analog meter

Electronic Instruments

Disadvantages of PMMC voltmeter

Low input impedance: Loading effectInsufficient sensitivity to detect low level signal

ApproachUtilized electronic devices such as BJT, FET or op amp to solve the above problems

Analog instrumentDigital instrument

Electronic voltmeters

Basic PMMC

Voltmeter

Rm

Rs

Ammeter Ohmmeter

AC voltmeter

EB

R1

D Rm

Basic Electronicvoltmeter

AmmeterRS

DVoltmeter Ohmmeter

AC voltmeter

EB R1

R1

R2

Electronicvoltmeter

Electronicvoltmeter

Electronicvoltmeter

Electronicvoltmeter

Loading Effect

Circuit before measurement

10 V

R1100kΩ

R2100kΩ

5 V

5 V V

V

10 V

100kΩ

100kΩ

6.7 V

3.3 V 100kΩ

10 V

100kΩ

100kΩ

6 V

4 V 200kΩ

Circuit under measurement

V 3.3V 10100//100100

100//100=

+=measV

V 0.4V 10100//200100

100//200=

+=measV

V

10 V

100kΩ

100kΩ

5.2 V

4.8 V 1000kΩ

V 8.4V 10100//1000100

100//1000=

+=measV

V

Example Find the voltage reading and % error of each reading obtained with a voltmeter on (i) 5 V range, (ii) 10 V range and (iii) 30 V range, if the instrument has a 20 kΩ/V sensitivity, an accuracy 1% of full scale deflection and the meter is connected across Rb

Loading Effect

SOLUTION The voltage drop across Rb with output to the voltmeter connection

50 V

Ra45kΩ

Rb5kΩ

Rm

Loading Effect

± 6.10± 0.35± 0.3-0.054.9530

± 4.40± 0.22± 0.1-0.124.8810± 5.36± 0.27± 0.05-0.224.785

% errorTotal error (V)

Meter error (V)

Loading error (V)

Vb .(V)

Range (V)

Transistor Voltmeter: Emitter Follower

Emitterfollower

increase input resistance reduceoutput resistance

Vin

+

-

Vin

VBE

IB

VinRi = IB

Rs

Rm

+

-

VCC

IE = Im

Basic concept

Voltage to bemeasured

PMMCVoltage drop across meter: m in BEV V V= −

Meter current: in BEm

s m

V VIR R

−=

+

Transistor base current:

where VBE is base-emitter voltage ~ 0.7 V for Si

EB

FE

IIh

≈ hFE = Transistor current gain (Typical values ~ 100-200

Schematic diagram of emitter follower


Circuit input resistance: ( )in ini FE FE s m

B E

V VR h h R RI I

= ≈ ≈ +

Example The simple emitter-follower circuit has VCC = 20 V, Rs+Rm = 9.3 kΩ, Im = 1mA at full scale, and transistor hFE = 100

(a) Calculate the meter current when Vin = 10 V(b) Determine the voltmeter input resistance with and without the transistor.

SOLUTION

+

-

Vin

VBE

IB

VinRi = IB

Rs

Rm

+

-

VCC

IE = Im


*The base-emitter voltage drop (VBE) introduces some limitations in using emitter follower as a voltmeter:

•The circuit cannot measure the input voltage less than 0.6 V•a non-proportional deflection: error

From the above experiment, if we apply Vin with 5 V, the meter should read half of full scale I.e. Im = 0.5 mA. But, the simple calculation shows that Im = 0.46 mA

Q2

RsVE1 VE2

V

VPVin

Q1

RmR2 R3 R6

R5

R4

I2 I3-VEE

+VCC

Practical emitter-follower voltmeter using second transistor Q2 and voltage divider R4, R5and R6 to eliminate VBE error in Q1

Use negative supply also to measure Vin < 0.6 VPMMC

Bridge configuration

1 2m E EV V V= −

1 1E in BEV V V= − 2 2E P BEV V V= −where

Zero adjust


At the condition of Vin = 0, Vp should be set to give zero meter reading, Vm = 0. Therefore, the potentiometer R5 is for the zero adjust. If transistors Q1 and Q2 are identical, VBE1 = VBE2

