JP3233074B2 - Half-wave rectifier - Google Patents

Description

DETAILED DESCRIPTION OF THE INVENTION

[0001]

The present invention relates to an operational amplifier,
The present invention relates to a half-wave rectifier that performs analog signal processing by combining a diode that is a nonlinear element.

[0002]

2. Description of the Related Art FIG. 10 is a circuit diagram showing a conventional half-wave rectifier circuit described in, for example, "Design of OP Amplifier Circuit" by Okamura, page 186. In FIG. The input resistors Rs, 2b are feedback resistors Rf, 3a,
3b is a diode, 4 is an input terminal, and 5 is an output terminal.

Next, the operation will be described. When the voltage Vin at the input terminal 4 is a positive voltage, all the current Vin / Rs flowing through the input resistor 2a flows into the output terminal of the operational amplifier 1 via the diode 3a. The non-inverting input of the operational amplifier 1 is grounded, and the inverting input and the non-inverting input have the same potential.
Since no current flows through the feedback resistor 2b, the voltage drop of the feedback resistor 2b is equal to zero. From the above, the output terminal 5
Is given by the following equation. Vout = 0

When the voltage Vin at the input terminal 4 is a negative voltage,
All the voltages Vin / Rs flowing through the input resistor 2a flow through the feedback resistor Rf and do not flow through the diode 3a. Therefore, the voltage Vout of the output terminal 5 is given by the following equation. Vout = −Vin × (Rf / Rs) From the above, the relationship between Vin and Vout is as shown in the graph of FIG. 11, and it can be seen that the circuit operates as a half-wave rectifier.

[0005]

The above description has been made on the assumption that all components used in the conventional half-wave rectifier have ideal characteristics. That is, it is assumed that the amplification degree of the operational amplifier 1 is infinite, the input offset voltage and the input bias current are zero, and the reverse current Ir of the diodes 3a and 3b is zero.

Next, the influence of the reverse current Ir of the diode will be considered. When the voltage Vin at the input terminal 4 is a negative voltage, the voltage at the output terminal of the operational amplifier 1 is Vout + Vf obtained by adding the forward voltage Vf of the diode 3b to the voltage at the output terminal 5. Since the voltage at the inverting input terminal of the operational amplifier 1 is 0 V,
A reverse voltage (Vout + Vf) is applied to the diode 3a. Since the reverse current Ir of the diode 3a flows through Rf,
The voltage Vout of the output terminal 5 is given by the following equation. Vout = {− Vin × (Rf / Rs)} − Rf × Ir

Since the reverse current Ir of the diode 3a increases as the reverse voltage increases, the effect of the reverse current of the diode 3a appears as shown in the graph of FIG. In the figure, 6 is an ideal characteristic, and 7 is a characteristic including the influence of the reverse current of the diode 3a. The effect of the reverse current of the diode 3a becomes a problem that cannot be ignored when high signal processing accuracy is required, when the feedback resistance Rf is large, and when the reverse current is large.

The present invention has been made to solve the above problems, and has as its object to provide a half-wave rectifier in which the influence of the reverse current of the diode 3a is extremely small.

[0009]

SUMMARY OF THE INVENTION Half-wave rectification according to the present invention
Input terminal, non-inverting input terminal and output
An operational amplifier having a terminal and an inverting input of the operational amplifier
An input resistor connected in series with the terminal and one end
A first diode connected to the output terminal of the band, and one end
Is connected to the inverting input terminal of the operational amplifier.
And a series circuit consisting of
A series circuit comprising second and third diodes;
Is connected between the second diode and the third diode.
Is continued, the flow front Stories second diode third diode
The reverse current that is about to flow into the ground
A resistor flowing to the non-inverting input terminal of the width unit.

Further, in the half-wave rectifier according to the present invention,
Is the resistance connected between the second and third diodes.
Has the other end connected to the ground point and the third diode
Has a resistance value smaller than the reverse resistance of.

Further, in the half-wave rectifier according to the present invention,
Is the resistance connected between the second and third diodes.
Has the other end connected to the ground point.
Is equal to the parallel resistance value of the feedback circuit and the input resistance.
A non-inverting input of the operational amplifier is connected to a second resistor having a high resistance value.
And the second resistor is grounded.

