JPH05121959A - Low frequency range correction amplifier circuit - Google Patents

Abstract

PURPOSE:To make the capacitance of a coupling capacitor small and to compensate a low frequency range without addition of a capacitor with respect to the low frequency range correction amplifier circuit for sag countermeasure for an amplifier or the like. CONSTITUTION:An output voltage V4 of an operational amplifier OP1 is fed back negatively to an inverting input terminal of the operational amplifier OP1 via a resistor R11. Moreover, feedback is applied to an inverting input terminal with a voltage resulting from dividing a load voltage V5 with resistors R12, R13 and R14.

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 amplifier circuit for taking measures against sag in an amplifier or the like.

[0002]

2. Description of the Related Art It is desirable to increase the high cutoff frequency in order to improve the rising of the waveform of an amplifier or the like. However, if the frequency characteristic of the amplifier is lowered in the low range, the top of the rectangular wave is attenuated. A phenomenon called sag that tilts in a vertical direction occurs. It is necessary to improve the waveform transfer characteristic by subjecting this sag.

FIG. 4 shows a circuit diagram of a conventional low-frequency correction amplifier circuit. In FIG. 4, the AC input signal e _i from the AC source 2 is input to the non-inverting input terminal (+) of the differential amplifier (hereinafter referred to as “op amp”) OP1 via the capacitor C ₁ . The reference voltage obtained by dividing the voltage Vcc from the DC power supply 3 by the resistors R ₁ and R ₂ is input to the inverting input terminal (−) 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, the inverting input terminal of the op-amp OP1 is fed back by resistors R ₅ and R ₆ from the output terminal, and an AC output signal from the output terminal to the load R _L (75Ω) via the coupling capacitor C _3. e _O is output.

Here, when the sag countermeasure is taken as described above
Is the coupling capacitor C₃Increase the capacity of (eg
470 μF or more), or as shown in FIG.
R_Five, R ₆Connection point and coupling capacitor C₃And load R_L
A low-frequency correction capacitor C with frequency characteristics at the connection point_FourTo
Connecting. For example, a correction capacitor C_FourIf you connect
Coupling capacitor C₃The capacity is about 100μF and vertical
Countermeasures for sag of sync signal.

Namely, in the circuit of FIG. 4 (R ₅ · R
₆ ) / (R ₅ + R ₆ ) >> 1 / Jω C 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 e _O = R ₅ / R ₃ (1) Further, the output voltage e _a of the operational amplifier OP1 is, e _a = become _{(R 5 / R 3) ·} {1 / (1 + JωC 3 R L)} ... (2). Therefore, the load voltage e ₀ is irrelevant to the frequency and the coupling capacitor C ₃ , and the output voltage e _a becomes R in the low frequency region even if the capacitance of the coupling capacitor C ₃ is reduced.
It comes close to ₅ / R ₃ and is compensated in the low range.

[0007]

However, when the capacitor C ₄ for low frequency correction is not used, the coupling capacitor C ₃ is used.
Must have a large capacity, and when the low-frequency correcting capacitor C ₄ is used, the coupling capacitor C ₃ can have a small capacity, but a separate capacitor C ₄ is required. Therefore, it is difficult to reduce the size of the device, and it is not possible to reduce the cost.

Therefore, the present invention has been made in view of the above problems, and an object of the present invention is to provide a low-frequency correction amplifier circuit which does not add a capacitor and uses a coupling capacitor as a small capacity to perform low-frequency compensation.

[0009]

SUMMARY OF THE INVENTION In a low-frequency correction amplification circuit for outputting an AC signal from a operational amplifier to a load via a coupling capacitor, the above-mentioned problem is obtained by connecting one input terminal of the operational amplifier to the operational amplifier. First feedback from the output side
This is solved by providing the second feedback means for feedback from the load side to the input terminal together with the second feedback means.

[0010]

As described above, the feedback is applied to the one input terminal of the operational amplifier by the first feedback means. This feedback ensures the DC bias of the operational amplifier. Further, feedback is applied to the input terminal by the second feedback means.

As a result, the voltage gain of the operational amplifier is independent of the coupling capacitor connected to the output terminal. Further, the output voltage of the operational amplifier approaches the 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 capacitance of the coupling capacitor and lower the low frequency region without adding a capacitor for low-frequency correction, and it is possible to reduce the size and cost. ..

[0013]

DETAILED DESCRIPTION OF THE PREFERRED EMBODIMENT A preferred embodiment of the present invention will be described below with reference to the drawings. FIG. 1 shows a circuit diagram of an embodiment of the present invention. The same components as those in FIG. 4 are designated by the same reference numerals. The low-frequency correction amplification circuit of FIG.
The constant current Ic ₁ flowing from the DC power source 3 of the voltage Vcc to the P-type transistor Q1 and the PNP-type transistors Q2 and Q3.
In the reference voltage formed by the collector current Ic _{2 of the} transistor Q2 and the resistor R ₁₀ which form the current mirror with
The AC signal from which the DC component has been removed by the capacitor C ₁ from the AC source (e _i ) 2 is superimposed and input to the non-inverting input terminal (+) of the operational amplifier OP1 which is an operational amplifier.

Further, the collector current I of the transistor Q3
c₃Is the resistor R which is the first feedback means ₁₁, And resistance R₁₂To
It is shunted by and is input to the inverting input terminal of operational amplifier OP1.
To be done. Resistance R₁₁Is from the output terminal of the operational amplifier OP1
Apply negative feedback to the inverting input terminal.

