Inductance and Capacitance Measurements
Capacitance Measurements
I=
Voltage
source
C
I
V
= V (2π fC )
XC
Example: C = 10 pF, V = 10 V (limited by breakdown
voltage of the capacitance), I = 100 mA; give f = 1600
MHz
parasitic components are dominate the
measurements
Simple
Not practical in the real applications
Residual components
1 

Z = R + j ω L −

ωC 

Equivalent circuit of a capacitor at high frequency
Impedance (Ω)
Capacitance dominate
Inductance
dominate
Ideal
capacitor
0.1 µF
Frequency (Hz)
0.01 µF
0.001 µF
Inductance and Capacitance Measurements
AC
amplifier
C
Voltage
source
AC
voltmeter
R
V =A
RVin
R 2 + X C2
=A
RVin
 1 
R2 + 

 2π fC 
2
Where A = Amplifier gain
Non-linear relationship between V and C
It is hard to keep the amplifier gain constant over
the large input voltage
Inductance and Capacitance Measurements
Limiting
amplifier
C
Voltage
source
phase
detector
R
DC
voltmeter
In this scheme, we measure
the phase difference between
the input voltage, Vin and the
voltage across the series
resistance.
Here, the magnitude of the amplifier is not a critical factor, its gain should be large
enough to give an output that can be detected by the phase detector.
V=A
jX CVin
X ( X + jR)Vin
= A C 2C
R + jX C
R + X C2
Taylor’s Series: θ = arctan(2π fRC ) = 2π fRC −
 R
θ = arctan 
 XC

 = arctan(2π fRC )

