JPH0466878A - Apparatus and method for measuring electrostatic capacitance, resistance and inductance - Google Patents

Abstract

PURPOSE:To enable fast and simultaneous measurement of C, R and L components between two terminals of a device (DUT) to be tested by applying a signal waveform to the DUT being worked originally to sample a response signal at a proper time. CONSTITUTION:A C-R-L series equivalent circuit is given between two terminals of a DUT. In this case, when a current signal iS (t) for measurement is inputted into the DUT1, a voltage v(t) is generated between the two terminals as response signal. Here, the v(t) at the final point of the signal iS is sampled to detect a voltage vZ. If so, an electrostatic capacitance CX is determined by dividing a time integrated value of the signal iS by the voltage vZ. In addition, the v(t) is sampled at the time T/2 using as the current iS a signal which has one peak point and is symmetrical horizontally with the peak as center with the time integration having a value other than zero and the value at the final point and a gradient being zero to detect a voltage v1. Thus, the resistance RX can be determined by a specified formula.

Description

[Detailed Description of the Invention] [Industrial Application Field] The present invention provides a
, R, and L at high speed and simultaneously.

[Prior Art] Conventionally, as a method for measuring C, R, and L between two terminals of a DUT, the so-called instantaneous value method, which uses a step function as a measurement signal, is widely known.

In this method, a step function is input between two terminals of a DUT, and the voltage between the two terminals is measured.

For example, when measuring the capacitance CX using a step voltage, the measurement circuit shown in FIG. 4(A) is used.

In the figure, the voltage signal aVS is connected to both terminals of a capacitor CX via a known resistor R, and a voltmeter (2) is connected to both terminals.

In this case, as shown in FIG. 4(B), a voltage signal source (5) with an amplitude (2) is applied to a series circuit of a resistor R and a capacitor CX. When a step voltage is applied manually, the terminal voltage v (t) of the capacitance C changes transiently based on .

Therefore, the voltage between the terminals of the capacitance CX at T1 at an arbitrary time within the transient time (for example, a time constant C11R is adopted as this time) is sampled, and this is set to 0.
Solving for 0 gives the following equation, and the capacitance cx between the two terminals of the DUT can be found.

It is also possible to use a current signal as a step function. The measurement circuit for capacitance CX in this case is shown in Figure 5 (A
), a current signal source i is connected to both terminals of the capacitance CX.
5 and a voltmeter ■ are connected.

In this case, as shown in the same figure (B), the capacitance CX is
The amplitude I from the current signal source is. When a step current of is input, the terminal voltage v (t) of the capacitance CX changes based on.

Therefore, if we sample the voltage V between the terminals of the capacitance Cx at an arbitrary time T, and solve this for CX, we get (2).

Therefore, the capacitance C between the two terminals of the DUT can be obtained.

The conventional methods described above provide fast and relatively accurate measurements when the DUT is a pure capacitance (i.e., a capacitor) or can be considered equivalent to a pure capacitance. possible.

[Problem to be solved by the invention]

However, there are few cases in which the DUT is equivalent to pure capacitance or capacitance, and in reality, it is often represented by a C-R-L series or parallel equivalent circuit.

For this reason, the conventional method for measuring the capacitance CX of a DUT has the disadvantage that the influence of the resistance R and the inductance L appears as a measurement error.

In order to eliminate this inconvenience, it is necessary to measure the resistance R and inductance L using a separate method and calculate the capacitance CX using these values R and L as correction factors, which leads to the problem of complicating the measurement procedure. occurs.

Also known is a method of directly measuring only capacitance by using an intermittent sine wave as a measurement signal to alleviate the effects of resistance and inductance.

In this method, an intermittent sine wave current is input to a DUT, a response voltage to this current is subjected to Fourier analysis, and the capacitance of the DUT is measured by focusing only on the capacitance component of the voltage.

However, even with this measurement method, there is a problem that measurement accuracy decreases when the value of resistance or inductance is large.

