JPH0361394B2 - - Google Patents

Description

[Detailed Description of the Invention] [Summary] Consists of three mirror circuits for supplying current from the ground side to the B line and three mirror circuits for drawing current from the A line to the battery side. , a resistor and a capacitor are connected so that it has high impedance for differential signals and low impedance for common mode signals, and a stabilized power supply is connected to the third mirror circuit on the battery side. This is a power supply circuit with a configuration that eliminates the influence of power supply noise and enables large-scale integration.

[Industrial application field]

The present invention relates to a power supply circuit in a subscriber circuit of an exchange.

A power supply circuit having a function of supplying communication current to a subscriber is made smaller by using an electronic circuit instead of having a large structure including a transformer or the like. Recently, there has been a demand for further miniaturization and cost reduction for such electronic power supply circuits.

[Conventional technology]

As a conventional electronic subscriber circuit, for example, the configuration shown in FIG. 7 or FIG. 8 is known. In each figure, OP ₀ and OP ₁ are operational amplifiers,
Q ₀ , Q ₁ are transistors, C ₀ , C ₁ , C _AB are capacitors, Ra ₀ , Rb ₀ , Re ₀ , Ra ₁ , Rb ₁ , Re ₁ are resistors,
_VBB is the power supply voltage, and A and B are subscriber lines.

The power supply resistance in the power supply circuit shown in Fig. 7 is B
On the line side, Z _B = (Ra ₁ + Ra ₁ / Ra ₁ ) · Re ₁ ... (1), and on the A line side, Z _A = (Ra ₀ + Rb ₀ / Ra ₀ ) · Re ₀ ... It is expressed as (2). Therefore, the power supply resistance is determined by the absolute accuracy of each resistor.

In addition, capacitors C ₀ and C ₁ are connected to the non-inverting input terminals of operational amplifiers OP ₀ and OP ₁ , and in the frequency band used, Rb ₀ ≫1/jωC ₀ and Rb ₁ ≫1/
In the case of the condition jωC ₁ , the AC signal is bypassed by the capacitors C ₀ and C ₁ , so the AC signal at the non-inverting inputs of the operational amplifiers OP ₀ and OP ₁ can be regarded as almost 0. I can do it.

The output terminals of the operational amplifiers OP ₀ , OP ₁ are connected to the bases of the transistors Q ₀ , Q ₁ , and the emitters of the transistors Q ₀ , Q ₁ are connected to the bases of the operational amplifiers OP ₀ , OP ₁ .
The resistors Re ₀ and Re ₁ are connected between the emitter and the power supply or ground. Therefore, if the voltage at the non-inverting input terminals of operational amplifiers OP ₀ and OP ₁ is V ₊ , then transistors Q ₀ ,
As the emitter current of Q ₁ , V ₊ /Re ₀ , V ₊ /Re ₁
current flows. If the current amplification factor h _FE of the transistors Q ₀ and Q ₁ is large compared to 1 (h _FE ≫ 1), and the non-inverting inputs of the operational amplifiers OP ₀ and OP ₁ are approximately 0 as described above, then , transistors Q ₀ , Q ₁
No alternating current flows through.

Therefore, in this power supply circuit, resistances Rb ₀ and Rb ₁ appear in response to the AC signal, and from the condition that Rb ₀ and Rb ₁ ≫1,
It becomes a high impedance circuit. In this way, by forming a high impedance circuit for AC signals, there is no attenuation for speech signals, and noise from the power supply V _NBB is reduced by A,
Assuming that the impedance terminated on the B line is Z _L ,
It can be attenuated to the relationship of {Z _L /(Z _L +Rb ₀ +Rb ₁ )}V _NBB .

