Question

The op-amp (ideal except \(I_b=1\,\text{nA}\) input bias current and \(V_{\!os}=10\,\mu\text{V}\) input offset) is the inverting amplifier with \(R_{\text{in}}=1~\text{k}\Omega\) and \(R_f=100~\text{k}\Omega\). Find the worst-case output offset voltage \((\pm~\mu\text{V})\) (rounded to the nearest integer).

Hint:

In inverting amps: \(V_{o,\;V_{\!os}}=(1+R_f/R_{\text{in}})V_{\!os}\) and \(V_{o,\;I_b}\approx I_b R_f\) (with non-inverting input at 0~\(\Omega\), its bias term is negligible).

Answer 1

Correct Answer: 1110

Step 1: Output due to input offset voltage.
Noise gain (for \(V_{\!os}\)) in an inverting amplifier: \(1+\dfrac{R_f}{R_{\text{in}}}=1+100=101\).
\(\therefore V_{o,\;V_{\!os}} = 101\times 10~\mu\text{V}=1010~\mu\text{V}\).
Step 2: Output due to input bias current.
With the non-inverting input grounded, the inverting node is at virtual ground. The bias current through \(R_f\) sets \(V_{o,\;I_b}= I_b\,R_f = (1~\text{nA})(100~\text{k}\Omega)=100~\mu\text{V}\).
Step 3: Worst case.
Taking the same sign so they add: \(|V_o|_{\max}=1010+100=1110~\mu\text{V}\).
Final Answer: \(\pm 1110~\mu\text{V}\)