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 $=\frac{1}{{10}^N}\times rangeofvoltage$

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

$V_0=-\frac{1}{RC}\underset{0}{\overset{t}{\int }}{V}_{i}dt$

$=-\frac{1}{100\times {10}^3\times 1\times {10}^{-6}}\underset{0}{\overset{1}{\int }}1dt=-10V$

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 $=\pm \frac{0.5}{20}\times 100=\pm 2.5\mathrm{\%}$

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, $T_2=\frac{6.8}{10}\times 10=6.8ms$

T₁ = nT_s

$\Rightarrow 150clks=n\times \frac{1}{f_s}$

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

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

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

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 $=\frac{1}{100\times {10}^{-6}}=10k\mathrm{\Omega }/v$

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

$\begin{array}{l}{V}_{rms}=\frac{V_m}{\sqrt{3}}=\frac{4.98}{\sqrt{3}}=2.87V\\ \mathrm{\%}error=\frac{2.77-2.87}{2.87}\times 100=-3.48\mathrm{\%}\end{array}$