Analog Electronic Volt-Ohm

Analog Electronic
Volt-Ohm-Milliammeters
4
Objectives
You wiU be able to:
l. Sketch various transistor analog voltmeter circuits, and explain the operation of each
circuit. Calculate circuit currents, voltages, and input resistance.
2. Sketch an input aUenuator circuit as used with an electronic voltmeter. Explain its op­
eration, and define the circuit input resistance.
3. Using illustrations, explain the problems that can occur with electronic voltmeter
ground terminals when measuring voltages in a circuit.
4. Draw the circuit diagrams of various op-amp analog voltmeters. Explain the operation
of each circuit.
5. Draw series, shunt, and linear ohmmeter circuits as used in electronic instruments, and
explain the operation of each circuit.
6. Sketch the circuit diagrams of various ac electronic voltmeters, and explain their oper­
ation.
7. Sketch a circuit to show how current is measured by an electronic voltmeter. Explain
the circuit operation.
8. Draw the front panel of a typical analog electronic voltmeter showing the various con­
trols and meter scales. State typical performance specifications for the instrument, and
discuss its applications.
Introduction
Voltmeters constructed of moving-coil instruments and multiplier resistors (see Chap­
ter 3) have some important limitations. They cannot measure very low voltages, and their
resistance is too low for measurements in high-impedance circuitry. These restrictions
86
are overcome by the use of electronic circuits that offer high input resistance, and which
amplify low voltages to measurable levels. When such circuits are used, the instrument
becomes an electronic voltmeter.
Electronic voltmeters can be analng instruments, in which the measurement is indi­
cated by a pointer moving over a calibrated scale, or digital instruments, which display
the measurement in numerical form (see Chapter 5). As well as amplification, transistor
and operational amplifier circuits offer advantages in the measurement of resistance, di­
rect current, and alternating current.
4-1 TRANSISTOR VOLTMETER CIRCUITS
Emitter-Follower Voltmeters
Voltmeter loading (see Section 3-4) can be greatly reduced by using an emitter follower.
An emitter follower offers a high 4input resistance to voltages being measured, and pro­
vides a low output resistance to drive current through the coil of a deflection meter. The
basic emitter-follower voltmeter circuit illustrated in Figure 4-1 shows a PMMC instru­
ment and a multiplier resistance (Rs) connected in series with the transistor emitter. The
dc supply is connected-positive to the transistor collector and the negative to the deflec­
tion meter. The positive terminal of voltage E (to be measured) is supplied to the transis­
tor base, and its negative is connected to the same terminal as the power supply negative.
The transistor base current in Figure 4-1 is substantially lower than the meter cur­
rent.
where hF£ is the transistor current gain. Thus, the circuit input resistance is
,------0+
+o----------~---~
F"lgUre 4-1 An emitter follower offers a
high input resistance to a measured voltage,
and a low output resistance to a dellection
voltmeter circuil VSE introduces an error in
the measurement.
Sec. 4-\
Transistor Voltmeter Circuits
87
E
R·=­
•
18
which is much larger than the meter circuit resistance (Rs + Rm).
Example 4-1
The simple emitter-follower voltmeter circuit in Figure 4-1 has Vee = 20 V, Rs + Rm =
9.3 ill, 1m = 1 rnA at full scale, and transistor hF£ = 100.
(a) Calculate the meter current when E = 10 V,
(b) Detennine the voltmeter input resistance with and without the transistor.
Solution
VE = E - VBi';= to V - 0.7 V
(a) =9.3 V
9.3 V
VE
Im= Rs+Rm
= 9.3 kfl
=1 rnA
IB= ~
(b) With the transistor.
hF/,
=
1 rnA
100
= 10 fJ-A
R=.!i=~
• IB
to fJ-A
= I Mfl
Without the transistor.
The transistor base--emitter voltage drop (VBd introduces an error in the simple
emitter-follower voltmeter. For example, when E is 5 V in the circuit in Example 4-1, the
meter should read half of full-scale, that is, 0.5 rnA. However, as a simple calculation
shows, the meter current is actually 0.46 rnA. The error can be eliminated by using a po­
tential divider and an additional emitter follower, as illustrated in Figure 4-2.
The practical emitter-follower circuit in Figure 4-2 uses a plus-and-minus. or dual­
polarity supply (typically, ±12 V). Transistor Q\ has its base biased to ground via resistor
Rio and a potential divider (R 4 , Rs. and R6 ) provides an adjustable bias voltage (Vp) to the
base of transistor Q2' Resistors R2 and R3 connect the transistor emitter terminals to the
negative supply voltage (-V£d, and the meter circuit is connected between the transistor
emitters. The circuit input resistance is R \ in parallel with the input resistance at the tran­
sistor base.