The present invention has been proposed in order to solve the above-mentioned problems, and in the case where the DUT is represented by a C-R-L series or parallel equivalent circuit, the present invention has the advantage of high speed, which is an advantage of the conventional instantaneous value measurement method. It is an object of the present invention to provide a measuring device and a measuring method for capacitance, resistance, and inductance, which enable high-accuracy and simple measurement while maintaining the following.

[Means to solve the problem]

In measuring capacitance, the inventor believes that (1) measurement errors occur because step functions are used as voltage and current signals, and as a result, the effects of resistance and inductance inevitably appear; (1) On the other hand, an electric needle? In jIII, we focused on the fact that sine waves are often used to measure the electrical characteristics of various elements. and,
Based on the above (■) and (■), if you devise the waveforms of the voltage signal and current signal and perform sampling between the two terminals of the DUT at an appropriate time, you can easily adjust the capacitance C1 resistance R and inductance L with extremely high precision. The present invention was completed based on the knowledge that measurement can be performed.

That is, the capacitance, resistance, and inductance measurement device of the present invention is capable of measuring (1) a measurement signal that satisfies the conditions that the time integral has a value other than O and the end point value and slope are O, or that satisfies these conditions. A signal source that supplies the device under test with a measurement signal that has one peak point and is symmetrical with respect to the peak point, or (2) the sum of time integrals and the value of the end point are 0.
and a signal source for supplying a measurement signal having an end point gradient other than O to the device under test, and means for detecting a response signal due to the measurement signal. .

In the measuring device of the present invention, the measurement signal of the signal source has an initial phase of (3) 90° with the amplitude being DC biased.
(4) One period of a sine wave signal with an initial phase of 0°.

Furthermore, the method for measuring capacitance and inductance of the present invention includes (5) transmitting the measurement signal of (1) or (3) above to the DU
T is input to a device under test represented by a C-R-L series or parallel equivalent circuit, a response signal at the end point of the measurement signal is sampled, and the value of the time integral is divided by the sampled value. (6) Input the measurement signal of (2) or (4) above into the device, sample the response signal at the end point of the measurement signal, and convert the sampling signal to the gradient at the end point of the response signal. (7) The above (1) or (3)
A measurement signal of is input to the device, a response signal at the peak point and an end point of the measurement signal is sampled, and the difference between the sampling value at the peak point and 1/2 of the sampling value at the end point is calculated as It is characterized by dividing by the peak value of the measurement signal.

[Effect]

The present invention will be described in detail below. In the circuit between the two terminals of the DIJT, the capacitance C8, resistance RX, and inductance L1 are often not independent electric elements, but for the sake of explanation, , these components will be explained by replacing them with individual elements.

In addition, the present invention is used when the two terminals of I) UT are represented by (, -RL series or parallel equivalent circuits, but the two terminals of DUT are C, , RxL, Even if it does not contain the element, set the value of the element that is not included to 0 or (
Considering 1), it can be treated as a C-R-L series circuit or a parallel circuit. Therefore, hereinafter, the connection between the two terminals of the DUT will be described as a C-RL series circuit or a parallel circuit.

In the present invention, the measurement signal of the signal source is
If the #A terminal is a (, -R-L series equivalent circuit), it is a current signal, and if it is a C-R-L parallel equivalent circuit, it is a voltage signal.

First, as the above measurement signal, a signal that satisfies the conditions that the time integral has a value other than 0 and the end point value and slope are 0, or in addition to these conditions, has one peak point and the peak point A case will be described in which capacitance, resistance, and inductance are measured using a left-right symmetrical signal as a reference.

When DOT is represented by a C-R-L series equivalent circuit, when measuring capacitance and resistance, the current signal 1. , (t) are used.

If a current signal 1s(t) is input to two terminals of the DUT, a voltage v as a response signal is generated between the two terminals of the DUT.
(t) occurs.

The voltage of C, , R, , L is vc, ■1. If vl, y, I=R, lIs...(2) t holds true, so v(t) becomes v(t)=vc+vR+Vl.

