4. Analog Meter for measring current and voltage PDF

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Electronic Instruments Disadvantages of PMMC voltmeter Low input impedance: Loading effect Insufficient sensitivity to detect low level signal Approach Utilized electronic devices such as BJT, FET or op amp to solve the above problems Electronic voltmeters Analog instrument Digital instrument

Rm Voltmeter R1 EB Rs

Basic PMMC

Ammeter

Ohmmeter D

Rm

AC voltmeter

Electronic voltmeter

RS Ammeter R1

R2

Electronic voltmeter

Basic Electronic voltmeter Voltmeter

EB

R1

Ohmmeter D Electronic voltmeter

AC voltmeter

Electronic voltmeter

Loading Effect R1

100kΩ Ω

5V

6.7 V

100kΩ Ω

10 V

10 V R2

100kΩ Ω

5V

3.3 V V

100kΩ Ω

100kΩ Ω

Vmeas =

Circuit under measurement

Circuit before measurement

100kΩ Ω

100 // 100 10 V = 3.3 V 100 + 100 // 100

100kΩ

6V

5.2 V

10 V

10 V 100kΩ Ω

Vmeas =

4V

200 // 100 10 V = 4.0 V 100 + 200 // 100

V

100kΩ

200kΩ Ω

V meas =

4.8 V V

1000 // 100 10 V = 4.8 V 100 + 1000 // 100

1000kΩ

Loading Effect 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

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

Ra 45kΩ

50 V Rb 5kΩ

V Rm

Loading Effect

Range (V)

Vb . (V)

Loading error (V)

Meter error (V)

Total error (V)

% error

5

4.78

-0.22

± 0.05

± 0.27

± 5.36

10

4.88

-0.12

± 0.1

± 0.22

± 4.40

30

4.95

-0.05

± 0.3

± 0.35

± 6.10

Transistor Voltmeter: Emitter Follower +

Basic concept

IB

+ Voltage to be Vin measured

VBE

Emitter follower

IE = Im Rs

Vin

V CC

Rm V Ri = in IB

reduce output resistance

increase input resistance

Voltage drop across meter:

Vm = Vin − VBE

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

Transistor base current:

Vin − VBE Rs + Rm I IB ≈ E hFE

-

PMMC

Schematic diagram of emitter follower

Im =

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

Transistor Voltmeter: Emitter Follower Circuit input resistance:

Ri =

Vin V ≈ hFE in ≈ hFE ( Rs + Rm ) IB IE

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 + IB

+ VBE

IE = Im Rs

Vin

VCC

Rm V Ri = in IB

-

-

Transistor Voltmeter: Emitter Follower *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 Bridge configuration

+VCC R4

Q1

V

Q2 VP

V in R2

VE1

Rs

Rm

I2

V E2

Vm = VE1 − VE 2

R5

R3

where R6

Zero adjust

I3 -VEE

PMMC

VE1 = Vin − VBE1 VE 2 = VP − VBE 2 Use negative supply also to measure Vin < 0.6 V

Practical emitter-follower voltmeter using second transistor Q2 and voltage divider R4, R5 and R6 to eliminate VBE error in Q1

Transistor Voltmeter: Emitter Follower 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

Vm = VE 1 − VE 2 = Vin − VBE 1 − (Vp − VBE 2 ) = Vin − V p 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

800k

Ra 5V

Voltage to be measured

100k

1V

Input Range Switch 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

Rb 10V

E 60k

To meter

Rc 25V

40k

Vin

Rd

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

Example On the 5 V range:

Rb + Rc + Rd Ra + Rb + Rc + Rd 100 k Ω + 60 k Ω + 40 k Ω = 5 V× 800 kΩ + 100 kΩ + 60 kΩ + 40 kΩ =1V

Vin = 5 V ×

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

FET input stage

Emitter follower

+VCC 800k

Ra

1V

R4

5V

100k

Rb 10V

E 60k

Rc 25V

40k

Q1

EG VG S

V

Q2

R5 VP

VS I2

R2

Rs+Rm

R3

R6

I3

Rd

-VEE

PMMC

A FET Input Voltmeter V m = V E1 − V E 2

where VE1 = EG − VGS − VBE1

VE 2 = VP − VBE 2

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

Input attenuator

FET input stage

Emitter follower

+V CC 800k

Ra

100k

Rb

1V

R4

5V

10V

E 60k

Rc 25V

40k

Rd

EG

Q1

VG S

V

Q2

R5 VP

VS I2

R2

Rs+R m

R3

R6

I3 - VEE

On the 10 V range:

Operational Amplifier Voltmeter Op-Amp Amplifier Voltmeter Non-inverting amplifier

Vout = (1 +

meter circuit

+V CC

R4 )E R3

The voltage gain

+ E

Av = (1 +

-

I4

R4 Vout

-V EE IB

Rs +Rm

R4 ) R3

Selection of R3 and R4 I3

R3

R3 =

E I3

and

R4 =

Vout − E I3

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.

