JPH0635503A - Industrial instrument output circuit - Google Patents

Abstract

PURPOSE:To attain the automatic switching of two types of standard signals by means of a single industrial instrument by using a control means which decides the operating state of each switch part in accordance with the both-side voltage of a detecting resistance and the output voltage level of main circuit which are set in the OFF states of two switch parts. CONSTITUTION:A differential amplifier 1 receives the input/output voltage E1 and E2 at a non-inverted input terminal (+) via the resistances 6 and 12. The transistors TR 2 and 3 output the voltage of the same place as the output of the amplifier 1, and a detecting resistance 10 leads the output of the TR 3 to the output OUT. Then the output of the TR 3 is fed back to an inverted input terminal (-) after the division of voltage. An electronic switch 4 performs the ON/OFF control of connection between the input terminal (-) and a common line COM. Meanwhile an electronic switch 5 performs the ON/OFF control of connection between an output terminal OUT and the input terminal (+). A deciding circuit 13 turns off both switches 4 and 5 and receives the voltage of the resistance 10 and the output voltage ED to reset the switches 4 and 5 according to the level of the voltage ED.

Description

Detailed Description of the Invention

[0001]

BACKGROUND OF THE INVENTION 1. Field of the Invention The present invention relates to an industrial instrument used in process control, and more particularly to an output circuit of an industrial instrument capable of handling a plurality of standard signals in one device.

[0002]

2. Description of the Related Art An output signal of an industrial instrument is DC 1V
There is a voltage signal of ~ 5V and a current signal of 4mA ~ 20mA. The industrial measuring instrument has an output circuit corresponding to the type of the output signal, and is manufactured as a different model. 5 and 6 show an output circuit of a conventional industrial instrument. FIG. 5 shows an output circuit of an industrial instrument for outputting a voltage signal and FIG. 6 shows an output circuit of an industrial instrument for outputting a current signal.

The output circuit shown in FIG. 5 comprises a differential amplifier 20.
Input voltage E to the input terminal IN_iIs not inverted
Amplified output E supplied to the input terminal (+)_OIs the output terminal
It is led to OUT and negatively returns to the inverting input terminal (-).
It has been returned. Therefore, this differential amplifier 20 is a buffer
Functions as an amplifier, E_O= E_iAnd the input voltage E _i
If the output voltage is 1-5V, the output voltage E_OIs also 1 to 5V.

The output circuit shown in FIG. 6 has a differential amplifier 21.
, NPN type transistor 22, PNP type transistor 23, and seven resistors 24 to 30. The input voltage E _i applied to the input terminal IN is supplied to the non-inverting input terminal (+) of the differential amplifier 21 through the resistor 24, and the inverting input terminal (−) of the differential amplifier 21 is passed through the resistor 25. Connected to the common line COM.

The output of the differential amplifier 21 is the transistor 2
2 is supplied to the base terminal, and a resistor 26 is provided between the collector terminal of the transistor 22 and the power supply line (+ B).
A resistor 27 is placed between the emitter terminal and the common line COM.
Are connected. The collector voltage of the transistor 22 is supplied to the base terminal of the transistor 23,
The collector voltage of the transistor 23 is led to the output terminal OUT via the resistor 28 and becomes the output voltage E _O. In addition,
The emitter terminal is directly connected to the power supply line (+ B).
Also, the collector terminal of the transistor 23 and the differential amplifier 2
The resistor 29 is connected between the first inverting input terminal (-) and the non-inverting input terminal (+) of the differential amplifier 21.

In the circuit of FIG. 6 having such a configuration,
Output current I derived from output terminal OUT_OAnd the input end
Input voltage E supplied to child IN_iExplain the relationship with.
However, the equivalent resistance between the output terminal OUT and the common line COM is
Anti (load resistance) R_O(= E _O/ I_O) And the resistance 2
Resistance values of 4, 25, 29 and 30 are all R₁, Resistor 28
Resistance value is R₂Will be described.

Since the resistances of the resistors 24 and 30 are equal, the potential of the non-inverting input terminal (+) of the differential amplifier 21 is
Intermediate value between input voltage E _i and output voltage E _O (E _i + E _O ) /
It is 2. Therefore, the potential of the inverting input terminal (−) of the differential amplifier 21 is also (E _i + E _O ) / 2, and since the resistance values of the resistors 29 and 25 are equal, the collector potential of the transistor 23 is (E _i + E _O). ).