88 Analog Electronic Volt-Ohm-Millimeters
Ql
Chap. 4
r-------------------------------~-------.----~+voc
t
Rs
E
1
RIO I-+-------V---------~
~----------------------------~~-----.----~ -VEE
F"lgure 4-2 Practical emitter-follower voltmeter cffi;uit using a second transistor (Q2)
and a potential divider (R4 • Rs. and R 6 ) to eliminate the Vs£error produced by Qt.
When no input voltage is applied (E =0 V). the base voltage of Q2 is adjusted to give
zero meter current. This makes Vp =0 V. VEl =VEl =-0.7 V. and (meter circuit voltage) V=
o V. Now suppose that a 5 V input is applied to the Q\ base. The meter voltage is
V= VEl - VE2
= (E - VBE1 ) - VE2
=(5 V -0.7 V)-(-0.7 V)
=5V
Thus. unlike the case of the simple emitter-follower voltmeter, all of the voltage to be
measured appears across the meter circuit; no part of it is lost as transistor VBE.
Example 4-2
An emitter-follower voltmeter circuit such as that in Figure 4-2 has R2 = R3
Vcc=±l2 V.
(a) Determine 12 and 13 when E =0 V.
(b) Calculate the meter circuit voltage when E = I V and when E =0.5 V.
=3.9 kfi and
Solution
(a) V R2 = VR3 = 0 V - VBE
-
VEE =OV-O.7V-(-12V) = 11.3 V
[_/_ V R2 _ Il.3V
2 - 3R z - 3.9 kfl
= 2.9mA
Sec.4-1
Transistor Voltmeter Circuits
89
(b) When E= 1
V.
VEl
=E- V8g =1 V -0.7 V
=0.3 V
VEl
= Vp- V8g ==0 V -0.7 V
=-0.7V
V= Vgl
-
VE2 =0.3 V -(-0.7 V)
==IV
WhenE=O.5 V.
VIOl ==E- VBg =O.5 V -0.7 V
=-0.2V
VEl == Vp - VBg '" 0 V - 0.7 V
=-0.7V
V= VIOl - VE2 == -0.2 V - (-0.7 V)
=0.5 V
Ground Terminals and Floating Power Supplies
The circuit in Figure 4-2 shows the input voltage E as being measured with respect to
ground. However, this may not always be convenient. For example, suppose that the volt­
age across resistor Rb in Figure 4-3(a) were to be measured by a voltmeter with its nega­
tive terminals grounded. The voltmeter ground would short-circuit resistor Rc and seri­
ously affect the voltage and current conditions in the resistor circuit. Clearly, the
voltmeter should not have one of its terminals grounded.
For the circuit in Figure 4-2 to function correctly, the lower end of RI must be at
zero volts with respect to +Vcc and -Vee. The + and - supply voltage may be derived
from two batteries [Figure 4-3(b)] or from two de power supply circuits [Figure 4-3{c)J.
[n both cases, the negative terminal of the positive supply is connected to the positive ter­
minal of the negative supply. For ±9 V supplies, Vee is +9 V with respect to the common
terminal, and Vee is -9 V with respect to the common terminal. [n many electronic cir­
cuits. the power supply common terminal is grounded. In electronic voltmeter circuits.
this terminal is not grounded, simply to avoid the kind of problem already discussed.
When left without any grounded terminal. the voltmeter supply voltages are said to be
floating. This means that the common terminal assumes the absolute voltage (with respect
to ground) of any terminal to which it may be connected. An inverted triangular symbol
is employed to identify the common terminal or zero voLtage tenninaL in a circuit [see
Figure 4-3(b),(c)].
Although the electronic voltmeter supply voltages are allowed to float, some in­
stlUments have their common terminal connected to ground via a capacitor. usually 0.1
fLF. If batteries are used as supply, the capacitor is connected to the chassis. Where a
90
Analog Electronic Volt-Ohrn-Millimeters
Chap. 4
Voltmeter
+£0-----,
R"
Short circuit
(a) A voltmeter with one of
it's tenninals grounded can
short..d rcuit a component
in a circuit in which voltage
is being measured .
