Electrical and Electronic Measurements Test 6

In a 3 ½ digit digital multimeter, there are three full digits and one half digit.

A full digit counts from 0 to 9 whereas an half digit counts from 0 to 1.

Concept:

The resolution in a N bit DVM is given by \(= \frac{1}{{{{10}^N}}} \times range\;of\;voltage\)

Where N is the number of full digits.

In a DVM, a full digit counts 0 to 9 and half digit counts from 0 to 1.

Calculation:

The full-scale reading = 9.99 V

It is a 3-digit voltmeter.

Range of voltmeter = 10 V

Resolution for the given DVM is \(= \frac{1}{{{{10}^3}}} \times 10 = 0.01\)

Resistance (R) = 100 kΩ

Capacitance (C) = 1 μF

Integrator input voltage (V_i­) = 1 V

For an integrator type DVM, output voltage is given by

\({V_0} = - \frac{1}{{RC}}\mathop \smallint \limits_0^t {V_i}dt\)

\( = - \frac{1}{{100 \times {{10}^3} \times 1 \times {{10}^{ - 6}}}}\mathop \smallint \limits_0^1 1dt = - 10\;V\)

We know that,

Average value of triangular wave = \(\frac{{{V_m}}}{2}\)

rms value of triangular wave \(= \frac{{{V_m}}}{{\sqrt 3 }}\)

\(\begin{array}{l} \Rightarrow \frac{{{V_m}}}{{\sqrt {30} }} = 60\\ \Rightarrow {V_m} = 60\sqrt 3 \end{array}\)

Analog instrument:

Voltage error = ±2% of 25 V = ±0.5 V

Error \( = \pm \frac{{0.5}}{{20}} \times 100 = \pm 2.5\% \)

Digital instrument:

For 20 V displayed on 3 ½ digit display, 1 digit = 0.01 V

Voltage error = ±(0.6% of reading + 1 count)

= ±(0.12 + 0.01) = ±0.13 V

Concept:

For a dual-slope analog to digital converter,

De-integration time, \({T_2} = \frac{{{V_i}}}{{{V_{Ref}}}} \times {T_1}\)

Where Vi = input voltage

VRef = reference voltage

T1 is the first integration time

Total conversion time = first integration time (T2) + de-integration time.

Calculation:

Given that, input voltage (Vi) = 6.8 mV

Reference voltage (VR) = 10 mV

 Integration time (T1) = 10 ms

De-integration time, \({T_2} = \frac{{6.8}}{{10}} \times 10 = 6.8\;ms\)

T₁ = nT_s

\(\Rightarrow 150\;clks = n \times \frac{1}{{{f_s}}}\)

\(\Rightarrow 150 \times {T_{clk}} = n \times \frac{1}{{{f_s}}}\)

\(\Rightarrow 150 \times \frac{1}{{{f_{clk}}}} = n \times \frac{1}{{{f_s}}}\)

\(\Rightarrow {f_{clk}} = \frac{{{f_s} \times 150}}{n}\)

For maximum clock frequency, the value of n should be i.

Let the voltage is V_S and the internal resistance is R

For 5 V scale:

Internal resistance = full scale voltage × sensitivity

= 5 V × 20 kΩ/V = 100 kΩ

Reading = 4.5 V

\( \Rightarrow {V_S} \times \left( {\frac{{100}}{{R + 100}}} \right) = 4.5\)

For 10 V scale:

Internal resistance = full scale voltage × sensitivity

= 10 V × 20 kΩ/V = 200 kΩ

Reading = 6 V

\( \Rightarrow {V_S} \times \left( {\frac{{200}}{{R + 200}}} \right) = 6\)

By solving the above two equations, we get

The dc sensitivity \(= \frac{1}{{100 \times {{10}^{ - 6}}}} = 10k{\rm{\Omega }}/v\)

For a halfwave rectifier,

Ac sensitivity = 0.45 × dc sensitivity

= 0.45 × 10 = 4.5 kΩ/v

Series multiplier = (S_ac) (V) - R_m - R_d

= (4.5)(20) – 2 – 0.5

The meter uses a full wave rectifier circuit it indicates value of 2.77 V the form factor for full wave rectified sinusoidal waveform is 1.11

Average value of voltage, V_avg = 2.49 V

For triangular wave shape, peak value of voltage V_m = 2 V_avg = 4.98 V

\(\begin{array}{l} {V_{rms}} = \frac{{{V_m}}}{{\sqrt 3 }} = \frac{{4.98}}{{\sqrt 3 }} = 2.87\;V\\ \% \;error = \frac{{2.77 - 2.87}}{{2.87}} \times 100 = - 3.48\;\% \end{array}\)