Further, in the half-wave rectifier according to the present invention,
A second input resistor is connected to the non-inverting input terminal of the operational amplifier.
Connected and connected between the second and third diodes.
The other end of the resistor is connected to the second input resistor.

Further, in the half-wave rectifier according to the present invention,
Means that the resistance value of the second input resistor is equal to that of the feedback circuit
It is equal to the parallel resistance of the input resistance.

Further, according to the half-wave rectifier of the present invention,
Connected to the inverting input terminal of the operational amplifier.
It has a plurality of input resistances, and a plurality of
An input signal is input.

[0015]

DESCRIPTION OF THE PREFERRED EMBODIMENTS Embodiment 1 FIG. 1 is a circuit diagram of a half-wave rectifier according to this embodiment. In FIG. 1, 1 is an operational amplifier, 2a is an input resistor connected to the inverting input side of the operational amplifier, 2b
Is a feedback resistor having one end connected to an input resistor and the other end connected to a diode 3b described later, and 2c is a ground resistor connected at one end to a diode 3a described later and the other end is grounded. In FIG. 1, the resistance values of the respective resistors are Rs, Rf, and Rg.
Indicated by.

3a, the anode side is connected to the input resistor 2a,
A diode having a cathode connected to the ground resistor 2c;
Reference numeral 3b denotes a diode whose anode side is connected to the output side of the operational amplifier 1 and which is connected to the feedback resistor 2b. Reference numeral 4 denotes an input terminal connected to the input resistor 2a, and reference numeral 5 denotes an output terminal connected to the feedback resistor 2b and the cathode of the diode 3b. The non-inverting input side of the operational amplifier 1 is grounded.

Next, the operation will be described. When the voltage Vin at the input terminal 4 is a positive voltage, all the current Vin / Rs flowing through the input resistor 2a flows into the output terminal of the operational amplifier 1 via the diodes 3a and 3c. At this time, diode 3
Since the voltage of the cathode a is equal to the forward voltage −Vf, a current of Vf / Rg flows through the ground resistor 2c.

This current merges with the current flowing through the diode 3a and flows into the output terminal of the operational amplifier 1 via the diode 3c. The non-inverting input of the operational amplifier 1 is grounded, and the inverting input and the non-inverting input have the same potential. Since no current flows through the feedback resistor 2b, the voltage drop of the feedback resistor 2b is equal to zero. From the above, the voltage Vo of the output terminal 5
ut becomes 0v.

When the voltage Vin at the input terminal 4 is a negative voltage,
All the current Vin / Rs flowing through the input resistor 2a flows through the feedback resistor 2b and does not flow through the diode 3a. Therefore, the voltage Vout of the output terminal 5 is given by the following equation. Vout = −Vin × (Rf / Rs)

Next, the effect of the reverse current Ir of the diode will be considered. When the voltage Vin at the input terminal 4 is a negative voltage, the voltage at the output terminal of the operational amplifier 1 is Vout + Vf obtained by adding the forward voltage Vf of the diode 3b to the voltage at the output terminal 5.

The voltage at the inverting input terminal of the operational amplifier 1 is 0 V
Therefore, the voltage at the output terminal of the operational amplifier 1 is equal to the sum of the reverse voltages applied to the two diodes 3a and 3b. When the resistance value Rg of the ground resistor 2c is set to a value sufficiently smaller than the reverse resistance of the diode 3a, the reverse current I
r almost flows through the ground resistance 2c. That is, the reverse current that is going to flow into the diode 3a is bypassed to the ground resistor 2c. Neglecting the reverse current flowing through the diode 3a, the voltage at the cathode of the diode 3a is Ir × Rg, which is the reverse voltage of the diode 3a.

In practice, since the reverse current Ir of the diode is very small, the value of Ir × Rg is
It is sufficiently smaller than the sum of the reverse voltages of a and 3b. That is, the reverse voltage applied to the diode 3c is substantially equal to Vout + Vf, and the reverse voltage applied to the diode 3a is equal to much lower Ir × Rg. As a result, the reverse current flowing through the diode 3a, that is, the current flowing through the diode 3a to the inverting input of the operational amplifier 1 becomes extremely small. As a result, a half-wave rectifier having a small influence of the reverse current can be obtained.

The diode 3 in this embodiment is
c and the ground resistor 2c form a reverse current bypass circuit. The same applies to embodiments after this point.