(4) Here, sampling of v (t) at the end point (time T, ) of is is performed by a desired detection means.

However, since the value and slope of the end point of is are O, ①a in the above equation (2) and vl in the equation (3) are respectively 0.

Therefore, the sampling voltage v z (v (T
−) ) becomes equal to vc, and becomes .

If the integral of the above equation is known, the capacitance C, is 1sd
t Cx-...(5) 2 can be easily obtained without being influenced by R, spear, etc.

In particular, as is, a signal is used in which the time integral has a value other than 0, the value and slope of the end point are 0, and the signal has one peak point and is symmetrical with respect to the peak point. In some cases, R, as well as CX can be measured at the same is.

In this case, prior to sampling v (t) at time T, time T, <'2 (i.e., current signal i
Sampling of v (t) is also performed at the peak point of 5 (t)).

Since the slope of the current signal is is 0 at the peak point, vl in the above equation (3) is 0, and the sampling voltage v + (v (T-/2)) at the peak point is 1/2 of .

Here, the sampling voltage at the peak point ■.

and the sampling voltage v2 at the end point of 172, which is (6) - (7)/2, is calculated as vl-
−v2=R,Ii, , (T,/2). i, (T
, /2) is the peak value (1,), from the above equation, we get:

Also, since the value and slope of the current signal i at the end point are 0, vl in the above equation (2) and (1) in the equation (3) are respectively 0, and the sampling voltage vz(v(T-)
) becomes .

Incidentally, since i is symmetrical with respect to the peak point, the first term on the right side of equation (6) can be used to obtain the right side of equation (7).

When measuring inductance and resistance in a case where the DUT is represented by a (,-R-L parallel equivalent circuit), a voltage signal V having the same shape as the above 1s(t) is used as the measurement signal.
Use yo (1).

Then, the response signal (circuit current 1(t)) of the DUT to this 5 is sampled at time T (when resistance is also measured, at time T and T,/2).

In this case, based on "circuit duality", 5vsdt L,=...(5)' can be easily derived. Here, i+, iz are time T, 5T
, the sampling value of i (t) at /2, ■8 is V
.

is the wave height value.

In addition, in the present invention, the measurement signal jS+v is a sine wave signal whose initial phase is 90° and whose amplitude is DC biased, i s = I o (1-cosa +, t ) Vs =
One cycle of V6 (I C03 (1111t)) is also used.

In this case, equation (5) is (5)′ formula is ■. T.

It becomes simple as L, = (where 1., V. is the amplitude of the sine wave signal, and ω is 2π/T).

Next, a case will be described in which capacitance and inductance are measured using a signal in which the sum of time integrals and the value of the end point are 0, and the slope of the end point has a value other than 0. .

In the case where the DUT is represented by a C-R-L series equivalent circuit, when measuring inductance, the current signal 1s(t) is used as the measurement signal.

If a current signal (1 Jt) is input to two terminals of the DUT, a voltage v (
t) occurs.

Then, sampling is performed at the end point (time T,) of the current signal.

At the end point, the sum of the time integrals of the current signal is and the value at the end point are 0, so the equation (1) is ``■''. and ■ in equation (2)
, becomes O, and the sampling voltage v z (v
(T-) ) becomes , and becomes .

When measuring capacitance in a case where the DUT is represented by a C-R-L parallel equivalent circuit, a voltage signal vS(t) having the same shape as the above-mentioned 1Jt) is used as the measurement signal.

Also in this case, Cx is derived as follows based on "circuit duality". However, 12 is a sampling current at the end point of the voltage signal.

Furthermore, as the measurement signals is and vs, the initial phase is O.
One period of the sine wave signal of o, i s=I osinω1It vs0Vos+nω*t is used.

When using these sinusoidal functions, (9).

Equation (9)′ is simplified as follows.

That is, equation (9) is (9)′ formula is becomes.

(Example〕 The present invention will be explained in detail below.

Figure 1 shows the capacitance CX of the circuit between the two terminals of DUTI,
FIG. 2 is a simplified circuit diagram illustrating an example of a circuit for measuring resistance RX and inductance L. FIG.

