Section 5. Means of extending the limits of measurement. Shunts and additional resistances

Expanding the measurement limits of devices is an important technical and economic task, the purpose of which is to reduce the volume of the enterprise’s instrumentation fleet without compromising the metrological support for product testing and process control. With the means to expand the measurement range, it is possible to use the same usually expensive instrument to measure quantities of different sizes. In specific situations, it may be necessary to change the measurement limit in the direction of increasing the upper measurement limit, i.e., reduce the sensitivity of the device, and in other cases, on the contrary, increase the sensitivity, i.e., change the measurement limit in the direction of decreasing the upper measurement limit. There are two options for solving this problem.

In the first version, the means of expanding the measurement limits are built into the measuring device, which is equipped with a manual limit switch. Such a device is multirange, and the metrological characteristics of this device may differ at different limits. Then they are normalized for each measurement limit separately. The consumer is informed about this by inscriptions on the scale or in the accompanying documentation.

In the second option, external means of expanding the measurement limits are used. This option is used where measurements at one selected limit are performed over a long period of time, such as in process control systems.

Such an external means of expanding the limits of measurement is nothing more than a scaling linear measuring transducer , which does not change the form of the measured value, but only its scale. These converters are produced by the industry as autonomous measuring instruments. Each group of such transducers has unified properties, connecting dimensions and metrological characteristics. Therefore, when they are connected to a single-limit measuring device, in fact, a new device is obtained, the metrological characteristics of which must be calculated from the metrological characteristics of the connected components.

As external means of expanding the limits of measurement, the following are used:

– shunts – to expand the limits of current measurement in the direction of increasing the maximum value;

– voltage dividers and additional resistances – to expand the limits of voltage measurement in the direction of increasing the maximum value;

– current and voltage amplifiers – to expand the limits of current or voltage measurement in the direction of decreasing the maximum value of the measured value;

– current and voltage measuring transformers – can be used to expand the limits of measuring current or voltage in both directions, but most often they are used to expand the limits of measurement in the direction of increasing the maximum value of the measured value.

5.1. Shunts

The connection diagram of a single-limit ammeter with a shunt is shown in fig. 5.1.

The shunt has four clamps. A pair of clamps L 1 , L 2 are called current clamps , a line with a measured current is connected to them. The other two clamps P 1 , P 2 are potential , an ammeter is connected to them, the own resistance of which is shown in fig. 5.1 and denoted by .

Potential clamps are rigidly connected between certain points of the shunt by welding or other methods, providing high stability of the location of these points and negligible and stable transition resistance from these points to potential clamps. Direct connection of the ammeter to the current clamps is unacceptable, since in this case the instability of the resistance of the contacts in the current clamps due to various forces at the screw connection, the ingress of dirt and dust at high current strength will cause a corresponding instability of the voltage drop across these contacts and a measurement error that does not may be guaranteed by the ammeter and shunt manufacturers and cannot be determined by measurement.

The resistance of the shunt between the points of connection of potential clamps is indicated by (Fig. 5.1, but ).

Let be – the current of the full deviation of the needle, corresponding to the upper limit of the measuring range of the ammeter A; is the voltage drop across the ammeter resistance at this current ( ); – the upper limit of the current measurement range, which is desirable to provide with a shunt.

Obviously, at this current strength, the equality . If the shunt is considered as a current divider with a division factor , then its resistance

In a two-limit ammeter (Fig. 5.1, b ), if we accept , shunt resistance for limits and are respectively equal to:

, (5.1)

where are shunting coefficients.

By jointly solving (5.1), we can determine the shunt resistances:

.

Similarly, one can calculate the resistances for a multi-limit stepped shunt.

5.2. Additional resistances

To expand the limits of voltage measurement, voltage dividers and additional resistances can be used. However, due to the fact that the voltage divider must consume a current from the object that exceeds the current of the voltmeter’s own consumption, in practice additional resistances are used to expand the measurement limits of voltmeters (Fig. 5.2).

Additional resistance connected in series with a voltmeter. To change the voltage measurement limit with before magnitude at a given value of the total deflection current of the voltmeter needle determined from expressions

where – coefficient of expansion of the measurement limit of the voltmeter (scale multiplier).

To ensure compatibility of the additional resistance and the voltmeter to which it is connected, the documentation for the voltmeter and, as a rule, on its scale indicates the full deflection current of the needle. A suitable additional resistance is selected according to the following criteria:

– according to the expansion coefficient of the measurement limit;

– according to the maximum allowable current through , which should not be greater than so that the additional resistance does not overheat with this current;

– according to the characteristics of the instrumental error of the new voltmeter created in this way, which will be the sum of the intrinsic error of the voltmeter and the error of the additional resistance, including that resulting from overheating by the current flowing through it.

