JPS63314060A - power supply circuit - Google Patents

Abstract

PURPOSE:To obtain a simplified power feeding circuit by coupling respective control input terminals of first and second mutual conductance amplifiers with each other by a capacitor. CONSTITUTION:Mutual conductance amplifiers 10 and 20 which can be operated as constant current sources respectively are provided correspondingly to lines A and B, and respective control input terminals CNT1 and CNT2 of mutual conductance amplifiers 10 and 20 on the line A side and the line B side are coupled with each other by a capacitor 30. Thus, a prescribed DC equivalent resistance required for feeding DC, a high impedance for a differential signal, and a low impedance for an in-phase signal (AC inducting signal) are simultaneously obtained, and complicated feedback control circuit and VBB/2 power source are eliminated.

Description

[Detailed description of the invention] [Summary] A power supply circuit provided in a subscriber circuit, which is a so-called A
The control input terminals of the transconductance amplifiers on the A-line and B-line sides are connected to each other by a capacitor. A predetermined DC equivalent resistance necessary for DC power supply, high impedance for differential signals, and low impedance for in-phase signals (AC induced signals) can be obtained at the same time by simply doing this.

[Industrial application field]

The present invention relates to a power supply circuit, and particularly to a power supply circuit provided in a subscriber circuit of an exchange.

Generally, the power supply circuit is placed on the exchange side, and its main purpose is to supply direct current to each subscriber's telephone to operate it through the telephone line. Since this power supply circuit is connected to the ends of the AvA and B lines forming the telephone line, it must be designed so that the input impedance for AC signals seen from the A and B lines becomes a predetermined value. There are two types of input impedance (Znr
= Z Cy), and must satisfy the following conditions.

(2) AC impedance Z0 for differential signals (audio signals) is high; and (2) AC termination impedance ZCt for in-phase signals (harmful AC signals such as AC induction) is low. In addition, (1) The DC power supply resistance must have a value that can supply the DC current required by the telephone, for example several hundred ohms.

Two AC termination impedances for the above differential signals □
is desirably high so as not to attenuate it, and on the other hand, the AC termination impedance ZCT for the above-mentioned common mode signal is desirably low so as to attenuate it as much as possible.

Furthermore, in recent years, the conditions for common-mode signals have become stricter, and it is required that the audio signal not be distorted even if the common-mode signal current becomes larger than the DC power supply current.

[Conventional technology]

FIG. 6 is a diagram showing an example of a conventional power supply circuit.

In this figure, the feeder circuit consists of a first transconductance amplifier 10, a second transconductance amplifier 20, and a feedback control circuit 50, and is connected to the subscriber's telephone equipment via feeder terminals 0UT1 and 0UT2, so-called A line 4.
DC power is supplied to the 1 and B wires 42. Both the first and second transconductance amplifiers 10 and 20 have the same configuration and consist of operational amplifiers 16 and 26 and various resistor groups. In addition, bias input terminals B1 and B2, control input terminals CNTl and CNT2, and the above-mentioned power supply terminal 0U
TI and 0UT2.

Note that the various resistors mentioned above include the first to fifth resistors (11 to 15) for the first mutual conductance amplifier 10;
Regarding the second transconductance amplifier 20, these are the first to fifth resistors (21 to 25).

The power supply circuit shown in Fig. 6 is based on the literature (IEEE JOllR
NALOF 5OLID-3TATE C11? C1I
ITS, VOL, 5C-16, NO, 4, AIJGIJ
ST1981), page 262, and includes a pair of transconductance amplifiers (10, 20) and a feedback control circuit (50) as its basic elements. The transconductance amplifier is described in F on the same page of the same document.
Transconductance am to ig, 3
Although shown as a plifier, its function is equivalent to a constant current source. Therefore, the above-mentioned condition (2) (high impedance for differential signals) and condition (2) (predetermined DC power supply resistance) are satisfied. However, the above-mentioned condition (2) (low impedance with respect to the common-mode signal) cannot be satisfied by the transconductance amplifier alone. Therefore, in order to have a low impedance with respect to the common mode signal, a feedback control circuit 50 is introduced, and its output is connected to the control input terminal CN described above.
It will be added to Tl and CNT 2 (Fig. 6).

In addition, as a conventional example, Japanese Patent Application Laid-Open No. 57-42263
Although it satisfies the aforementioned conditions (■, ■, and ) cannot cope with it. because,
The power supply circuit shown in the same publication is a P
NPN whose collector is connected to the NP transistor and the A line
The former PNP l-transistor only sends direct current from its collector, and the latter N
This is because a PN transistor can only absorb direct current from its collector, and cannot conduct direct current in the opposite direction. On the other hand, the operational amplifier (16, 2
6) can inherently send or absorb current from its output, so it does not matter if the in-phase signal current is larger than the DC supply current.

