Question: I need to terminate an unused op amp and plan to use the [recommended circuit](https://youtu.be/gwQiFuckMz0?t=727) below with +3.3V. How do I choose the resistor values? The op amp I'm using is the [MCP6002](https://www.microchip.com/en-us/product/MCP6002) ([datasheet](http://ww1.microchip.com/downloads/en/DeviceDoc/MCP6001-1R-1U-2-4-1-MHz-Low-Power-Op-Amp-DS20001733L.pdf)).

[![](https://i.stack.imgur.com/gY6Ji.png)](https://i.stack.imgur.com/gY6Ji.png)

I understand too low allows a lot of current, wasting power. Too high and it's not enough current to drive the load. In this case the load is just the op amp. Here is where I get lost. What determines a reasonable but low amount of current to put through the op amp?

`Output Short-Circuit Current` is +/-6mA @ 1.8V, so something like < 3mA @ 3.3V would be very safe for the high end, but I want the low end.

10K ohms for both resistors would be 0.33mA. I guess that would be fine, but can I go lower?

Answer: 
The op-amp has rail-to-rail input capabilities hence, you can tie the unused input to either: -

* 3V3 or
* 0 volts

Noting this from the DS: -

[![enter image description here](https://i.stack.imgur.com/T7p9M.png)](https://i.stack.imgur.com/T7p9M.png)

I'd probably tie IN+ to 0 volts.

> 
> *What determines a reasonable but low amount of current to put through
> the op amp?*
> 
> 
> 

Well, the graph above gives some indication but, unfortunately, the DS doesn't give an equivalent graph when the input is close to the positive rail.

If you insist on using a potential divider, take note that the input bias current is about 1 nA over temperature and, if your resistors were 1 MΩ roughly, that would mean a 1 mV drop across them so, I'd go as high as possible because it won't really matter that much.
The general way to do this is to look at the op-amp's input current specification (not the output current). You want to choose a resistance that (A) won't generate more than about 10% of the power supply voltage when it's drawn through the bias network, and (B) is well below the board's self-leakage resistance.

I'm just going to trust AndyAka's answer that the input bias current is 1nA -- in that case, (A) is satisfied with \$\frac{0.33\mathrm V}{1 \mathrm{nA}} = 330 \mathrm M\Omega\$\*. This doesn't come *close* to satisfying (B) -- if you have a pair of pads for an 0603 resistor on a board and you swipe your finger across them you'll have less resistance than that. My "I don't worry about this" resistance is \$1 \mathrm M\Omega\$, so that's what I'd use.

I'd also add a feedback resistor, from output to the \$V\_-\$ input. This won't change the circuit operation (it's *supposed* to do nothing), but if you're in the middle of prototyping and you discover that you need an op-amp for something, the resistor in the layout will make it much easier to use that op-amp.

\* I'm answering for free, so I refuse to pull out a calculator. This answer is exactly correct, unless I'm off by a piddly little factor of 1000 somewhere.