable range of output voltages is 1 volt to 4 volts.

But, you should also verify that the package dissipation isn't being exceeded under these conditions. For instance, if you are sourcing 100 mA at an output level of 2 volts from a 5 volt supply, the internal power dissipation will be based on 3 volts (5 volts - 2 volts) x 100 mA. This is a power of 300 mW and may cause too much overheat.

  • \$\begingroup\$ Thank you for your answer. So, for the line items mentioned in the document, it is mentioned as OUTPUT HIGH only. So, it is safe to assume that the output will be 4.9V or 4.85V as mentioned in the table right? \$\endgroup\$
    – user220456
    Commented Oct 28, 2022 at 10:08
  • \$\begingroup\$ Correct and, I'm adding the bit about load effects on output voltage so bear with me.... \$\endgroup\$
    – Andy aka
    Commented Oct 28, 2022 at 10:09
  • \$\begingroup\$ Yes, with 100mA, the package dissipation is very high. \$\endgroup\$
    – user220456
    Commented Oct 28, 2022 at 10:24
  • \$\begingroup\$ If I want the output to swing only between 0-2V, with an output current of 100mA, then is it not possible, right? \$\endgroup\$
    – user220456
    Commented Oct 28, 2022 at 10:28
  • \$\begingroup\$ Thank you for your answer. Also, if the short circuit current is 120mA or 200mA, at that time, when can calculate the power dissipation, the junction temperature exceeds, right? So,won't the device get damaged? Then what is the purpose of providing the short circuit current? Could you please clarify ? \$\endgroup\$
    – user220456
    Commented Oct 28, 2022 at 10:39

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