JPS5915357A - Current supply circuit - Google Patents

Abstract

PURPOSE:To prevent deterioration of an exchange function between 2-wire and 4-wire circuits, by securing the active voltage of a feeding transistor TR by performing the constant voltage supply at a region of a small loop current. CONSTITUTION:A telephone set 4 is connected to a 2-wire circuit 3 and receives feed from a power supply 9 via a feeding circuit consisting of amplifiers 7 and 8 and feeding TR5 and 6. The call signal is converted by a 2-wire/4-wire converter consisting of amplifiers 10, 13 and 14 and delivered to the 4-wire side through a terminal 11 to be fed into the 2-wire side through a terminal 12. Conductance amplifiers 21 and 22 applied with reference voltages Vref1 and Vref2 monitor potentials of contacts 15 and 16. When the call current is reduced and the voltage of each contact becomes higher than the reference voltage, the feeding circuit has a control to assure the supply of constant voltage. When the call current exceeds a prescribed level, diodes 23 and 24 are cut off. Thus the feeding circuit functions as a power supply having an internal resistance of a prescribed level.

Description

DETAILED DESCRIPTION OF THE INVENTION The present invention relates to a current supply circuit that does not impair the 2-wire/4-wire conversion function even at low loop currents and is suitable for integrated circuit implementation.

FIG. 1 shows an example of a current supply circuit in a conventional electronic subscriber circuit, in which 1 and 2 are two-line side terminals, which are connected to a terminal device 4 such as a telephone via a subscriber line transmission line 3. 5
and 6 are Darlington configuration transistors, 7 and 8
is an operational amplifier (hereinafter referred to as an operational amplifier).
, and resistors R3 and R2 are connected in series between terminal 1 and ground.

A resistor R3 is connected between the emitter of the transistor 5 and ground, and the base and collector of the transistor 50 are connected to the output terminal and terminal l of the operational amplifier 7, respectively, and the negative phase input of the operational amplifier 7 is connected to the emitter of the transistor 5, while the positive phase input is connected to the emitter of the transistor 5. The input is connected to the connection point between resistors R and R2. The circuit connected to terminal 2 is configured similarly, but
The difference is that a series circuit of resistors R4 and R5 is connected to the terminal 2 and the negative power supply 9, while a resistor R6 is connected between the emitter of the transistor 6 and the negative power supply 9.

The operation of this circuit is such that the positive-phase input potential and the negative-phase input potential of the operational amplifier 7 are equal, so that 2.sup.+ ""-R3'3 R, +R2 IL=I3+I.

However, Vl... Potential of terminal 1 11...
・Current flowing through R1+R2■3 ・・・・・・R
The current 11 flowing through the circuit 3 becomes a loop current and is equivalent to a circuit of a resistor (Rx) represented by R, +R2, that is, having a value of R2/R3+1.

Similarly, the circuit connected to the terminal 2 is equivalent to the circuit of the resistor (RY) connected to the power supply 9.

As a result, the power supply circuit has an output resistance (RX+BY), a power supply 9
It will work as the voltage (vBB).

However, in order to operate transistors 5 and 6 of the Turlington configuration for output in a non-saturated state, a voltage of approximately 1.5 V or more is required as a collector-emitter voltage. As shown in FIG. The characteristics deviate from the broken line.

On the other hand, when looking at audio signals, terminal devices such as telephones 4
The signals input to terminals 1 and 2 from the operational amplifier 10
The signal is converted into an unbalanced signal and output to the four-wire signal output terminal ll. The signal input from the 4-wire signal input terminal 12 is converted into a balanced signal by the inverting amplifiers 13 and 14,
It is applied to the positive phase inputs of operational amplifiers 7 and 8 via nodes 15 and 16, modulates the collector currents of transformers 7 and 5 and 6, and outputs signals to two-wire terminals 1 and 2.

Note that 17, 18, 19 and 20 in the figure are capacitances for compling. Further, since the circuit for suppressing the wraparound is not directly related to this circuit, its explanation will be omitted here.

However, as mentioned earlier, since transistors 5 and 6 perform saturation operation in the area circuit shown in Figure 2, a problem arises in that the 4-wire input signal is not transmitted to the 2-wire, and the 2-wire 4-wire conversion function is completely disabled. The problem was that it didn't work.

The present invention was made to eliminate these drawbacks, and it is an object of the present invention to provide a current supply circuit that does not impair the 2-wire/4-wire conversion function even when the 2-wire load resistance is large and the power supply current is small. In order to ensure the active voltage of the transistor in the region where the loop current is small (1), constant voltage power supply is performed at a value 1. lower than the voltage VBB of the power supply 9. explain.