In the power supply circuit shown in Fig. 8, the capacitor C _AB is connected between the non-inverting input terminals of operational amplifiers OP ₀ and OP ₁ , and the DC power supply characteristics are similar to the power supply circuit shown in Fig. 7. , and a differential signal (speech signal) is applied to the capacitor C _AB in opposite phase, so the non-inverting inputs of the operational amplifiers OP ₀ and OP ₁ become almost 0, as shown in Fig. 7. Like the power supply circuit shown in , it has high impedance. Furthermore, for common-mode signals, since the common-mode voltage is applied across capacitor C _AB , the situation is the same as when capacitor C _AB is not connected. Therefore, for common mode signals, the impedance is reduced to a level equal to the feed resistances Z _A and Z _B.

[Problem that the invention seeks to solve]

In the conventional power supply circuit shown in FIG. 7, since the power supply resistors Z _A and Z _B each have high impedance independently, when common mode noise is superimposed on the line, it is not attenuated and the transistors Q ₀ , The collector-emitter voltage V _CE of Q ₁ will rise and fall together according to the level of its common mode noise. Therefore,
Due to the high level of noise, the collector
When the emitter voltage V _CE is saturated, a normal differential signal (speech signal) is clipped and becomes a no-signal state while the emitter voltage V CE is saturated, which has the drawback of significantly degrading the transmission characteristics. Furthermore, since the precision of the power supply resistors Z _A and Z _B depends on the absolute precision of each resistor, there is a drawback that it is not easy to integrate these resistors into an integrated circuit.

In addition, in the conventional power supply circuit shown in Fig. 8, as mentioned above, the power supply resistance Z _A ,
The impedance is reduced to be equal to Z _B. Therefore,
Common mode noise will be attenuated. However, there is a problem in that it is greatly affected by the noise V _NBB added from the power supply side. The noise V _NBB is
If Ra ₁ , Ra ₀ ≫1/jωC _AB , then the operational amplifier
The non-inverting input terminals of OP ₀ and OP ₁ each have V ₊₀ .
= (1-k)・V _NBB and V ₊₁ =k・V _NBB (however, 0
<k<1) is input.

Therefore, the current flowing through transistors Q ₀ and Q ₁ is
I _Re0 = V ₊₀ /Re ₀ and I _Re1 = V ₊₁ /Re ₁ , respectively. Here, if Rb ₁ =Rb ₀ >>1, the current flowing through the transistors Q ₀ and Q ₁ becomes I _Re0 =I _Re1 .

Furthermore, if Re ₀ = Re ₁ , when k = 1/2, and the subscript is omitted, I _Re = V _NBB /2・Re
Then, the impedance terminated on A and B lines is
If Z _L , then due to the noise V _NBB from the power supply,
A noise voltage of Z _L・V _NBB /2・Re appears at the terminal.
Therefore, the power supply circuit has no noise suppression effect.

Countermeasures against such noise have been proposed, but since they require external circuit elements, they have the drawback of being difficult to implement into large-scale integrated circuits.

An object of the present invention is to improve the above-mentioned conventional drawbacks and provide a power supply circuit suitable for large-scale integrated circuit LSI.

[Means for solving problems]

_The power supply circuit of the present invention will be explained with reference to FIG. Convert voltages V B and _{V A} between the first to third mirror circuits B0 to B2, A0 to A2 on the side and battery side, and the B line and the earth and between the A line and _the battery into currents. A pair of resistors connected between the resistors Rb ₁ and Rb ₀ and the input terminals of the third mirror circuit B2 and A2
The output terminal of the first mirror circuit _B0 on the earth side is connected to the _B line _, and the input terminal is connected to the third mirror circuit A2 on the battery side.
connect the output terminal of the first mirror circuit A0 on the battery side to the A line, connect the input terminal to the output terminal of the third mirror circuit B2 on the earth side,
A pair of resistors Rc ₁ and Rc ₀ are connected to the output terminals of the second mirror circuits B1 and A1, whose input terminals are connected to the resistors Rb ₁ and Rb ₀ , respectively, and the input terminals of the third mirror circuits B2 and A2. cross-connect through the battery side third
A stabilized power supply (voltage V _Z ) is connected to the mirror circuit A2.

