JPS6362038B2 - - Google Patents

Description

DETAILED DESCRIPTION OF THE INVENTION The present invention relates to a displacement converter that converts displacement corresponding to various process variables such as pressure, flow rate, tension, etc. into an electrical output signal.

Such displacement converters are used to convert various process quantities into output signals and transmit them to a remote receiving unit, but in differential pressure flowmeters, etc. There is a relationship Q ² = P between the obtained displacement P, and after converting the displacement P into an electric signal, a square rooter is used to make the relationship between the measured quantity and the output signal linear. It was necessary to do so.

For this reason, in the past, a flattener was required in addition to the displacement transducer, which resulted in disadvantages such as an increase in the mounting space for the equipment and a complicated configuration of the entire device.
As a countermeasure to this problem, recently, a square root type displacement converter has been proposed in which a square root calculation function is added to the displacement converter itself.

FIG. 1 is a circuit diagram of a square root type displacement converter disclosed in "Displacement Converter" (Japanese Patent Application No. 51-9096) filed separately by the present applicant, and the details are clear from the same application. , its outline is as follows.

That is, a high frequency excitation signal is applied by the excitation power supply G to a pair of variable capacitances C ₁ and C ₂ that are differentially varied according to the displacement to be detected, and then the variable capacitances C _{1 and} C 2 are A detection circuit detects the signal current passing through C ₂ .
In addition to detecting with DET, variable capacitance C ₁ ,
A voltage E ₁ corresponding to the difference between C ₂ and a voltage E ₂ corresponding to the sum of the capacitances C ₁ and C ₂ are obtained, and at the same time when applying the voltage E ₁ to the non-inverting input of the comparator CP , _the voltage _{E 2}
is applied to the inverting input of comparator CP.

Furthermore, the switching circuits of multipliers ML ₁ and ML ₂ operate according to changes in the output of comparator CP, repeating charging and discharging of the integration circuit, and switching circuits of multipliers ML 1 and ML 2.
Since the output of the comparator CP changes according to the integrated voltage of the integrating circuit in ML ₂ , the multipliers ML ₁ and ML ₂
If the duty ratio for charging and discharging the switching circuit in is D, the voltage applied to the inverting input of the comparator CP is E ₂ ·D ² , and furthermore,
From the relationship that both input voltages of comparator CP are equal,
E ₁ = E ₂・D ² holds true, and based on this relationship, the duty ratio when the output of the comparator CP changes is D = √
E ₁ /E ₂ is obtained, and the square root operation is performed based on this.

Furthermore, if the amplitude of the excitation power supply G is E, the angular frequency is ω, the initial capacitance of variable capacitances C ₁ and C ₂ in a non-displacement state is C ₀ , and the capacitance that changes according to displacement is ΔC, then the following equation is obtained. holds true.

C ₁ =C ₀ +ΔC, C ₂ =C ₀ −ΔC …(01) E ₁ =ω・E・C ₁ −ω・E・C ₂ =ω・E(C ₁ −C ₂ ) …(02) E ₂ =ω・E・C ₁ +ω・E・C ₂ =ω・E(C ₁ +C ₂ ) …(03) By substituting equation (01) into equations (02) and (03), E ₁ = 2・ω・E・ΔC …(04) E ₂ =2・ω・E・C ₀ …(05) Therefore, In other words, since a result corresponding to the square root of the variable capacitance ΔC is obtained, a square root calculation is performed at the same time as the displacement is converted into an electrical signal, and this signal is passed through the input resistor R ₀₁ to the differential output circuit of the output circuit. Applied to the non-inverting input of amplifier A.

The line terminals L ₁ and L ₂ of the output circuit are connected to a power source E _{S and a load resistor R L of} the receiving section etc. via a two-wire line, and the voltage applied by the power source E _S is
It is stabilized by the constant current circuit CC and constant voltage diode ZD and supplied as the power supply voltage for each part, while the reference voltage as a reference value determined by resistors R ₀₂ and R ₀₃ and the voltage of the non-inverting input are , a differential amplifier A amplifies the difference between both voltages, controls a transistor Q, and determines a line current value as an output signal passed to a load resistor R _L via line terminals L ₁ and L ₂ .

