Question

3-8 An operational amplifier to be used in a current follower has an open-loop gain $A$of $1\times {10}^6$and an input bias current of $2.5\text{\hspace{0.17em}nA}$.

(a) Design a current follower that will produce a $1.0\text{V}$output for a$10.0\mu \text{A}$ input current.

(b) What is the effective input resistance of the current follower designed in part (a)?

(c) What is the percent relative error for the circuit designed in part (a) for an input current of $25\mu \text{A}$?

Short answer

The effective input resistance is$0.1\Omega$

Step-by-step solution

Part (a)  Step 1. Given information

Open loop gain A = 1 x 10⁶

Input bias current i_b = 2.5 nA

Output voltage V₀ = 1.0 V

Input current i­_I = 10.0 µA

Explanation

Operational amplifiers can be used to measure or process currents by connecting them in the current follower mode. This mode provides a nearly zero resistance load to the current source and prevents it from being loaded by a measuring device or circuit.

Here,

V₊, V_- = input voltages

V_s = Input difference voltage

V₀ = output voltage

i_b = input bias current

i_f = feedback current

i_i = input current

R_f = feedback resistor

Also, ${\text{R}}_f=\frac{{\text{V}}_0}{{\text{i}}_i}$

So,${\text{R}}_f=\frac{1.0\text{V}}{10.0\times {\text{10}}^{-6}\text{A}}=1\times {10}^5\Omega$

Part (b).  The effective input resistance Ri=RfA

The effective input resistance

${\text{R}}_{\text{i}}\text{=}\frac{{\text{R}}_{\text{f}}}{\text{A}}$

${\text{R}}_{\text{i}}\text{=}\frac{{\text{1×10}}^{\text{5}}\text{Ω}}{{\text{1×10}}^{\text{6}}}\text{=0.1Ω}$

Part (c) . The relative error percentage −RiRL+Ri×100

The relative error percentage

$\frac{-R_i}{R_L+R_i}\times 100=\frac{-4\times {10}^{-2}\Omega }{R_L+4\times {10}^{-2}}\times 100=-\frac{4}{R_L+4\times {10}^{-2}}\%$