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

This is exactly my practical circuit:

Non-inverting amplifier with TL084

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

(Datasheet: TL084)

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2  
Use a better op-amp. The TL084 is old, and frankly a bit rubbish. –  Connor Wolf Jun 18 '12 at 3:55
9  
@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$. –  hkBattousai Jun 18 '12 at 4:00
3  
@FakeName, come be an engineer in Turkey, you will understand hkBattousai. –  abdullah kahraman Jun 18 '12 at 7:07
2  
Can't you get free samples sent to Turkey from manufacturers? It's super easy to get free samples sent to Germany and it is rather not complicated to get them in Lithuania, which is in a "blacklist" of manufacturers as east europe. –  miceuz Jun 18 '12 at 11:09
2  
@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. –  abdullah kahraman Jun 24 '12 at 11:24
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2 Answers 2

up vote 9 down vote accepted

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.

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 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.

enter image description here

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.


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.

enter image description here

enter image description here

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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.

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