Embodiment 2 FIG. FIG. 2 is a circuit diagram of the half-wave rectifier in this embodiment. In this circuit, the diodes 3a, 3b, 3c are arranged with polarities opposite to those of the first embodiment.

Next, the operation will be described. When the voltage Vin at the input terminal 4 is a negative voltage, all the current Vin / Rs flowing through the input resistor 2a is supplied from the output terminal of the operational amplifier 1 via the diodes 3a and 3c. At this time, the voltage of the anode of the diode 3a is equal to the forward voltage Vf.
A current of Vf / Rg flows through the ground resistor 2c.

The current flowing through the ground resistor 2c and the diode 3
The current flowing through a is supplied from the output terminal of the operational amplifier 1 via the diode 3c. The non-inverting input of the operational amplifier 1 is grounded, and the inverting input and the non-inverting input have the same potential. Since no current flows through the feedback resistor 2b, the voltage drop of the feedback resistor 2b is equal to zero. As described above, the voltage Vout of the output terminal 5 becomes 0V.

When the voltage Vin of the input terminal 4 is a positive voltage,
All the current Vin / Rs flowing through the input resistor 2a flows through the feedback resistor 2b and does not flow through the diode 3a. Therefore, the voltage Vout of the output terminal 5 is given by the following equation. Vout = −Vin × (Rf / Rs) Next, the effect of the reverse current Ir of the diode will be considered. When the voltage Vin at the input terminal 4 is a positive voltage, the voltage at the output terminal of the operational amplifier 1 becomes Vout-Vf obtained by subtracting the forward voltage Vf of the diode 3b from the voltage at the output terminal 5.

The voltage at the inverting input terminal of the operational amplifier 1 is 0 V
Therefore, the voltage at the output terminal of the operational amplifier 1 is equal to the sum of the reverse voltages applied to the two diodes 3a and 3b. When the resistance value Rg of the ground resistor 2c is set to a value sufficiently smaller than the reverse resistance of the diode 3a, the reverse current Ir of the diode 3c
Almost flow through the ground resistance 2c. That is, the diode 3
The reverse current that is about to flow into a is bypassed to the ground resistor 2c. Neglecting the reverse current flowing through the diode 3a, the voltage at the anode of the diode 3a is -Ir * Rg, which is the reverse voltage of the diode 3a.

Actually, since the reverse current Ir of the diode is very small, the value of Ir × Rg is determined by the two diodes 3
It is sufficiently smaller than the sum of the reverse voltages of a and 3b. That is, the reverse voltage applied to the diode 3c is substantially equal to Vout-Vf, and the reverse voltage applied to the diode 3a is equal to much lower Ir × Rg.

As a result of the lower reverse voltage, the diode 3a
, That is, the current flowing through the diode 3a to the inverting input of the operational amplifier 1 becomes very small. As a result, a half-wave rectifier having a small influence of the reverse current is obtained.

Embodiment 3 FIG. 4 is a circuit diagram of the half-wave rectifier in this embodiment. This embodiment is based on the operational amplifier 1 of the first embodiment.
Are grounded via a resistor 2d. The resistance value Rc of the resistor 2d is determined by the feedback resistor 2b and the input resistor 2
The effect of the bias current of the operational amplifier 1 can be reduced by making the resistance value of the parallel resistor a equal.

Embodiment 4 FIG. FIG. 5 is a circuit diagram of the half-wave rectifier in this embodiment. This embodiment is based on the operational amplifier 1 of the second embodiment.
Are grounded via a resistor 2d. The resistance value Rc of the resistor 2d is determined by the feedback resistor 2b and the input resistor 2
The effect of the bias current of the operational amplifier 1 can be reduced by making the resistance value of the parallel resistor a equal.

Embodiment 5 FIG. FIG. 6 is a circuit diagram of a half-wave rectifying adder according to this embodiment. In this embodiment, a plurality of resistors 2e, 2f, and 2g are connected in parallel to the inverting input terminal of the operational amplifier 1 of the first embodiment, and one end of each resistor is connected to the input terminals 4a, 4b, and 4g.
c is a half-wave rectifying adder. In FIG. 6, the resistance values of the respective resistors are indicated as R1, R2, and R3.