FIG. 3 shows the circuit configuration of one embodiment of the current supply circuit according to the present invention. The circuit surrounded by the broken line is the part added to the circuit in FIG. 1, and 21 and 22 are conductance amplifiers, 23 and 24 is a diode, 25 and 26 are capacitors,
R7 to R10 represent resistance.

This circuit inputs the potential of the node 15 through the resistor R7 to the positive phase of the conductance amplifier 21, inputs the reference voltage Vrefl to the negative phase, and outputs the voltage from the diode 23t=
A is connected to the opposite phase of the operational amplifier 7, a resistor R8 is connected between the emitter of the power/dimetal 5 and the opposite phase of the operational amplifier 7, and a capacitor 25 is connected between the conductance amplifier 21 and the ground.

The polarity of the output current of the conductance amplifier 21 is as follows:
The direction in which the current is sunk is positive, and an output current that is gm times the differential input voltage can be obtained, and gm is the positive side.

The operation of this circuit is, from a DC perspective, when the potential of the node 15 is 15, when v15 <'ref I, the output of the conductance amplifier 21 forces a current, but no current flows due to the diode 23'. The power supply circuit performs the same operation as described in FIG.

On the other hand, vI5>outside. conductance amplifier 21 in fl
The output of sinks current through diode 23 and through resistor R8.

The power supply characteristics at this time are V15---R3 (13
+ ix) -Rs Ix Engineering x-(V+5-Vref
l ) gmI, = 13-V, /(R,4-R2
) However, el is the output value of the conductance amplifier 21.

The relationship holds, and therefore, if we approximate it by assuming that the box is large enough, then we get. That is, constant voltage power is supplied without depending on the loop current (1), and its value is given by the right side of equation (2). FIG. 4 is a diagram showing this relationship.

In terms of alternating current, the capacitor 25 removes the alternating current component, so the conductance amplifier 21 does not operate, and the dashed circuit in FIG. 3 has no effect on the alternating current operation in FIG. 1 in terms of alternating current. .

Note that the operations within the broken line in the lower half of FIG. 3 are also the same as those described above, so their explanation will be omitted.

As mentioned above, the reference voltages vref+ and Vref2
By choosing appropriately, the loop current 11. Constant voltage power supply is possible in a small area of transistors 5 and 6.
operating voltage can be ensured.

FIG. 5 shows a circuit diagram for explaining the present invention in detail.

This circuit detects a 2-wire differential voltage and sends it to a 4-wire output.
It performs the same function as a circuit consisting of an operational amplifier 10 and an operational amplifier 10.

The potential at terminal 1 is converted into a voltage/current by a current mirror circuit consisting of a diode 27 and a transistor 28 and a resistor R1□ and inputted to the positive phase of an operational amplifier 29. The voltage/current is converted by the mirror circuit and resistor R1□ and inputted to the opposite phase of the operational amplifier 29, and the resistor R43 and R14
The voltage is converted again using the value determined by , and the 4-wire signal output terminal 11
It is a circular wire that outputs a voltage proportional to the differential voltage between terminals 1 and 2.

At this time, in the conventional power supply circuit shown in FIG. 1, when the loop current is small, the potential value IV11 of terminal 1 becomes small, and the current mirror circuit consisting of the diode 27 and transistor 28 in FIG. Operating voltage(
In this example, approximately 0.6V) cannot be secured, but the
By using the power supply circuit shown in the figure, the value of IVll is maintained at a constant value even with a small loop current, so the operating voltage of the current mirror circuit shown in Figure 5 can be secured and signal transmission from 2 wires to 4 wires is possible. becomes.

FIG. 6 shows the circuit configuration of another embodiment of the current supply circuit according to the present invention, which mainly uses current mirror circuits, including current mirror circuits 32, 33, 34 .
35, 36 and 37, resistors R15 and R16, diodes 38 and 39, and constant current sources 40 and 41.

Since this circuit is symmetrical in the top and bottom of the figure, the upper half will be explained below.

The other end of the resistor R15, one end of which is connected to terminal 1, is connected to the input end of the current mirror circuit 32, and the output of the current mirror circuit 32 is input to the current mirror circuit 33, and the output of the circuit 33 is 2. One side is input to the current mirror circuit 34, and the other side is connected to the anode of the diode 38. The output of the current mirror circuit 34 is connected to the terminal 1, the cathode of the diode 38 is connected to the input of the current mirror circuit 32, and the constant current source 4o is connected to the anode of the diode 38.

Although a simple basic circuit is shown as a current mirror circuit here, it is clear that there are various other modifications.

Next, as the mirror ratio of the current mirror circuit, for example, the current mirror circuits 32 and 33 have I:1 and 1:1:1. J4 is set to n.