+ Vee 0 - - - - - - - - - '
+Vcc~+
CirCUilSymbol~
forcommom tenninal
Common
terminal
Common terminal
+
-VEE~-
- VEE o-----~
(c) ~ supply using power supplies
(b) : supply using batteries
Serious measurement errors can result when a grounded voltmeter terminal is incor­
rectly connected to a circuit. When a circuit has a plus-and .. minus supply voltage, the voltmeter com­
mon terminal should always be connected to the common terminal of the supply.
Figure 4-3
liS V power supply is included in the voltmeter. the chassis and the capacitor are
grounded. Thus. when measuring voltage levels in a transistor circuit, for example. the
common terminal introduces a capacitance to ground wherever it is connected in the cir­
cuit. To avoid any effect on conditions within the circuit (oscillations or phase shifts).
the voltmeter common terminal should always be connected to the transistor circuit
ground or zero voltage terminal. All voltages are then measured with respect to this
point..
Sec. 4.1
Transistor Voltmeter Circuits
b
91
Voltmeter Range Changing
The potential divider constituted by resistors Rm Rb , Re, and Rd in Figure 4-4 allows large
input voltages to be measured on an emitter-follower voltmeter. This network, called an
input attenuator; accurately divides the voltage to be measured before it is applied to the
input transistor. Calculation shows that the Q3 input voltage (Ed is always I V when the
maximum input is applied on any range. For example, on the 5 V range,
EG = 5 V
X
Rb + Rc + Rd Ra + Rb + Rc + Rd =5 V x
looko'+60 ko' +40 ko'
800 kfl + 100 kfi +60kfl+40 kfi
=lV
The input resistance offered by this circuit to a voltage being measured is the total
resistance of the attenuator, which is 1 Mo'. A 9 Mfl resistor could be induded in series
with the input terminal to raise the input resistance to 10 MO. This would further divide
the input voltage by a factor of 10 before it is applied to the gate terminal of Q3'
FET.Input Voltmeter
The input resistance of the transistor voltmeter circuit can be increased further by using
an additional emitter follower connected at the base of Q. in Figure 4-2. However, the use
1
1 iii
1
1
I
1
Input
attenuator
~loE
•
FET
input
stage
1
~111i
I
1
1
1
1
1
1
800k
IV
Emitter-follower
voltmeter
~
+
5V
R4
Vee
100 k
E
Rs
60k
R6
40k
VEE
Figure 4-4 A voltmeter input attenuator is simply a potential divider that accurately divides the
voltage to be measured. The FET input stage (Q) gives .he emitter follower a very high input resist­
ance.
92
Analog Electronic Volt-Ohm-Millimeters
Chap. 4
of a FET source follower (Q3)' as illustrated in Figure 4-4 gives a higher input resistance
than can be achieved with a bipolar transistor. The PET source terminal is able to supply
all of the base current required by Q" while the input resistance at the FEr gate is typi­
cally in ex.cess of 1 MO..
.
Consider the voltage levels in the circuit of Figure 4-4. When E = 0 V, the PET gate
is at the zero voltage level. But the gate of an n-channel PET must always be negative
with respect to its source terminal. This is the same as stating that the source must be pos­
itive with respect to the gate. [f Vos is to be -5 V, and Eo = 0 V, the source te~inal volt­
age must be +5 V. This means that the base t~al of Q. is at +5 V, and, since Q2 base
voltage must be equal to Q. base voltage, Q2 base must also be at +5 V. As in the circuit
of Figure 4-2, Rs in Figure 4-4 is used to zero the meter when the input voltage is 0 V.
Now consider what occurs when a voltage to be measured is applied to the circuit
input. With the attenuator shown, Ea will be a maximum of 1 y.This causes the PET
source terminal to increase until Vas is again -5 Y. That is, ~ goes from +5 to -t{i V to
maintain Vas equal to -5 V. The ~ increase of 1 V is also a 1 V increase in the base volt­
age of Q •. As already ex.plained, all of this (1 V) increase appears across the meter cir­
cuit.
Example 4-3
Determine the meter reading for the circuit in Figure 4-4 when E = 7.5 V and the meter is
set to its 10 V range. The PET gate-source voltage is -5 V, Vp = +5 V, Rs + Rm = 1 ka,
and 1m = I rnA at full scale.
Solution
On the 10 V range: .
=7.5 V x
60 kfl +40 kfl
800 kfl + tOO kfl + 60 kfl + 40 kfl
=0.75 V
Vs=EG- VGs=0.75 V -(-5 V).