[Problem that the invention seeks to solve]

In the conventional power supply circuit shown in FIG.
is introduced, and first, the midpoint potential between the A line 41 and the B line 42 is detected by the resistors 51 and 52. The differential signal is the 6th
Waveform d1 in the figure. Since the differential signal component appears as d2, the differential signal component is not detected as a midpoint potential. However, in the same figure, the in-phase signal appears as waveforms cl and c2, and the midpoint potential is detected as waveform m. . This midpoint potential m is defined as Vl
l/2 (v□ is the power supply voltage for amplifiers 16 and 26)
is applied to the non-inverting input of the positive phase amplifier 53 which has an inverting input, and its output is applied to the control input terminals CNT 1 and CNT 2. This allows the common mode signal current, which is AC induction from the outside, to be transferred to the mutual conductance amplifier 16.
, 26 alternately.

As mentioned above, in the conventional power supply circuit (Fig. 6), the resistors 51 and 52 for midpoint potential detection, the positive phase amplifier 53, and the V
A large-scale feedback control circuit 50 consisting of an a++/2 power supply is required, and there is a problem that the circuit becomes large and complicated.

The present invention has been made in view of these problems, and it is an object of the present invention to provide a power supply circuit that substantially eliminates the need for a feedback control circuit.

[Means for solving problems]

FIG. 1 is a diagram showing the basic configuration of a power supply circuit according to the present invention. As shown in this figure, the conventional feedback control circuit (50 in Fig. 6) is eliminated, and in its place, the control input terminal CNT1
A capacitor 30 is provided which couples the control input terminal CNT2 to the control input terminal CNT2.

[For production]

Generally speaking, the capacitor 30 appears to be absent for common-mode signals, and the desired low impedance for common-mode signals is achieved.

Moreover, this state is the same as when looking at the DC power supply current. On the other hand, for differential signals, this is equivalent to apparently dividing the capacitor 30 into two and connecting the midpoint to virtual ground, thereby achieving the desired high impedance for differential signals.

In this way, by simply providing one capacitor, it is possible to eliminate the need for a conventional complicated feedback control circuit and a Vts/2 power supply (the details of the operating principle will be described later).

〔Example〕

FIG. 2 is a diagram showing an embodiment of the present invention. The circuit configuration is basically the same as that shown in FIG. 1, but the operational amplifier 16,
Buffer circuits 60 and 65 consisting of emitter followers are provided at each of the outputs 19 and 29 of 26.
S, Patent 4.387.273 (Fig. 6)), so that even if a common mode signal considerably larger than the power supply current intrudes, it can be adequately dealt with. Also, the bias input terminal Bl.

A predetermined voltage divided by three voltage dividing resistors 71, 72 and 73 is applied to B2. Although the input/output portion of the audio signal can be selected as appropriate, this figure shows an example in which a transformer 74 is provided at the illustrated position and the audio signal V is input/output via this.

Looking at the buffer circuits 60 and 65 above, the former has a PNP transistor 61 and an NPN transistor 62.
(mutually emitter-coupled) and a resistor 63, the latter buffer circuit 65 consists of a PNP transistor 66, an NPN
It consists of a transistor 67 (emitter coupled to each other) and a resistor 68. If the common-mode signals C1 and C2 that are larger than the DC power supply current at a certain instant are represented by hatching in the figure, c
l is PNP) By Ranjistaro 1, C2 is PNP
) are respectively absorbed by Ranjistaro 6. When the polarity of the common mode signals cl and C2 switches at the next instant,
Next, a current corresponding to cl is injected from the NPN l-rangistaro 2, and a current corresponding to C2 is injected from the NPN l-rangistaro 7, respectively. In addition, the original DC power supply current is a-h l) -* from GND (OV) in the figure.
C-1d −+ 6−+ V RB (−48V) always flows.

Next, the principle of operation of the power supply circuit according to the present invention shown in FIG. 1 will be explained with reference to FIGS. 3, 4, and 5.

FIG. 3 is a DC equivalent circuit of the power supply circuit according to the present invention,
This shows that the desired DC power supply resistance can be obtained. In addition, for the resistance value R of the resistor in the figure, RI5<<RI2 + Rz
(The subscripts 11, 12, and 15 are the resistances 11 and 15, respectively.
, 12, and 15. (hereinafter, the same applies) is set in advance so as to satisfy the following relationship. For DC power supply, the capacitor 30 (FIG. 1) is as good as not present. The absence of capacitor 30 means that resistance 14.24
It is equivalent to not having (Figure 1). Here, looking at the first transconductance amplifier 10 side, the direct current I flowing into it is expressed as follows. The reason why the current ■1 is expressed by equation (1) is as follows.

In FIG. 1, for direct current, the capacitor 30 is
Since it shows infinite impedance, it is equivalent to nothing.
(OPEN), the resistor 14 (because it is connected to an infinite impedance) is in the state of 0PEN at one end.

Therefore, the resistor 14 has no effect on the circuit, so it is virtually nonexistent. Therefore, only the resistor 13 is connected between the output terminal and the inverting input terminal of the operational amplifier. The resistor 13 is
The output signal is fed back as is to the inverting input terminal, but since the input current of the operational amplifier is extremely small, the voltage drop across the resistor 13 can be ignored (current is zero and only the voltage signal is fed back).

Therefore, the value of the resistor 13 may be 0Ω (short circuit), resulting in a so-called voltage follower configuration.