Each of the voltages at the input terminals 4a, 4b, 4c is V
1, V2 and V3, V1 / R1 + V2 / R2 + V
When 3 / R3> 0, the currents flowing through the resistors 2e, 2f, and 2g are all transmitted through the diodes 3a and 3c.
Flows into the output terminal. The non-inverting input of the operational amplifier 1 is grounded, and the inverting input and the non-inverting input have the same potential. Since no current flows through the feedback resistor 2b, the feedback resistor 2b
The voltage drop of b is equal to zero. As described above, the voltage Vout of the output terminal 5 is given by the following equation. Vout = 0

V1 / R1 + V2 / R2 + V3 / R3 <0
In the case of, the sum of the currents flowing through the input resistors 2e, 2f, 2g (V1 / R1 + V2 / R2 + V3 / R3) all flows through the feedback resistor 2b and does not flow through the diode 3a. Therefore, the voltage Vout of the output terminal 5 is given by the following equation. Vout = −Rf × (V1 / R1 + V2 / R2 + V3 / R3) From the above, it can be seen that this embodiment operates as a half-wave rectifying adder.

Embodiment 6 FIG. FIG. 7 is a circuit diagram of a half-wave rectifying adder according to this embodiment. In this embodiment, a plurality of resistors 2e, 2f, and 2g are connected to the inverting input terminal of the operational amplifier 1 according to the second embodiment, and one end of each resistor is used as a half-wave rectifying and adding terminal 4a, 4b, or 4c. It is a vessel. The resistance value of each resistor is R1, R
2 and R3.

The voltages at the input terminals 4a, 4b, 4c are
1, V2 and V3, V1 / R1 + V2 / R2 + V
When 3 / R3 <0, the currents flowing through the resistors 2e, 2f, and 2g are all transmitted through the diodes 3a and 3c.
Is supplied from the output terminal of. The non-inverting input of the operational amplifier 1 is grounded, and the inverting input and the non-inverting input have the same potential. Since no current flows through the feedback resistor 2b, the voltage drop of the feedback resistor 2b is equal to zero. As described above, the voltage Vout of the output terminal 5 is given by the following equation. Vout = 0

V1 / R1 + V2 / R2 + V3 / R3> 0
In this case, the sum of the currents flowing through the input resistors 2e, 2f, and 2g (V1 / R1 + V2 / R2 + V3 / R3) all flows through the feedback resistor Rf and does not flow through the diode 3a. Therefore, the voltage Vout of the output terminal 5 is given by the following equation. Vout = −Rf × (V1 / R1 + V2 / R2 + V3 / R3) From the above, it can be seen that the circuit in this embodiment operates as a half-wave rectifying adder.

Embodiment 7 FIG. 8 is a circuit diagram of the half-wave rectifier in this embodiment. This embodiment is based on the operational amplifier 1 of the first embodiment.
The input resistor 2h is connected to the non-inverting input terminal of the non-inverting input terminal, and the terminal on the opposite side of the resistor is an input terminal 4d. The connection point between the diodes 3a and 3c and the input terminal 4d are connected by a resistor 2i. In addition, resistance 2h, 2i
Are represented by R4 and R5, respectively.

The voltages at the input terminals 4 and 4d are V1 and V2, respectively.
If V1−V2> 0, all the current flowing through the input resistor 2a flows into the output terminal of the operational amplifier 1 via the diodes 3a and 3c.

If the input bias current of the operational amplifier 1 is ignored, the voltage at the non-inverting input terminal is equal to the input voltage V2,
The inverting input and the non-inverting input have the same potential. Since no current flows through the feedback resistor 2b, the voltage drop of the feedback resistor 2b is equal to zero. From the above, the voltage Vout of the output terminal 5
Is given by the following equation. Vout = V2

When V1−V2 <0, all the current (V1−V2) / Rs flowing through the input resistor 2a flows through the feedback resistor Rf and does not flow through the diode 3a. Further, as described above, the voltage at the non-inverting input terminal of the operational amplifier 1 is equal to V2. Therefore, the voltage Vout of the output terminal 5 is given by the following equation. Vout = V2−Rf × (V1−V2) / Rs = (1 + Rf / Rs) V2− (Rf / Rs) V1 From the above, it can be seen that this embodiment operates as a half-rectifying addition subtractor.

