# Feedback deviation

If a measuring device is built into an apparatus, the original reality changes. The measuring device influences the physical quantity to which the measurement applies. Its effect on the measured value leads to a measurement error that in the metrology basic standard 1319 DIN -1 feedback deviation is called.

## Measurement of temperature

To measure the temperature of a gas flowing through a pipe, z. B. installed a resistance thermometer . The measuring insert and the thermometer protection tube have a different heat dissipation than the original tube wall. In the equilibrium between this increased heat dissipation on the one hand and the supply of heat through the gas on the other hand, a temperature is created in the thermometer that deviates from the gas temperature. The temperature difference to the environment is regularly determined to be too small.

Given the large number of parameters such as installation depth, diameter and flow velocity, the deviation cannot be specified quantitatively. At most, there are empirical values ​​as to when the measurement deviation becomes negligibly small.

## Measurement of electrical quantities

In electrical circuits, one speaks of the fact that a measuring device exerts an influence on the circuit if its insertion changes the quantity to be measured. The influence and reaction-free measurement of the electrical current is only possible with an ideal current source or the electrical voltage with an ideal voltage source or with ideal measuring devices. A measurement deviation must be expected in every real case. The cause is self-consumption through its measuring connections. If parameters are known for electrical measuring devices, the feedback deviation can often be determined mathematically. Such marks come into question

• the internal resistance ${\ displaystyle R_ {i}}$ • the maximum voltage or current consumption:
• in the case of an ammeter, the voltage drop at the end of the measuring range ,${\ displaystyle U_ {I \ mathrm {(MB)}}}$ ${\ displaystyle I _ {\ mathrm {MB}}}$ • in the case of voltmeter, the current consumption at full scale value or the voltage-related resistance${\ displaystyle I_ {U \ mathrm {(MB)}}}$ ${\ displaystyle U _ {\ mathrm {MB}}}$ ${\ displaystyle \ varrho}$ • occasionally (most likely with measuring devices for variable quantities) the power consumption at full scale value.

In the ideal case (no voltage drop on the ammeter with ) or (no current consumption of the voltmeter with ). Otherwise the measured value always contains a systematic measurement deviation with a negative sign. How big this turns out is not a property of the measuring device alone, but always the result of its interaction with the circuit. ${\ displaystyle U_ {I \ mathrm {(MB)}} = 0}$ ${\ displaystyle R_ {i} = 0}$ ${\ displaystyle I_ {U \ mathrm {(MB)}} = 0}$ ${\ displaystyle R_ {i} = \ infty}$ The application of the indicators for self-consumption should be shown in examples. - The calculated deviation cannot be verified experimentally by repeating the measurement with the same measuring device.

example 1

In the circuit shown here, = 24 V; = 10 kΩ. The control of the transistor should be set so that = becomes. It is assumed that the transistor is independent of , an approximation acceptable above about 2V. The voltmeter gives: = 15 V and = 10 kΩ / V. The question is how far the measured and set voltage is also the voltage that results when the measuring device is removed after the setting. ${\ displaystyle U_ {0}}$ ${\ displaystyle R_ {a}}$ ${\ displaystyle U_ {CE}}$ ${\ displaystyle {\ tfrac {1} {2}} U_ {0}}$ ${\ displaystyle I_ {C}}$ ${\ displaystyle U_ {CE}}$ ${\ displaystyle U _ {\ mathrm {MB}}}$ ${\ displaystyle \ varrho}$ At = 15 V, = 0.1 mA flows through the measuring device . With an actually applied voltage of 12 V, the current is lower in the ratio 12:15, i.e. = 0.08 mA. During the measurement the current flows through it . After removing the measuring device, only the unchanged current flows through it , the voltage drop on is = 0.8 V lower. Correspondingly increases to 12.8 V. The absolute feedback deviation is = - 0.8 V, the relative deviation = - 0.8 / 12.8 = - 6%. ${\ displaystyle U _ {\ mathrm {MB}}}$ ${\ displaystyle I_ {U \ mathrm {(MB)}} = 1 / \ varrho}$ ${\ displaystyle I_ {U}}$ ${\ displaystyle R_ {a}}$ ${\ displaystyle I_ {C} + I_ {U}}$ ${\ displaystyle R_ {a}}$ ${\ displaystyle I_ {C}}$ ${\ displaystyle R_ {a}}$ ${\ displaystyle R_ {a} \ cdot I_ {U}}$ ${\ displaystyle U_ {CE}}$ ${\ displaystyle F}$ ${\ displaystyle f}$ If the setting were made using a measuring device with 1 kΩ / V, = 0.8 mA and accordingly the voltage change to 8 V when the measuring device was removed. The setting of the transistor control can be omitted in such a measuring device after this preliminary planning. ${\ displaystyle I_ {U}}$ ${\ displaystyle R_ {a}}$ Example 2