[0008] From the above relationship, the voltage across the resistor 28 is Motomari and _{(E i + E O) -E} O from E _i, the voltage across the resistor 30 _{(E i + E O) /} 2-E O from (E _i -E _O )
/ 2 is obtained. Since the output current I _O is the sum of the current flowing through the resistor 28 and the current flowing through the resistor 30, I _O =
Become _{E i / R 2 + (E} i -E O) / (2R 1). Substituting the relationship of E _O = I _O R _O into this equation, I _O = (2R ₁ + R ₂ ) E _i / {R ₂ (2R ₁ +
R ₀ )}, and the relationship between the input voltage E _i and the output current I _O is obtained.
Since the resistance value R ₁ is now sufficiently large (that is, R ₁ >>
R ₂ , R _O ), the above equation is simplified to I _O ≈E _i / R ₂ . That is, from the output circuit of FIG. 5, the input voltage E _i
An output current I _O proportional to is output and the input voltage is E _i =
In the case of 1 to 5 V, the output current may be I _O = 4 to 20 mA.

[0009]

As described above, the output circuit of the conventional industrial instrument is made as a different model depending on whether it outputs a voltage signal or a current signal.
There is a disadvantage in that one industrial instrument cannot output two types of standard signals. In addition, since there is no compatibility between each model, there was an inconvenience such as needing a spare machine separately and having to replace it with another model when local modification is required.

The present invention has been made in view of this problem, and an object thereof is to provide an industrial instrument output circuit capable of appropriately switching and outputting two kinds of standard signals by one industrial instrument. To do.

[0011]

In order to achieve the above object, an industrial instrument output circuit according to the present invention has a first resistor and a second resistor whose resistance values are equal to each other. A differential amplifier that is received by the non-inverting input terminal via
An amplifier section that receives the output of this differential amplifier and outputs an amplified voltage of the same phase as this, a detection resistor that leads the output of this amplifier section to the output terminal of this circuit, and an output of the amplifier section In an industrial instrument output circuit comprising a resistance voltage dividing circuit for dividing the voltage by two resistors having the same voltage and feeding back to the inverting input terminal of the differential amplifier, the voltage dividing circuit is provided in the resistance voltage dividing circuit. A first switch unit that controls ON / OFF of a connection with a reference potential point and a second resistor that are connected in series to control ON / OFF of a connection between an output terminal of the circuit and the non-inverting input terminal. The second switch section and the first and second switch sections are turned off, the voltage across the detection resistor in this state and the output voltage of the circuit are received, and the first switch is operated according to both voltage values.
And a control means for determining the operating state of the second switch section.

[0012]

The differential amplifier has an input voltage E _i and an output voltage E _{o of} this circuit.
Are received by the non-inverting input terminal via the first and second resistors having the same resistance value. Therefore, ignoring the input resistance of the differential amplifier, the potential at the non-inverting input terminal is (E _i + E _o ) /
It becomes 2. The amplification section receives the output e of the differential amplifier and outputs an amplified voltage Ae having the same phase as the output e.

The detecting resistor guides the output of the amplifying section to the output terminal of this circuit. The resistance value of the detection resistor is R ₂ . The resistance voltage dividing circuit divides the output Ae of the amplification section by two resistors having the same resistance value and feeds it back to the inverting input terminal of the differential amplifier. Therefore, the potential of the inverting input terminal of the differential amplifier is Ae / 2, and the amplification factor of the differential amplifier is very large, so that the relationship of Ae / 2 = (E _i + E _o ) / 2 is established. That is, the output voltage Ae of the amplification unit
Is given by E _i + E _o .

The first switch section is provided in the resistance voltage dividing circuit and controls ON / OFF of the connection between the inverting input terminal of the differential amplifier and the reference potential point. Therefore, the voltage dividing circuit functions only when the first switch unit is in the ON state, and when the first switch unit is in the OFF state, the potential of the inverting input terminal of the differential amplifier is equal to the output voltage of the amplifying unit. Will be equal.

The second switch section is connected in series with the second resistor and controls ON / OFF of the connection between the output terminal of the circuit and the non-inverting input terminal of the differential amplifier. Therefore, the potential of the non-inverting input terminal of the differential amplifier is (E _i + E _o ) / 2 when the second switch section is in the ON state, while the non-inverting input terminal is in the OFF state when the non-inverting input terminal is in the OFF state. The potential of both the terminal and the inverting input terminal is E _i .

The circuit operation will be summarized and described below by classifying the operating states of the first and second switch sections. [When the first and second switch parts are in the ON state] In this case, the voltage across the test resistor is E calculated from Ae-E _o.
i, and the voltage across the second resistor is (E _i −
_Eo ) / 2. Therefore, the resistance value of the inspection resistor is R
₂ and the resistance value of the second resistor is R ₁ , the output current I
_{o is, I o = E i / R} 2 + (E i -E o) / (2R 1)
Is. Here, if the load resistance is rearranged as R _o , then I _o
= (2R ₁ + R ₂ ) E _i / {R ₂ (2R ₁ + R ₀ )} ≈
It becomes E _i / R ₂ . That is, when the first and second switch parts are in the ON state, this circuit functions as a circuit for outputting a current signal.