Further, the first mirror circuits B0 and A0 include transistors Q ₁ and Q ₀ , operational amplifiers OP ₁ and OP ₀ , and resistors Ra ₁ ,
It is composed of Ra ₀ and is supplied with power to the B line and A line via transistors Q ₁ and Q ₀ .

[Effect]

The voltages on the B line and A line are converted into current by the resistors Rb ₁ and Rb ₀ , and the current flows to the output terminals of the second mirror circuits B1 and A1. Correspondingly, current flows through the output terminals of the third mirror circuits B2 and A2, and current is supplied from the first mirror circuits B0 and A0. At that time, the differential signal (call signal) is bypassed by the capacitor C _AB , so it is not input to the third mirror circuit B2, A2, and the differential signal is input to the first mirror circuit B0, A
Since 0 supplies current regardless, it presents a high impedance to differential signals.

In addition, with respect to the common-mode signal (inductive noise), the circuit operates in the same way as a circuit without the capacitor C _AB connected, and an alternating current component due to the common-mode signal flows through the third mirror circuit B2, A2. Therefore, the first
Since an alternating current component also flows through the mirror circuits B0 and A0, they exhibit low impedance with respect to the common mode signal.

Furthermore, by connecting a stabilized power supply to the common terminal of the third mirror circuit A2, power supply noise can be attenuated.

〔Example〕

Embodiments of the present invention will be described in detail below with reference to the drawings.

FIG. 1 is a block diagram of an embodiment of the present invention.
The first to third mirror circuits B0, B1, B2 for supplying current to the B line from the earth side, and the A to the battery side.
First to third mirror circuits A0, A1, A2 for drawing current from the line, and resistors Rb ₁ , Rb ₀ , Rc ₁ ,
It is composed of Rc ₀ and a capacitor C _AB , and the battery voltage V _BB is, for example, -48V. In addition, the voltage V _Z of the stabilized power supply is connected to the common terminal of the third mirror circuit on the battery side.
has been added.

The mirror circuits B0 and A0 on the earth side and battery side are as follows:
As in the conventional example shown in FIGS. 7 and 8, operational amplifiers OP ₁ , OP ₀ , transistors Q ₁ , Q ₀ , resistors Ra ₁ ,
Composed of Ra ₀ , Re ₁ , Re ₀ , operational amplifier
By using the non-inverting input terminals of OP ₁ and OP ₀ as input terminals and the collectors of transistors Q ₁ and Q ₀ as output terminals, it becomes equivalent to a normal mirror circuit. In addition,
This circuit has an input-to-output current gain of Ra ₁ /
Re ₁ , Ra ₀ /Re ₀ .

Also, connect the output terminal of the first mirror circuit B0 on the earth side to the B line, connect its input terminal to the output terminal of the third mirror circuit A2 on the battery side, and connect the output terminal of the first mirror circuit B0 on the battery side to the B line.
Connect the output terminal of the mirror circuit A0 to the A line, connect its input terminal to the output terminal of the third mirror circuit B2 on the earth side, and connect the output terminal of the mirror circuit A0 on the earth side to the second mirror circuit B1 on the earth side.
A resistor Rb ₁ is connected between the input terminal and the B line, and its output terminal is connected to the input terminal of the third mirror circuit A2 on the battery side via the resistor Rc ₀ .
A resistor is connected between the input terminal of the mirror circuit A1 and the A line.
Rb ₀ is connected, and its output terminal is connected to the input terminal of the third mirror circuit B2 on the earth side through the resistor Rc ₁ , and the input terminal of the third mirror circuit B2 on the earth side and the battery side are connected. A series circuit of resistors Rc ₁ and Rc ₀ and a capacitor C _AB is connected between the input terminal of the third mirror circuit A2 and the input terminal of the third mirror circuit A2.

