JPH0351350B2 - - Google Patents

Description

[Detailed description of the invention] 〔Technical field〕 The present invention relates to a subscriber circuit for public telephones.

[Prior art]

Conventionally, this type of subscriber circuit has been constructed using mechanical elements such as a power supply relay 5 and polarity reversing relays 3 and 4 as shown in Figure 1, which has been an obstacle to miniaturization and electronicization of equipment. was.

[Purpose of the invention]

The purpose of the present invention is to eliminate the need for mechanical elements such as relays by digitizing the power supply function circuit and the talk current polarity reversal function circuit in the subscriber circuit, thereby reducing the size of the circuit and achieving high-density packaging. The object of the present invention is to provide a subscriber circuit for public telephones that makes it possible to use public telephones.

[Structure of the Invention] The present invention relates to a subscriber circuit for supplying a talking current to a subscriber line, which is provided with a means for sending out a polarity reversal signal by reversing the polarity of the talking current. The circuit comprises two feeder circuits each connected with opposite polarity to the subscriber line, each of the feeder circuits detecting a voltage on the subscriber line and converting it into a current. a conversion circuit, a current mirror circuit having a series circuit of a switch and a capacitor in parallel and inputting the output of the voltage/current conversion circuit; and inputting the output of the current mirror circuit and sending a current to the subscriber line. It comprises a current supply circuit and a control terminal that controls the operation and stop of the power supply circuit and the opening and closing of the switch according to an input signal from the outside, and each of the two power supply circuits is operated independently by an external signal applied to the control terminal. , is characterized in that it is configured to perform a stop and send out the polarity reversal signal.

〔Example〕

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

FIG. 2 is a circuit configuration diagram showing an embodiment of the present invention. In Figure 2, 1 and 2 are subscriber lines, 3
a and 3b are power supply circuits having the same configuration, respectively, having control terminals 4a and 4b, and connecting subscriber lines 1 and 2.
are connected with opposite polarity to each other. below,
They are called a normal power supply circuit 3a and a reverse power supply circuit 3b, respectively.

Normal power supply circuit 3a (reverse power supply circuit 3b)
have high input impedance voltage/current conversion circuits 5a, (5b) that detect subscriber line voltages of subscriber lines 1, 2 and convert them into currents, and common terminals 6a, (6).
b) and input the outputs of the voltage/current conversion circuits 5a, (5b) into current mirror circuits 7a, 7b.
and current supply circuits 8a, (8b) which input the outputs of the current mirror circuits 7a, (7b), suck the communication current from the subscriber lines 2, (1), and discharge the communication current to the subscriber lines 1, 2. Furthermore, switches 9a, (9b) and capacitors 10a, 10a are connected in parallel with the current mirror circuits 7a, (7b).
(10b) are connected in series.

Further, the common terminals 6a and 6b are connected to the power supply _VBB .

Next, the circuit operation will be explained.

During normal power supply, the normal power supply circuit 3a is turned on by a signal from the control terminal 4a from the outside, the reverse power supply circuit 3b is turned off by a signal from the control terminal 4b, and the communication current is supplied to the current supply circuit 8a.
It flows in the direction of the arrow in FIG. 2, that is, flows out through the subscriber line 1 to a terminal (not shown), and flows from the subscriber line 2 into the same circuit 8a. In order to turn on and off the power supply circuit, the following configuration may be used, for example. That is, as shown in FIG. 2, the switches 11a, (11b) and 9a, (9b) are configured to operate in response to signals from the control terminals 4a, (4b). Here, switch 11
When a, (11b) are open, there is no input to the current supply circuits 8a, (8b), so the current supply is stopped. At this time, the reverse power feeding circuit 3b has a high impedance with respect to the subscriber lines 1 and 2, so that it does not affect the above operation.

On the other hand, during reverse power feeding, the control terminals 4a, 4
The signal b is inverted, and current flows from the current supply circuit 8b in the direction of the arrow (dashed line) in the figure, that is, from the subscriber line 2 via the terminal and back to the subscriber line 1, the circuit 8
Flows into b. Further, when both the normal and reverse power supply circuits 3a, 3b are off, the signals of the control terminals 4a, 4b are both off, so the supply of communication current to the terminal is stopped.