=5.75 V
VEl
= Vs - V 8E =5.75 V -0.7 V = 5.05 V
Va= Vp - V BE =5 V -0.7 V
=4.3 V
V=
VEl -
VEl:;:
5.05 V -4.3 V
=0.75 V =EG
[=
m
V
Rs+Rm
0.75 V
=lkfi
== 0.75 rnA (75% of full scale)
. Sec. 4-1
Transistor Voltmeter Circuits
93
On the 10 V range, full scale represents 10 V. and 75% offull scale would be
read as 7.5 V.
Difference Amplifier Voltmeter
The instruments discussed so far can measure a maximum of around 25 V. This could
be extended further, of course, simply by modifying the input attenuator. The minimum
(full-sCale) voltage measurable by the electronic voltmeter circuits already considered is
I V. This too can., be altered to perhaps a minimum of 100 mV by selection of a meter
that will give FSD when 100 mV appears across Rs + Rm. However, for accurate meas­
urement of low voltage levels, the voltage must be amplified before it is applied to the
meter.
Transistors QI and Q2 together with RLI , Ru, and RE in Figure 4-5(a) constitute a
differential amplifier, or emitter-coupled amplifier. The circuit as a whole is known as a
difference amplifier voltmeter. This is because when the voltage at the base of Q2 is zero,
and an input voltage (E) is applied to the Q. base, the difference between the two base
voltages is amplified and applied to the meter circuit.
When a small positive voltage is applied to the base of QI in Figure 4-5, the current
through QI is increased, and that through Q2 is decreased. An increase in lei causes
fetR LI to increase and thus produces a fall in voltage Vet. Similarly, a decrease in fez pro­
duces a rise in Vez. The consequence of this is that the voltage across the meter circuit in­
creases positively at the right-hand side and negatively at the left. This meter voltage (\I)
is directly proportional to the input voltage (E).
+Vcc
Ru
Ru
Rs
ICl
t
fa
V
VCl
VC2
~
~
-­
-----
i
,----o+vcc
Rm
Ru
R2
(b) Zero control
1£2
1£1
RE
-VEE
(a) Voltmeter circuit
Figure 4-5 A difference amplifier voltmeter amplifies low-level input voltages for measurement on
the deflection voltmeter circuit.
94
Analog Electronic Volt-Ohm-Millimeters
Chap. 4
Potentiometer R3 in Figure 4-5(b) is an alternative method of providing meter-zero
adjustment. Q2 base control, as in Figure 4-4, could also be used in the circuit of Figure
4-5. When the movable contact of R3 is adjusted to the right, the portion of R3 added to
RLI is increased and the portion of R3 added to Ru is reduced. When the contact is moved
left, the reverse is true. Thus, VC\ and VC2 can be adjusted differentially by means of R 3 •
and the meter voltage can be set to zero.
4-2 OPERATIONAL AMPLIFIER VOLTMETER CIRCUITS
Op-Amp Voltage-Follower Voltmeter
The operational amplifier voltage-follower voltmeter in Figure 4-6 is comparable to the
simple emitter-follower circuit. However, unlike the emitter-follower, there is no base­
emitter voltage drop from input to output. The voltage-foLLower also has a much higher
input resistance and lower output resistance than the emitter-follower. The voltage­
foLLower input (£8) is applied to the op-amp noninverting input terminaL, and the
feedback from the output goes to the inverting input The very high internal voltage
gain of the operational amplifier, combined with the negative feedback, tends to keep
the inverting terminal voltage exactly equal to that at the noninverting terminal. Thus.
the output voltage (Vo) exactly follows the input. As discussed earlier, the attenuator
selects the voltmeter range.
I
I
10(
I
I
I
I
lnput
attenuator
~
I
I 0(
Voltage follower
I
I
I
I
:o---~~----,
I
I
I
I
--_~+I""o(--
I
I
I
I
I
I
I
I
I
I
I
Meter
.
. _
CirCUit
I
I
I
I
I
I
I
I
I
I
I
I
I
I
Figure 4·6 An Ie operational amplifier voltage-follower voltmeter is similar to the
emitter·follower voltmeter. except that the voltage-fOllower input resistance is much
higher than that of the emitter follower. and there is no base~miner voltage drop.