1
1
3
5
2
π
fRC
+
(
)
( 2π fRC ) …
3
5
If 2πfRC < 0.1; that gives θ ~ 2πfRC within 0.3 % error
θ ≈ 2π fRC
Inductance and Capacitance Measurements
Example: to cover C from 0-100 pF with a source frequency of 1 MHz; how can
we select the series resistance.
0.1 = 2π fRC = 2π × 1 MHz × R × 100 pF
0.1
R=
= 1590 Ω
−4
2π ×10
Solve for R:
Inductance Measurements
L
Voltage
source
R
Limiting
amplifier
V=A
phase
detector
RVin
R( R − jX L )Vin
=A
R + jX L
R 2 + X L2
2π fL
X 
θ = arctan  L  = arctan(
)
R
 R 
DC
voltmeter
If 2πfL/R < 0.1; that gives θ ~ 2πfL/R
within 0.3 % error
θ ≈ 2π fL / R
Example: to cover L from 0-1µH with a source frequency of 1 MHz; the series
resistance for full scale deflection is 62.8 Ω
Source of Errors
Parasitic components in the instruments
Lead
inductatnce
Generator
resisance
Limiting
amplifier
R
Voltage
source
C
phase
detector
DC
voltmeter
Amplifier
Input capacitance
Non-ideal characteristic of the device under test
R
Voltage
source
ESR
C
Limiting
amplifier
phase
detector
DC
voltmeter
Harmonic distortion of the signal source
Therefore, This type of
measurement is suitable
for High Q inductance and
Low D capacitance
Q Meter
Q-meter is an instrument designed to measure the Q factor of a coil and for
measuring inductance, capacitance, and resistance at RF.
Basic Q-meter Circuit
•The basic operation based on the well-known characteristics of series resonant circuits.
VL
R
Coil
V
VL
XL
XC
IR
Series resonant circuit
VC
X L = XC
VC = VL = IX C = IX L
IXL
V
VC
At resonance:
I
V = IR
By the definition of Q:
Q=
X L X C VC
=
=
R
R
V
Therefore, if V is a known constant, a voltmeter
connected across the capacitor can be calibrated in
term of the circuit Q
Q Meter
25
R=5Ω
20
L = 10H
E
C
VC (Volt)
sin10t
15
VC =
10
VC
E
(ω RC )
2
+ (ω 2 LC − 1)
2
5
0
.0001
.001
.01
Capacitance (F)
.1
Q Meter
Example: when the below circuit is in the resonance, V = 100 mV, R = 5 Ω, and XL= XC
= 100 Ω
(a) Calculate the coil Q and the voltmeter indication.
(b) Determine the Q factor and voltmeter indication for another coil that R = 10 Ω, and
XL= XC = 100 Ω at resonance
L
Signal
generator
V
Unknown coil
Multiply
Q by
R
Solution (a)
Circuit
C
V Q meter
Tuning
capacitor
I=
V 100 mV
=
= 10 mA
R
10 Ω
VL = VC = IX C = 10 mA × 100 Ω = 1 V
Q=
VC
1V
=
= 10
V 100 mV
V 100 mV
=
= 20 mA
R
5Ω
VL = VC = IX C = 20 mA × 100 Ω = 2 V
Q=
(b) For the second coil:
I=
VC
2V
=
= 20
V 100 mV
Measurement Method: Direct Connection
The circuit can be adjusted to the resonance by
V
Multiply
Q by
Circuit
C
V Q meter
Tuning
capacitor
X L = X C and L =
Practical Q meter:
V
2
100
Q
(a) Capacitor voltmeter calibrated
to monitor Q
1
0.6
2
7
Multi-Q-by
( 2π f )
2
C
Henry
100
150
mV
30
0
20
50
3
V
10
VC
1
100
(b) Supply voltmeter calibrated as
a multiply-Q-by meter
pF
200 H
µ
50
0
1
9 Preset the source frequency, and
then vary the tuning capacitor
9 Preset the tuning capacitor and
then adjust the source frequency
20
Signal
generator
R
50
L
0
Unknown coil
(c) Capacitance dial
calibrated to indicate coil
inductance
Measurement Method: Series Connection
For low-impedance components: low value resistors, small coils, and large capacitors
Shorting
strap
Z
Signal
generator
Low
impedance
L
For the first measurement: short Z
R
Work coil
Tuning
capaitor
C1,C2
X L = X C or ω L =
Circuit
V Q meter
Thus
Q meter for a low-impedance component
For the second measurement:
The capacitor is tuned again to the new resonance point.
X S = X C 2 - X L or X S =
1
1
ω C2 ω C1
Xs is inductive if C1 > C2 and capacitive if C2 > C1
XS =
C1 − C2
ω C1C2
Q1 =
1
ω C1
ωL
1
=
R ω C1 R
Measurement Method: Series Connection
The resistive components of the unknown impedance Z.
R1 =
RS = R2 − R1 =
1
1
−
ω C1Q1 ω C2Q2
X1
X
and R2 = 2
Q1
Q2
RS = R2 − R1 =
Thus
Case I: Z = purely resistive, C1 = C2
RS = R2 − R1 =
C1Q1 − C2Q2
ω C1Q1C2Q2
Q1 − Q2
∆Q
=
ω C1Q1Q2 ω C1Q1Q2
LS =
C1 − C2
ω 2C1C2
X s ( C1 − C2 ) Q1Q2
QS =
=
RS
C1Q1 − C2Q2
Case III: Z = capacitive, C1 < C2 CS =
C1C2
C2 − C1
QS =
Case II: Z = inductive, C1 > C2
X s ( C1 − C2 ) Q1Q2
=
RS
C1Q1 − C2Q2
Measurement Method: Parallel Connection
For high-impedance components: high value resistors, Large coils, and small capacitors
L
R
For the first measurement: open switch
Work coil
Signal
generator
XP
High
impedance
RP
X L = X C or ω L =
Circuit
C1,C2
V Q meter
Thus
Q meter for a high-impedance component
For the second measurement:
The capacitor is tuned again to the new resonance point.
X X
X L = C2 P
XC2 + X P
If the unknown is inductive, XP = ωLP;
If the unknown is capacitive, XP = 1/ωCP;
1
ω ( C1 − C2 )
1
LP = 2
ω ( C1 − C2 )
XP =
CP = C1 − C2
Q1 =
1
ω C1
ωL
1
=
R ω C1 R
Measurement Method: Parallel Connection
In a parallel resonant circuit, the total resistance at resonance: Q2 = RT/XL
While X2 = XC1 , Therefore RT = Q2XL=Q2XC1 = Q2/C1
Where
RLP =
R
R 2 + ω 2 L2
1
1
1
=
+
RT RP RLP
1
1
1
ω C1
R
=
−
=
− 2
RP RT RLP
Q2
R + ω 2 L2
ω C1
1 ω C1  1 
1
1
=
− 
≈
−
RP
Q2  R  1 + ω 2 L2 / R 2
Q2
RQ12
since Q1 = 1/ ω C1 R
1 ω C1 ω C1
=
−
RP
Q2
Q1
QP =
RP =
Q1Q2
QQ
= 1 2
ω C1 (Q1 − Q2 ) ω C1∆Q
RP (C1 − C2 )Q1Q2
(C − C2 )Q1Q2
=
= 1
XP
C1 (Q1 − Q2 )
C1∆Q
Source of Error
ƒDistributed capacitance
Signal
generator
V
L
R
Multiply
Q by
Cd
f1, f2
C
Tuning
capacitor
Circuit
V Q meter
One simple method of finding the distributed capacitance (Cd) is to make
two measurements at different frequencies
f =
At first measurement: f1
f1 =
At first measurement: f2
f2 =
If we set f2 =2 f1
1
2π L ( C2 + Cd )
1
2π LC
1
2π L ( C1 + Cd )
1
2π L ( C2 + Cd )
=
2
2π L ( C1 + Cd )
Cd =
C1 -4C2
3
Source of Error
The effective Q of a coil with distributed capacitance is less than the true Q
 C + Cd 
True Q = Qe 

 C 
ƒResidual or insertion resistance: in the Q meter circuit can be an important
source of error when the signal generator voltage is not metered.
If RS is the source resistance, the circuit current resonance is
I=
V
R + Rs
Instead of
Also, the indicated Q factor of the coil is
Instead of the actual coil Q, which is
Thus Rs must be kept minimized.
I=
Q=
V
R
ωL
R + Rs
Q=
ωL
R
Source of Error
Example The self-capacitance of a coil is to be measured by using the procedure just
outlined. The first measurement is at f1 = 2 MHz, and C1 = 460 pF. The second
measurement, at f2 = 4 MHz, yields a new value of tuning capacitor, C2 = 100 pF. Find
the distributed capacitance, Cd.
Solution
Cd =
C1 − 4C2 460 − 4 × 100
=
= 20 pF
3
3
Example A coil with a resistance of 10 Ω is connected in the “ direction-measurement”
mode. Resonance occurs when the oscillator frequency is 1.0 MHz, and the resonating
capacitor is set at 65 pF. Calculate the percentage error in trounced in the calculated
value of Q by the 0.02 Ω insertion resistance.
Solution The effective Q of the coil equals
1
1
=
= 245
ω CR 2π × 106 × 65 × 10-12 × 10
1
Qe =
= 244.5
The indicated Q of the coil equals
ω C ( R + 0.02 )
Qe =
The percentage error is then
245 − 244.5
× 100% = 0.2%
245