It is necessary to set the DC resistance of both the power supply circuits 3a and 3b to a certain value when viewed from the terminal. Generally, it is matched to the conventional relay winding resistance. This resistance value is set by a DC feedback circuit inside the circuit. Next, this operation will be explained.

In normal power supply state, the voltage seen from subscriber line 2 to subscriber line 1 is V _L , the station battery voltage is V _BB (generally -48V), the ground voltage of subscriber line 1 is V ₁ , subscriber line 2 and If the voltage between the station batteries is V ₂ , the voltage/
The output current (denoted as I) of the current conversion circuit 5a is:
When the conversion coefficient is _K1 , I= _K1 ( _V1 + _V2 )=- _K1 ( _VBB + _VL ). This current I is the current mirror circuit 7a
The current is transmitted at a certain current transfer ratio. If this transmission ratio is K ₂ , then the input current of the current supply circuit 8a is K ₂ I. This current K ₂ I is the current supply circuit 8a.
is transmitted as a talking current _IL . Assuming that the conversion coefficient at this time is _K3 , the speech current I _L flowing through the subscriber lines 1 _{and 2} is I _L = _{K3K2I = -K1K2K3} ( _VBB + _VL ). Therefore, the equivalent direct current resistance (R _DC ) of the power supply circuit 3a is equal to the voltage drop (V ₁ +
V ₂ )=−(V _BB +V _L ) divided by the communication current I _L is expressed as R _DC =−(V _BB +V _L )/ _IL = 1/K ₁ K ₂ K ₃ . Setting of R _DC is performed by appropriately designing K ₁ , K ₂ , and K ₃ .

This point will be explained in detail again later.

The capacitors 10a, (10b) are connected to the subscriber line 1,
This is a bypass for removing AC signals such as audio signals applied between the two from the feedback circuit, and since no feedback is applied to AC signals, the impedance becomes high. An AC termination may be installed, for example, between subscriber lines 1 and 2, if necessary, but is not shown because it is not relevant here.

Next, a method of transmitting a so-called polarity reversal signal, which reverses the polarity of the power supply current to a public telephone, will be described.

FIG. 3 is a diagram showing the polarity reversal signal sending timing and the speech current waveform. The polarity reversal signal is sent to the power supply circuit 3.
It is transmitted by alternately turning on and off control signals applied to control terminals 4a and 4b of terminals a and 3b, respectively. This signal is sent out during a call and is used, for example, as a coin acceptance signal. In Fig. 3, the waveform of the communication current is slowed down (at t ₁ and t ₂ ).
For example, this is due to the effect of a filter (not shown) inserted in the subscriber line.

The polarity reversal noise suppression filter inserted into the subscriber line is generally constructed of a passive circuit consisting of inductance and capacitance in order to withstand lightning surges and the like. An example of this is shown in FIG. According to the figure, a low-pass filter composed of a transformer T and a capacitance C is inserted between the power supply circuits 3a, 3b and the terminal. This filter allows the power supply current of the power supply circuit to be smoothed at the time of polarity reversal. The impedance when looking at the subscriber line from the terminal is high impedance because the transformer T is connected in a differential mode, so the impedance is approximately equal to the input impedance of the 2-wire 4-wire circuit connected in parallel to the outside. In other words, the difference in subscriber line input impedance due to the presence or absence of this low-pass filter is small and within a range that does not affect the speech signal.

The reason why the communication current waveform becomes as shown in FIG. 3 when the power supply circuits 3a and 3b are turned on and off is as follows. Now, to simplify the explanation, we will explain the normal power supply circuit 3a and the reverse power supply circuit 3.
b is expressed by the equivalent DC resistance R _DC , and the control terminal 4
Consider replacing the on/off of the control signals a and 4b with a switch that switches the direction of the power supply current. This is shown in FIG. In other words, switch SW1, SW
2, the power supply current is N1 and N2 during normal power supply.
During reverse power supply, power is supplied to the public telephone TEL from the R1 and R2 terminals. When the control signal at the control terminal 4a is on, the switches SW1 and SW2 are connected to the terminals N1 and N2, respectively, and when the control signal at the control terminal 4a is on, the switches SW1 and SW2 are connected to the terminals R1 and R2, respectively. When both of these control signals are off, switches SW1 and SW2 are connected to the empty terminal A.
1, connected to A2.

