Why put a resistor in series with the positive terminal in a non inverting op-amp circuit?

I've tried to find the answer, but failed.

Why do we add R1 in series with the positive terminal for a non-inverting amplifier? What is the best value we can assume or choose for it?

• Is this actually about the 741 operation amplifier? It doesn't seem to be specifically about it, but just about opamps in general. If that's the case, we should remove that tag. If it is specifically about the 741, then this is very central and should be explained in the question text, I guess. – Marcus Müller 2 days ago
• ^ Abdallah, could you clarify? – Marcus Müller 2 days ago
• is this about opamps in general, or about the 741 (which is a special type of opamp) specifically? You used the 741-opamp tag. – Marcus Müller 2 days ago
• No, it's in general , I'm sorry for this thing , i want to remove it from tag now – Abdallah_98 2 days ago
• No need to apologize! – Marcus Müller 2 days ago

It's to compensate the effect of bias current in the non-inverting amplifier. The compensating resistor value equals the parallel combination of R2 and R3. The input current creates a voltage drop across R1 that offsets the voltage across the combination of R2 and R3.

• Input currents have higher temperature coefficients than input voltages, so the thermal DC errors are dominated by input resistance if those are not balanced. For a '741, resistive imbalance should be under 1k ohm for best performance. – Whit3rd yesterday
• I think it's worth to mention that in order for this "trick" to work, the op amp's input offset current should be less than an order of magnitude of bias current. :) – Long Pham 4 hours ago
• I wouldn't call it a trick so much as a design technique. I first came across it in some 1968/1969 op amp handbooks out of the old National Semiconductor, authored by some of the early pioneers in integrated analog circuit design like Bob Widlar and Robert Dobkins. – SteveSh 1 hour ago

In the "old times", it was for "impedance reasons" and "symmetry". The simplest value is $$\R_2 \parallel R_3\$$, so it should be: $$\20 \times 100/(100+20) \implies \$$ nearest $$\16.7 \text{ k}\Omega\$$.

The exact calculation is much more complicated, because you have to take into account variation of all offset characteristics versus temperature, power supplies and so on ... And sometimes, it can be used for noise reasons.

I have noted that the internal resistance of the generator is not shown. Some OPAs also need a resistance (50 ohm) to ground ... for RF impedance matching. That is not the case here.

Image to explain how microelectonics is ... see my comments lower. https://slideplayer.fr/slide/3251080/

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• What do you mean by "OPAs"? Op amps (operational amplifiers)? Or a particular series of (high-bandwidth) op amps whose part names start with "OPA"? Or something else? – Peter Mortensen 14 hours ago
• Symmetry of what? The influence of bias currents? Or something else? – Peter Mortensen 14 hours ago
• OPAs is what you say. Operational Amplifiers generally speaking. For symmetry, you must go into the amplifier and how is designed its internal topology, essentially based on differential amplifier, for some reason of temperature stability. Must see the schematic for understanding. – Antonio51 6 hours ago
• However, the topology described is not sufficient in terms of the diagram. It must also be at the level of the final achievement on the chip. Since this is microelectronics, you have to look at the most basic level of realization, and it's not that simple. Let us add the fact that a differential stage is not made up of two transistors but at least four placed in a crossed way! And placed in the 4 corners of a square so that the temperature (outside or inside released by the components themselves) has the least possible influence on the characteristics of the assembly! – Antonio51 6 hours ago
• Microelectronic technology is an architectural "marvel of ingenuity" in the true sense of the word ... not many people know the difficulty. Remember that a certain current component is like drawing plans for a city to the nearest mm with a square area of 100 km side! Sorry for this long comment ... – Antonio51 6 hours ago

Besides matching the impedance to null the bias current offset voltage, another reason is to limit currents in the case of an input over-voltage condition.

In the case of overvoltage (input beyond rails) most op amps can tolerate a few mA of input current through their internal rail clamping diodes without damage. A resistor of, e.g., 10 kohm hence allows the input voltage to exceed the rails by some 10s of volts.

Although some op amps do contain some serial input resistance, this is usually kept rather small for noise reasons. As a result, external current limiting is almost always necessary when inputs can leave the range between the supply rails.

Why put a resistor in series with the positive terminal in a non inverting op-amp circuit?

A compensation resistor is sometimes added in series to the non-inverting terminal of an op-amp when the non-inverting input signal comes from a low impedance source.

Although there are a variety of possible benefits that accrue from the presence of a compensation resistor, they all are related to the idea of balancing the impedance from each of the op-amp inputs to ground.

To achieve such balance, the value of $$\R1\$$ should be

$$R1 = (R2 || R3) - Rsource$$

Where $$\Rsource\$$ is the output impedance of the source driving the non-inverting input.

The input stage of any op-amp is some sort of differential pair. (It may not be a "pair", there is sometimes 4 or more transistors, but the idea is the same.)

The output of a differential pair is a highly non-linear function of the inputs. Op-amps rely upon negative feedback to attain linearity.

The gain of a differential pair varies according to how evenly the transistors conduct. If the pair is evenly balanced the gain is highest. When one transistor conducts much more than the other, the gain can be very low. While the gain of an op-amp based amplifier is set by the feedback network, the bandwidth of the amplifier will suffer if the open-loop gain of the op-amp is below its optimal.

One reason for using the compensation resistor, then, is to ensure the differential pair is operating at its maximum gain point, rather than off to the side where its gain is highly reduced, and by so doing, ensuring the maximum bandwidth is available for the op-amp.

Another reason for using the compensation resistor is to reduce input offset voltage.

There are yet other reasons that have been offered, and I cannot evaluate their significance apart from actual circuits with actual op-amps. But they all based upon balancing the differential pair.

What Mike said is true. I am providing more information for those who want to dig a little deeper. Write down the equation which provides the value of Vo, the output of the op-amp, as a function of Vin. Start with Vin = 1 Volt to make it easy.

In a perfect world, the input impedances of an op-amp are infinite. The output impedance of the op-amp is zero. Same is true of the voltage source Vin. Also, the differential voltage between + and - inputs to the op-amp is zero. Furthermore, the open loop gain of an op-amp is infinite. In the real world, none of this is true. The bias currents, which are necessary to activate the differential input circuits of the op-amp, are exceedingly small which means that what is said at the beginning are fair assumptions to make to address the present problem. The bias current flowing in the parallel combination of R2 and R3 produces a voltage at the inverting input of the op-amp which can be cancelled at the non-inverting input of the op-amp with the addition of R1 with the value of R2||R3. In other words, the error produced at the output of the op-amp caused by the input offset voltage, because of the bias currents, can be eliminated.

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