JP3180376B2 - Low-frequency correction amplifier circuit - Google Patents

Description

DETAILED DESCRIPTION OF THE INVENTION

[0001]

BACKGROUND OF THE INVENTION 1. Field of the Invention The present invention relates to a low-frequency correction amplifying circuit for taking measures against sag of an amplifier or the like.

[0002]

2. Description of the Related Art It is desirable to increase the high-frequency cutoff frequency in order to improve the waveform rise of an amplifier or the like. However, if the frequency characteristics of the amplifier decrease in the low frequency range, the top of the rectangular wave is attenuated. The phenomenon of sag that tilts gradually occurs. It is necessary to improve the waveform transmission characteristics by applying the sag target.

FIG. 4 shows a circuit diagram of a conventional low-frequency correction amplifier circuit. 4, to the non-inverting input terminal of the differential amplifier (hereinafter referred to as "operational amplifier") OP1 (+), the AC input signal e _i from the AC source 2 is inputted via the capacitor C _1. The inverting input terminal (-), the reference voltage of the voltage Vcc was pressurized by the resistors R ₁ and R ₂ min from the DC power supply 3 is input via the resistor R _3. The resistor R ₂ capacitor C ₂ are connected in parallel, the resistance R ₄ is connected between the resistor R _1, a connection point of the R ₂ and the non-inverting input terminal.

On the other hand, feedback is applied to the inverting input terminal of the op-amp OP1 from the output terminal by resistors R ₅ and R ₆ , and an AC output signal is supplied from the output terminal to a load R _L (75Ω) via a coupling capacitor C _3. e _O is output.

Here, when the sag countermeasures are taken as described above.
Is the coupling capacitor C_ThreeIncrease the capacity of (for example,
470 μF or more), or as shown in FIG.
R_Five, R ₆Connection point and coupling capacitor C_ThreeAnd load R_L
And low-frequency correction capacitor C with frequency characteristics_FourTo
Connecting. For example, correction capacitor C_FourWhen connected
If the coupling capacitor C_ThreeIs about 100μF and vertical
A countermeasure for sag of the synchronization signal is taken.

Namely, in the circuit of FIG. 4 (R ₅ · R
₆ ) / (R ₅ + R ₆ ) ≫1 / JωC ₄ , R ₄ ≫1 / Jω
C ₁ , (R ₁ · R ₂ ) / (R ₁ + R ₂ ) ≫1 / JωC ₂
Then, the load voltage e _O output to the load R _L is as follows: e _O = R ₅ / R ₃ (1) The output voltage e _a of the operational amplifier OP1 is as follows: e _a = (R ₅ / R ₃ ) · {1 / (1 + JωC ₃ R _L )} (2) Therefore, the load voltage e ₀ is the frequency, is independent of the coupling capacitor C _3, the output voltage e _a is R according also to reduce the capacitance of the coupling capacitor C ₃ becomes a low frequency range
₅ / R ₃ , which is compensated in the low frequency range.

[0007]

[SUMMARY OF THE INVENTION However, in the case of not using the capacitor C ₄ for the low frequency correction coupling capacitor C ₃
The must with mass, Although the case of using the capacitor C ₄ for the low frequency correction can be a coupling capacitor C ₃ and the small-capacity, and require a capacitor C ₄ separately. Therefore, it is difficult to reduce the size of the device, and there is a problem that the cost cannot be reduced.

SUMMARY OF THE INVENTION The present invention has been made in view of the above problems, and has as its object to provide a low-frequency correction amplifier circuit that performs low-frequency compensation without adding a capacitor and with a small capacity of a coupling capacitor.

[0009]

An object of the present invention is to provide a DC via.
The signal supplied to the non-inverting input terminal
In a low-frequency correction amplifier circuit that differentially amplifies the signal and outputs an AC signal to a load via a coupling capacitor, a resistor is used at the inverting input terminal of the operational amplifier from the output side of the operational amplifier .
And a second feedback means for performing feedback from the load side using a resistor is provided at the inverting input terminal, and a constant current is applied to the resistance of the second feedback means.
This problem is solved by flowing a current and performing a DC bias on the inverting input terminal .

[0010]

As described above, output to the inverting input terminal of the operational amplifier is performed.
Apply feedback from the force side by the first feedback means using a resistor.
And a second resistor using a resistor at the inverting input terminal from the load side.
Of the second feedback means.
Apply a constant current to apply a DC bias to the inverting input terminal.
U.