The line current also passes through the bias resistor R ₀₄ and the feedback resistor R ₀₅ of the transistor Q, and the voltage generated in the resistor R 05 is passed through the resistor R ₀₆ to the resistor R ₀₅ of the differential amplifier A. It provides negative feedback to the inverting input to stabilize the line current value.

However, in FIG. 1, the loop from the comparator CP to the multipliers ML ₁ and ML ₂ and the line terminals L ₁ and
Although individual negative feedback is provided by the loop from the _L2 side to the input side of the differential amplifier A, the negative feedback loop as a whole circuit is not configured, so the difference between the output of the comparator CP and If external noise etc. enters between the input of dynamic amplifier A and the output signal of comparator CP, this will be added to the output signal of comparator CP and converted to line current, causing an error in the line current value. It has caused drawbacks.

The present invention aims to fundamentally eliminate such drawbacks of the conventional technology, and provides a square root calculation function and eliminates the influence of external noise by applying negative feedback over the entire circuit via a squaring circuit. The present invention provides an extremely effective square-plane displacement transducer capable of

The details of the present invention will be explained below with reference to FIG. 2 and subsequent figures showing embodiments.

In the circuit diagram shown in Figure 2, a pair of variable capacitances C ₁ and C 2 constituting a differential capacitance sensor are used as variable impedance, and an excitation power supply B whose output amplitude can be controlled is connected to the variable capacitance C 1 and C ₂ . After applying an excitation signal, the signal current passing through the variable capacitors C ₁ and C ₂ is detected by the diodes D ₁ to _{D 4} , and the detected output of the diodes D ₁ and D ₄ is applied to the resistors R ₅ , R ₆ and A voltage is applied to the low frequency converter consisting of capacitors C ₃ and C ₄ , and a voltage corresponding to the value of variable capacitance C ₁ is supplied from resistor R ₆ .
e ₁ and variable capacitance from resistor R ₅
A voltage e ₂ is obtained according to the value of C ₂ .

Note that the positive half wave of the excitation signal is connected to the diode D ₁ ,
The negative half wave is connected to the load resistor _R7 via _the diodes _D2 and _D3 .

Further, the voltages e ₁ and e ₂ are given to a subtracter as a control circuit constituted by resistors R ₂ , R ₃ , R ₄ and a differential amplifier A ₁ , and here, variable capacitances C ₁ and C In addition to the signal voltage corresponding to the difference _between
The output amplitude of excitation power source B is controlled.

On the other hand, voltages e ₁ and e ₂ are applied to an adder of an output circuit constituted by resistors R ₈ and R ₉ and a differential amplifier A ₂ , where variable capacitances C ₁ and C ₂
The signal voltage becomes a signal voltage corresponding to the sum of , and is amplified as a difference from the reference voltage e ₀ as in FIG. 1, controls the transistor Q, and determines the value of the current I ₁ flowing thereto.

The configuration of the output circuit is the same as that of the first output circuit, except that negative feedback is not applied to the non-inverting input of differential amplifier _A2 .
It is similar to the figure.

In addition, the resistor generated by the line current I ₀
The voltage of R ₀₅ is applied to the inverting input of the differential amplifier A ₁ via a feedback circuit consisting of a bias voltage e ₃ , a squaring circuit SQ, and a resistor R ₁ , thereby increasing the variable capacitance. Negative feedback is applied to C ₁ and C ₂ in a direction that cancels out the signal voltage according to the difference.

Therefore, the output amplitude of excitation power supply B is controlled in the direction in which the sum of voltages e ₁ and e ₂ becomes equal to reference voltage e ₀ , and the output amplitude is corrected according to the difference between voltages e ₁ and e ₂ . angular frequency ω in the excitation signal
Fluctuations in line current I ₀ due to changes in I 0 are prevented.

Further, by negative feedback via the square circuit SQ, a change in the line current I ₀ can be obtained in accordance with a value obtained by square rooting the differential change in the variable capacitances C ₁ and C ₂ .

That is, the output amplitude of excitation power source B is E, the angular frequency is ω, and the initial values of variable capacitances C ₁ and C ₂ are C ₀ ,
The distance between the electrodes forming variable capacitance C ₁ and C ₂ is
If d ₀ is the change in the distance between the electrodes according to the displacement and Δ _d , then the following equation holds true.

