# Choosing resistor values for op amp termination

I need to terminate an unused op amp and plan to use the recommended circuit below with +3.3V. How do I choose the resistor values? The op amp I'm using is the MCP6002 (datasheet).

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?

• "the TI recommended circuit" is the well-known "voltage divider". If you use resistors with a few k resistances, the op-amp "load" is practically "nothing" (open circuit). "10K ohms" is fine. Commented Dec 26, 2021 at 14:00

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: -

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.

• 1) The chart X axis is confusing. Why negative and that range? 2) I like the voltage divider for efficiency, described in this video. Does the Input Bias Current (current that flows into the input pins to bias the transistors) of ~1nA mean that's all the current needed into the non-inverting input? 3) 1MΩ resistors gives me 3300nA (0.0033mA), if that's sufficient I'll likely use that since I don't have larger on hand. Just to be sure of my understanding, would 10MΩ (330nA) or 100MΩ (33nA) also work? Commented Dec 26, 2021 at 15:31
• (1) it's showing that even if you take the input slightly negative below the ground power pin of the op-amp, there isn't much current taken by the inputs. (2) That's all the current either input needs to work. (3) Either would work but, it's a bit pointless given that the op-amp's quiescent current is 100 microamps (0.1 mA) @NateS Commented Dec 26, 2021 at 16:12
• You're answer is not fully in line with answers to an older question of mine ( electronics.stackexchange.com/questions/576556/… ). Even with RR input, tying IN+ to VDD or VSS can make the output saturate due to input offset voltage..This in turn can lead to out of spec high supply current draw on many op amps. Would love to hear your opinion on this issue. Commented Dec 26, 2021 at 16:31
• @tobalt I wasn't answering your question; I was answering the one above. If you see discrepancies then please directly highlight them. I'm not about to look at both questions and make some form of compromise answer based on both of them so, do some work (or raise a new question) and explain what it is that is specifically troubling to you. I also note that you haven't left the same request of persons answering your question so, why not? Commented Dec 26, 2021 at 20:32

The voltage divider has the purpose of guaranteeing a voltage on the IN+ that is within both the input common-mode voltage range and the output voltage range. Mid-supply is a safe bet but not a must.

If you use resistors, use the largest resistor present in your BOM. That way, the least amount of current from the supply rail will be wasted. The "op amp load" has nothing to do with these resistors, it has no load being connected only to one MOSFET gate.

The datasheet section 4.5. recommends an alternative solution with no additional components but this one is not recommended as it can cause increasing supply current (out-of-spec even), (more discussion here).

Instead, another solution that is safer and also requires no components, is to connect the output of the used amp in the package to the IN+ of the unused one. This assures that the unused one is always kept at a safe voltage, assuming the used amp's circuit is proper.

simulate this circuit – Schematic created using CircuitLab

• It seems the latest circuit solution is the most original. In addition, it gives a "free" buffer for the working op-amp:-) Commented Dec 26, 2021 at 21:47

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.