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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 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
Transistor Voltmeter: 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
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
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
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
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
Operational Amplifier Voltmeter
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
Operational Amplifier Voltmeter
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
Electronicvoltmeter(1.5 V range)
+
-
R1
Rx E
A
B
1kΩ
100Ω
10Ω
100kΩ
1MΩ
EB1.5V
standardresistor
rangeswitch
Electronic Ohmmeter: Parallel Connection
Electronicvoltmeter(1.5 V range)
+
-
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 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
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