The operation will be explained based on FIG. First, considering when the normal power supply circuit 3a is turned on, the power supply current
_IL is smoothed by a low-pass filter and supplied to the terminal TEL via terminals N1 and N2. terminal
The current I _T flowing through TEL has a smooth waveform due to the capacitance C of the low-pass filter and the inductance of the transformer T.

This is because many high-frequency components are included in the steep rise portion of the communication current I _L , and these are attenuated by the low-pass filter.

Next, consider when the normal power supply circuit turns off. At this time, the reverse power supply circuit also remains off.

The terminal current I _T smoothly approaches zero until the discharge current from the capacitance C disappears.

Next, consider when the reverse power supply circuit is turned on. At this time, the normal power supply circuit remains off. The circuit operation at this time is the same as when the normal power supply circuit described above is on, only that the direction of the terminal current I _T is reversed.

Corresponding to Fig. 3, during the period _t1 when the normal power supply circuit is off, the capacitance C is
During the period t ₂ when the reverse power supply circuit is turned off, the terminal TEL is also discharged from the capacitance C in the opposite direction to the normal time, so the terminal current I _T smoothly approaches zero. At the moment the power supply circuit is turned on, current flows to the terminal while charging the capacitance and being blocked by the inductance of the transformer T, so the terminal current I _T
rises smoothly. In other words, the terminal current I _T has a smooth waveform due to the low-pass characteristics of the filter.
This rounding of the waveform has the effect of preventing pulse noise from being heard by the subscriber's ears, and is important for speech quality.

Here, the functions of the switches 9a and (9b) shown in FIG. 2 will be described. If the polarity reversal signal is sent with the switches 9a and (9b) closed, the power supply circuit 3
While capacitors 10a and 3b are off, capacitors 10a and 10b are charged by the output offset currents of voltage/current conversion circuits 5a and 5b.
The moment the control signal is turned on, the current mirror circuit 7 is connected to the capacitors 10a and (10b).
A, (7b) a discharge occurs, and the current mirror circuits 7a, (7b) until the discharge ends.
An output current flows through the subscriber lines 1 and 2, and as a result, an overshoot current momentarily flows through the subscriber lines 1 and 2 as well. This situation is shown in FIG. This deteriorates the noise characteristics.

That is, since the separation switches 9a, (9b) of the capacitors 10a, (10b) are turned on and off at the same time as the control signals 4a, (4b) are turned on and off, the charging voltage of the capacitors 10a, (10b) is turned on and off. It remains constant regardless. This capacitor is used to bypass communication signals (alternating current) and does not require charging or discharging. Figure 8 shows the waveform. The reverse power supply circuit side is also similar to the normal power supply circuit side. Note that the charging voltage is determined by current mirror circuits 7a, (7b).
saturates at a certain voltage limit determined by the input side circuit configuration. For example, current mirror circuits 7a, (7b)
If the format is as shown in Fig. 9, the charging voltage V _C of the capacitors 10a and (10b) is as follows, ignoring the base current, where V _BE is the voltage between the base and emitter of the input transistor, and R10 is the emitter resistance. Saturation occurs at V _C =R10×I+V _BE .

Switches 9a, (9b) used in this embodiment
is provided to eliminate this drawback.
That is, when the control signal is turned off, the switches 9a,
(9b) is left open and the capacitors 10a and (10
This is to prevent the charging of the switch 9a, 9b, and close the switches 9a and 9b at the same time as the control signal is turned on, so that the communication current cannot flow immediately after the control signal is turned on.

FIG. 5 is a diagram showing a specific example of the circuit shown in FIG. 2, and members having the same reference numerals as those shown in FIG. 2 have the same functions.

Here, FIG. 5 shows a specific example of the normal power supply circuit 3a in FIG. 2.

