Transcript 14-Maxwell-Bridgesx

Maxwell Bridges
Hamzeh AlRashdan
Joseph Twall
Types of Maxwell bridges
Maxwell bridges consist of two types
1. Maxwell’s inductance bridge
2. Maxwell’s inductance-capacitance
bridge (Maxwell-Wien Bridge)
Types of Maxwell bridges
Maxwell’s inductance bridge
Maxwell-Wien inductance capacitance
bridge
Why is there more than one type?
For a balancing condition
𝑍1 ∡𝜙1 . 𝑍3 ∡𝜙3 = 𝑍2 ∡𝜙2 . 𝑍4 ∡𝜙4
When measuring an
inductance, its 90 ° phase
shift can be compensated by
two methods
How to compensate the 90° phase
shift?
1. A known impedance with an equal
positive phase angle (i.e. inductance) in
either of the adjacent arms as in
Maxwell’s inductance bridge
2. A Known impedance with an equal
negative phase angle(i.e. Capacitance)
in opposite arm as in Maxwell-Wien
bridge
Maxwell’s inductance bridge
Maxwell’s inductance bridge
In this figure
𝑍1 = 𝐿1 + 𝑅1 is
unknown
R₂,R₃,Z ₄ are known
Maxwell’s inductance bridge
L₄ is a variable
inductance
The Bridge is
balanced by
varying L₄ and one
of the resistances
R₂,R₃
Maxwell’s inductance bridge
The balance condition is
that
Z₁ Z ₃ =Z ₂ Z ₄
(𝑅1 + 𝑗𝜔𝐿1 )𝑅3 = (𝑅4 + 𝐽𝜔𝐿4 )𝑅2
Maxwell’s inductance bridge
Equating the real and
imaginary parts on
both sides , we have
1. R ₁ R ₃= R ₂ R ₄
2. L ₁ R ₃= L ₄R ₂
Maxwell’s inductance bridge
By rearranging:
1.
2.
𝑅2 𝑅4
𝑅1 =
𝑅3
𝐿4 𝑅2
𝐿1 =
𝑅3
Maxwell’s inductance bridge
limitations
It is important to note the Maxwell’s inductance bridge
cannot be used with high quality inductors.
1 < 𝑄 − 𝑓𝑎𝑐𝑡𝑜𝑟 < 10
The quality of an inductor is given by the Q-Factor
𝑊𝐿
𝑄 − 𝑓𝑎𝑐𝑡𝑜𝑟 =
𝑅
For High quality inductors we use Hay’s Bridge
Example.
The Arms of an a.c Maxwell bridge are
arranged as follows: AB and BC are
resistors of 100 Ω each, DA is a
standard variable reactor 𝐿1 of
resistance 32.7 Ω, and CD comprises a
standard variable resistor R in series of
a coil of unknown impedance.
Balance was obtained with
𝐿1 = 47.8 mH
R=1.36
Find the resistance and indcuctance
of the coil.
Since the products of
resistances of opposite
arms are equal:
32.7 ∗ 100 = 1.36 + 𝑅4 100
𝑅4 = 32.7 − 1.36 = 𝟑𝟏. 𝟑𝟒𝜴
Since the products of
resistances of opposite
arms are equal:
𝐿1 ∗ 𝑅3 = 𝐿4 ∗ 𝑅2
47.8 𝑚𝐻 ∗ 100𝛺
𝐿4 =
100𝛺
𝐿4 = 47.8 𝑚𝐻
Maxwell-Wien Bridge
L3= unknown inductance
R3=effective resistance of
inductor L3
R1, R2, R4=known non
inductive resistances
C=variable standard
capacitor.
the positive phase angle
of an inductive
impedance may be
compensated by the
negative phase angle of a
capacitive impedance
put in the opposite arm.
The unknown inductance
then becomes known in
terms of this capacitance.
The balance condition is
Z₁ Z ₃ =Z ₂ Z ₄
𝑅2 𝑅4 = 𝑅3 + 𝑗𝐿3
𝑅1
(
)
1+𝑗𝐶𝑅1
 separating and
rearranging the real
and imaginary
terms, we get
𝑅3 =
𝑅2𝑅4
𝑅1
𝐿3 = 𝐶𝑅2𝑅4
Thus we have two
variables R1 and C which
appear in one of the two
balance equations and
hence the two equations
are independent.
The expression for Q factor
𝑄 =
𝜔𝐿3
𝑅3
= ω𝐶𝑅1
Advantages of Maxwell’s inductance
capacitance bridge:
The two balance equations are independent if we
choose R1 and C as variable elements.
The frequency does not appear in any of the two
equations.
This bridge yields simple expressions for L3 and R3 in
terms of known bridge elements.
Disadvantages of Maxwell’s inductance
capacitance bridge:
 This bridge requires a variable standard capacitor which may
be very expensive if calibrated to the high degree of
accuracy.
 The bridge is limited to the measurement of low Q coils
(1<Q<10). it is clear from the Q factor equation that the
measurement of high Q coils demands a large value of
resistance R1, perhaps 10^5 or 10^6 Ω. The resistance boxes
of such high values are very expensive. thus for values
of Q>10, the Maxwell’s bridge is unsuitable.
EXAMPLE:
The arms of an a.c. Maxwell
bridge are arranged as follows:
AB is a non-inductive resistance
of 1,000 Ω in parallel with a
capacitor of capacitance 0.5 μF,
BC is a non-inductive resistance
of 600Ω CD is an inductive
impedance (unknown) and DA is
a non-inductive resistance of 400
Ω. If balance is obtained under
these conditions, find the value of
the resistance and the
inductance of the branch CD.
Since R1R3=R2R4
R3=R2R4/R1
R3= 600*400/1000= 240 Ω
Also
L3=CR2R4
L3= 0.5 ∗ 10−6 ∗ 400 ∗ 600
=0.12 H