1 2 1 2( )m E E in BE p BE in pV V V V V V V V V= − = − − − = −

At Vin = 0 -> Vm = 0, give Vp = 0

Consequently, if Vp is set properly, Vm will be the same as Vin

Example An emitter-follower voltmeter circuit as shown in the previous picture has R2= R3 = 3.9 kΩ and supply with ±12 V. Calculate the meter circuit voltage when Vin = 1 V and when Vin = 0.5 V. Assume, both transistors have VBE = 0.7 V

SOLUTION when Vin = 1 V

when Vin = 0.5 V

Voltage Range Changing: Input Attenuator

E

800k

100k

60k

40k

Ra

Rb

Rc

Rd

1V

5V

10V

25V

Vin

Input Range Switch

Voltage to be measured To meter

The measurement point always sees a constant input resistance of 1 MΩ

The input attenuator accurately divides the voltage to be measured before it is applied to the input transistor. Calculation shows that the input voltage Vin is always 1 V when the maximum input is applied on any range

Example On the 5 V range:

5 V

100 k 60 k 40 k 5 V800 k 100 k 60 k 40 k

1 V

b c din

a b c d

R R RVR R R R

+ += ×

+ + +Ω + Ω + Ω

= ×Ω + Ω + Ω + Ω

=

FET Input Voltmeter

Q2

Rs+Rm

V

VP

Q1

R3 R6

R5

R4

I3

R2I2

E

800k

100k

60k

40k

Ra

Rb

Rc

Rd

1V

5V

10V

25V

VG SEG

-VEE

+VCC

VS

Inputattenuator

FETinput stage

Emitterfollower

The addition of FET at the input gives higher input resistance than can be achieved with a bipolar transistor

A FET Input VoltmeterPMMC

1 2m E EV V V= − where 1 1E G GS BEV E V V= − − 2 2E P BEV V V= −

In general, it is not simple to calculate VGS, for simplicity, we assume that VGS will be given.

FET Input Voltmeter

Example Determine the meter reading for the FET input voltmeter in the previous figure, when E = 7.5 V and the meter is set to its 10 V range. The FET gate-source voltage is –5 V, VP = 5 V, Rs+Rm = 1 kΩ and Im = 1 mA at full scale

SOLUTION On the 10 V range:

Q2

Rs+Rm

V

VP

Q1

R3 R6

R5

R4

I3

R2I2

E

800k

100k

60k

40k

Ra

Rb

Rc

Rd

1V

5V

10V

25V

VG SEG

-VEE

+VCC

VS

Inputattenuator

FETinput stage

Emitterfollower

Operational Amplifier Voltmeter

Rs+Rm

Non-invertingamplifier

metercircuit

+

-Vout

-VEE

+VCC

R4

R3

IB

I4

I3

E

Op-Amp Amplifier Voltmeter4

3

(1 )outRV ER

= +

4

3

(1 )vRAR

= +

The voltage gain

The non-inverting amplifier gives a very high input impedance and very low output impedance. Therefore, the loading effect can be neglected. Furthermore, it can provide gain with enabling to measure low level input voltage.

Selection of R3 and R4

33

ERI

= and 43

outV ERI−

=

Example Design an op-amp Voltmeter circuit which can measure a maximum input of 20 mV. The op-amp input current is 0.2 µA, and the meter circuit has Im = 100 µA FSD and Rm = 10 kΩ. Determine suitable resistance values for R3 and R4


SOLUTION To neglect the effect of IB, the condition of I4 >> IB must be satisfied. The rule of thumb suggested I4 should be at least 100 times greater than IB