Further, the resistance value R4 of the resistor 2h is changed to the feedback resistor 2
By making the resistance value of b and the parallel resistance of the input resistance 2a equal to each other, the influence of the bias current of the operational amplifier 1 can be reduced. Most of the reverse current of the diode 3c flows to the input terminal 4d via the resistor 2i. That is, a reverse current that is going to flow into the diode 3a is bypassed to the resistor 2i. Since the reverse current of the diode 3a is very small, it has the effect of reducing the effect of the reverse current as in the first embodiment. Note that the diode 3c in this embodiment is
And the resistor 2i constitute a reverse current bypass circuit. In this regard,
The same applies to the following embodiments.

Embodiment 8 FIG. FIG. 9 is a circuit diagram of a half-wave rectifier according to this embodiment. This embodiment is a half-wave rectifying adder / subtractor in which a resistor 2h is connected to the non-inverting input of the operational amplifier 1 of the second embodiment, and the terminal on the opposite side of the resistor is an input terminal 4d. A connection point between the diodes 3a and 3c and the input terminal 4d is connected to a resistor 2i.
Connected with.

The voltages at the input terminals 4 and 4d are V1 and V2, respectively.
If V1−V2 <0, all the current flowing through the input resistor is supplied to the operational amplifier 1 via the diodes 3a and 3c.
Is supplied from the output terminal of. If the input bias current of the operational amplifier 1 is ignored, the voltage of the non-inverting input is equal to the input voltage V2.
And the inverting and non-inverting inputs are at the same potential.

Since no current flows through the feedback resistor 2b, the voltage drop of the feedback resistor 2b is equal to zero. As described above, the voltage Vout of the output terminal 5 is given by the following equation. Vout = V2

When V1−V2> 0, all the current (V1−V2) / Rs flowing through the input resistor 2a flows through the feedback resistor Rf and does not flow through the diode 3a. Further, as described above, the voltage at the non-inverting input terminal of the operational amplifier 1 is equal to V2. Therefore, the voltage Vout of the output terminal 5 is given by the following equation. Vout = V2-Rf (V1-V2) / Rs = (1 + Rf / Rs) V2- (Rf / Rs) V1 From the above, it can be seen that the circuit in this embodiment operates as a half-wave rectifying adder.

Further, the resistance value R4 of the resistor 2h is
By making the resistance value of b and the parallel resistance of the input resistance 2a equal to each other, the influence of the bias current of the operational amplifier 1 can be reduced. Most of the reverse current of the diode 3c flows to the input terminal 4d via the resistor 2i, and the reverse current of the diode 3a is very small. Therefore, the effect of reducing the influence of the reverse current is obtained as in the first embodiment.

In the embodiment described above,
The diode 3c and the ground resistor 2c provide a reverse current bypass circuit for bypassing the reverse current that is going to flow into the diode 3a.
Alternatively, the configuration is made up of the diode 3c and the resistor 2i, but this configuration is merely an example and is not necessarily limited to these configurations. Further, it is desirable to completely bypass the reverse current flowing into the diode 3a, but it is not necessary to completely bypass the reverse current. This is because even if a small amount of reverse current flows into the diode 3a, the characteristics of the conventional half-wave rectifier are improved.

[0050]

The present invention is configured as described above.
Therefore, the following effects are obtained. The invention
In the half-wave rectifier according to
An operational amplifier having a force terminal and an output terminal;
The input resistance connected in series to the inverting input terminal of the
A first diode having an end connected to the output terminal of the operational amplifier
And one end are connected to the inverting input terminal of the operational amplifier
And a series circuit consisting of a feedback circuit
A series of second and third diodes provided in a row
A circuit, one end of which is the second diode and the third diode.
Between the second diode and the second diode.
The reverse current flowing into the third diode is supplied to the ground point or
Comprises a resistor flowing to a non-inverting input terminal of the operational amplifier.
Therefore, the characteristics of the half-wave rectifier
Deterioration can be prevented.

In the half-wave rectifier according to the present invention,
Is the resistance connected between the second and third diodes.
Has the other end connected to the ground point and the third diode
Has a resistance value smaller than the reverse resistance of the third
The reverse current that is about to flow into the
To the resistor connected between the diode and the third diode.
Can be shed.

In the half-wave rectifier according to the present invention,
Is the resistance connected between the second and third diodes.
Has the other end connected to the ground point.
Is equal to the parallel resistance value of the feedback circuit and the input resistance.
A non-inverting input of the operational amplifier is connected to a second resistor having a high resistance value.
Because the second resistor is grounded, the third die
Bypass reverse current that tries to flow into the Aether to ground
In addition to the bias voltage of the operational amplifier.
The effect of the flow can be reduced.