As shown in FIG. 2a, the mirror circuit has an input terminal IN, an output terminal OUT, and a common terminal C, and has the configuration shown in FIG. 2b or c. b
In this circuit, the bases of the transistor Qa connected to the input terminal IN and the transistor Qb connected to the output terminal OUT are commonly connected to the input terminal IN, and the emitters of the transistors Qa and Qb are connected to the resistor.
It is connected to the common terminal C via R ₁ and R ₂ .
In addition, the circuit shown in c connects a transistor Qc for compensating the base currents of transistors Qa and Qb to the circuit shown in b, and further connects transistors Qa and Qb to the circuit shown in b.
A transistor Qd is connected to compensate for the shortage of the current amplification factor h _FE and to suppress fluctuations due to the Early effect of the transistor Qb. The second and third mirror circuits B1, B2, A1, and A2 on the earth side and battery side in FIG. 1 can use the circuits shown in b or c in FIG. 2, and are as follows. A case where the current ratio is 1 will be explained.

The voltage V _{B of} the B line is
A current of V _B /Rb ₁ will be passed through the second mirror circuit B1, and if the internal resistance value of this second mirror circuit B1 is included in the resistor Rb ₁ and the collector-emitter voltage of the internal transistor is ignored, then Since the current ratio of the current I _Rb1 = V _B /Rb ₁ flowing to the input terminal is 1, the same current also flows to the output terminal, and is connected to the third mirror circuit on the battery side via the resistor Rc ₀ . It flows to the input terminal of A2. Since this third mirror circuit A2 also has a current ratio of 1, the current flows from its output terminal to the resistor Ra ₁ of the first mirror circuit B0 on the earth side. Therefore, the power supply resistance Z _B on the B line side becomes = Rb ₁ ·Re ₁ /Ra ₁ . Similarly, for the power supply resistance Z _A on the A line side, Z _A =Rb ₀ ·Re ₀ /Ra ₀ .

The differential signal (call signal) of voltages va and vb is connected to the A line,
When applied to the B line, currents of i _Rb1 = vb/Rb ₁ and i _Rb0 = vb/Rb ₀ flow through the resistors Rb ₁ and Rb ₀ , respectively. This alternating current flows through resistors Rc ₀ and Rc ₁ similarly to the above-mentioned direct current. Since the input terminal of the mirror circuit is considered to be a type of diode, the input impedance of the third mirror circuit B2, A2 is Z ₃
Assuming that Z ₃ <<Rc ₁ , Rc ₀ , then in a circuit where the capacitor C _AB is not connected, AC voltages of i _Rb1 ×Rc ₀ and i _Rb0 ×Rc ₁ are generated across the resistors Rc ₀ and Rc ₁ .

Since the capacitor C _AB is connected between the resistors Rc ₀ and Rc ₁ and the alternating current voltages generated across the resistors Rc ₀ and Rc ₁ have opposite phases, an alternating current flows through the capacitor C _AB . Therefore, the capacitor C _AB and the resistors Rc ₁ and Rc ₀ constitute a kind of high-frequency blocking filter, and the alternating current flowing to the input terminals of the third mirror circuits B2 and A2 is almost zero. becomes. Therefore, since no AC component is input to the second mirror circuits B1 and A1, no AC component is input to the first mirror circuits B0 and A0, and as in the conventional example shown in FIG. High impedance to dynamic signals.

In addition, for the in-phase signal, since the in-phase AC voltage is applied to both ends of the capacitor C _AB , it is equivalent to the case where the capacitor C _AB is not connected, so the third mirror circuit B2, A2 An alternating current flows through the input terminal of the first mirror circuit B.
Since AC components also flow through 0 and A0, the impedance for the common-mode signal is the power supply impedance.
Z _B and Z _A.

