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I am trying to solve section d in the following question by substituting a current source of -Ios/2 at the '+' and +Ios/2 at the '-', as the resistor(s) found in section c would nullify the output for all the other offsets, namely IB and Vos. However, I obtain a result which is different than that asserted in the formal solution and wished to confirm my method here.

For a 150 kΩ resistor connected to '+' and IB flowing away from the ground the output would be zero, except that now Ios is added. Hence, using superposition:

For the -Ios/2 source connected to '+' and flowing away from ground: (Ios/2) × 150 kΩ = Vo

For the +Ios/2 source connected to '-' and flowing away from ground:

Vo/100 kΩ = (Ios/2)

This yields 1.25 mV, whereas the official answer is -1 mV.

Where's my mistake?

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Sorry if I wasn't clear. In part C, I hope you put a 100K resistor to ground on the positive input, so that the effect of Ib is cancelled; so the offset current is left (10na). The capacitor blocks DC, so the current has to flow through the feedback resistor. 10nA * 100K = 1mv. Because bias current flowing out of each input is the same, and the DC impedance becomes identical when you add the resistor, the bias current component is cancelled out and lacking offset current you would have zero at the output. So the offset current component appears across one of the 100 k resistors but not the other. Hence 10nA * 100K = 1mv.

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  • \$\begingroup\$ You should explain to him why you use bias current and not the offset current he favors. \$\endgroup\$ Jun 30, 2017 at 20:22
  • \$\begingroup\$ @JohnBirckhead And what about Vos? Where was that taken into account in your analysis? Furthermore, as I pointed out in my question, the resistor on the positive input should be either 50k or 150k, based on the direction of the current \$I_B\$. Why couldn't I simply add -Ios/2 to the positive input and +Ios/2 to the negative? \$\endgroup\$
    – peripatein
    Jun 30, 2017 at 20:56

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