Static capacitance-to-voltage converter and converting method

Details Static capacitance-to-voltage converter and converting method Static capacitance-to-voltage converter and converting method

1999-09-23

2001-12-18

Metjahic, Safet (Department: 2858)

Electricity: measuring and testing

Impedance, admittance or other quantities representative of...

Lumped type parameters

C324S658000, C324S683000, C324S686000

Reexamination Certificate

active

06331780

ABSTRACT:

TECHNICAL FIELD
The present invention relates to a static capacitance-to-voltage converter and an associated converting method which are capable of highly accurate conversion of a static capacitance into a corresponding voltage by eliminating the influence of a stray capacitance occurring on a signal line for connecting the static capacitance to an operational amplifier.
BACKGROUND ART
FIG. 1
generally illustrates the configuration of a static capacitance-to-voltage converter described in Laid-open Japanese Patent Application No. 61-1457. This static capacitance-to-voltage converter has been proposed to solve a problem of the prior art which suffers from the inability of accurate voltage conversion due to the fact that a stray capacitance of a cable used to connect an unknown static capacitance is superimposed on the unknown static capacitance, and that these static capacitances may vary due to movements and bending of the cable or the like. As illustrated in
FIG. 1
, an unknown capacitance Cx is connected between an alternating current (AC) signal generator OS and an operational amplifier OP with connection cables covered with shielding lines s to reduce the influence of stray capacitances Cs
1
, Cs
2
, Cs
3
. Specifically, an output and an inverting input of the operational amplifier OP are connected through a feedback circuit formed of a parallel circuit including a resistor Rf and a capacitor Cf. The unknown capacitance Cs has one end connected to the inverting terminal of the operational amplifier OP through a shielding line s, and the other end connected to the AC signal generator OS through another shielding line s. Both of the shielding lines and a non-inverting input of the operational amplifier OP are grounded.
With the configuration described above, since substantially no voltage difference exists between the two ends of the unknown capacitance Cx, the stray capacitance Cs
2
is not charged. Also, since the stray capacitance Cs
3
is regarded as a coupling capacitance of both the shielding lines s, the stray capacitance Cs
3
can be eliminated by grounding the shielding lines s. In this way, the influence exerted by the stray capacitances of the cables for connecting the unknown capacitance Cx is reduced by using the shielding lines s, so that a charge equal to that induced on the unknown static capacitance Cx is induced on the capacitor Cf of the feedback circuit, resulting in an output proportional to the unknown static capacitance Cx produced from the operational amplifier OP. Stated another way, assuming that an output voltage of the AC signal generator OS is Vi, an output voltage Vo of the operational amplifier OP is expressed by −(Cx/Cf)Vi, so that the converter of
FIG. 1
may be used to convert the unknown static capacitance Cx into the voltage Vo from which the unknown static capacitance Cx can be derived together with the known values Cf and Vi.
SUMMARY OF THE INVENTION
The static capacitance-to-voltage converter illustrated in
FIG. 1
, however, implies a problem that as the unknown static capacitance Cx is smaller, the influence of stray capacitances becomes prominent, so that the static capacitance Cx cannot be accurately converted into a voltage. In addition, since the feedback circuit of the operational amplifier OP is formed of a parallel circuit including the resistor Rf and the capacitor Cf, separate steps are required to form a resistor and a capacitor for actually integrating necessary components into a converter in a one-chip form, causing disadvantages of a complicated manufacturing process and an increased chip size. Furthermore, since the capacitor cannot be applied with an AC signal when one electrode of the static capacitance Cx is being biased at a certain voltage, a conversion of the static capacitance into an output voltage cannot be performed.
To solve the problem as mentioned, the applicant has proposed a static capacitance-to-voltage converter constructed as illustrated in FIG.
2
. In the following, this static capacitance-to-voltage converter will be described in detail with reference to FIG.
2
. An operational amplifier
21
has a voltage gain extremely larger than a closed loop gain. A gain seems to be almost infinity. A feedback resistor
23
is connected between an output terminal
22
and an inverting input terminal (−) of the operational amplifier
21
to form a negative feedback for the operational amplifier
21
. The operational amplifier
21
has a non-inverting input terminal (+) connected to an alternating current (AC) signal generator
24
and the inverting input terminal (−) connected to one end of a signal line
25
which has the other end connected to one electrode
26
1
of a capacitor
26
having an unknown or known static capacitance. The other electrode
26
2
of the capacitance
26
is grounded, clamped to a fixed direct current (DC) bias voltage or not grounded. The other electrode
26
2
may be applied with an AC bias. In this case, the bias current may have a frequency identical to or different from the frequency of an AC signal output from the AC signal generator
24
.
The signal line
25
is surrounded by a shielding line
27
for preventing unwanted signals such as noise from being induced into the signal line
25
from the outside. The shielding line
27
is not grounded but is connected to the non-inverting input terminal (+) of the operational amplifier
21
.
Since the operational amplifier
21
is formed with a negative feedback through the feedback resistor
23
, and the operational amplifier
21
has a voltage gain extremely larger than a closed loop gain, the operational amplifier
21
is in an imaginary short-circuit state, and a gain seems to be almost infinity. In other words, an electric potential difference between the inverting input terminal (−) and the non-inverting input terminal (+) of the operational amplifier
21
is substantially zero. Thus, since the signal line
25
and the shielding line
27
are at the same voltage, it is possible to cancel a stray capacitance possibly occurring between the signal line
25
and the shielding line
27
. This holds true irrespective of the length of the signal line
25
, and also holds true irrespective of movements, bending, folding and so on of the signal line
25
.
Assume now that an AC output voltage of the AC signal generator
24
is Vi; its angular frequency is &ohgr;; the static capacitance of the capacitor
26
is Cx; a current flowing through the capacitance
26
is i
1
; the resistance of the feedback resistor
23
is Rf; a current flowing through the feedback resistor
23
is i
2
; a voltage at the inverting input terminal of the operational amplifier
21
is Vm; an output voltage of the operational amplifier
21
is V, the voltage Vm at the inverting input terminal (−) is at the same voltage as the AC signal output voltage Vi of the AC signal generator
24
since the operational amplifier
21
is in an imaginary short-circuit state, as mentioned above. That is, the following equation is satisfied:
Vi=Vm
In addition, the following equations are also satisfied:
i
1
=−Vm/(1/j&ohgr;Cx)=−Vi/(1/j&ohgr;Cx)
i
2
=(Vm−V)/Rf=(Vi=V)/Rf
Here, since i
1
=i
2
, the output voltage V of the operational amplifier
21
is expressed by the following equation:
V=Vi(1+j&ohgr;Rf·Cx)
This equation indicates that the output voltage V of the operational amplifier
21
includes an AC component proportional to the static capacitance Cx. It is therefore possible to derive a DC voltage proportional to the static capacitance Cx by appropriately processing the output voltage V.
As described above, since the operational amplifier
21
is in an imaginary short-circuit state so that a stray capacitance occurring between the signal line
25
and the shielding line
27
will not appear between the inverting input terminal (−) and the non-inverting input terminal (+) of the operational amplifier
21
, the equation representing the output vol

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