Thus, the voltage gain of the operational amplifier becomes independent of the coupling capacitor connected to the output terminal. Further, the output voltage of the operational amplifier approaches an input voltage proportional to the voltage gain as the frequency becomes lower even if the capacitance of the coupling capacitor is reduced.

That is, it is possible to reduce the low-frequency region by adding a low-frequency correction capacitor and reducing the capacitance of the coupling capacitor, thereby making it possible to reduce the size and cost. .

[0013]

DESCRIPTION OF THE PREFERRED EMBODIMENTS Preferred embodiments of the present invention will be described below with reference to the drawings. FIG. 1 shows a circuit diagram of one embodiment of the present invention. The same components as those in FIG. 4 are denoted by the same reference numerals. The low-frequency correction amplifier circuit shown in FIG.
Type transistor Q1 constant current Ic ₁ flowing from the DC power supply 3 of the voltage Vcc of the PNP transistor Q2, Q3
In the collector current Ic ₂ and the reference voltage formed by resistors R ₁₀ of the transistor Q2 which form a current mirror,
AC source (e _i) 2 are superimposed removed AC signal DC component from the capacitor C _1, is input to the non-inverting input terminal of the operational amplifier OP1 is an operational amplifier (+).

The collector current I of the transistor Q3 is
c_ThreeIs a resistor R which is a first feedback means. ₁₁And the resistance R₁₂To
Shunted and input to the inverting input terminal of the operational amplifier OP1
Is done. Resistance R₁₁From the output terminal of the operational amplifier OP1
Apply negative feedback to the inverting input terminal.

Output voltage V of operational amplifier OP1_FourIs the join
Capacitor C_ThreeThe DC component is removed by the
_FiveAs load R_LSupplied to This load voltage V_FiveThe
Anti-R ₁₃, R₁₄The resistance R₁₂But
Connected. These resistors R₁₂, R₁₃, R₁₄Due to the load
R_LFeedback to the inverting input terminal of the operational amplifier OP1 from the
This constitutes a second feedback means. In this case, the resistance R₁₁, R
₁₂To balance the operational amplifier OP1.
Anti-R₁₂Ensures the DC bias of the operational amplifier OP1
Is what you do.

Next, the operation of the above circuit will be described.
Here, the signal is input to the non-inverting input terminal of the operational amplifier OP1.
Input voltage to V_Two, Input voltage input to the inverting input terminal
To V _Three, Resistance R₁₁The current flowing through₁, Resistance R₁₂The flow
Current I_TwoAnd Also, the inverting input of the operational amplifier OP1
The current flowing through the force terminal is i_-, The current flowing through the non-inverting input terminal
Flow i₊And

In this case, the balance condition of the DC amplifier is as follows: V ₂ = Ic ₂ · R ₁₀ (3) V ₃ = I ₂ [R ₁₂ + ｛R ₁₄ · (R ₁₃ + R _L ) / (R ₁₄ + R ₁₄ + R _L )｝] (4) Ic ₃ = I ₁ + I ₂ (5)

The voltage gain A of the operational amplifier OP1 is represented by A
≫1, offset voltage is 0, R ₁₂ ≫ ｛R ₁₄ · (R ₁₃ + R
_{L) / (R 14 + R} 13 + R L)}, i + «Ic 2, i - «
Ic ₃ .

[0019] Now, the voltage at the inverting input terminal when the open state since the Ic ₃ = I ₂ When V ₃₀ resistance R ₁₁ is not _{connected, V 30 = R 12 · I} 3 ... (6 ). If V ₂ −V ₃₀ = ΔVin (7), ΔVin = R ₁₀ · Ic ₂ −R ₁₂ · Ic ₃ (8)

Therefore, the resistance R₁₁And apply negative feedback
And V_Two= V_ThreeSince it operates to become
c_Two, Ic_ThreeAnd resistance R_Ten, R₁₂Raw by the variation of
Tying V_TwoAnd V_ThreeOutput voltage of the operational amplifier OP1
Force voltage V_FourIs V_Four= ｛-(R₁₁/ R₁₂) / ΔVin｝ + V_Two ... (9) In this case, the resistance R₁₁When not connecting R₁₁
= ∞ and the output voltage V _FourIs on the ground side or + power supply side
It does not work properly after swinging. That is, the resistance R ₁₁Ha
Plays the role of securing the DC bias of the pair amplifier OP1
I have.