In the half-wave rectifier according to the present invention,
A second input resistor is connected to the non-inverting input terminal of the operational amplifier.
Connected and connected between the second and third diodes.
Since the other end of the resistor is connected to the second input resistor,
3 Reverse current flowing into the diode is passed through the resistor.
Can be detoured.

In the half-wave rectifier according to the present invention,
Means that the resistance value of the second input resistor is
Since the parallel resistance of the input resistor is equal to the
Reverse current that is about to flow into the
Of the operational amplifier bias current
The effect can be reduced.

Further, in the half-wave rectifier according to the present invention,
Connected to the inverting input terminal of the operational amplifier.
It has a plurality of input resistances, and a plurality of
Characteristics caused by reverse current because input signal is input
It is possible to construct an addition type half-wave rectifier that prevents deterioration of
it can.

[Brief description of the drawings]

FIG. 1 is a circuit diagram of a half-wave rectifier circuit according to a first embodiment.

FIG. 2 is a circuit diagram of a half-wave rectifier circuit according to a second embodiment.

FIG. 3 is a graph showing input / output characteristics of a half-wave rectifier circuit according to a second embodiment.

FIG. 4 is a circuit diagram of a half-wave rectifier circuit according to a third embodiment.

FIG. 5 is a circuit diagram of a half-wave rectifier circuit according to a fourth embodiment.

FIG. 6 is a circuit diagram of a half-wave rectifying adder according to a fifth embodiment.

FIG. 7 is a circuit diagram of a half-wave rectifying adder according to a sixth embodiment.

FIG. 8 is a circuit diagram of a half-wave rectifying subtractor according to a seventh embodiment.

FIG. 9 is a circuit diagram of a half-wave rectification adder / subtractor according to an eighth embodiment.

FIG. 10 is a circuit diagram of a conventional half-wave rectifier circuit.

FIG. 11 is a graph showing input / output characteristics of a conventional half-wave rectifier circuit.

FIG. 12 is a graph showing the influence of reverse current of a diode in a conventional half-wave rectifier circuit.

[Explanation of symbols]

1 operational amplifier, 2a input resistance, 2b feedback resistance, 2
c Ground resistance, 2d-2i resistance, 3a-3c diode, 4, 4a-4d input terminal, 5 output terminal, 6
Ideal characteristics, 7 Including the effect of diode reverse current

Claims (6)

(57) [Claims] 1. An inverting input terminal, a non-inverting input terminal and an output.
An operational amplifier having a terminal, and an input connected in series to an inverting input terminal of the operational amplifier.
A resistor and a first die having one end connected to the output terminal of the operational amplifier
And one end is connected to the inverting input terminal of the operational amplifier.
And a second and a third diode provided in parallel with the series circuit.
Between a series circuit consisting of over de one end and said second diode and said third diode
Connected to the third diode through the second diode.
The reverse current that is about to flow into the
And a resistor flowing to a non-inverting input terminal of the operational amplifier . 2. A power supply connected between the second and third diodes.
The other end of the resistor is connected to the ground point,
Having a resistance smaller than the reverse resistance of the diode.
The half-wave rectifier according to claim 1, wherein: 3. A power supply connected between the second and third diodes.
The other end of the resistor is connected to the ground point.
Ri, further connecting the second resistor having the feedback circuit and the parallel resistance value equal to the resistance value of said input resistor to the noninverting input terminal of the operational amplifier, characterized by being grounded second resistor claims Item 1. A half-wave rectifier according to Item 1 . 4. A non-inverting input terminal of the operational amplifier is connected to a second
Connect an input resistor and connect between the second and third diodes
And the other end of the resistor that is, a half-wave rectifier according to claim 1, characterized in that the connection between the second input resistor. 5. The half-wave rectifier according to claim 4 , wherein a resistance value of the second input resistor is equal to a parallel resistance value of the feedback circuit and the input resistor. 6. The operational amplifier according to claim 1, further comprising a plurality of input resistors connected to an inverting input terminal of the operational amplifier, wherein a plurality of input signals are input through the plurality of input resistors. 5. The half-wave rectifier according to any one of 5 .