In multi-limit voltmeters (Fig. 5.2, b ) use the stepped inclusion of resistors and for the corresponding voltage measurement limits at a given full frame deflection current the resistance of additional resistors is calculated by the formulas

or ;

or ,

where are the expansion coefficients of the limits.

Additional resistors can be internal (up to 600 V) and external (up to 1500 V). External additional resistors, in turn, can be individual and interchangeable for rated currents of 0.5; one; 3; 7.5; 15 and 30 mA.

5.3. Typical examples for the calculation of shunts and additional resistors

Example 5.1. Determine the limits for measuring currents I 1 and I 2 in the circuit of a two-limit milliammeter (Fig. 5.1, b ) with a current of total deflection of the measuring mechanism frame I A u003d 50 μA, internal resistance R A u003d 1.0 kOhm. The resistance values of the resistors of the stepped shunt R 1 u003d 0.9 Ohm; R 2 u003d 0.1 Ohm.

Solution . The current I A flowing through the milliammeter is related to the measured current I by the dependence

From here .

At the current measurement limit I 1 R w1 u003d R 1 + R 2 , and at the current measurement limit I 2 , the resistor R 1 is connected in series with R A , and R 2 serves as a shunt. From here

Example 5.2. To expand the measuring range of the ammeter by 50 times with an internal resistance R A = 0.5 Ohm, you must connect a shunt. Determine the resistance of the shunt, the total deviation current of the device and the maximum value of the current at the extended limit, if the voltage drop across the shunt is U n = 75 mV.

Solution .Shunt resistance Ohm.

Instrument Total Deflection Current

Maximum current value at extended limit

Example 5.3. An ammeter with a measurement limit of 100 A has an external shunt with a resistance R W = 0.001 Ohm. Determine the resistance of the instrument’s measuring coil if the total deflection current I A = 25 mA. Determine the maximum power consumed by the ammeter.

Solution . Measuring coil resistance of the instrument

R A u003d R w ( n – 1) u003d 0.001 Ohm ( I / I A – 1) u003d 0.001 [(100 A / 25 10 3 ) – 1] = 4 ohms.

Power consumption of the ammeter

P A = ,

where R is the equivalent resistance of R A and R w connected in parallel, calculated by the formula

Then the power consumption P A =

Example 5.4. Determine the resistance values of additional resistors R 1 , …, R 4 in the circuit of a multi-limit magnetoelectric voltmeter (see Fig. 5.2, b ), which is designed to measure voltage in four ranges with upper limits U 1 u003d 30 V, U 2 u003d 50 V, U 3 u003d 100 V and U 4 u003d 200 V, if the current of the total deflection of the frame of the measuring mechanism of the voltmeter is 10 mA, and the frame resistance is 400 ohms .

Solution . The value of the additional resistor is calculated by the formula

R d u003d R V ( n – 1), where n u003d U / ( I V R V ) .

For the first measurement range (30 V ) R d1 u003d R 1 ; for the second measurement range (50 V) R d2 u003d R 1 + R 2 ; for the third measurement range (100 V) R d3 u003d R 1 + R 2 + R 3 ; for the fourth measurement range (200 V) R d4 u003d R 1 + R 2 + R 3 + R 4 ;

n 1 = 30 B/( n 2 u003d 50 B / 4 B u003d 12.5; n 3 u003d 100 B / 4 B u003d 25; n 4 u003d 200 B / 4 B u003d 50.

Hence R 1 u003d R d 1 u003d R V ( n 1 – 1) u003d 400 (7.5 – 1) u003d 400 Ohm; R 2 u003d R d 2R d 1 u003d 400 (12.5 – 1) – 2600 u003d 4600 – 2600 u003d 2000 Ohm; R 3 u003d R d 3R d 2 u003d 400 (25 – 1) – 4600 u003d 9600 – 4600 u003d 5000 Ohm; R 4 u003d R d 4R d 3 u003d 400 (50 – 1) – 9600 u003d 19600 – 9600 u003d 10000 Ohm.

Example 5.5. The measurement limit of the voltmeter is 7.5 V with an internal resistance R V u003d 200 Ohms. Determine the additional resistance that must be included to expand the measurement limit to 600 V.

Solution .

R d u003d R V ( n – 1); n = 600 V / 7.5 V = 80.

R d u003d 200 (80 – 1) u003d 15.8 kOhm.

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