In the same figure, between the two terminals a and a' of 0UTI, Cx
A series equivalent circuit of -RX-L is formed, and a and a' are current signal sources capable of outputting a measurement signal for one cycle of a sine wave signal whose initial phase is 90° and whose amplitude is DC biased. 2 and a voltmeter 3 that detects the response signal of the DUT (in this case, the voltage between a and a') are connected in parallel.

This will be explained below with reference to the waveform diagram in FIG.

First, from the current signal source 2, a current signal for measurement i 5 (t)
-] o (1-CO5(+3.t) (however, 0≦
t≦T1. ω, -2π/T, , Io is the amplitude, 1<0. Input i,=o) to DUTI (i in Figure 2).
(See the waveform diagram of s).

At this time, the voltage VCX(t) of Cx is =R*-1o(
1-cosω, 1) t -L , l' I o'ωa-sin (ω, 1) (see the waveform diagram of v+tx(t), v(1) in FIG. 2).

Next, the voltage v+ (i.e.,
v (T,/2) ).

At time T, /2, the differential coefficient of is (i.e.,
Since the slope) is 0, VLX is 0.

Therefore, vl is v c, (T, /2) and VIIX (T@
/2), v l=v c-(T-/2) + v wiX(T-
72) (see the waveform diagram of vcx(t) in FIG. 2).

Further, the voltage v, 1W(t) of R, and the voltage vlX of L, respectively, are v, 1x(t) - R, 1, 1s(t) 2 C1l.

Note that in FIG. 1, for convenience of explanation, a case is shown in which a voltmeter 3 is used to measure the voltage v (t), but
In reality, although not shown, the voltage v (t) is measured using a sample-and-hold circuit, an A/D converter, etc.
t) is sampled, and this sampling voltage is stored in a register or the like.

Furthermore, the voltage between terminals a and a' at time T, (i.e., the end point of current signal i) vz (i.e., v (T,)
) to sample.

At time T, the value and differential coefficient of i, , are 0, so VIIX and vt+i are 0.

Therefore, the sampling voltage is So, the capacitance C8 becomes 1, T.

z It is required as.

This calculation of CX is performed by an arithmetic device (not shown) via the registers and the like.

In this way, the capacitance CX can be extremely easily determined only from the sampling voltage v2 at the end point of is.

On the other hand, the resistance RX also has an influence on R1 by calculating v, -v, /2 -02) using the peak point sampling voltage ■ expressed by the equation 00) and the sampling voltage Vz at the end point. It can be found very easily without being

co), from formula 00, and using the current signal is to measure C1,
Resistance R8 can be determined without being influenced by C8 or L8.

Note that the waveform diagram of v (t) in FIG. 2 shows that the difference between ■ and ■ 1/2 of 2 is 2Rx-1°.

In practice, it is often sufficient to simply find the capacitance C and the resistance R8, but according to the measuring method of the present invention, the inductance L can also be found.

The value of L is not determined from Equation 10 (00), but the waveform of the current signal is changed, and the current signal is of one cycle of the sine wave signal is used.

In other words, the signal current of i, -1osinω*t (where 0≦t≦T, ω, = 2π/TII, Io is the amplitude, t<O, t>T, and i s=O) is shown in Figure 1. Enter the DUTI of

Then, at time T, (i.e., the end point of is), a, a
The voltage v2' between the terminals of ' is sampled.

The sine wave signal has a sum of time integral values of i and an end point of 0.
Therefore, the voltage cost of the capacitance C8 and the voltage V of the resistor R are O, and 2' becomes equal to the voltage V of the inductance.

= L x' I o-o)
It can be determined very easily without being influenced by the above (see VIN in Figure 3).

Further, in the above embodiment, the case where the initial value of the voltage VCX of the capacitance C is 0 is described as a pre-capture, but the voltage VCX of the capacitance CX. If the initial value of is not 0, the present invention can be applied by sampling the voltage v (o) at 1=0 and performing a correction calculation (that is, subtracting ■(0) from the measured voltage).

