# Is this resistor needed in this opamp circuit?

I'm looking at the datasheet of the MCP6241. There's this schematic:

The op-amp has a input bias current of 1 pA, and $10^{ 13}\Omega$ input impedance. Is resistor Rz still needed then?

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No it is not needed, but not for the reason you think. In applications where every bit of offset voltage matters, you try to present the same impedance to each opamp input so that the input bias current doesn't create a differential voltage between the two inputs.

Whether a explicit resistance is needed depends on the impedance at the other input, the bias current, and how much you care about the input offset voltage. For example, with 1 pA bias current, a 1 MΩ resistor would drop only 1 µV. That's not going to matter since the opamps inherent input offset voltage is much larger than that. Unless you have a very large impedance, trying to match the impedances in a case of very low bias current opamps is silly.

However, the real reason Rz is not needed is because exactly the same effect can be accomplished with a different choice of Rx and Ry. The output impedance of the Rx,Ry divider is the parallel combination of Rx and Ry while the divider fraction is governed by the ratio of the two. It is therefore possible to chose Rx and Ry so that they have both the desired divider fraction and output impedance.

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 Looks like my answer isn't needed. +1 – stevenvh Aug 3 '12 at 17:44 So, if Rx = Ry = 100k and Rz = 50k, I can delete Rz if I take 200k for Rx and Ry. Correct? – Federico Russo Aug 4 '12 at 9:39 @Federico: Yes, because in both cases you have a voltage source of 1/2 Vdd with 100 kOhm impedance. – Olin Lathrop Aug 4 '12 at 11:35

It's there to match the impedance on both inputs, which minimises offset voltage in addition to the inherent input offset (to be thorough, the resistor in question is not actually necessary if the divider resistor values are chosen correctly, as Olin notes in his answer)

Since the same current flows through each input*, if the impedances are matched then the same voltage drop will be caused on each input and cancel out.
*This only holds true for opamps with matched input currents, which is not always the case. An excellent reference which discusses this and much more is Opamp Applications by Walt G. Jung.

To give an example, if we take a simple opamp non-inverting buffer, the opamp having an input impedance of 1 Megaohm (to exaggerate the effect, though you do get opamps with similar input resistances)
Vin is at 1V:

R1 is the input impedance at 500k. Often you see buffers with no Rf, just the output wired straight to the inverting input. However to match the offset voltage correctly we need an Rf equal to the input impedance.
To show the offset effect, we sweep Rf from 1 ohm to 500 kOhm:

Note how with an Rf of 1 ohm, Vout is ~500mV offset from Vin. As Rf increases toward 500k, we can see the offset heads to zero.

If you look on page 13 (4.7) you will see an explanation of this with the same circuit used as an example.
As Olin notes, at typical 1pA input current, unless you have huge impedances there's no point doing this as the effect will be tiny compared to the inherent offset. It doesn't hurt to get into the habit of thinking about it though.

However, a high temperatures the input bias current can increase quite dramatically, in which case the effect can become more noticeable again. For the MCP6421 the current increases to 1100pA at 125degC. So make sure you take all this into account when deciding what is necessary.

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+1 for the simulation and graph – justing Aug 3 '12 at 17:58
Matching the impedance of the signals driving the two opamp inputs is not about minimizing the effect of the offset voltage. That voltage is there all the time regardless of impedance. Impedance matching is to keep the bias current from creating additional offset voltage. – Olin Lathrop Aug 3 '12 at 21:03
Yes, maybe it's badly worded, I'll mention the inherent offset (was trying to just focus on the relevant effect) – Oli Glaser Aug 3 '12 at 22:01
Thanks, I hadn't looked further in the datasheet yet. Does this mean tht all those billions of buffer amps without feedback resistor are wrong? – Federico Russo Aug 4 '12 at 9:30
No, it just means to be aware of it, as some of them may be "not ideal". It all depends on the opamp, conditions and circuit function, for example if it's a simple (usually AC coupled) audio buffer then a few mV DC offset doesn't matter. – Oli Glaser Aug 4 '12 at 10:57