The subscriber line voltage is detected by calendar mirror circuits 27 and 24 whose input terminals are connected to the subscriber lines 1 and 2, respectively, and after converting it into a current, the current is supplied to an adding circuit 25 and added by the adding circuit 25. The current flows into the input terminal of the current mirror circuit 7a. The output current of the current mirror circuit 7a becomes the input current of the current mirror circuit 26, and the current mirror circuit 26 becomes the input current of the current mirror circuit 26.
The output current of circuit 26 flows through resistor R ₄ and the sum of the input and output currents of circuit 26 flows out of the common terminal and flows through resistor R ₃ . The current flowing through resistance R ₃ is resistance
Since the current is larger than the current flowing through R ₄ , the value of resistor R ₃ is set smaller than the value of resistor R ₄ so that the voltage drops across resistors R ₃ and R ₄ are equal.
Operational amplifier 20 and PNP transistor 22, operational amplifier 21 and NPN transistor 2
3 constitute a portage follower, and the current equal to the voltage across the resistors R ₃ and (R ₄ ) divided by the resistors R ₁ and (R ₂ ) connected to the emitters is connected to the subscriber line. 1, (2). In this example,
A current supply circuit is composed of the current mirror circuit 26 and two voltage followers.

To explain in detail, the voltage-current conversion circuits 5a, (5b) shown in FIG. 2 are specifically composed of the current mirror circuits 27, 24 and the adder circuit 25 shown in FIG. Furthermore, the addition circuit 25
As an example, as shown in FIG. 10, it is composed of current mirror circuits 28 and 29.

Furthermore, an example of the configuration of the current mirror circuit is shown in the 11th section.
12. FIG. 11 shows an input current discharge type, and FIG. 12 shows an input current sink type. In these figures, as is well known, the input currents I ₁ , I ₂ and the output currents I ₁ ′, I ₂ ′ are the emitter resistors R5, R7 and R6,
Using R8, there are the following relationships: I ₁ '/I ₁ = R5/R6 (1) I ₂ '/I ₂ = R7/R8 (2). If this ratio is n ₁ , then I ₁ ′=n ₁ I ₁ …(3) I ₂ ′=n ₁ I ₂ …(4) (n ₁ = R5/R6=R7/R8 …(5)) . The input voltage of the current mirror circuit is V ₁ ,
Since V ₂ , ignoring the base current, I ₁ = (V ₁ − V _BE1 )/R5 ……(6) I ₂ = (V ₂ − V _BE2 )/R7 ……(7) Here, V _BE1 ( V _BE2 ) is the base-emitter voltage of the input transistor.

In subscriber circuits, generally V ₁ , V ₂ >>V _BE1 , V _BE2 , so the above equation can be set as I ₁ =V ₁ /R5 (8) I ₂ =V ₂ /R7 (9).

In FIG. 10, the current mirror circuit 27,
From equations (1) and (8), the output current of 24 is I ₁ ′=V ₁ /R6 ……(10) From equations (2) and (9), I ₂ ′=V ₂ /R8 ……(11) Become.

The input/output transmission ratio of the Karen mirror circuit 29 is 1:1.
Then, the output current becomes I ₁ ″=I ₁ ′. Therefore,
The input current of the current mirror circuit 28 is given by the sum of two currents, I _S = I ₁ ″ + I ₂ ′ = I ₁ ′ + I ₂ ′ ... (12) If the input/output transfer ratio is 1:n ₂ , the output The current is
Using equations (10), (11), and (12), I=n ₂ I _S = n ₂ (I ₁ ′+I ₂ ′)=n ₂ (V ₁ /R6+V ₂ /R8)…
…(13) becomes. If R8=R6 is set here, the above equation becomes I=(n ₂ /R6)(V ₁ +V ₂ ) (14). Here, if the subscriber line voltage seen from subscriber line 2 to subscriber line 1 is V _L , then from Fig. 10, V ₂ +V _L +V ₁ = -V _BB (∵V _BB <0)... ) Therefore, V ₁ +V ₂ =-(V _BB +V _L )...(16). Therefore, equation (14) is I = (n ₂ / R6) (V ₁ + V ₂ ) = - (n ₂ / R6) (V _BB + V _L )
...(18)

Therefore, the formula I=K ₁ (V ₁ +V ₂ )=-K ₁ (V _BB +V _L ) ……(19 ) and Equation (18), they agree based on the relationship K ₁ = n ₂ /R6 (20).

