JPH0143271B2 - - Google Patents

Description

[Detailed description of the invention] [Industrial application field]

The present invention relates to an improvement in a method for measuring the capacitance of a capacitive element.

[Conventional technology]

Conventionally, a method for measuring the capacitance of a capacitive element has been proposed as described below with reference to FIG. That is, the reference AC power supply 1 from which the reference AC voltage e ₀ is obtained is represented by a series circuit of a capacitor C _x and a resistor R _x , and therefore, Z _x = R _x + 1/(jωC _x ) ......(1 ) A capacitive element 2 to be measured having an impedance Z _x (where ω is an angular frequency) is connected through a resistor 3 . In this case, when the value of the resistor 3 is Rs, the value _Rs of the resistor 3 is selected so that the relationship R _s ≫Z _x (2) is obtained, and therefore, the capacitive element 2 has An alternating current i ₀ (where I ₀ is the amplitude) flows, which is substantially in phase with the alternating voltage e ₀ and is represented by i ₀ = I ₀ sinωt (3), and the capacitive element 2 changes accordingly. At both ends, an AC voltage e _x expressed as e _x = i ₀ Z _x = I ₀ R _x sinωt + {I ₀ / (ωC _x )} sin (ωt − π / 2) ......(4) is obtained. It is configured as follows. Also, the AC voltage e _x obtained across the capacitive element 2
is supplied to the positive input terminal 4a of the operational amplifier 5 whose negative input terminal 4b is grounded, and from the output terminal 4c of the amplifier 5, e ₁ =K ₁ e _x =I ₀ [R _x sinωt −{1/(ωC _x )}cosωt] (5) The configuration is such that an alternating current voltage e ₁ (where K ₁ is the gain of the amplifier 5) is obtained. Further, the AC voltage e ₁ obtained from the operational amplifier 5 is supplied to one input terminal 6 a of the multiplier 7 . Further, the AC voltage e _s obtained across the resistor 3 described above and expressed as e _s =−i ₀ R _s =−I ₀ R _s sinωt (6) is supplied to the polarity inversion amplifier 8, The polarity inverting amplifier 8 is configured to obtain an alternating current voltage e ₂ expressed as e ₂ = _−es = I ₀ R _s sinωt (7). Furthermore, the AC voltage e ₂ obtained from this polarity inversion amplifier 8 connects the positive polarity input terminal 10a to ground through the input series capacitor 9, and connects the feedback resistor 11 between the output terminal 10c and the negative polarity input terminal 10b. The negative polarity input terminal 10 of the operational amplifier 12
b, and from its operational amplifier 12, the capacitor 9
When the value of is C ₁ and the value of resistor 11 is R ₁ , e ₃ = e _s ωC ₁ R ₁ = −(I ₀ R _s ωC ₁ R ₁ ) sin (ωt + π/2) ………(8) The configuration is such that an AC voltage e ₃ expressed as is obtained. Further, the AC voltage e ₃ obtained from this operational amplifier 12 is supplied to the other input terminal 6b of the multiplier 7, and therefore, from the multiplier 7, e ₄ = e ₁ · e ₃ = {(-K ₁ I ₀ ² R _x R _s ωC ₁ R ₁ )/2}・(cos(
π/2)−cos(2ωt+π/2)} +{(K ₁ I ₀ ² R _s ωC ₁ R ₁ )/(2ωC _x )}・{sin(2ωt
+π/2)+sinπ/2} (9) The multiplication output voltage e ₄ is obtained as follows. Also, the multiplier output voltage obtained from this multiplier 7
e ₄ is supplied to a low-pass filter 13, the output of which is connected to the positive input terminal 14 of an operational amplifier 15, which connects the negative input terminal 14b to ground.
Therefore, from the output terminal 14c of the operational amplifier 15, E ₁ =(K ₁ K ₂ I ₀ ² R _s C ₁ R ₁ )/(2C _x ) ......(10) The configuration is such that a DC output voltage E ₁ (where K ₂ is the gain of the operational amplifier 15) can be obtained. Further, the DC output voltage E ₁ obtained from the operational amplifier 15 is configured to be supplied to the reference voltage input terminal 17 a of the digital voltmeter 16 . In addition, the AC voltage obtained from the polarity inversion amplifier 8
e ₂ is supplied to the two input terminals 19a and 19b of a multiplier 18 similar to multiplier 7, and thus from the output terminal 19c of that multiplier 18, e ₅ =e ₂ ² =I ₀ ² R _s ² {(1-cos2ωt)/2} = {(I ₀ ² R _s ² )/2} − {(I ₀ ² R _s ² cos2ωt)/2} ......(11) Squared output voltage It is configured so that e ₅ can be obtained. Furthermore, the squared output voltage e ₅ obtained from this multiplier 18 is supplied to a low-pass filter 20, and the output of the low-pass filter 20 is connected to an operational amplifier 22 whose positive polarity input terminal 21a is connected to ground. is supplied to the negative polarity input terminal 21b of the operational amplifier 22.
A DC output ^voltage E ₂ (where K ₃ is the gain of _the operational _amplifier ²² ) _is obtained from _the output terminal 21c of the It is configured so that it can be used. Further, the DC output voltage E ₂ obtained from the operational amplifier 22 is supplied to the unknown voltage input terminal 17b of the digital voltmeter 16, and therefore, the voltmeter 16
E ₂ /E ₁ = {(K ₃ I ₀ ² R _s ² )/2}・{(2C _x )/K ₂ K ₁ I ₀ ² R _s
C ₁ R ₁ )} = {(2C _x R _s ) / (C ₁ R ₁ )}・{K ₃ / (K ₁ / K ₂ )}...
