7 Variable conversion elements

 7.1.2 Deflection-type d.c. bridge

A deflection-type bridge with d.c. excitation is shown in Figure 7.2. This differs from the Wheatstone bridge mainly in that the variable resistance R_v is replaced by a fixed resistance R₁ of the same value as the nominal value of the unknown resistance R_u. As the resistance R_u changes, so the output voltage V₀ varies, and this relationship between V₀ and R_u must be calculated.

This relationship is simplified if we again assume that a high impedance voltage measuring instrument is used and the current drawn by it, I_m, can be approximated to zero. (The case when this assumption does not hold is covered later in this section.) The analysis is then exactly the same as for the preceding example of the Wheatstone bridge, except that Rv is replaced by R₁. Thus, from equation (7.1), we have:

                    V₀ = V_i(R_u/(R_u + R₃) - R₁/(R₁ + R₂)            (7.3)

When R_u is at its nominal value, i.e. for R_u = R₁, it is clear that V₀ = 0 (since R₂ = R₃). For other values of R_u, V₀ has negative and positive values that vary in a non-linear way with R_u.

Example 7.1

A certain type of pressure transducer, designed to measure pressures in the range 0–10 bar, consists of a diaphragm with a strain gauge cemented to it to detect diaphragm

deflections. The strain gauge has a nominal resistance of 120 Ω and forms one arm of a Wheatstone bridge circuit, with the other three arms each having a resistance of 120 Ω. The bridge output is measured by an instrument whose input impedance can be assumed infinite. If, in order to limit heating effects, the maximum permissible gauge current is 30 mA, calculate the maximum permissible bridge excitation voltage. If the sensitivity of the strain gauge is 338 mΩ /bar and the maximum bridge excitation voltage is used, calculate the bridge output voltage when measuring a pressure of 10 bar.

Solution

This is the type of bridge circuit shown in Figure 7.2 in which the components have the following values:

                            R₁ = R₂ = R₃ = 120 Ω

Defining I₁ to be the current flowing in path ADC of the bridge, we can write:

                           V_i = I₁/(R_u + R₃)

At balance, R_u = 120 and the maximum value allowable for I₁ is 0.03 A.

Hence:

                         V_i = 0.03(120 + 120) = 7.2 V

Thus, the maximum bridge excitation voltage allowable is 7.2 volts.

For a pressure of 10 bar applied, the resistance change is 3.38 Ω, i.e. R_u is then equal to 123.38 Ω.

Applying equation (7.3), we can write:

V₀ = V_i{R_u/(R_u + R₃) - R1/(R₁ + R₂)} = 7.2 {123.38/243.38 – 120/240} = 50 mV

Thus, if the maximum permissible bridge excitation voltage is used, the output voltage is 50 mV when a pressure of 10 bar is measured.

The non-linear relationship between output reading and measured quantity exhibited by equation (7.3) is inconvenient and does not conform with the normal requirement for a linear input–output relationship. The method of coping with this non-linearity varies according to the form of primary transducer involved in the measurement system.

One special case is where the change in the unknown resistance R_u is typically small compared with the nominal value of R_u. If we calculate the new voltage V’₀ when the resistance R_u in equation (7.3) changes by an amount R_u, we have:

V’₀ = V_i{(R_u + R_u)/(R_u + R_u + R₃) - R₁/(R₁ + R₂)}                         (7.4)

The change of voltage output is therefore given by:

            V₀ = V’₀ - V₀ = Vi Ru/(R_u + R_u + R₃)

If Ru << R_u, then the following linear relationship is obtained:

                           V₀/ Ru = V_i/(R_u + R₃)                                        (7.5)

This expression describes the measurement sensitivity of the bridge. Such an approximation to make the relationship linear is valid for transducers such as strain gauges where the typical changes of resistance with strain are very small compared with the nominal gauge resistance.

However, many instruments that are inherently linear themselves at least over a limited measurement range, such as resistance thermometers, exhibit large changes in output as the input quantity changes, and the approximation of equation (7.5) cannot be applied. In such cases, specific action must be taken to improve linearity in the relationship between the bridge output voltage and the measured quantity. One common solution to this problem is to make the values of the resistances R₂ and R₃ at least ten times those of R₁ and R_u (nominal). The effect of this is best observed by looking at a numerical example.

Consider a platinum resistance thermometer with a range of 0° –50°C, whose resistance at 0°C is 500 Ω and whose resistance varies with temperature at the rate of 4 Ω/°C. Over this range of measurement, the output characteristic of the thermometer itself is nearly perfectly linear. (N.B. The subject of resistance thermometers is discussed further in Chapter 14.)

Taking first the case where R₁ = R₂ = R₃ = 500 Ω and V_i = 10 V, and applying equation (7.3):

                At 0°C; V₀ = 0

At 25°C;    R_u = 600 Ω and V₀ = 10(600/1100 – 500/1000) = 0.455 V

At 50°C;    R_u = 700 Ω and V₀ = 10(700/1200 – 500/1000) = 0.833 V

This relationship between V₀ and R_u is plotted as curve (a) in Figure 7.3 and the non-linearity is apparent. Inspection of the manner in which the output voltage V₀ above changes for equal steps of temperature change also clearly demonstrates the non-linearity.

