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. The circuit becomes: R2 = 10 ohm, R1 = short circuit, potentiometer = 10 kohm linear. 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. If R1 and R2 form a ~~ 1000:1 divider then R1 will be about 10 ohm x 1000 = 10 kohm. Maxim manage to say this in fewer words in the diagram below. If Rf is, say, 10 kohm then a value of R2 = 10 ohm causes an error of 10/10,000 = 0.1%. If the resistance of R2 is small relative to Rf then minimal errors are caused. The effective resistance of the R1, R2 divider is R1 and R2 in parallel or about = R2 for large division ratios.
If, for example, a +/- 15 mV change is required then the ratio of R1:R2 can be about 15 V:15 mV = 1000:1. R1 and R2 divide down changes in potentiometer voltage by the ratio R2/(R1+R2). For a small error the potentiometer value would need to be small or Rf would need to be reduced by an equal amount.įor small offset voltage adjustments the adjustment of the potentiometer becomes difficult and most of the potentiometer range is not used.Īdding R1 and R2 overcomes both these problems. The equivalent resistance of Rf (equal to Rf/4) will add to Rf and cause gain errors. If R1 is a short circuit and R2 an open circuit the whole change in potentiometer voltage is applied to the end of Rf. 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. 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. 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. This injects a small current into the node which causes an offset voltage.Ĭurrent injection effectively occurs at a high impedance point and voltage adjustment at a low impedance point, but both methods are functionally equivalent. Or a say 100 kohm resistor from the op-amp inverting input can be fed by a 10 kohm potentiometer connected to +/- 15 V. The ease of use of this method is improved by adding one two-resistor divider to the potentiometer voltage, as explained below. The methods described below can easily be applied to your circuit byĪdding a divider and potentiometer at the ground end of your R2. Or vary the voltage of a node which a circuit element connects to. The best method to use varies with the application circuit, but all eitherĪpply a variable current to a circuit node Adjust the wiper until the output offset voltage is reduced to zero.There are a range of methods which can be used to provide offset voltage compensation. Adjustment of this pot will null the output.Ģ) By varying the position of the wiper on the 10kΩ potentiometer, we are trying to remove the mismatch between inverting and non-inverting terminals of op-amp. Method to achieve compensation:ġ) To offset this input voltage we have offset null pins in 741 op-amp, hence connect 10kΩ potentiometer across offset null pins 1 and 5 and a wiper be connected to negative supply pin 4 as shown in Figure 1. Need of input offset voltage compensation:ġ) For 741C $V_$ to zero, the op-amp is then said to be nulled or balanced.
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