C ₁ = C ₀ d ₀ / d ₀ −Δ _d , C ₂ = C ₀ d ₀ / d ₀ + Δ _d …(1) Also, for convenience of explanation, R ₁ = R ₂ = R ₃ = R ₄ , R ₅ =
Let R ₆ = R, R ₈ = R ₉ , the current flowing through the reference potential point COM to the resistor R ₀₅ is I ₂ , and the resistance value of the resistor R ₀₅ is
If R _f , the following equations hold true.

e ₁ =E・ω・C ₁・R …(2) e ₂ =E・ω・C ₂・R …(3) (e ₁ +e ₂ /2−e ₀ )μ _t =I ₁ …(4) However, μ _t is the total amplification factor of the differential amplifier A ₁ and the transistor Q. Also, differential amplifier A ₁
If the amplification factor of is μ ₁ , then I ₀ = I ₁ + I ₂ …(5) {e ₁ /2−e ₂ + (R _f・I ₀ −e ₃ ) ² /2} μ ₁ = E …( 6) By transforming equation (6), {(e ₁ −e ₂ )−(R _f・I ₀ −e ₃ ) ² }=2・E/μ ₁ …(7) (4), (5) From the formula, I ₀ = (e ₁ + e ₂ / 2 - e ₀ ) μ _t + I ² ∴e ₁ + e ₂ = 2・e ₀ + 2I ₁ − I ₂ / μ _t …(8) (2), (3) , From equation (8), E・ω・R(C ₁ +C ₂ )=2・e ₀ +2I ₀ −I ₂ /μ _t ∴E=2・e ₀ /ω・R(C ₁ +C ₂ ) +2I ₀ −I ₂ /ω・R(C ₁ +C ₂ ) μ _t …(9) Furthermore, from equations (2), (3), and (7), [{2・e ₀ /ω・R(C ₁ +C ₂ )+2I ₀ −I ₂ /ω・R(
C ₁ + C ₂ ) μ _t }ω・R (C ₁ − C ₂ ) − (R _f・I ₀ −e ₃ ) ² ] μ ₁ /2=2・e ₀ /ω・R (C ₁ + C ₂ ) +2I ₀ −I
₂ /ω・R(C ₁ +C ₂ )μ _tIf we transform this, we get (2・e ₀ /C ₁ +C ₂ +2I ₀ −I ₂ /(C ₁ +C ₂ )μ _t )・(C
₁ −C ₂ ) − (R _f・I ₀ −e ₃ ) ² = 4・e ₀ /ω・R (C ₁ +C ₂ )μ ₁ +4I ₀ −I ₂ /ω
・R(C ₁ + C ₂ ) μ _t・μ ₁ …(10) Here, if μ _t and μ ₁ are sufficiently large, then 2・e ₀ /C ₁ +C ₂ (e ₁ − e ₂ ) −(R _f・I ₀ −e ₃ ) ² = 0 ∴(R _f・I ₀ −e ₃ ) ² = 2・e ₀ C ₁ −C ₂ /C ₁ +C ₂ …(11) Go to equation (11) Introducing equation (1), we get Therefore, as shown in equation (12), the line current I ₀ is proportional to the square root of the change Δ _d in the distance between the electrodes according to the displacement to be detected.

FIG. 3 is a circuit diagram showing another embodiment, in which Rosemount sensors are used as the variable capacitances C ₁ and C ₂ and the primary winding P and two secondary windings are
Variable capacitance by transformer T with S ₁ and S ₂
In addition to applying an excitation signal to C ₁ and C ₂ , the common electrodes of the same capacitances C ₁ and C ₂ are connected to a common potential by a capacitor C ₅ , and the sum voltage of voltages e ₁ and e ₂ is applied to a resistor. The reference voltage e0 is taken out from the amplifier _R10 and applied to the inverting input of the differential amplifier _A2 , while the reference voltage _e0 is applied to the non-inverting input of the same amplifier _A2 . Further, the power supply voltage V is divided by the resistors R ₁₁ and R ₁₂ to obtain the bias voltage e ₃ , but other aspects are the same as in FIG. 2, and the same operational results as in FIG. 2 can be obtained.