                    For the temperature change from 0 to 25°C, the change in V₀ is

                                   (0.455 – 0) = 0.455 V

                   For the temperature change from 25 to 50°C, the change in V₀ is

                                   (0.833 - 0.455) = 0.378 V

If the relationship was linear, the change in V₀ for the 25–50°C temperature step would also be 0.455 V, giving a value for V₀ of 0.910 V at 50°C.

Now take the case where R₁ = 500 Ω but R₂ = R₃ D 5000 Ω and let V_i = 26.1 V:

       At 0°C;     V₀ = 0

       At 25°C;     R_u = 600 Ω and V₀ = 26.1(600/5600 – 500/5500) = 0.424 V

       At 50°C;      R_u = 700 Ω and V₀ = 26.1(700/5700 – 500/5500) = 0.833 V

This relationship is shown as curve (b) in Figure 7.3 and a considerable improvement in linearity is achieved. This is more apparent if the differences in values for V₀ over the two temperature steps are inspected.

                              From 0 to 25°C,    the change in V₀ is 0.424 V

                               From 25 to 50°C, the change in V₀ is 0.409 V

The changes in V₀ over the two temperature steps are much closer to being equal than before, demonstrating the improvement in linearity. However, in increasing the values of R₂ and R₃, it was also necessary to increase the excitation voltage from 10 V to 26.1 V to obtain the same output levels. In practical applications, V_i would normally be set at the maximum level consistent with the limitation of the effect of circuit heating in order to maximize the measurement sensitivity (V₀/ R_u relationship). It would therefore not be possible to increase V_i further if R₂ and R₃ were increased, and the general effect of such an increase in R₂ and R₃ is thus a decrease in the sensitivity of the measurement system.

The importance of this inherent non-linearity in the bridge output relationship is greatly diminished if the primary transducer and bridge circuit are incorporated as elements within an intelligent instrument. In that case, digital computation is applied

to produce an output in terms of the measured quantity that automatically compensates for the non-linearity in the bridge circuit.

Case where current drawn by measuring instrument is not negligible

For various reasons, it is not always possible to meet the condition that the impedance of the instrument measuring the bridge output voltage is sufficiently large for the current drawn by it to be negligible. Wherever the measurement current is not negligible, an alternative relationship between the bridge input and output must be derived that takes the current drawn by the measuring instrument into account.

Thevenin’s theorem is again a useful tool for this purpose. Replacing the voltage ´ source V_i in Figure 7.4(a) by a zero internal resistance produces the circuit shown in Figure 7.4(b), or the equivalent representation shown in Figure 7.4(c). It is apparent from Figure 7.4(c) that the equivalent circuit resistance consists of a pair of parallel resistors R_u and R₃ in series with the parallel resistor pair R₁ and R₂. Thus, RDB is

given by:

                         R_DB = - R₁R₂/(R₁ + R₂) + R_uR₃/(R_u + R3)           (7.6)

The equivalent circuit derived via Thevenin’s theorem with the resistance ´ R_m of the measuring instrument connected across the output is shown in Figure 7.4(d). The open-circuit voltage across DB, E₀, is the output voltage calculated earlier (equation 7.3) for the case of R_m = 0:

                  E₀ = V(Ru/(R_u + R₃) - R₁/(R₁ + R₂)                      (7.7)

If the current flowing is I_m when the measuring instrument of resistance R_m is connected across DB, then, by Ohm’s law, I_m is given by:

                           I_m = E₀/(R_DB + R_m)                                   (7.8)

If V_m is the voltage measured across R_m, then, again by Ohm’s law:

                         V_m = I_mR_m = E₀R_m/(R_DB + R_m)                    (7.9)

Substituting for E₀ and R_DB in equation (7.9), using the relationships developed in equations (7.6) and (7.7), we obtain:

V_m = V_i [Ru/(R_u + R₃) - R1/(R₁ + R₂)] Rm /R₁R₂/(R₁ + R₂) + R_uR₃/(R_u + R₃) + R_m

Simplifying:

V_m = V_iR_m(R_uR₂ - R₁R₃) / R₁R₂(Ru + R₃) + R_uR₃(R₁ + R₂) + R_m(R₁ + R₂)(R_u + R₃)   (7.10)

Example 7.2

A bridge circuit, as shown in Figure 7.5, is used to measure the value of the unknown resistance Ru of a strain gauge of nominal value 500 Ω. The output voltage measured across points DB in the bridge is measured by a voltmeter. Calculate the measurement sensitivity in volts/ohm change in Ru if

(a) the resistance R_m of the measuring instrument is neglected, and

(b) account is taken of the value of R_m

Solution

For R_u = 500 Ω, V_m = 0.

To determine sensitivity, calculate V_m for R_u = 501 Ω.

(a) Applying equation (7.3):    V_m = V_i [R_u/(R_u + R₃) - R₁/(R₁ + R₂)]

Substituting in values: V_m = 10 (501/1001 – 500/1000) = 5.00 mV

Thus, if the resistance of the measuring circuit is neglected, the measurement sensitivity is 5.00 mV per ohm change in R_u.

(b) Applying equation (7.10) and substituting in values:

V_m = 10 × 10⁴ × 500(501 – 500) / 500² (1001) + 500 × 501(1000) + 10⁴ × 1000 × 1001 = 4.76 mV

Thus, if proper account is taken of the 10 kΩ value of the resistance of R_m, the true measurement sensitivity is shown to be 4.76 mV per ohm change in R_u.