Chapter 3.7
AC analog Meters
• The D’Arsonval movement responds only to DC currents.
• Can we use it to measure AC values?
• Consider the following circuit:
D1
R1
AC
A
• On the +ve half cycle of the AC input, D1 will conduct once Vin is > 0.6V.
– Current will then flow through the PMMC movement.
• On the -ve half cycle, D1 is reversed biased
– No current flows except for a very small leakage current.
• This meter therefore responds to half of the sinusoidal input.
• In fact the meter responds to the average of the half-wave rectified signal:
AC signal
Amplitude
1
0.5
0
-0.5
-1
0
0.005
0.01
0.015
0.02
0.025
Time (s)
Half-wave rectified AC signal
0.005
0.01
Amplitude
1
0.03
0.035
0.03
0.035
0.5
0
-0.5
-1
0
0.015
0.02
Time (s)
0.025
Signal at PMMC
2 × Vrms
= 0.45Vrms
π
π
Note that the total circuit current is restricted by R1 to <= IFSD
For a very sensitive movement, this current may be too small to cause D1 to conduct.
We need to increase the current through D1 without increasing the current through
the PMMC movement.
Consider the next circuit:
Vm =
•
•
•
•
EE11A Handouts: Chapter 3.7
Prepared by: Mr. Fasil Muddeen
Vp
=
1
© 2001
D1
R1
AC
RSH
A
• Here, the meter movement is shunted by RSH. Since the parallel combination of RSH
and Rm is < Rm, the total circuit resistance drops and therefore the circuit current
through D1 increases.
• The arrangement still has a problem:
– The meter and D1 have to handle the reverse half cycle voltage.
• Consider this circuit:
D1
R1
D2
AC
RSH
A
• Here, D2 will conduct heavily during the - ve half-cycle, effectively bypassing the
PMMC and D1.
Example
• In the last configuration, if Rm is 100Ω, IFSD is 1mA, RSH is 100 Ω and the on
resistance of both D1 and D2 is 400 Ω, what is the value of R1 and the sensitivity of
the meter for the 10V range.
Solution:
Rm = RSH ;
∴ I SH = I FSD = 1mA Max
∴ Max Source current I T = 2 mA
• On the 10 V range the maximum input is 10V. The circuit conducts only on the +ve
half cycle. Therefore the average DC voltage causing the current flow in the meter is:
VDC = 0.45Vrms = 4.5V
4.5
= 2.25kΩ
2mA
RT = R1 + RD1 + Rm RSH
∴ RT =
R1 = 2.25k − 400 − 50 = 1800Ω
• Sensitivity on the 10V range is:
EE11A Handouts: Chapter 3.7
Prepared by: Mr. Fasil Muddeen
2
© 2001
RT 2250
=
= 225Ω / V
VT
10
This is typical for AC meters based on this principle, the sensitivity is lower than that
of a DC meter using the same movement.
S=
Alternative AC meter
Coil 1
Coil 2
• Consider the above arrangement. Here, the permanent magnet has been replaced by 2
coils. Recall:
T = BNIA sin α
∴T ∝ IB
but B ∝ I
∴T ∝ I 2
• Note that all coils are in series
– Because of this, current flow changes direction in all together
• The torque is therefore always in one direction
• This type of meter movement is called an electrodynamometer
– It can measure both DC and AC
– It has a higher coil resistance than the PMMC meters
– It has increased power consumption
– It therefore has a lower sensitivity than PMMC meters
– The magnetic fields are weaker than those for a permanent magnet
• this also causes lower sensitivity.
• The circuit responds to the average squared current, the scale can be graduated to
read the square root of this
– This gives a direct reading of the rms. values.
• Useful for low frequency measurements < 125Hz
– Higher frequencies mean increased coil impedance
• With slight modifications can be made to read watts, VARS or power factor.
EE11A Handouts: Chapter 3.7
Prepared by: Mr. Fasil Muddeen
3
© 2001