[Case where the first and second switch parts are in the OFF state] In this case, a negative feedback loop is formed by the amplifying part and the resistor in the voltage dividing circuit, and the voltage dividing circuit does not function.
Therefore, the potentials of the non-inverting input terminal and the inverting input terminal of the differential amplifier are E _i , and the output of the amplifying unit is also E _i.
Is equal to Therefore, the output voltage E _O of this circuit is E _o = R _o E _i / (R _o + R ₂ ) ≈E _i , and the circuit functions as a voltage signal output circuit.

The control means operates, for example, when the power of the present circuit is turned on to turn off the first and second switch parts, and the voltage across the detection resistor and the output voltage of the present circuit in this state. In response to both voltage values, the first
And the operating state of the second switch section. A specific example of the operation will be described. Since the load is connected when the power is turned on, the load resistance value R _o is calculated from the voltage E ₂ across the detection resistor and the output voltage E _{o of} this circuit, and the value is calculated. When is large, the first and second switch parts are maintained in the OFF state, while when the value is small, the first and second switch parts are changed to the ON state.

As described above, when the switch circuit is in the ON state, it has the characteristic of I _O ≈E _i / R ₂ , and when it is in the OFF state, it has the characteristic of E _o ≈E _i. By the initial operation when the power is turned on, the circuit corresponding to the load resistance value is automatically set.

[0020]

The present invention will be described in more detail based on the following examples. FIG. 1 is a circuit diagram showing an output section of an industrial instrument according to an embodiment of the present invention. This circuit includes a differential amplifier 1, an NPN type transistor 2, a PNP type transistor 3, transistors 4 and 5 which are electronic switches,
It is composed of seven resistors 6 to 12, a discrimination circuit 13, and an LED 14 showing the discrimination result.

The input voltage E _i applied to the input terminal IN is supplied to the non-inverting input terminal (+) of the differential amplifier 1 through the resistor 6, and the inverting input terminal (−) of the differential amplifier 1 is a resistor. 7 and the electronic switch 4 are connected to a common line COM via a series circuit. The output of the differential amplifier 1 is supplied to the base terminal of the transistor 2, and a resistor 8 is provided between the collector terminal of the transistor 2 and the power supply line (+ B).
However, a resistor 9 is connected between the emitter terminal and the common line COM.

The collector voltage of the transistor 2 is supplied to the base terminal of the transistor 3, and the collector voltage of the transistor 3 is led to the output terminal OUT via the resistor 10 and becomes the output voltage E _O. The transistor 3
The emitter terminal of is directly connected to the power supply line (+ B). Further, a resistor 11 is connected between the collector terminal of the transistor 3 and the inverting input terminal (−) of the differential amplifier 1 to connect the output terminal OUT and the non-inverting input terminal (+) of the differential amplifier 1. A series circuit of the resistor 12 and the electronic switch 5 is connected between them.

The determination circuit 13 determines the voltage E ₂ across the resistor 10.
And a output voltage E _O, and outputs a control voltage to the gate terminals of the electronic switches 4 and 5. Then, the electronic switches 4 and 5 are turned on or off according to the level of the control voltage. The LED 14 also functions as an indicator lamp that indicates whether the circuit is outputting a voltage signal or a current signal.

Further, in the circuit of FIG. 1, the resistors 6, 7, 1
The resistance values of 1 and 12 are all R ₁ [Ω], and the resistance value of the resistor 10 is R ₂ [Ω]. Note that FIG. 3 is an external view of the industrial measuring instrument, in which the LED 14 is shown. FIG. 2 is a flow chart for explaining the operation procedure of the discrimination circuit 13. Hereinafter, the circuit operation of FIG. 1 will be described with reference to the flowchart of FIG.

When the power is turned on, the discriminating circuit 13 first outputs a control voltage for controlling the electronic switches 4 and 5 in the OFF state (step ST1). When the electronic switches 4 and 5 are turned off, the circuit of FIG. 1 operates equivalently as the circuit of FIG. 4, and the collector potential of the transistor 3 matches the input voltage E _i . Where resistance 1
The current flowing through 0 is common with the current flowing through the load resistance R _O , so that the relationship of R _O = R ₂ E _O / E ₂ is established. The resistance value of the resistor 10 was R ₂ , and the voltage across it was E ₂ .