Sec. 4-2
Operational Amplifier Voltmeter Circuits
95
Op-Amp Amplifier Voltmeter
Like a transistor amplifier, an IC operational amplifier circuit can be used to amplify low
voltages to levels measurable by a deflection instrument. Figure 4-7 shows a suitable op­
amp circuit for this purpose. Input voltage E is applied to the op-amp noninverting input,
the output voltage is divided across resistors R3 and R4 , and VR3 is fed back to the op-amp
inverting input terminal. The internal voltage gain of the op-amp causes VR3 to always
equal E. Consequently, the output voltage is
v
o
R-'..4
= E_R..:;..3_+_
(4-1)
R3
The circuit is known as a noninverting ampLifie 1; because its output is positive when a positi ve
input voltage is applied, and negative when the input is a negative quantity. The noninverting
amplifier has a very high input resistance, very low output resistance, and a voltage gain of
(4-2)
1
1
1
1
Noninverting ------...,..~:f.o(f__- Meter ------:
amplifier
1
circuit
10(
1
1
1
1
t
E
Figure 4-7 An operational amplifier noninverting amplifier can be used to amplify low
input voltages to a level suitable for the deflection meter circuit. The voltmeter gain is
(R3 + R4 )IR 3·
96
Analog Electronic Volt-Ohm-Millimeters
Chap. 4
An op-amp noninverting amplifier voltmeter is very easily designed. Current 14
through R3 and ~ is first selected very much larger than the op-amp input bias current
(I8)' Then the resistors are calculated as
and
Example 4-4
An op-amp voltmeter circuit as in Figure 4-7 is required to measure a maximum input of
20 mV. The op-amp input current is 0.2 JLA. and the meter circuit has I". = 100 fLA FSD
and Rm = 10 kil. Determine suitable resistance values for R3 and R4 •
Solution
Select
[4
= 1000 X In = 1000 x 0.2 !-LA
=0.2mA
At full scale,
1m = 100 V-A
and
=IV
= 100 n
R4 = Vo - E = I V 14
20 mV
0.2mA
=4.9 kil
Voltage-to-Current Converter
The circuit shown in Figure 4-8 is essentially a noninverting amplifier. as in Figure 4-7.
However. instead of connecting the meter between the op-amp output and ground. it is
substituted in place of resistor R4 (in Figure 4-7). Once again. VR1 remains equal to the
input voltage. and as long as IRl is very much greater than lB. the meter current is
(4-3)
Sec. 4-2
Operational Amplifier Voltmeter Circuits
97
=
E
1
Figure 4-8 Voltmeter circuit using an
op-amp voltage-to-current converter. The
meter current is F/R3'
Example 4-5
Calculate the value of R3 for the circuit in Figure 4-8 if E = 1 V is to give FSD on the
meter. The moving-coil meter has 1= 1 rnA at full scale and Rnr = 100 n. Also detennine
the maximum voltage at the operational amplifier output terminaL
Solution
From Equation 4-3,
E
R) = - I(FSD)
= -1 V- = 1 kfl
1 rnA
Vo = I(R 3 + Rm)
= 1 mA(l kfl + 100 fl)
=1.1 V
Many electronic rnultirange instruments do not have any current-measuring facili­
ties. Those that do measure current generally have very low-level current ranges, and
some have relatively high resistances when operating as ammeters. For example, the
meter resistance on one instrument is specified as 9 kfl when operating on a 1.S fJ.A
range. This must be taken into account when the instrument is connected in series with a
circuit in which the current is to be measured. The instrument terminal voltage drop when
used as an ammeter is termed the burden voltage. For a 9 kfl resistance on a 1.5 ILA
range, the burden voltage is
VB =9 kn x 1.S ILA= 13.5 mV
Other typical burden voltage specifications are 250 mV max, 2 Von a 10 A range. and
6 m VIrnA. These voltages drops mayor may not be important, depending on the circuit
under test.
PROBLEMS
4-1 A simple emitter-follower voltmeter circuit as in Figure 4-1 has Vee = 12 V, Rm =
1 kfl. a 2 rnA meter, and a transistor with hF£ = 80. Calculate a suitable resistance
for R. to give full scale deflection when E = 5 V. Also, determine the voltmeter
input resistance.
4-2 An emitter-follower voltmeter circuit, as in Figure 4-2, has the following compo­
nents: R, = 12 kfl, R2 =R3 = 2.1. kG, R4 = 14 = 3.3 kfl, Rs =SOO n, and R. + Rm =
10 kG. A 100 f1A meter is used, the supply voltage is ±9 V, and the transistors have
hF£ = 7S. Determine Vp ' [81, [82, [2. [3, and 14 when E = O. Also, calculate the range
of adjustment for Vp.