FIG. 4 is a circuit diagram showing an example of the squaring circuit SQ, in which a resistor R ₂₁ is connected to the non-inverting input of the comparator CP.
The input voltage Vi from the input terminal IN is applied through the input terminal IN, and positive feedback is provided by the resistor _R22 , so that the operation of the comparator CP exhibits hysteresis characteristics with respect to the input level.

In addition, the current from the power supply E _B to the capacitor C ₂₁ passes through the transistor Q ₂₁ and the resistor R ₂₃ , which are turned on and off according to the change in the output of the comparator _CP .
While the capacitor C ₂₁ is charged by the charging characteristic according to the time constant of R ₂₃ and the capacitor C ₂₁ , when the transistor Q ₂₁ is turned off, the output change of the comparator CP is applied via the inverter IV. Q ₂₂ turns on, discharging the charge in capacitor C ₂₁ via resistor R ₂₄ .

Therefore, a feedback loop is formed between the change in the terminal voltage of the capacitor _C21 applied to the inverting input of the comparator CP and the corresponding change in the output of the comparator CP. The duty ratio D at which the operation is automatically repeated and the output of the comparator CP repeats high and low levels is:
It is determined depending on the input voltage Vi, and when the proportionality constant is _K1 , the following formula holds true.

D=V _i ·K ₁ (21) On the other hand, the input voltage V _i is applied to transistors Q ₂₃ , Q ₂₄ , resistor R ₂₅ and It is applied to a circuit consisting of capacitor C ₂₂ , transistor Q ₂₃ is connected to transistor Q ₂₁ and transistor Q ₂₄
turns on and off at the same time as the transistor Q ₂₂ , the input voltage V _i is stepped according to the duty ratio D, and the capacitor C ₂₂ is charged and discharged via the resistor R ₂₅ . If the constant is _K2 , the output voltage _V0 appearing at the output terminal OUT is expressed by the following equation.

V ₀ = D・V _i・K ₂ …(22) By substituting equation (21) into equation (22), V ₀ =V _i・K ₁・V _i・K ₂ =V _i 2・K ₁・K ₂ ...(23), and the output voltage is proportional to the square of the input voltage V _i
V ₀ is obtained.

However, in addition to what is shown in FIG. 4, any circuit that exhibits a square calculation function, such as a cascaded circuit of multipliers, can be used.

In addition, variable inductance, variable resistance, etc. may be used instead of variable capacitances C ₁ and C ₂ .
The same effect can be achieved even if only one of a pair of impedances is variable and the other is a fixed reference impedance, and the configuration of each circuit in FIGS. 2 and 3 can be selected according to the conditions. can be freely modified in various ways.

As is clear from the above explanation, according to the present invention, negative feedback as a whole is performed via a squarer circuit, so the influence of external noise that enters between each circuit is eliminated, and an error-free output is generated. A signal can be obtained, and an output signal corresponding to a value obtained by square root calculation of a given displacement can be obtained, and it is highly effective as a displacement converter for various uses requiring a square root calculation function.

[Brief explanation of drawings]

Fig. 1 is a circuit diagram of a conventional example, Figs. 2 and 3 are circuit diagrams showing an embodiment of the present invention, and Fig. 4 is a circuit diagram showing an example of the squarer circuit used in Figs. 2 and 3. It is. C ₁ , C ₂ ... Variable capacitance (variable impedance), B ... Excitation power supply, D ₁ - _{D 4} ... Diode,
A ₁ , A ₂ ...Differential amplifier, SQ ...Squaring circuit,
_R02 to _R05 , _R1 to _R12 ...Resistor, _C3 to _C5 ...Capacitor, Q...Transistor, T...Transformer.

Claims (1)

[Claims] 1: a pair of variable impedances, at least one of which changes in accordance with the displacement to be detected; an excitation power source whose output amplitude is controllable for applying an excitation signal to each of the variable impedances; and a signal corresponding to the difference between the variable impedances. a control circuit that calculates and then controls the output amplitude of the excitation power source; an output circuit that calculates a signal corresponding to the sum of the variable impedances and converts the difference from the reference value into an output signal; A square root type displacement converter comprising a feedback circuit that feeds back the output signal via a squaring circuit in a direction that cancels the signal according to the difference.