The current of the constant current power supply 40 is I. , the current flowing through the resistor R15 is 1, the input current of the current mirror circuit 32 is 2, and the current flowing through the diode 38 is 13, then IJ, -■
, +n■2 I , −-V/R,5 I,, +13=1゜I3-0 ...... (I2〉Io) +3-I
o-I2...(I2<■.) holds true. (
However, although the potential drop of the diode at the input stage of the current mirror circuit 320 has been ignored to simplify the formula, even if it is taken into consideration, it does not affect the effects of the present invention in any way. ) Therefore, it becomes.

That is, in the region where the loop current is large, this circuit has RI5/
It looks equivalent to a resistor with a value of (1+n), and in a region where the loop current is small, it works as a constant voltage circuit with a value of -R15·Io.

FIG. 7 shows this characteristic. By appropriately selecting the value of -RIs IO, the operating voltage of the diode in the input stage of the current mirror circuit 32 can be secured, and the operating voltage of the output transistor of the current mirror circuit 34 can be maintained. The same effect as explained in FIG. 3 can be obtained.

FIG. 8 shows an example of a circuit configuration that can be used as the conductance amplifiers 21 and 22 in FIG. !l:
rlnl'l transistors, the former has a current sink type output, and the latter has a current force type output.

Although the above explanation is based on the case where a bipolar transistor is used, it is obvious that it may be replaced with an FET element, and also, as a characteristic in a place where the loop current is large, only a type with a constant output resistance is used. Although the present invention is described as an example, the present invention is not limited to this, and can also be applied to constant current power supply or other devices having intermediate characteristics.

As explained above, the current supply circuit of the present invention can secure the operating voltage of the output transistor in the power supply circuit even in the region of small loop current, and also ensures the operating voltage of the current mirror that converts the two-wire voltage signal into current. The function of the 2-wire and 4-wire conversion circuit is not impaired because the 2-wire and 4-wire conversion circuits can also be secured. This means that 2
This means that the circuit operates even if the load resistance on the line side is large, and the P1 circuit has the advantage of being able to extend the line length and the internal resistance of the terminal equipment to be high, so the range of applications of the P1 circuit is wide. Chiru. Furthermore, since the power supply characteristics are switched depending on the magnitude of the loop current value (or loop voltage value) and the threshold current (or threshold voltage), there is also the advantage that there is no need for software control.

[Brief explanation of the drawing]

FIG. 1 is a diagram showing an example of a current supply circuit in a conventional electronic subscriber circuit, FIG. 2 is a diagram explaining the characteristics of the circuit in FIG. 1,
FIG. 3 is a circuit configuration diagram of one embodiment of the current supply circuit according to the present invention, FIG. 4 is a characteristic diagram of the circuit of FIG. 3, and FIG. 5 is a circuit configuration diagram for detailed explanation of the present invention. FIG. 6 is a circuit configuration diagram of another embodiment of the present invention, FIG. 7 is a characteristic explanatory diagram of the circuit in FIG. 6, and FIG. 8 is an example of a circuit configuration that can be used as the conductance amplifier in FIG. 3. FIG. 1, 2 ...... 2 g tu++ terminal, 3
- Subscriber line transmission line, 4...terminal equipment, 5.
6.28.31...Transistor, 7.8. +
0.29.42.44 Operational amplifier, 9 ・
・・Power supply, 11・・ 4-wire signal output terminal, 12
・・・・・・4-wire signal input terminal, 13.14
・Pa. Inverting amplifier, +5. +6 ° node, 17~2
0゛・Capacity for cambric link, 2], 22・・・・・
・Conductance amplifier, 23,2i, 27.30
．． 38.39 ・Diode, 25.26-
Capacity, 32-37... Current mirror circuit, 40.41... Constant current source, 43 ・
npn transistor, 45... pnpl
・Randister, R1-R16... Resistor. Figure 4 111-V+ Figure 5 1 Figure 6

Claims (1)

[Claims] (In a current supply circuit using active elements that supplies power to a 02-wire 4-wire conversion circuit, there is a circuit that monitors the voltage or current supplied to the 2-wire side, and a circuit that determines the current supply characteristics. The current supply characteristic is determined depending on whether the output of the monitored circuit is higher or lower than a predetermined threshold. A current supply circuit characterized in that it operates either a circuit that has a voltage supply characteristic or a circuit that has a characteristic other than a constant voltage supply characteristic. Claim (1) characterized in that a rectifying element is used as a changeover switch for operating either a circuit that provides constant voltage power supply characteristics or a circuit that provides characteristics other than constant voltage characteristics. Current supply circuit as described.