Calculate the input bias current at each input of an op-amp having specified values of \(I_{\mathrm{IO}}=4 \mathrm{nA}\) and \(I_{\mathrm{IB}}=20 \mathrm{nA}\).

Based on the question, two parameters have been provided: \(I_{IO}=4 \, \mathrm{nA}\) and \(I_{IB}=20 \, \mathrm{nA}\). Here, \(I_{IO}\) is the input offset current, and \(I_{IB}\) is the input bias current.

The values of \(I_{B1}\) and \(I_{B2}\) can be obtained from the following established op-amp formulas: \(I_{IB} = \frac{I_{B1} + I_{B2}}{2}\) and \(I_{IO} = |I_{B1} - I_{B2}|\). These will be used to solve the values of \(I_{B1}\) and \(I_{B2}\) respectively.

From the given \(I_{IB} = 20 \, \mathrm{nA}\), and using the first equation \(I_{IB} = \frac{I_{B1} + I_{B2}}{2}\), we can infer that the sum of \(I_{B1}\) and \(I_{B2}\) is equal to \(2 * I_{IB} = 40 \, \mathrm{nA}\), giving the possibility that \(I_{B1}\) and \(I_{B2}\) could be any values that add up to \(40 \, \mathrm{nA}\). However, using the second equation \(I_{IO} = |I_{B1} - I_{B2}|\) with the given \(I_{IO} = 4 \, \mathrm{nA}\), this narrows down the possible combinations to \(I_{B1} = 18 \, \mathrm{nA}\) and \(I_{B2} = 22 \, \mathrm{nA}\), or \(I_{B1} = 22 \, \mathrm{nA}\) and \(I_{B2} = 18 \, \mathrm{nA}\), since in both cases, the difference in absolute terms equals \(4 \, \mathrm{nA}\) and the sum equals \(40 \, \mathrm{nA}\).