In addition, the current supply circuits 8a and 8b include operational amplifiers 20 and 21, transistors 22, and
23, resistors R1, R2, R3, R4, and a current mirror circuit 26. This part is the 13th
Extracted and shown in the figure. In FIG. 13, the input/output transmission ratio of the current mirror circuit 26 is 1:
It is set as 1. If the input current of the current mirror circuit 26 is I ₀ , the output current is I ₀ ′=I ₀ (21) The common current is the sum of the input current and output current, and I ₀ ″=I ₀ +I ₀ ′=2I ₀ ...(22) Here, if the ratio of resistors R3 and R4 is set to R3:R4=1:2...(23), the voltage drop V across resistor R3 is (22), using equation (23), V=I ₀ ″R3=2I ₀ R3=I ₀ R4 …(24) On the other hand, the voltage drop V' across resistor R4 is (21)
Using equation (24), V'=I ₀ 'R4=I ₀ R4=V (25), which is equal to the voltage drop across resistor R3. Resistor R1,
As mentioned above, the voltage drop across R2 is equal to the voltage drop across resistors R3 and R4, and the resistance values are equal: R1=R2 (26), so the current I _L flowing out from subscriber line 1 is equal to the voltage drop across resistors R3 and R4. Ignoring the base current, I _L = V/R1...(27) Based on the same assumption, I _L ' flowing from subscriber line 2 is I _L ' = V'/R2 = V/R1 = I _L ... ...(28) Thus, current is supplied from this power supply circuit to the terminal in so-called balanced mode.

From formulas (27) and (24), I _L = V/R1 = (R4/R1) I ₀ ... (29) Therefore, the transmission ratio K ₃ is obtained from I _L = K ₃ I ₀ ... (30) K ₃ = R4/R1 ...(31).

The input/output transmission ratio of the current mirror circuit 7a is 1:
If K ₂ is assumed, the output current I ₀ becomes from the input current I as follows with reference to FIG. 14: I ₀ =K ₂ I (32).

To summarize the above matters, the output current I of the voltage-current conversion circuit is I = K ₁ (V ₁ + V ₂ ) ... (19) The output current I _L of the power supply circuit is I _L = K ₃ I ₀ ... (30 ), and by substituting equations (32) and (19) here, I _L =K ₃ K ₂ K ₁ (V ₁ +V ₂ ) ...(33).

Now, the equivalent DC resistance of the power supply circuits 3a and (3b)
R _DC is the value obtained by dividing the circuit voltage drop (V ₁ + V ₂ ) by the power supply current I _L flowing through this power supply circuit, so the first
Referring to Figure 5, R _DC = V ₁ / I _L + V ₂ / I _L = (V ₁ + V ₂ ) / I _L ... (34) Substituting it into equation (33) gives R _DC = 1/K. ₁ K ₂ K ₃ ...(35)

As mentioned above, these coefficients are determined by the circuit constants (resistance ratio, current mirror ratio) (K ₁ is the equation (20), K ₂ is the equation (32), and K ₃ is the equation (31)).

Regarding each portion in FIG. 5 corresponding to K ₁ , K ₂ , and K ₃ in FIG. 2, K ₁ corresponds to the current mirror circuits 27 and 24 and the addition circuit 25.

K ₂ corresponds to the current mirror circuit 7a and is the input/output current transmission ratio.

_K3 is the current mirror circuit 26, resistors R1, R
2, R3, R4, operational amplifiers 20, 21, and transistors 22, 23.

As mentioned above, the definitions of K ₁ , K ₂ , and K ₃ are equations (20), (32), and (31), respectively.

In order to re-explain the effects of the switches 9a and (9b) already mentioned, let us assume that the switches 9a,
Let us consider the case where (9b) does not exist. The normal power supply circuit 3a is turned off, and the reverse power supply circuit 3b is turned off.
Assuming that is on, the potential of subscriber line 1 is lower than the potential of subscriber line 2 (reverse power feeding), and the voltage between subscriber line 1 and ground is V ₁
and the voltage _V2 between the subscriber line 2 and the power supply _VBB are both higher than during normal power supply. Therefore, as is clear from equation (19), in the normal power supply circuit, the voltage-current conversion circuit output current I is larger than during normal power supply. Therefore, the saturation voltage V _C of the capacitor 10a also increases as shown by the already mentioned formula V _C =R10×I+V _BE (period of time t ₃ in FIG. 16).