In Figure 3, when the voltage of ■4 is applied to the circuit, 0UT
The voltage VX between I and the non-inverting input is RovA. Since the voltage of the non-inverting input appears as is from the output of the voltage follower, VX is applied to both ends of the resistor 15. Now, since R+s<< R+z+R1+, it becomes.

Using equation (1), it becomes. As is clear from equation (2), the three resistors 11
By appropriately setting the resistance values of 12 and 15, the desired resistance RA (for example, several hundred Ω) can be obtained (under the conditions
). Further, the accuracy of the resistor R7 is also improved by increasing the accuracy of the resistance values of these resistors.

On the other hand, the DC equivalent resistance Rs can be similarly set on the second transconductance amplifier 20 side and is expressed by the following equation.

FIG. 4 is a differential signal equivalent circuit of the power supply circuit according to the present invention. For differential signals, it appears equivalent to two narrow plane rows of capacitors 30' connected around the virtual ground E for alternating current components. The capacity of each capacitor 30' is
If the capacitance of the capacitor 30 in FIG. 1 is Cab, then C1
bx2. The reason for such an equivalent circuit is that the differential signals di and d2 of the A line and B line fluctuate vertically symmetrically in the figure. Here, for the resistance values RI41R24 of resistors 14 and 24, if C and b are chosen sufficiently large so that (r is the lowest frequency of the differential signal), the voltage applied to each capacitor 30' becomes almost zero. Become. Using this, the voltage at the point indicated by ■+ in the figure becomes V+ = V- due to a loose imaginary short, so the voltage ■- at the inverting input (-) is (The numerator 10 on the right side indicates that the voltage across the capacitor 30' becomes zero as described above.)

From this equation (6), it becomes. This equation (7) and the above equation (5) (■10= are expressed as follows. Here, Rth = R, 4 and R
I 2 ” RI :Resistors 11 and 12 so that l
, 13 and 14 are set in advance, then ◯ in the above equation (8) becomes Ia=O. 1A-〇 means that a high impedance is achieved for the differential signal (the above-mentioned condition ① is met).

FIG. 5 is a common-mode signal equivalent circuit of the power supply circuit according to the present invention. When the in-phase signal (cl, c2) enters, the voltage at the non-inverting input (+) of the operational amplifier 16 in the first transconductance amplifier 10 + and the voltage at its inverting input (-)
- changes with the waveform shown. This change is exactly the same on the second transconductance amplifier 20 side. Then, no voltage will appear across the capacitor 30 with respect to the common mode signal. In other words, for the common mode signal, it is as if the capacitor 30 does not exist.

As a result, the common-mode signal equivalent circuit becomes the same as the DC equivalent circuit shown in Figure 3, and the DC power supply resistance (for example, several hundred ohms)
The corresponding impedance becomes the impedance for the common-mode signal, and a low value is obtained (meeting the condition (2) described above).

〔Effect of the invention〕

As explained above, according to the present invention, a power feeding circuit that is simpler than the conventional one is realized.

[Brief explanation of the drawing]

FIG. 1 is a diagram showing the basic configuration of a power supply circuit according to the present invention, FIG. 2 is a diagram showing an embodiment of the present invention, FIG. 3 is a DC equivalent circuit of the power supply circuit according to the present invention, and FIG. FIG. 5 is a diagram showing an example of a differential signal equivalent circuit of a power feeding circuit according to the present invention. FIG. 5 is a diagram showing an in-phase signal equivalent circuit of a power feeding circuit according to the present invention. FIG. 6 is a diagram showing an example of a conventional power feeding circuit. lO...first transconductance amplifier, 11 to 15
...Resistor, 16...Operation amplifier, 20...
2nd transconductance amplifier, 21 to 25... Resistor, 26... Operational amplifier, 30... Capacitor, 41... Al, 42... B line, OUT 1, OUT 2... Feeding end , CNT 1
, CNT 2...control input terminal, Bl, B2...
・Bias input terminal.

Claims (1)

[Claims] 1. Line A (41) connected to the subscriber's telephone (TEL)
and B line (42), respectively at the feed end (OUT
1, OUT2) are connected to the first transconductance amplifier (10) and the second transconductance amplifier (2
0), and the first and second transconductance amplifiers (10, 20) have non-inverting inputs (17, 2), respectively.
7), inverting input (18, 28) and output (19, 29
), an operational amplifier (16, 26) having a bias input terminal (B1, B2) and the non-inverting input (1
a first resistor (11, 21) connected between the non-inverting input (17, 27) and the power supply end (OUT
1, OUT2) connected between the second resistor (12, 22
), the output (19, 29) and the inverting input (18, 2
8), the third resistor (13, 23) connected between the control input terminals (CNT1, CNT2) and the inverting input (1
8, 28), and a fourth resistor (14, 24) connected between the power supply terminals (OUT1, OUT2) and the output (1
9, 29), the power supply circuit includes a fifth resistor (15, 25) connected between the control input terminal (CNT1) and the control input terminal (CNT1).
A power supply circuit characterized in that a capacitor (30) is used to connect between the two terminals (NT2).