[0021] Then, the DC gain A _{VD C} between the inverting input terminal and the non-inverting input terminal of the operational amplifier _{OP1, A VDC = ∂V 4 / ∂V} 2 = ∂V 4 / ΔVin = R 11 / R 12 ... (10 ). That is, the DC gain of the operational amplifier OP1 is determined only by the resistor R _11, R _12.

Next, the AC characteristics will be described. FIG.
FIG. 2 is a diagram for explaining the AC characteristics of FIG.
The circuit diagram of (A) is extracted from FIG.

[0023] Now, in the use frequency band of the circuit in FIG _{1, R 2 »1 / JωC 3} , R 11 »R 12, R 12 · i -
≪e _i . In FIG. 2A, the voltage e ₆ and the load voltage e ₅ divided by the resistors R ₁₃ and R ₁₄ are e ₆ = e _i (11) e ₅ = e _i ｛1+ (R ₁₃ / R ₁₄ )｝ (12), and the voltage gain A _V in this case is as follows: A _V = e ₅ / e _i = 1 + (R ₁₃ / R ₁₄ ) (13) (13) In this case, regardless of a fixed value and the capacitor C _3.

At this time, the output voltage e _{4 of the} operational amplifier OP1
Is R _L {R ₁₃ + {(R ₁₂ · R ₁₄ ) / (R ₁₂ + R ₁₄ )}
_{When, e 5 = e 4 · R} L / {(1 / JωC 3) + R L} ... a (14). Therefore, e ₄ = e ₅ · {(1 / JωC ₃ ) + _RL } / _RL = e ₅ · {(1 / JωC ₃ R _L ) +1} = A _V · e _i {(1 / JωC ₃ R _L) ) +1｝ (15) Therefore, e ₄ / e _i = A _V {1 / (1 + JωC ₃ R _L )} (16) An example of the frequency characteristic in this case is shown in FIG. Here, at a frequency of 10 Hz or less, the load voltage e ₅
Decreases because the operational amplifier OP1 is saturated.

As described above, according to the equations (10) and (16), even if the capacitance of the coupling capacitor C ₃ is reduced in the low frequency region (ω becomes smaller), e ₄ / e _i becomes equal to the voltage gain A _V. As they approach, low frequencies are compensated. Therefore, it is possible to reduce the size and cost of the amplification device by taking measures against sag.

FIG. 3 is a circuit diagram of another embodiment of the present invention.
Is shown. FIG. 3 shows the transistor Q in FIG.₁~ Q_Threeof
Instead of a current mirror, a resistor R₂₀, R_{twenty one}Added
Things. That is, the non-inverting input of the operational amplifier OP1
The reference voltage input to the terminal is₂₀, R_TenVoltage
The input to the inverting input terminal is_{twenty one}, R₁₁, R
₁₂It is not a constant current as shown in FIG.
Can perform the same operation as described above.

In the embodiment shown in FIG. 1, the PNP
Although the case where the constant current is supplied by the current mirror by the transistors Q1 to Q3 of the type is shown, the same operation can be performed even if the polarity of the power supply voltage is reversed by using the NPN type transistor.

[0028]

As described above, according to the present invention, the feedback is performed from the output side to the inverting input terminal of the operational amplifier by the first feedback means using a resistor, and the second feedback means using the resistor. There line feedback from the load side, the second feedback means
By applying a constant current to the resistor and applying a DC bias to the inverting input terminal, it is possible to perform low-frequency compensation without adding a capacitor and reducing the capacity of the coupling capacitor to take measures against sag. Can be reduced in size and cost.

[Brief description of the drawings]

FIG. 1 is a circuit diagram of one embodiment of the present invention.

FIG. 2 is a diagram for explaining the AC characteristics of FIG.

FIG. 3 is a circuit diagram of another embodiment of the present invention.

FIG. 4 is a circuit diagram of a conventional low-frequency correction amplifier circuit.

[Explanation of symbols]

 DESCRIPTION OF SYMBOLS 1 Operational amplifier 2 AC source 3 DC power supply 4 Current source

Claims (1)

(57) [Claims] 1. A power supply for a non-inverting input terminal which is DC-biased.
In a low-frequency correction amplifier circuit for differentially amplifying a supplied signal with an operational amplifier and outputting an AC signal to a load via a coupling capacitor, a resistor is connected to an inverting input terminal of the operational amplifier from an output side of the operational amplifier. provided with a first feedback means for performing feedback using, to the inverting input terminal, a second feedback means for performing feedback by the resistance from the load side is provided, a constant current to the resistor of the second feedback means And the reverse
A low-frequency amplifier circuit characterized in that a DC bias is applied to a force terminal .