Rs+Rm

Non-invertingamplifier

metercircuit

+

-Vout

-VEE

+VCC

R4

R3

IB

I4

I3

E

Select I4 = 1000 x IB = 1000 x 0.2 µA = 0.2 mA

At full scale: Im = 100 µA


Op-Amp Amplifier Voltmeter: voltage to current converter

33

mEI IR

= =

Rs+Rm

EB

+

-

-VEE

+VCC

R3

Im

VR3I3

IB

3

mm

RV ER

=

Since I3 >> IB, therefore Im= I3

Meter current

Meter voltage

if Rm > R3, voltage E is amplified by the ratio of Rm/R3

Current Measurement with Electronic Voltmeter

Rs+Rm

+

-

+VC

C

R3

-VEE

RS+ -+ -

Ammeterterminals

I

E

Electronicvoltmeter

An electronic voltmeter can be used for current measurement by measuring the voltage drop across a shunt (Rs). The instrument scale is calibrated to indicate current.

Electronic Ohmmeter: Series Connection

Electronicvoltmeter(1.5 V range)

+

-

R1

Rx E

A

B

1kΩ

100Ω

10Ω

100kΩ

1MΩ

EB1.5V

standardresistor

rangeswitch

Series Ohmmeter for electronic instrument

Ohmmeter scale for electronic instrument

Meter fu

ll

scale

Rx = 0 Rx = ∞

1

xB

x

RE ER R

=+

Suppose that R1 is set to 1 kΩ1 k1.5 V 0.75 V

1 k 1 kE Ω

= × =Ω + Ω

At Rx = ∞ or open circuit, the voltmeter indicate full scale defection (E = 1.5 V) and Rx = 0 or shorted circuit, since E = 0, no defection is observed. At other values of resistance, the battery voltage EB is potentially divided across R1 and Rx, given by

(50% defection)

Thus if Rx = R1, half scale will be indicated

R1

Electronic Ohmmeter: Series Connection

Example For the electronic ohmmeter in the Figure, determine the resistance scale marking at 1/3 and 2/3 of full scale

SOLUTION From1

xB

x

RE ER R

=+

1

1x

B

RR EE

=−

Rearrange, give us

At 1/3 FSD; E = EB/3

1 1

3 21x

B

B

R RR EE

= =×

−

At 2/3 FSD; E = 2EB/3

1123 1

2

xB

B

RR REE

= =×

−Mete

r full

scale

Rx = 0 Rx = ∞

R1 2R1R1/2


+

-

R1

Rx E

A

B

1kΩ

100Ω

10Ω

100kΩ

1MΩ

EB1.5V

standardresistor

rangeswitch

Electronic Ohmmeter: Parallel Connection


+

-

R2 Rx E

A

B

+

-

6V

R14kΩ

1.33kΩ

Shunt Ohmmeter for electronic instrument

At Rx = ∞ or open circuit,

2

1 2

1.33 k 6 V 1.5 V4 k 1.33 k

BRE E

R R=

+Ω

= × =Ω + Ω

Therefore, this circuit give FSD, when Rx = ∞

When, Rx = 0 Ω, E = 0 V, therefore, the meter gives no defection.

At any value of Rx2

1 2

||||

xB

x

R RE ER R R

=+

So, the meter indicates half-scale when Rx = R1|| R2

AC Electronic Voltmeter

Classification:Average respondingPeak respondingRMS responding (True rms meter)

Most ac measurements are made with ac-to-dc converter, which produce a dc current/voltage proportional to the ac input being measured

Principle

ac to dc converterVin dc meter

periodic signal only

any signal

AC Electronic Voltmeter

The scale on ac voltmeters are ordinarily calibrated in rms volts

Average responding meter

ac to dc converterVin dc meter

Form factor is the ratio of the rms value to the average value of the wave form

Form Factor rms

average

VV

=

It should be noted that the rms value is calculated from Vin, while the average value is calculated from the output of ac-dc converter.

Peak responding meter

Form factor is the ratio of the peak value to the rms value of the wave form

Crest Factor peak

rms

VV

=

Average-Responding Meter

In this type of instrument, the ac signal is rectified and then fed to a dc millimeter.In the meter instrument, the rectified current is averaged either by a filter or by the ballistic characteristics of the meter to produce a steady deflection of the meter pointer.

