Chapter 3, Problem 3.8QAP

Interpretation Introduction

(a)

Interpretation:

Current follower that will produce a 1.0 V output for 10.0 µA input current should be designed.

Concept introduction:

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,

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

Interpretation Introduction

(b)

Interpretation:

Effective input resistance of the current follower designed in part (a) should be calculated.

Concept introduction:

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

R_i = effective input resistance

A = open loop gain

Also,

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

Interpretation Introduction

(c)

Interpretation:

The percent relative error for the circuit designed in part (a) for an input current of 25 µA should be calculated.

Concept introduction:

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

R_i = effective input resistance

A = open loop gain

Here,

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

Also,

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

Thus,

Relative error percentage = $\frac{-R_i}{R_L+R_i}\times 100$