How do I correct offset voltage of Opamps which have no explicit offset-null pins?

Question

Not all opamps have explicit offset-null support, but all opamps have offset voltage.

This is exactly my practical circuit:

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

(Datasheet: TL084)

Answer 1

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 pot at the ground end of your R2.
  The ease of use of this method is improved by adding 1 2 resistor divider to the pot voltage, as explained below.

• Or a say 100k resistor from the opamp inverting input can be fed by a 10k pot connected to +/- 15V. This injects a small current into the node which causes 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. ie Injecting a current causes it to flow in related circuitry and causes a voltage change, and adjusting a voltage causes current flows to alter.

To compensate for 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.

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

  The effective resistance of the R1,R2 divider is R1 & 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 = say 10k then a value of R2 = 10 Ohms causes an error of 10/10,000 = 0.1%.

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

If R1 & R2 form a ~~ 1000:1 divider then R1 will be about 10 Ohms x 1000 = 10k.
Use of a say 50 k pot will result in an equivalent resistance of about 12k5 at the mid point and this can be used in place of R1.
The circuit becomes: R2 = 10 Ohms, R1 = short circuit, pot = 10k 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.

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Added:

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.

Answer 2

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.