Further, the voltage _VZ applied to the common terminal of the third mirror circuit A2 on the battery side is approximately the same value as the voltage _VBB , but is stabilized. For example, the third
It is supplied from a stabilized power source, an example of which is shown in the figure. In the figure, Q ₂ , Q ₃ , and Q ₄ are transistors, D ₀ is a diode, R ₃ and R ₄ are resistors, and Dn is a Zener diode connected in series, forming a modified mirror circuit. There is. Corresponding to the current flowing in resistor R ₃ , transistor Q ₃ and diode D ₀ by supply voltage V _BB , Zener diode Dn,
Current flows through the transistor Q ₂ and the resistor R ₄ , and a constant voltage is applied to the base of the transistor Q ₄ by a plurality of cascade-connected Zener diodes Dn, and the stabilized voltage V _Z from the transistor Q ₄ is applied to the battery side. It is supplied to the common terminal of the third mirror circuit A2.

When noise is superimposed on the power supply voltage V _BB , a current I _N = V _NBB /Rb ₀ flows through the resistor Rb ₀ due to the noise V _NBB , and this noise current I _N flows to the output terminal of the second mirror circuit A1. . This noise current I _N is a resistance
As mentioned above, _{Rc 1 =} Rc ₀ ≫
Because of the relationship (1/jωC _AB ), this noise current I _N flows through the third mirror circuits B2 and A2, respectively, with the same current of 1/2· _IN . This current finally flows to the input terminals of the first mirror circuits B0 and A0, and the transistors Q ₁ and Q ₀ have
I _BN = 1/2・I _N・Ra ₁ /Re ₁ , I _AN = 1/2・I _N・
A noise current of Ra ₀ /Re ₀ flows. This noise current only generates a common mode voltage on the A line and B line, and does not become differential noise. Therefore, it will not affect the call signal.

Also, the noise current flowing through the A line through the resistor Rb ₀ is the main noise source, but its value is _equal to the resistance
Rb ₀ and Rb ₁ attenuate in proportion to the value of Z _L /(Rb ₀ +Z _L +Rb ₁ ), resulting in a sufficiently small value.

Further, as the resistors Rb ₁ and Rb ₀ , for example, a current limiting circuit shown in FIG. 4 can be used. In the figure, r _c and r _e are resistors, and Q ₅ and Q ₆ are transistors. Transistor Q ₅ to which base current is supplied via resistor r _c is in an on state, and this transistor Q ₅
The on-resistance and resistance r _e appear as the resistance of the current limiting circuit. When the voltage applied across this circuit increases, the voltage _Vre across resistor r _e increases, which causes the base potential of transistor Q ₆ to rise, causing transistor Q ₆ to start operating. Ru. Therefore, the current I _C flowing to the collector of the transistor Q ₅ is limited to a value expressed by I _C =V _BE6 / _re , where the base-emitter voltage of the transistor Q ₆ is V _BE6 . FIG. 5 is an explanatory diagram of the characteristics of this current limiting circuit, in which the current I
rise is limited. Therefore, when the line resistance is small, it is possible to limit the power supply current.

In the circuit configuration shown in FIG. 1, the A-line side and the B-line side are symmetrical, so the operation and characteristics of the B-line side will mainly be supplementarily explained.

When a voltage V _B is applied to the B line, a current of V _B /Rb ₁ is input to the mirror circuit B1 through the resistor Rb ₁ .
Here, for the sake of simplicity, the input resistance of the mirror circuit B1 is assumed to be 0. Note that although the actual input resistance of the mirror circuit B1 is not 0, since it is connected in series with the resistor Rb ₁ in terms of the circuit, it can be considered to be included as a part of the resistor Rb ₁ . Also, the output resistance of the mirror circuit is very large and can be made infinite for simplicity.