...(13) It is configured so that the measurement expressed as follows is performed. The above is the conventionally proposed method for measuring the capacitance of a capacitive element. According to this method of measuring the capacitance of a capacitive element, the measurement expressed by equation (13) is performed in the voltmeter 16. . In this case, in equation (13), everything except the capacitance C _x is a constant. Therefore, in the voltmeter 16, the capacitive element 2
It is clear that measurements of the capacitance C _x can be obtained. Furthermore, in this case, equation (13) does not include terms for the amplitude and angular frequency of the AC output obtained from the reference AC power supply 1. Therefore, according to the conventional method for measuring the capacitance of a capacitive element shown in _FIG . ,Obtainable,
It has the following characteristics. However, in the case of the conventional method for measuring the capacitance of a capacitive element shown in FIG. 1, the DC voltage E ₁ supplied to the voltmeter 16 is inversely proportional to the capacitance _C Despite this, the voltmeter 16
Although the DC voltage E ₂ is supplied to the reference side input terminal 17a of the voltmeter 16 and has a constant value as is clear from equation (12), the unknown side input terminal 17b of the voltmeter 16 is supplied to Therefore, the effective measurement range of capacitance C _x is now
On the voltmeter 16, check the full scale (e.g. rated 1000
display) to 1/10 scale (100 display), the value of the DC voltage _E1 supplied to the reference side input terminal 17a of the voltmeter 16 when the value of capacitance C _x full scale is reached (this is Let E _a ) be the capacity
DC voltage E ₁ when C _x has a value of 1/10 scale
is 1/10 of the value of (let this be E _b ). Therefore, by appropriately configuring the system for obtaining the DC voltage E ₂ , the value of the DC voltage E ₂ supplied to the unknown side input terminal 17b of the voltmeter 16 is set to a small value corresponding to the above-mentioned value E _a . There is a need to. Therefore, if the resolution of the measurement result of the capacitance C _x on the voltmeter 16 has the value when the capacitance C _x is at full scale, it is the value when the capacitance C _x is at 1/10 scale. It had the disadvantage that only an extremely low resolution could be obtained, which was 1/10 that of the conventional method. In addition, when the voltmeter 16 has a double integration type configuration, the value E _a of the DC voltage E ₁ when the capacitance C _x is full scale is the DC voltage when the capacitance C _x is 1/10 scale. Since the value of E ₁ is 1/10 of E _b , when the capacitance C _x is a full scale value, the capacitance C _x at the voltmeter 16 is smaller than when the capacitance C _x is 1/10 scale. It has the disadvantage that it requires an extremely long measurement time, which is 10 times longer. Therefore, the present invention aims to provide a novel method for measuring the capacitance of a capacitive element without the above-mentioned drawbacks, which will become clear from the detailed description below. FIG. 2 shows an embodiment of the method for measuring capacitance of a capacitive element according to the present invention. In FIG. 2, parts corresponding to those in FIG. 1 are designated by the same reference numerals, and detailed description thereof will be omitted. The method for measuring capacitance of a capacitive element according to the present invention shown in FIG. 2 is similar to the conventional method for measuring capacitance of a capacitive element described above in FIG. 1, except for the following points. That is, in the operational amplifier 5 in the system where the DC voltage _E1 is obtained, the output terminal 4c is connected to the ground through the series circuit of the resistors 31 and 32, and the input terminal 4b is connected to the ground. , through the switch circuit 34, the resistor 31
and 32 connection midpoints. The comparator 38 has one input terminal 36a connected to the output terminal 4c of the operational amplifier 5, and the other input terminal 36b connected to the other end of a reference power supply 37 whose one end is connected to ground. , and controls the switch circuit 34 to be turned on by the output obtained when the value of the AC voltage _e1 obtained from the operational amplifier 5 from the output terminal 36c becomes smaller than the voltage value of the power supply 37. Therefore, the gain K ₁ of the operational amplifier 5 is set to the value K _1a when the switch circuit 34 is off.
is configured to convert to a value K _1b larger than . Further, in the operational amplifier 22 in the system in which the DC voltage E ₂ is obtained, the output of the filter 20 is supplied to its input terminal 21b via the input series resistor 41, and the output terminal 21 of the operational amplifier 22 is