Operational Amplifier Voltmeter 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 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 Select I4 = 1000 x IB = 1000 x 0.2 µA = 0.2 mA meter

SOLUTION

Non-inverting amplifier

circuit

At full scale: Im = 100 µA

+VCC

+ E

-

I4

R 4 Vout

-VEE IB

I3

R3

Rs +Rm

Operational Amplifier Voltmeter Op-Amp Amplifier Voltmeter: voltage to current converter Since I3 >> IB, therefore Im= I3

+VCC

+ EB

-

Meter current

I m = I3 =

Meter voltage

Vm =

Im -V EE

Rs+R m

E R3

Rm E R3

IB

if Rm > R3, voltage E is amplified by the ratio of Rm/R3 VR 3

R3

I3

Current Measurement with Electronic Voltmeter Electronic voltmeter

+V C C

+ Rs +Rm -V EE E

R3

+

+

RS

-

-

I Ammeter terminals

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 range switch

Ω 1MΩ 100k Ω

R1

R1

1kΩ Ω

A

100Ω Ω

EB 1.5V

M sc ete al r f e ul l

standard resistor

Rx = 0

+

Ω 10Ω

Rx

E

Rx = ∞

Electronic voltmeter (1.5 V range)

-

Ohmmeter scale for electronic instrument

B

Series Ohmmeter for electronic instrument 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

E = EB

Rx R1 + R x

Suppose that R1 is set to 1 kΩ 1 kΩ = 0.75 V (50% defection) E = 1.5 V × 1 kΩ + 1 kΩ Thus if Rx = R1, half scale will be indicated

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 standard resistor

E = EB

SOLUTION From

range switch

1MΩ Ω

Rearrange, give us

100kΩ Ω

R 1 1k Ω

A

Rx =

Rx R1 + Rx R1 EB

− E 1

100Ω Ω

EB 1.5V

10 Ω

+

Rx

E

Electronic voltmeter (1.5 V range)

At 1/3 FSD; E = EB/3

-

Rx =

R1/2 R1 2R1 Rx = 0

M sc ete al r f e ul l

B

Rx = ∞

R1 R = 1 EB × 3 −1 2 EB

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

Rx =

R1 = 2 R1 EB × 3 −1 2 EB

Electronic Ohmmeter: Parallel Connection At Rx = ∞ or open circuit,

+

E = EB

R1 4kΩ Ω

A

6V R2 1.33k Ω

-

= 6 V×

+

Rx

E

R2 R1 + R2

Electronic voltmeter

1.33 k Ω = 1.5 V 4 k Ω+ 1.33 kΩ

(1.5 V range)

-

Therefore, this circuit give FSD, when Rx = ∞

B

Shunt Ohmmeter for electronic instrument

At any value of Rx

E = EB

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

R2 || Rx R1 + R2 || Rx

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

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

Vin

ac to dc converter

dc meter

Classification: Average responding periodic signal only Peak responding any signal RMS responding (True rms meter)

AC Electronic Voltmeter The scale on ac voltmeters are ordinarily calibrated in rms volts

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

Vin

ac to dc converter

dc meter

Form Factor =

Vrms Vaverage

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 =

V peak Vrms

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. + E

+VDD1

Input waveform

Vout

output waveform

+ Vm

Vout = E

V m = E −V D where VD = cut-in voltage ~0.6-0.7 for Si

For the negative cycle,

Vout = E Vm = 0

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

E

+VD-

output waveform

+

D1 Input waveform

Vm Vout

-

Conventional half-wave rectifier For the positive cycle,

+ -

-

precision rectifier For the positive cycle, Vout = Vm = E For the negative cycle,

Vout = 0

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

Average-Responding Meter

V2 Vin

V1

V2 V1

Vin

Average-Responding Voltmeter Voltage to current converter precision rectifier

precision rectifier

+V CC

D1

C1

+ E

R1

+ VF -

C1

Im

+ -

D3

+V CC Rs +Rm

D1 Rs +Rm -V EE meter current

E

R1

meter current

-VEE D2

D4

R3

Full-wave rectifier

Half-wave rectifier Meter peak current

Ip =

R3

Ep R3

Average meter current I = 1 I = 0.318I av p π p

Meter peak current

Ip =

Ep R3

2 Average meter current Iav = I p = 0.637I p π

Average-Responding Voltmeter 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. SOLUTION at FSD, the average meter current is 1 mA precision rectifier

+VCC C1

E

R1

+

+ VF -

-

D1 R s+Rm -VEE meter current

R3

Peak-Responding Voltmeter The primary difference between the peak-responding voltmeter and the averageresponding voltmeter is the use of a storage capacitor with the rectifying diode. dc amplifier

VD~0.7V Vin

C

+ VC -

Charge cycle

R

Vin

C

Discharge cycle

R the input impedance of the dc amp

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 at Vp – V D. The effect of discharging through R will be minimized if its value is large enough to yield that RC >> T.

Peak-Responding Voltmeter

VC tracks Vin VC

Vin

RMS-Responding Voltmeter Suitable for: low duty-cycle pulse trains voltages of undetermined waveform

RMS value definition: Mathematic Vin

x2

Vrms =

1T 2 v (t ) dt T ∫0

Vout

∫

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

Temp. rise ∝ Vrms Thermocouple

I

heating wire

TC output (mV)

Millivoltmeter

Non-linear Difficult to calibrate scale

Temp(oC)

RMS-Responding Voltmeter Null-balance technique: non-linear cancellation Compare the heating power generated by input voltage to the heating power generated the dc amplifier Measuring thermocouple

+ ac input voltage

ac Amplifier

dc Amplifier -

+ Balancing thermocouple

Vin

Heater & TC

+ A

Heater & TC

Vout

Feedback current

Negative Feedback VT1

Vin

Heater & TC

Ve

+

Vout

A

VT2

Heater & TC

Vout = Ve = A (VT 1 − VT 2 ) Let, VT1 = k Vin and VT2 = k V out 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

Vout = A ( kVin − kVout ) Vout Ak = Vin 1 + Ak

If A is large

Vout ≈ Vin

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...