Since the current input to the mirror circuit B1 has a mirror ratio of 1, the same output current as the input current is input to the mirror circuit A2 through the resistor _Rc0 . Similarly, in this mirror circuit A2, the same output current as the input current is input to the mirror circuit B0. Therefore, as mentioned above, the current flowing through the resistor Rb ₁ is I _Rb1 , the input current of mirror circuit B0 is I _B0i , its output current is I _B00 , the mirror ratio of mirror circuit B1 is β1, and the mirror ratio of mirror circuit A2 is Assuming α2, I _Rb1 = V _B /Rb ₁ I _B0i = I _Rb1 × β1 × α2 = I _Rb1 (However, β1 = 1, α2 = 1) I _B00 = I _B0i × Ra ₁ /Re ₁ = I _Rb1 × Ra ₁ /Re ₁ = (V _B /Rb ₁ ) x Ra ₁ /Re ₁ = V _B x Ra ₁ / (Rb ₁ x Re ₁ ) Here, the resistance Z _B of the power supply circuit on the B line side is the applied voltage _{Since it is obtained from the relationship between _ _ _ _} (Rb ₁ ×Re ₁ )]=Rb ₁ ×Re ₁ /Ra ₁ . Similarly, the resistance Z _A of the power supply circuit on the A line side can be found, Z _A ≒ V _A / I _A00 = V _A / [V _A × Ra ₀ / (Rb ₀ × Re ₀ )] = Rb ₀ × Re ₀ /Ra ₀ .

Furthermore, an AC signal such as an audio signal in the power supply circuit is applied in a form superimposed on a DC signal, and the mirror circuit operates with the same mirror ratio as for the DC signal with respect to this AC signal. For example, when an AC voltage of vb is applied to the B line, the current vb/Rb ₁ is the mirror circuit B1
input and output with a mirror ratio of 1. Similarly, if an AC voltage of va is applied to the A line, the current va/
Rb ₀ is input to the mirror circuit A1 and output with a mirror ratio of 1.

If the aforementioned AC voltages va and vb are in phase, the resistance
The currents flowing through Rc ₀ and Rc ₁ have the same value; one direction increases the direct current, and the other direction decreases it. Therefore, assuming that there is no capacitor C _AB , voltages of vb×Rc ₀ /Rb ₁ and va×Rc ₁ /Rb ₀ are generated at the connection point of capacitor C _AB , and these voltages have the same phase. Therefore, there is no alternating potential difference between both ends of capacitor C _AB . Since there is no alternating current potential difference,
Even if capacitor C _AB is connected, no current will flow through it, and the configuration will be the same as without capacitor C _AB , making it equivalent to a DC circuit. That is, the alternating current impedances zb and za between the B line side and the A line side are zb≒Rb ₁ ×Re ₁ /Ra ₁ za≒Rb ₀ ×Re ₀ /Ra ₀ .

Also, when the AC voltages va and vb are differential (opposite phase), assuming that there is no capacitor C _AB , the connection point of the capacitor C _AB has vb×Rc ₀ /Rb ₁ , respectively.
A voltage of va×Rc ₁ /Rb ₀ is generated differentially. here,
If the capacitance of capacitor C _AB is sufficiently large, its AC impedance Z _c is Z _c = 1/2πfC _AB ≪Rc ₀ ,
If Rc is ₁ , almost all of the alternating current output from the mirror circuits B1 and A1 will pass through the capacitor C _AB .

Therefore, the mirror circuit A is passed through the resistors Rc ₀ and Rc ₁ .
2, it will no longer be input to B2. That is, capacitor C _AB operates as a kind of AC removal filter. Also, since no alternating current is input to the mirror circuits A2 and B2, no alternating current is input to the mirror circuits A0 and B0.
AC is not output from. This shows that even if a differential voltage is applied, the current does not flow, and the impedance zb, za for the differential signal is zb=V _B / (I _B00 + I _Rb1 ) = V _B / (0 + V _B /Rb ₁ )=Rb ₁ >>1 za=V _A /(I _A00 +I _Rb0 )=V _A /(0+V _A /Rb ₀ )=Rb ₀ >>1. In other words, the impedance becomes extremely high.

Further, the voltage V _Z of the stabilized power supply does not include power supply noise, and the power supply noise is included in the voltage V _BB . There is a need for a function that prevents this power supply noise from being output as a differential signal to the B line and A line. Note that if the noise is in phase, there is no problem because the subscriber cannot hear it as noise. In addition, mirror circuit A
Even if the power supply voltage _VBB is applied to 0 and A1, noise does not mix there.