Next, when the reverse power supply circuit is turned off and then the normal power supply circuit is turned on, the voltages V ₁ and V ₂ return to their normal values, and therefore the voltage-current conversion circuit output current I also returns to its original value, and eventually the capacitor 10a The charging voltage V _C also returns to the saturation voltage during normal power supply. However, at the timing of this return start (period of time _t4 in FIG. 16), the overcharged charge is discharged.
This discharge current flows into the input side of the Kaletomirror circuit 7a. This is because the output impedance of the current mirror circuit (28 in FIG. 10) of the voltage-current conversion circuit is high and it is impossible to discharge to that side. As a result, the input current I ₀ to the current supply circuit 8a
overshoots, and from equation (30), the communication current I _L also overshoots.

The above phenomenon occurs at the moment when the normal power supply circuit is turned on from off, but overshoot also occurs at the moment when the reverse power supply circuit is turned on from off. The shaded area in FIG. 16 indicates overshoot. This overshoot cannot be removed even by a filter and is heard by subscribers as harsh noise. If this overshoot is to be removed by a filter, the order of the filter must be increased and the cut-off characteristic must be made steeper, which increases the circuit scale, which is not desirable. Switches 9a, (9b),
Capacitor 10 is turned off when power is not being supplied.
The above-mentioned overshoot can be prevented by not charging overvoltage to a and (10b).

As described above, when the power supply circuit is turned off, the AC bypass capacitors 10a, (1
0b) is prevented by opening the switches 9a and (9b), an overshoot current as shown in FIG. 4 does not flow, which has the effect of reducing polarity reversal noise.

〔Effect of the invention〕

As explained above, in the present invention, the power supply circuit is configured with electronic elements instead of conventional mechanical elements such as relays, so that it is possible to miniaturize the circuit and achieve high-density packaging. .

[Brief explanation of drawings]

FIG. 1 is a schematic circuit configuration diagram showing a conventional subscriber circuit, FIG. 2 is a circuit configuration diagram showing an embodiment of the present invention, and FIG. 3 is a polarity reversal signal sending timing for the circuit in FIG. 2. FIG. 4 is a diagram showing the communication current waveform without switches 9a and (9b), FIG. 5 is a diagram showing a specific example of the circuit shown in FIG. 2, and FIG. A diagram showing the connection relationship between a filter, a subscriber line, and a 2-wire/4-wire conversion circuit. Figure 7 is an equivalent circuit when sending a polarity reversal signal, and Figure 8 is a diagram of the switch 9a.
FIG. 9 is a diagram showing a specific example of the current mirror circuits 7a and 7b, FIG. 10 is a diagram showing an example of the voltage/current conversion circuit, FIGS. 12 is a diagram showing an example of a current mirror circuit, FIG. 13 is a diagram showing a current supply circuit, FIG. 14 is a diagram showing an example of the current mirror circuit 7a, FIG. 15 is an equivalent DC circuit diagram of the power supply circuit, and FIG. FIG. 16 is a diagram showing the charging voltage waveform and the talking current waveform of the capacitor 10a when the switch 9a is not provided. 1, 2...Subscriber line, 3a, 3b...Power supply circuit, 4a, 4b...Control terminal, 5a, 5b...Voltage/current conversion circuit, 6a, 6b...Common terminal, 7
a, 7b...Current mirror circuit, 8a, 8b...
...Current supply circuit, 9a, 9b...Switch, 10
a, 10b... Capacitor, 20, 21... Operational amplifier, 22, 23... Transistor, 24, 2
6, 27...Current mirror circuit, 25...Addition circuit.

Claims (1)

[Scope of Claims] 1. A subscriber circuit for a public telephone, which supplies a talking current to a subscriber line and is equipped with means for sending out a polarity reversal signal by reversing the polarity of the talking current. , comprising two power supply circuits each connected to the subscriber line with opposite polarity, each of the power supply circuits comprising a voltage/current conversion circuit that detects the voltage of the subscriber and converts it into a current; ,
a current mirror circuit which has a series circuit of a switch and a capacitor in parallel and inputs the output of the voltage/current conversion circuit; and a current supply circuit which inputs the output of the current mirror circuit and sends a current to the subscriber line. , and a control terminal for controlling the operation and stopping of the power supply circuit and the opening and closing of the switch in response to an external input signal, and the two power supply circuits are independently operated and stopped by an external signal applied to the control terminal. A subscriber circuit for a public telephone set, characterized in that it is configured to send out the polarity reversal signal.