Next, the discriminating circuit 13 determines the above-mentioned [R ₂ E _O / based on the output voltage E _O and the voltage E ₂ across the resistor 10.
E ₂ ], and the value of the load resistance R _O is calculated (step ST2). Then, this calculated value [R ₂ E _O / E ₂ ]
Is compared with a specified value to determine whether the load is a current signal output load or a voltage signal output load (step ST3).

If the calculated value [R ₂ × E _O / E ₂ ] is larger than the specified value, a voltage signal may be output. Therefore, the discrimination circuit 13 outputs the voltage signal to the electronic switches 4 and 5.
Is controlled to an OFF state (step ST4). However, since the electronic switches 4 and 5 are in the OFF state after the power is turned on, there is no change in the circuit operation and the output voltage E _O is still E _O = R _O E _i / (R ₂ + R _O ). (See Figure 4). Here, the load resistance R _O is sufficiently larger than the resistance value R ₂ of the resistor 10, so that R ₂ << R _O , E _O ≈
The relational expression of E _i holds.

On the contrary, when the calculated value is smaller than the specified value, the discrimination circuit 13 controls the electronic switches 4 and 5 to be in the ON state so as to output the current signal (step ST5).
When the electronic switches 4 and 5 is turned ON, since the circuit of Figure 1 is comprised of the same connected to the circuit of Figure 6 described above, I _O
= (2R ₁ + R ₂ ) E _i / {R ₂ (2R ₁ + R ₀ )} holds, and from the relationship of R ₁ >> R ₂ and R _O , I _O ≈E
_{The i} / R ₂ relationship holds.

As described above, the circuit connection is appropriately switched after the determination in step ST3, and the voltage signal E _O ≈E _i or the current signal I _O ≈E _i / R ₂ is output. That is,
Assuming that the input voltage E _i is 1 to 5 V, the voltage output is E
_O = 1 to 5 V, if the current output becomes I _O = 4 to 20 mA. The circuit for switching between the voltage output and the current output for output is not limited to the circuit having the configuration of FIG.
Other circuits having the same function may be used. Further, although the electronic switches 4 and 5 are used in FIG. 1, the invention is not limited to this, and may have a mechanical contact such as a relay. Further, since the LED is an auxiliary member for displaying the operation status of the circuit of FIG. 1, it may be omitted.

[0030]

As described above, the industrial instrument output circuit according to the present invention automatically switches the circuit connection when the power is turned on and thereafter outputs the reference signal according to the load state. There is. Therefore, as compared with the conventional device, it is possible to save the trouble of distinguishing and switching the voltage / current output, and there is no risk of erroneous switching. Further, even when both a voltage output type and a current output type are used in one plant, it is convenient to eliminate the need for preparing both types as spare parts.

[Brief description of drawings]

FIG. 1 is a circuit diagram of an industrial instrument output circuit according to an embodiment of the present invention.

FIG. 2 is a flow chart for explaining the operation of the circuit of FIG.

FIG. 3 illustrates an external view of the circuit of FIG.

FIG. 4 is an equivalent circuit for explaining an operating state of the circuit of FIG.

FIG. 5 is a circuit diagram of a conventional industrial instrument output circuit that outputs a voltage signal.

FIG. 6 is a circuit diagram of a conventional industrial instrument output circuit that outputs a current signal.

[Explanation of reference numerals] 1 differential amplifier 2,3 amplifying transistor (amplifying section) 10 detecting resistor 11,7 resistance voltage dividing circuit 4 first switch section 5 second switch section 13 determination circuit (control means)

Claims (1)

[Claims] 1. A differential amplifier for receiving an input voltage and an output voltage of the present circuit at a non-inverting input terminal via first and second resistors having the same resistance value, and an output of the differential amplifier. An amplifying section that outputs an amplified voltage of the same phase, a detection resistor that leads the output of this amplifying section to the output terminal of this circuit, and an output of the amplifying section are divided by two resistors having the same resistance value. An industrial instrument output circuit comprising: a resistance voltage divider circuit that compresses the voltage and feeds it back to the inverting input terminal of the differential amplifier, wherein the connection between the inverting input terminal and a reference potential point is turned on provided in the resistance voltage dividing circuit. A first switch unit for controlling ON / OFF, a second switch unit connected in series to the second resistor for ON / OFF controlling the connection between the output terminal of the circuit and the non-inverting input terminal, Turn off the first and second switches. And a control unit that receives the voltage across the detection resistor and the output voltage of the circuit in this state, and determines the operating states of the first and second switch units according to the two voltage values. And industrial instrument output circuit.