For example, there is noise between the A line and the power supply with voltage V _BB .
When V _NBB is present, the resistor Rb ₀ causes the mirror circuit A1 to
A noise current I _N of V _NBB /Rb ₀ is input. The mirror circuit A1 is not directly affected by the noise contained in the power supply voltage V _BB , but if the input current contains noise, it will output a current proportional to it, and the noise current I _N is passed through the resistor Rc ₁ . Mirror circuit B
Trying to enter 2.

However, since the AC impedance of the capacitor C _AB is smaller than that of the resistors Rc ₀ and Rc ₁ , the noise current _IN is shunted to the resistors Rc ₀ and Rc ₁ and sent to the mirror circuit A.
2. Input in B2. At that time, the voltages across the capacitor C _AB will be in phase, so the voltages caused by the noise V _NBB generated on the B and A lines due to the noise current I _N will be in the same phase, and as mentioned above, this will not be a problem for the subscriber. It turns out.

Note that if the voltage V _Z of the stabilized power supply contains noise, the noise current will flow through the capacitor C _AB , and therefore the capacitor C _AB
This poses a problem because the voltages at both ends of the line are differential, and the voltages caused by noise generated on the B and A lines are differential.
That is, the stabilized power supply needs to have a configuration that contains as little noise as possible.

Also, if the resistors Rb ₀ and Rb ₁ are replaced with a current limiting circuit as shown in Fig. 4, this current limiting circuit becomes
Since it has a function of increasing the resistance value when a certain voltage or more is applied, when the voltage exceeds a certain value, it is equivalent to increasing the resistances Z _B and Z _A of the power supply circuit. That is, as mentioned above, since Z _B ≒ Rb ₁ ×Re ₁ /Ra ₁ Z _A ≒Rb ₀ ×Re ₀ /Ra ₀ is expressed, increasing the resistances Rb ₀ and Rb ₁ means that the resistance Z _A , Z _B increases. Considering these characteristics as a subscriber circuit, when the line resistance is small (when the subscriber is close), the voltage drop due to the line resistance is small, and therefore the voltages V _A and V in Figure 1 This is a state in which _B increases,
In this state, the resistances Rb ₀ and Rb ₁ increase, so that the power supply resistance increases, thereby suppressing an increase in the power supply current.

FIG. 6 is a block diagram of an embodiment of the present invention,
Telephone TEL, terminal impedance Z, 2 wire-4
A case is shown in which a hybrid circuit H that performs line conversion is added, and the output terminals of the second mirror circuits B1 and A1 and the A line and B line are connected to the hybrid circuit H. Note that the same reference numerals as in other FIG. 1 indicate the same parts.

Regarding common-mode impedance, the variation between the power supply circuit parts on the A line and B line side is called the unbalanced attenuation to ground, and it is one of _the important characteristics. The second mirror circuit B1, which is the latter stage of the feedback loop,
By being connected between the output terminals of A1,
It is possible to cancel out the influence of variations among the circuit elements connected before the capacitor C _AB , the precision of the resistors Rb ₁ and Rb ₀ , and the second mirror circuits B1 and A1.

In addition, the ground unbalance is determined by the relative ratio of the resistors Rc ₁ and Rc ₀ connected after the connection point of capacitor C _AB , and the third
mirror circuit B2, A2, first mirror circuit B
It is determined only by the relative ratio of 0 and A0, and the mirror circuit is
It is easy to improve the relative accuracy, and the resistance
Since it is easy to improve the relative accuracy of Rc ₁ and Rc ₀ when integrated circuits, it is possible to reduce the ground unbalance.

Also, since the power supply circuit is characterized by having a high impedance with respect to differential signals (normal call signals), as a subscriber circuit, the AC terminal impedance Z is connected between the B line and the A line. has been done. That is, the voice signal sent out from the telephone will generate a voltage across this termination impedance Z. This audio signal is directly input to the hybrid circuit H and sent out to the 4-wire line.