+VD-

E Inputwaveform

outputwaveform

D1

+-

Vout

Vm

+

-

E Inputwaveform

outputwaveform

D1

+-

Vm

+

-

+VD-

Vout

precision rectifierConventional half-wave rectifierFor the positive cycle,

m DV E V= −outV E=

where VD = cut-in voltage ~0.6-0.7 for Si

For the negative cycle, outV E=

0mV =

Since Diode D1 is revered bias, no current flow through meter

For the positive cycle, out mV V E= =

For the negative cycle, 0outV =

Therefore, the voltage drop in the forward bias can be compensated by this configuration

Average-Responding Meter

V1

V2

Vin

V2

V1

Vin

Average-Responding Voltmeter

Rs+Rm

precisionrectifier

+

-

+VCC

R3

-VEE

+ VF -

D1

metercurrent

ER1

C1

Rs+Rm

precisionrectifier

+

-

+VCC

R3

-VEE

D1

metercurrent

D3

D4D2

Im

ER1

C1

Voltage to current converter

Half-wave rectifier Full-wave rectifier

Meter peak current3

pp

EI

R=

Average meter current 1 0.318av p pI I Iπ

= =

Meter peak current3

pp

EI

R=

Average meter current 2 0.637av p pI I Iπ

= =

Example The half-wave rectifier electronic voltmeter circuit uses a meter with a FSD current of 1 mA. The meter a coil resistance is 1.2 kΩ. Calculate the value of R3 that will give meter full-scale pointer deflection when the ac input voltage is 100 mV (rms). Also determine the meter deflection when the input is 50 mV.

Average-Responding Voltmeter

SOLUTION at FSD, the average meter current is 1 mA

Rs+Rm

precisionrectifier

+

-

+VCC

R3

-VEE

+ VF -

D1

metercurrent

ER1

C1

Peak-Responding Voltmeter

The primary difference between the peak-responding voltmeter and the average-responding voltmeter is the use of a storage capacitor with the rectifying diode.

dcamplifier

Vin C R

VD~0.7V

C RVin VC

+

-

In the first positive cycle: VC tracks Vin with the difference of VD, until Vin reaches its peak value. After this point, diode is reversed bias and the circuit keeps VC atVp – VD. The effect of discharging through R will be minimized if its value is large enough to yield that RC >> T.

Charge cycle Discharge cyclethe input impedanceof the dc amp

Peak-Responding Voltmeter

VC

Vin

VC tracks Vin

RMS-Responding Voltmeter

Suitable for: low duty-cycle pulse trainsvoltages of undetermined waveform

RMS value definition: Mathematic 2

0

1 ( )T

rmsV v t dtT

= ∫

RMS value definition: Physicalrms voltage is equivalent to a dc voltage which generates the same amount of heat power in a resistive load that the ac voltage does.

x2 ∫Vin Vout

Millivoltmeter

Thermocouple

heating wireI Temp(oC)

TC o

utpu

t (m

V)

Temp. rise ∝ VrmsNon-linear

Difficult to calibrate scale

RMS-Responding Voltmeter

acAmplifier

ac inputvoltage

dcAmplifier

+

+-

-

Measuring thermocouple

Balancingthermocouple

Feedbackcurrent

Null-balance technique: non-linear cancellation

Compare the heating power generated by input voltage to the heating power generated the dc amplifier

Heater & TC

Heater & TC

+-

AVin Vout

Negative Feedback

VeHeater & TC

Heater & TC

+

-AVin Vout

VT1

VT2

( )1 2out e T TV V A V V= = −

( )out in outV A kV kV= −

Let, VT1 = k Vin and VT2 = k Vout where k is proportional constant of the heater and TC in the system. Note that k may depend on the level of the input signal

1out

in

V AkV Ak

=+

If the amplifier gain is very large, Vout is equal to Vin, this means that the dc voltage output is therefore equal to the effective, or rms value of the input voltage

out inV V≈If A is large

analog meter

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