Output voltage V of the operational amplifier OP1_FourJoins
Capacitor C₃DC component is removed by the additional voltage V
_FiveAs load R_LIs supplied to. This load voltage V_FiveTo
Anti-R ₁₃, R₁₄At the connection point divided by₁₂But
Connected. These resistance R₁₂, R₁₃, R₁₄Due to the load
R_LFeedback from the side to the inverting input terminal of operational amplifier OP1
It constitutes a second feedback means. In this case, the resistance R₁₁, R
₁₂The op amp OP1 is balanced by
Anti-R₁₂Secures the DC bias of the operational amplifier OP1
To do.

Next, the operation of the above circuit will be described.
Here, it is input to the non-inverting input terminal of the operational amplifier OP1.
Input voltage V₂, Input voltage input to the inverting input terminal
To V ₃, Resistance R₁₁The current flowing through₁, Resistance R₁₂Flow through
Current I₂And Also, the inverting input of the operational amplifier OP1
The current flowing through the force terminal is i_-, The current flowing to the non-inverting input terminal
Flow i₊And

The balance condition of the DC amplifier in this case is: V ₂ = Ic ₂ · R ₁₀ (3) V ₃ = I ₂ [R ₁₂ + {R ₁₄ · (R ₁₃ + _RL ) / (R ₁₄ + R ₁₄ + _RL )}] (4) Ic ₃ = I ₁ + I ₂ (5).

Further, the voltage gain A of the operational amplifier OP1 is set to A
>> 1, the offset voltage _{0, R 12 »{R 14} · (R 13 + R
_L ) / (R ₁₄ + R ₁₃ + R _L )}, i ₊ << Ic ₂ , i _- <<
Ic ₃ .

Now, assuming that the voltage of the inverting input terminal when the resistor R ₁₁ is not connected and is in the open state is V ₃₀ , Ic ₃ = I _2, and therefore V ₃₀ = R ₁₂ · I ₃ (6 ) Becomes. Further, if V ₂ −V ₃₀ = ΔVin (7), then ΔVin = R ₁₀ · Ic ₂ −R ₁₂ · Ic ₃ (8).

Therefore, the resistance R₁₁And apply negative feedback
And V₂= V₃Because it operates as
c₂, Ic₃And resistance R_Ten, R₁₂Raw due to variations in
V₂And V₃Of the operational amplifier OP1 due to the voltage difference between
Force voltage V_FourIs V_Four= {-(R₁₁/ R₁₂) / ΔVin} + V₂ … (9). In this case, the resistance R₁₁R when not connected₁₁
= ∞ and output voltage V _FourOn the ground side or + power supply side
It shakes off and does not operate normally. That is, the resistance R ₁₁Oh
Plays the role of securing the DC bias of the amplifier OP1
There is.

[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 ) Becomes. 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. Figure 2
2 is a diagram for explaining the AC characteristics of FIG. 1, and 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 divided by the resistors R ₁₃ and R ₁₄ and the load voltage e ₅ are as follows: e ₆ = e _i (11) e ₅ = e _i {1+ (R ₁₃ / R ₁₄ )} ... (12) and the voltage gain a _V in this _{case, a V = e 5 / e} i = 1 + (R 13 / R 14) ... a (13). In this case, the equation (13) has a constant value regardless of 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 RL} ) +1} = A _V · e _i {(1 / JωC _{3 RL} ) +1} (15) Therefore, e ₄ / e _i = A _V {1 / (1 + JωC _{3 RL} )} (16) An example of frequency characteristics in this case is shown in FIG. Here, at a frequency of 10 Hz or less, the load voltage e ₅
Is decreased 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 small), e ₄ / e _i becomes the voltage gain _AV . It comes closer and the low range is compensated. Therefore, it is possible to reduce the size and cost of the amplification device by taking sag measures.

Next, FIG. 3 is a circuit diagram of another embodiment of the present invention.
Indicates. FIG. 3 shows the transistor Q in FIG.₁~ Q₃of
Resistor R instead of current mirror₂₀, R_{twenty one}Added
It is a thing. That is, the non-inverting input of the operational amplifier OP1
The reference voltage input to the terminal is the resistance R₂₀, R_TenTo the voltage of
In addition, the input to the inverting input terminal is a resistor R_{twenty one}, R₁₁, R
₁₂It is a voltage corresponding to, and not a constant current as shown in Fig. 1.
Also, the same operation as described above can be performed.

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

[0028]

As described above, according to the present invention, one input terminal of the operational amplifier is fed back from the output side by the first feedback means and fed back from the load side by the second feedback means. As a result, it is possible to perform the low-frequency compensation for the sag by reducing the capacitance of the coupling capacitor without adding a capacitor, and it is possible to reduce the size and cost of the device.

[Brief description of drawings]

FIG. 1 is a circuit diagram of an 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]

 1 operational amplifier 2 alternating current source 3 direct current power source 4 current source

Claims (1)

[Claims] 1. A low-frequency correction amplification circuit for outputting an AC signal to a load from an operational amplifier via a coupling capacitor, wherein feedback is provided to one input terminal of the operational amplifier from the output side of the operational amplifier. And a second feedback means for feeding back from the load side to the input terminal.