Zero and Span Circuit

 Zero and Span Circuits

Zero and Span circuits consist of two inverting amplifiers which can be used to convert DC signal levels in order to match levels you want to provide to the controller, display, or computer. The circuits is shown in Figure 1 below.

Figure 1. Zero and Span Circuits

Figure 2 mx + b definition

The signal to be conditioned is a single input, E_in. It receives a gain of –R_f/R_i. A reference voltage, ±V, is the other output.It receives a gain of –R_f/R_os.So the output from IC₁ is

                             E_out1 = (–R_f/R_i) E_{in –} (–R_f/R_os) V . . . . . . . . . . (1)

This signal is then fed through an inverting amplifier with a gain -1.

E_out2 = (R_f/R_i) E_{in +} (R_f/R_os) V . . . . . . . . . . . . (2)

Compare this to the equation of a straight line,

                             y = mx + b   . . . . . . . . . . . . . . . . . . . . . . . . (3)

Where y is the dependent variable and x is the independent variable. For a plot of output versus input voltage, as in Figure 2,

                             m = R_f/R_i      --à slope or gain or span

                             b = (R_f/R_os) V  à y intercept or offset or zero

·    Application Zero and Span Circuits

The circuits can be applied such as when a temperature sensor has outputs range between  2.48 V – 3.90 V (for minimum to maximum temperature). The output is fed to an A/D converter which has input range 0 – 5 V. To provide maximum resolution, zero and span circuit can be used so that it fills the entire range of the converter.

Solution:

The circuit in Figure 1 should be used. The required gain is

                             m       =  E_{out /} E_in

                                                                = (E_{out(max) -} E_out(min) )/( E_{in(max) -} E_in(min))

                                      = (5 V – 0V) / (3.90V – 2.48V)

                                      = 3.52

But the gain is set by

                             m       = R_f/R_i      

Pick R_f relatively large, so that a smaller  R_i   will not load down the sensor.

                             R_f       = 330 kΏ

                             R_i      = R_f/m   = 330 kΏ/ 3.52       = 93.7 kΏ

Pick R_i    as a 47 kΏ fixed resistor with series 100 kΏ multiturn potensiometer. This set the gain (span). To get the offset (zero), substituse the values you have into the circuit equation at one point.

y = mx + b  

or

                             E_out = mE_in  + b

At E_in = 2.48 V, E_ot  = 0 V,

                             0 V     = (3.52)(2.48V) + b

                             b        = -(3.52)(2.48V)       = - 8.73V

but

                             b        = (R_f/R_os) V 

Since you need negative offset, select V = -12 V (The negative power supply voltage).

                             R_os     = (R_fV)/ b

                                      = ((330 kΏ)(-12V))/-8.73V    = 454 kΏ

Pick R_os as a fixed 220 kΏ resistor with a 500 kΏ multiturn potentiometer. To find R_comp,

                             R_comp = R_f||Ri ||R_os = 62.9 kΏ        (pick R_comp = 56 kΏ)

For the resistor in the second stage should be in the kΏ range. This lower offset without loading the first stage. Pick R = 2.2 kΏ. So R/2 = 1.1 kΏ. To check your design it’s work at the other specified point, use the equation 2.

                             E_out2   = (R_f/R_i) E_{in +} (R_f/R_os) V

                                       = (330 kΏ/93.7 kΏ)(3.90V) + (330 kΏ/454 kΏ)(-12V)

                                      = 13.72 V – 8.72 V   = 5 V.

(Source: Industrial Control Electronics Applications and Design)