Also, the signal from the 4-wire line is sent to the hybrid circuit H.
The mirror circuit B0, A0 is converted into 4 wires and 2 wires by
is input. The output terminals of the mirror circuits B2 and A2 are also connected to the output terminal of the hybrid circuit H, but since their output impedances are very large as described above, they can be omitted. Further, the output impedance of the hybrid circuit H is also large.

Signals from the 4-wire line are sent to mirror circuits B0 and A0.
Then, the signals are multiplied by Ra ₁ /Re ₁ and Ra ₀ /Re ₀ and output. This output current value is usually the same value, and the voltage generated between the B line and the A line is determined by the subscriber line resistance RL and the output current of the hybrid circuit H.
If I _N , then V _B −V _A = [(Z×RL)/(Z+RL)]×Ra ₁ /Re ₁ × _IN .

〔Effect of the invention〕

As explained above, the present invention provides the first
The first to third mirror circuits B0, B1, B2 and the first to third mirror circuits A0, A1, A2 on the battery side
, resistors Rb ₁ , Rb ₀ , Rc ₁ , Rc ₀ and capacitors
C _AB and the desired power supply resistance Z _B , depending on the relative accuracy of the resistor and mirror circuit.
Z _A can be obtained, and high impedance can be obtained for differential signals (speech signals) and low impedance for common mode signals (noise signals). Furthermore, regarding power supply noise,
This can be suppressed by connecting a stabilized power supply to the common terminal of the third mirror circuit A2. The desired power supply characteristics can be obtained depending on the relative accuracy of each part, and while it is not easy to obtain absolute accuracy of resistance values etc. when implementing LSI, it is easy to obtain relative accuracy. Therefore, the power supply circuit of the present invention has the advantage that it can be easily integrated into an LSI.

[Brief explanation of drawings]

FIG. 1 is a block diagram of an embodiment of the present invention, FIGS. 2 a to c are mirror circuits, FIG. 3 is a stabilized power supply, and FIG.
The figure shows a current limiting circuit, Fig. 5 is a diagram explaining current limiting characteristics, Fig. 6 is a block diagram of an embodiment of the present invention, and Fig. 7
8 and 8 show conventional power supply circuits. B0 to B2, A0 to A2 are the first to third mirror circuits, Ra ₁ , Ra ₀ to Re ₁ and Re ₀ are resistors, C _AB is a capacitor, OP ₁ and OP ₀ are operational amplifiers, and V _BB is the power supply voltage. , V _Z is the regulated power supply voltage.

Claims (1)

[Scope of Claims] 1. In a power supply circuit for supplying direct current for telephone calls in the subscriber circuit of an exchange and for passing voice signals, for supplying current from the ground to the B line. First to third mirror circuits B0, B1, B2 on the earth side; first to third mirror circuits A0, A1, A2 on the battery side for drawing current from the A line to the battery; and the B line. and resistors Rb ₁ and Rb ₀ for converting the voltage between the ground air and the voltage between the A line and the battery into current, and the third mirror on the ground air side and the battery side. The first mirror circuit B0 on the ground air side includes a pair of resistors Rc ₁ and Rc ₀ and a capacitor C _AB connected between the input terminals of the circuits B2 and A2, and the output terminal of the B line is connected to the first mirror circuit B0. is connected to the output terminal of the third mirror circuit A2 on the battery side, and the input terminal of the first mirror circuit A0 on the battery side, whose output terminal is connected to the A line, is connected to the ground. a resistor Rb ₁ connected to the output terminal of the third mirror circuit B2 on the air side and for converting the voltage into a current;
the second mirror circuits B1 and A1 on the earth side and the battery side, respectively, with Rb ₀ connected to the input terminal;
The output terminals of are cross-connected to the input terminals of the third mirror circuits B2 and A2 on the earth side and the battery side, respectively, via the pair of resistors Rc ₁ and Rc ₀ , and the output terminals of the battery side A power supply circuit characterized in that a stabilized power supply is connected to a common terminal of the third mirror circuit A2.