In the above embodiment, a current signal is used to determine the capacitance CX and resistance Rx between two terminals of the DUT forming the C-RL series circuit.
, the inductance LX has been described, but if "circuit duality" is considered, CX, R, , LX between the two terminals of the DtJT forming the L-R-C parallel circuit can be found.

(Effects of the Invention) According to the measurement method of the present invention, the waveform of the measurement signal that the signal source supplies to the DUT is devised, and the response signal is sampled at an appropriate time, resulting in the following effects. can be played.

In other words, even if capacitance, resistance, and inductance coexist, if the DUT is represented by a C-R-L series or parallel equivalent circuit, the instantaneous values of each element do not affect each other. Measurement using the value method becomes possible.

As a result, when measuring capacitance, for example, it is no longer necessary to measure the values of resistance or inductance by a separate method and perform correction calculations based on the measured values.

Furthermore, capacitance, resistance, and inductance can be measured with extremely high precision and in a simple manner while maintaining the high speed of the instantaneous value method.

[Brief explanation of drawings]

FIG. 1 is a circuit diagram showing an example of a measuring circuit for carrying out the measuring method of the present invention. FIG. 2 is a waveform diagram showing the current waveform of the current signal of the measuring circuit of FIG. 1, the voltages at both terminals of the C-R-L series circuit, and the waveforms of the voltages at various parts due to the current signal waveforms. FIG. 3 is a waveform diagram showing a current signal and a voltage waveform of the inductance of the C-R-L series circuit, which are different from the current waveform shown in FIG. 2. Figure 4 (A) is a circuit diagram showing a measurement circuit applied to the conventional measurement method using a step voltage signal source, and Figure 4 (B) shows the step voltage signal and its response in the circuit of Figure 4 (A). FIG. FIG. 5(A) is a diagram showing a measurement circuit applied to a conventional measurement method using a step current signal source, and FIG. 5(B) is a diagram showing a step current signal and its response in the circuit of FIG. 5(A). FIG. ■...DUT (device under test) 2...Current signal source 3...Voltmeter

Claims (7)

[Claims] (1) A measurement signal that satisfies the conditions that the time integral has a value other than 0 and the end point value and slope are 0, or that has one peak point in addition to these conditions and is symmetrical with respect to the peak point. A capacitor comprising: a signal source that supplies a measurement signal to a device under test; and means for detecting a response signal of the device due to the measurement signal.
Resistance and inductance measurement equipment. (2) a signal source that supplies a measurement signal to a device under test in which the sum of time integrals and the value of the end point are 0 and the gradient of the end point is a value other than 0; and A device for measuring capacitance and inductance, comprising means for detecting a response signal. (3) The measurement signal of the signal source according to claim (1) is a sine wave signal whose initial phase is 90° and whose amplitude is DC biased.
A device for measuring capacitance, resistance and inductance, characterized in that it measures periodicity. (4) An apparatus for measuring capacitance and inductance, wherein the measurement signal of the signal source according to claim (2) is one cycle of a sine wave signal with an initial phase of 0°. (5) The measurement signal of claim (1) or (3) is
-L input to a device under test represented by a series or parallel equivalent circuit, sample the response signal of the device at the end point of the measurement signal, and divide the value of the time integral of the measurement signal by the sampling value. A method for measuring capacitance and inductance, characterized by: (6) The measurement signal according to claim (2) or (4) is
-R-L input to the device under test represented by a series or parallel equivalent circuit, sample the response signal of the device at the end point of the measurement signal, and divide the sampled value by the slope of the end point of the measurement signal; A method for measuring capacitance and inductance, characterized by: (7) The measurement signal according to claim (1) or (3) is
- R-L input to the device under test represented by a series or parallel equivalent circuit, sample the response signal of the device at the peak point and end point of the measurement signal, and sample the sampling value at the peak point and the sampling at the end point. A method for measuring resistance, characterized in that the difference between the value and 1/2 is divided by the peak value of the measurement signal.