The voltage source is ideal; the load resistance is ohmic with = 3.0 Ω. ${\ displaystyle R_ {L}}$ The ammeter has the measuring ranges 1 | 3 | 10 | 30… 1000 mA; Self-consumption indicator = 0.6 V in all areas. ${\ displaystyle U_ {I \ mathrm {(MB)}}}$ In the measuring range = 300 mA it just shows full deflection. ${\ displaystyle I _ {\ mathrm {MB}}}$ With an internal resistance = 2.0 Ω one determines = 1.5 V. ${\ displaystyle R_ {i} = U_ {I \ mathrm {(MB)}} / I _ {\ mathrm {MB}}}$ ${\ displaystyle U_ {0} = I (R_ {L} + R_ {i})}$ If you switch the ammeter to the measuring range 1 A, then = 0.6 Ω, and the current increases to 1.5 V / 3.6 Ω = 0.42 A; that is 40% more than first measured. Such a discrepancy is a sure sign of a defective measuring device (which can be excluded with this consideration) or of a feedback deviation. The current after removing the ammeter is even higher at = 0.50 A. The current measured at the beginning differs by −40% compared to the current without the measuring device. ${\ displaystyle R_ {i}}$ ${\ displaystyle U_ {0} / R_ {L}}$ Despite the larger systematic measurement deviation in the 300 mA range, you should not switch to the 1 A range. Because in the 1 A range the error limits to be considered due to a class symbol are greater than in the 300 mA range. The deviation due to retroactive effects is easy to calculate and correctable; with error limits, the effort to correct the measured value would be much higher.

Example 3

The circuit and meter are the same as before. Only now = 70 V and = 68 kΩ are given, both with 1% relative error limit. ${\ displaystyle U_ {0}}$ ${\ displaystyle R_ {L}}$ If the expected current is slightly more than 1 mA, select the 3 mA measuring range, which includes an internal resistance = 0.6 V / 3 mA = 0.2 kΩ. ${\ displaystyle R_ {i}}$ The displayed current is . The right current is . ${\ displaystyle I_ {a} = U_ {0} / (R_ {L} + R_ {i})}$ ${\ displaystyle I_ {r} = U_ {0} / R_ {L}}$ According to the definition of the relative measurement error is

{\ displaystyle {\ begin {aligned} f & = {\ frac {I_ {a} -I_ {r}} {I_ {r}}} = {\ frac {I_ {a}} {I_ {r}}} - 1 \\ & = {\ frac {R_ {L}} {R_ {L} + R_ {i}}} - 1 = {\ frac {-R_ {i}} {R_ {L} + R_ {i}} } = {\ frac {-0 {,} 2} {68 {,} 2}} = - \; 0 {,} 3 \; \% \ end {aligned}}} According to the rules of error propagation , the current has an error limit of 2%. This means that the feedback deviation of the current is significantly less than its error limit and cannot be taken into account here.

Example 4

The current from a battery with 1.2 V is measured with a digital ammeter at 90% of the full scale value. If the measuring device drops 0.20 V at the end of the measuring range, the consumer is missing 0.18 V. That is 15% of the battery voltage; the consumer receives 85%. After removing the measuring device, the voltage at the consumer increases in the ratio 100/85 ≈ 1.18, i.e. an additive of 18%. With an ohmic load , the current increases to the same extent; it flows 18% higher than measured.