4-3 Calculate the meter deflections for the circuit in Probl~m 4-2 when the input volt­
age levels are 0.6 V, 0.75 V, and 1 V.
4-4 A 3.S V input (E) is applied to the input attenuator shown in Figure 4-4. Calculate
the voltage EG on each range selection.
4-5 The FET input voltmeter circuit in Figure 4-4 has the following components: R, =
6.8 kG, R2 = R3 = 4.7 kil, R4 = 1.5 kG. Rs = 500 il, R6 = 3.3 kG, R. + Rm = 20 kG.
The meter full-scale current is 50 ILA, the supply voltage is ±10 V, the transistors
Problems 115
•
have hFE = 80, and the FET gate-source voltage is VGS = -3 V. Determine Vp ' In 12,
13 , and 14 when E = O. Also, calculate the range of adjustment for Vp.
4-6 Calculate the meter deflectio(ls for the circuit in Problem 4-5 when the attenuator is
set to its 5 V range, and the input voltage levels are I V, 3 V, and 4 Y.
4-7 The difference amplifier voltmeter in Figure 4-5(a) has the following compo­
nents: Rl = R2 = 15 kfl, RLi = RL2 = 3.9 kfl, R£ = 3.3 kfl, Rs = 33 kfl, and Rm =
750 fl. The meter full-scale current is 50 ~A, and the supply voltage is ±12 Y.
Calculate the transistor voltage levels when E = o.
4-8 The circuit in Problem 4-7 has transistors with hFE = 100 and hi~ = 1.2 kfl. Deter­
mine the input voltage (E) that will give full-scale deflection on the meter.
4-9 An op-amp voltage-follower voltmeter, as in Figure 4-6, has Ra = 800 kfl, Rb =
100 kfl, Rc = 60 kfl, and Rd = 40 kfl. A 50 ~A meter is used with a resistance of
Rm = 750 fl. Determine the required resistance for Rs to give full-scale deflection
when E = 10 V and the range switch is as illustrated.
4-10 The noninverting amplifier voltmeter circuit in Figure 4-7 uses an op-amp with I B = .
300 nA, and a 50 /LA meter with Rm = 100 kfl. Determine suitable resistances for
R3 and R4 to give full-scale deflection when the input is 300 mY.
4-11 The voltage-to-current converter circuit in Figure 4-8 uses a 37.5 /LA (FSD) deflec­
tion meter with Rm = 900 fl. If R) = 80 kfl, determine the required input voltage
levels to give FSD and 0.5 FSD.
4-12 Determine the new resistance for R3 for the circuit in Problem 4-11 to give FSD
when E = 1 V. Also, calculate the op-amp output voltage.
4-13 Calculate the resistance scale markings at 25% and 75% of full scale for the series
ohmmeter circuit in Figure 4-9.
4-14 Determine the percentage meter deflection in the circuit of Figure 4-9 when the
100 kfl standard resistor is switched into the circuit and Rx = 166 kfl.
4-15 Calculate the meter deflection for the shunt ohmmeter circuit in Figure 4-10 when
Rx = 2 kO and when Rx = 300 O.
4-16 A 16.67 kO resistor is substituted for RE in the linear ohmmeter circuit in Figure
4-11. Calculate the measured resistance when the meter indicates 3.9 V.
4-17 The half-wave rectifier electronic voltmeter in Figure 4-12(b) uses a 500 ~A deflec­
tion meter with a 460 fl coil resistance. If Rs = 450 fl, calculate the rms input volt­
age required to give full-scale deflection.
4-18 The components used in Problem 4-17 are reconnected as in Figure 4-13(a) with
R3 = Rs. Determine the new rms input voltage required to give full-scale deflection.
Also determine the meter deft.ections when the input is 100 mV and 200 mY.
4-19 The ac electronic voltmeter circuit in Figure 4-12(c) uses the following compo­
nents: R, = 22 kfl, R2 = 2.25 kfl, R3 = 6.8 kfl, Rs + Rm = 1 kfl, and a 300 J.LA
meter. Calculate the rms input voltages for meter fun-scale deflection and for 0.5
FSD.
4-20 The full-wave rectifier voltmeter circuit in Figure 4-13(b) uses a 500 ~A meter
with Rm = 460 n together with R) = 450 n (as for Problems 4-17 and 4-18). Deter­
mine the rms input voltage for FSD on the meter.
116 Analog Electronic Volt-Ohm-Millimeters
Chap. 4