c is connected to the input terminal 21b through the switch circuit 42 and further through the feedback paper 43, and the switch circuit 42 is controlled to be turned on by the output of the comparator 38 mentioned above. It is configured to convert the gain K ₃ of the amplifier 22 to a value K _{3b that is greater than the value K 3a when} the switch circuit 42 is off. However, in this case, the value K _3b of the gain K ₃ becomes the value K _3a
and _the values _{K 1a and} K _{1b of} the gain K 1 _mentioned above, _the resistance is Values such as 41 and 43 are selected. The above are examples of the method for measuring capacitance of a capacitive element according to the present invention. The method for measuring the capacitance of a capacitive element according to the present invention is similar to the conventional method for measuring the capacitance of a capacitive element described above in FIG. 1, except for the matters mentioned above. I will omit the explanation, but
The value of the AC voltage _e1 from the operational amplifier 5 is
is greater than the voltage value of the operational amplifier 5
It operates with a gain K ₁ of K _1a and the operational amplifier 22 operates with a gain K ₃ of value K _3a . Further, when the value of the AC voltage e ₁ is smaller than the voltage value of the power supply 37, the operational amplifier 5 has a gain of the value K _1b.
K ₁ and operational amplifier 22 operates with a gain K ₃ of value K3 _b . And in this case the values K _1a , K _3a , K _1a and K _1b
Since there is a relationship between K _3a /K _1a = K _3b /K _1b = G (constant), the above equation (13)
The value of K ₃ /K ₁ does not change between the two operations described above. Therefore, the value of E ₂ /E ₁ expressed by equation (13) measured by the voltmeter 16 depends on the gain value of the operational amplifier 5, even if it takes the above-mentioned values K _1a and K _1b . do not. Therefore, in the case of the capacitive element capacitance measuring method according to the present invention shown in FIG. 2, as in the case of the conventional capacitive element capacitance measuring method described above in FIG.
Although it is possible to obtain the measurement result of the capacitance C _x of the capacitive element 2, if the reference side input terminal 17a of the voltmeter 16 is
Even if a DC voltage E ₁ having a value inversely proportional to the capacitance C _x is supplied, and a DC voltage E ₂ having a constant value is supplied to the unknown input terminal 17b, the conventional capacitive element shown in FIG. As described above in the case of the capacitance measurement method, when the effective measurement range of the capacitance C _x is from the full scale to 1/10 scale on the voltmeter 16, the operational amplifiers 5 and 22 The gains K ₁ and K ₃ take the values K _1a and K _3a when the capacitance C _x is a value of 1/10 scale, _and
When is at full scale, it takes the values of K _1b and K _3b . Therefore, the value E _a of the DC voltage E ₁ when the capacitance C _x is the full scale value is the same as the capacitance C _x Even if the DC voltage E ₁ is smaller than the value E _b when the value is 1/10 scale, by appropriately selecting the values of the resistors 31 and 32, for example, the value E _b can be reduced to 1/10. 10 ^1/2 ≒ 3.16, which is about 3.16 times that of the conventional capacitance measuring method of the capacitive element described above in FIG. Therefore, the value of the DC voltage E ₂ supplied to the unknown side input terminal 17b of the voltmeter 16 is set to approximately
It can be increased to a large value equivalent to 3.16 times. Therefore, the measurement result of the capacitance C _x with the voltmeter 16 when the capacitance C _x is the value at full scale is the same as that of the conventional capacitance measurement method of the capacitive element described above in FIG. It can be obtained with a high resolution of approximately 3.16 times. Furthermore, even if the voltmeter 16 has a double-integration type configuration, the value E _a of the DC voltage E ₁ when the capacitance C _x reaches full scale is different from that of the conventional capacitive element described above in FIG. For capacitance measurement method of approx.
Therefore, the measurement time with the voltmeter 16 in this case is approximately 3.16 times shorter than that in the conventional capacitance measuring method of the capacitive element described above in FIG. Note that in the above description, the operational amplifiers 5 and 22
We have described the case where the gain is configured so that it can take two values, but it is also possible to configure it so that the gain can take a value of 3 or more, or a system where the DC voltages E ₁ and E ₂ can be obtained. (For example, operational amplifiers 5 and 15, and 22 or 8) can be configured so that the gain can take a plurality of values, and the same effect as described above can be obtained.In short, the DC voltage
The gain of the system that obtains E ₁ and the gain of the system that obtains DC voltage E ₂ are controlled according to the value of DC voltage E ₁ to obtain the same effect as described above. You can also do that. In addition, without departing from the spirit of the invention,
Various modifications and changes may be made.

