Not all op-amps have explicit offset-null support, but all op-amps have an offset voltage.

This is exactly my practical circuit:

Non-inverting amplifier with TL084

How do I correct the offset voltage of TL084 in this circuit?

(Datasheet: TL084)

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    \$\begingroup\$ Use a better op-amp. The TL084 is old, and frankly a bit rubbish. \$\endgroup\$ Jun 18, 2012 at 3:55
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    \$\begingroup\$ @FakeName But it is easily found in local market, cheap, has high input impedance. I can't purchase opamp for my hobby projects; even the shipping cost is more than 40$. \$\endgroup\$ Jun 18, 2012 at 4:00
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    \$\begingroup\$ @FakeName, come be an engineer in Turkey, you will understand hkBattousai. \$\endgroup\$ Jun 18, 2012 at 7:07
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    \$\begingroup\$ @miceuz There are a few distributors but they are only taking big companies seriously. Last year, I wanted to order an ARM microcontroller, mailed all the distributors several times, but non of them replied me. Once they realize that you won't earn them money, they don't even speak to you (but yet they are playing the "kind company" role in the seminars very nicely). Actually, there is a company here which is not literally a distributor, they are just reselling Farnell products, but their waiting time and prices are very high. Because of these reasons, I'm working with local market products. \$\endgroup\$ Jun 18, 2012 at 17:29
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    \$\begingroup\$ @miceuz What is the meaning of the samples if you cannot afford importing them? Samples are valid for only the hobby projects in this case. For example, how hard do you think designing an SMPS in Turkey? Well, it is, really, hard. \$\endgroup\$ Jun 24, 2012 at 11:24

2 Answers 2


There are a range of methods which can be used to provide offset voltage compensation.

The best method to use varies with the application circuit, but all either

  • apply a variable current to a circuit node

  • or vary the voltage of a node which a circuit element connects to.

The methods described below can easily be applied to your circuit by

  • Adding a divider and potentiometer at the ground end of your R2.
    The ease of use of this method is improved by adding one two-resistor divider to the potentiometer voltage, as explained below.

  • Or a say 100 kohm resistor from the op-amp inverting input can be fed by a 10 kohm potentiometer connected to +/- 15 V. This injects a small current into the node which causes an offset voltage.

Current injection effectively occurs at a high impedance point and voltage adjustment at a low impedance point, but both methods are functionally equivalent. That is, injecting a current causes it to flow in related circuitry and causes a voltage change, and adjusting voltage causes current flows to alter.

To compensate for an offset voltage by injecting a current you can apply an adjustable voltage from a potentiometer via a high-value resistor to an appropriate circuit node. To adjust a "ground" voltage that a resistor connects to, you can connect it to a potentiometer which is able to vary either side of ground.

The diagram below shows one method. Here Rf would usually connect to ground.

enter image description here

If R1 is a short circuit and R2 an open circuit the whole change in potentiometer voltage is applied to the end of Rf. This causes two problems.

  • The equivalent resistance of Rf (equal to Rf/4) will add to Rf and cause gain errors. For a small error the potentiometer value would need to be small or Rf would need to be reduced by an equal amount.

  • For small offset voltage adjustments the adjustment of the potentiometer becomes difficult and most of the potentiometer range is not used...

Adding R1 and R2 overcomes both these problems.

R1 and R2 divide down changes in potentiometer voltage by the ratio R2/(R1+R2). If, for example, a +/- 15 mV change is required then the ratio of R1:R2 can be about 15 V:15 mV = 1000:1.

The effective resistance of the R1, R2 divider is R1 and R2 in parallel or about = R2 for large division ratios.

If the resistance of R2 is small relative to Rf then minimal errors are caused.

If Rf is, say, 10 kohm then a value of R2 = 10 ohm causes an error of 10/10,000 = 0.1%.

Maxim manage to say this in fewer words in the diagram below.

enter image description here

If R1 and R2 form a ~~ 1000:1 divider then R1 will be about 10 ohm x 1000 = 10 kohm.

Use of a, say, 50 kohm potentiometer will result in an equivalent resistance of about 12.5 kohm at the mid point and this can be used in place of R1.

The circuit becomes: R2 = 10 ohm, R1 = short circuit, potentiometer = 10 kohm linear.

The above circuit is taken from the useful Maxim Application note 803 - EPOT Applications: Offset Adjustment in Op-Amp Circuits which contains much other applicable information.

In his answer miceuz referred to NatSemi's AN-31 pages 6 & 7.

Not surprisngly, the circuits there apply the identical methods to what I describe above and to those in the Maxim app note, but the diagrams are more explanatory, so I've copied them here.

enter image description here

enter image description here

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    \$\begingroup\$ @Peter Mortenson - this post sat here for almost 3 years until you decided to mess with it. Useful technical edits are fine. Edits for style and preference, some of which are linguistically incorrect or questionable, are not appreciated. I add enough typos to keep the average pedant happy if that's their bent. Reformatting so IMHO information is lost or just stirred up does not feel like an optimum use of your time. I had a look at some of your answers - you seem to 'know your stuff' - I'd have thought you could do better spending your time on new answers than on rehashing my pontifications \$\endgroup\$
    – Russell McMahon
    Mar 29, 2015 at 13:01

This link has it sorted out. In general, you have to inject correcting voltage into one of inputs.

This PDF has it for inverting and noninverting configuration (Figures 18 and 19).


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