[Brief explanation of drawings]

FIG. 1 is a systematic connection diagram showing a conventional method for measuring capacitance of a capacitive element. FIG. 2 is a systematic connection diagram showing an embodiment of the method for measuring capacitance of a capacitive element according to the present invention. 1... Reference AC power supply, 2... Capacitive element to be measured,
3, 11, 31, 32, 41, 43...resistance,
5, 12, 15, 22... operational amplifier, 7, 18
... Multiplier, 8 ... Polarity inversion amplifier, 9 ... Input series capacitor, 13, 20 ... Low-pass filter, 16 ...
Digital voltmeter, 34, 42... switch circuit,
38... Comparator.

Claims (1)

[Claims] 1. A capacitive element to be measured is connected to a reference AC power source via a resistor having a sufficiently large value compared to its impedance, and from the AC voltage obtained from both ends of the resistor, 90 Obtain an AC voltage having a phase difference of °, and an AC voltage obtained from both ends of the capacitive element,
For the AC voltage obtained from both ends of the above resistor
Obtain the output voltage multiplied by the AC voltage with a phase difference of 90°, and for the AC voltage obtained from both ends of the above resistor,
Obtain the squared output voltage from the AC voltage having a phase difference of 90°, and calculate the ratio of the DC component of the multiplied output voltage to the DC component of the squared output voltage to express the capacitance of the capacitive element to be measured. In the method for measuring the capacitance of a capacitive element that obtains a measurement result, the gain of the system from which the DC component of the multiplication output voltage is obtained and the gain of the system from which the DC component of the multiplication output voltage is obtained are calculated by calculating the multiplication output A method for measuring capacitance of a capacitive element, characterized in that control is performed according to the value of a DC component of a voltage.