JPS62173808A - Buffer amplifier - Google Patents

Abstract

PURPOSE:To eliminate an offset voltage without increasing the current consumption and the occupied area by reducing an output current to 1/N to detect it and generating a base current on a basis of this detection current and multiplying the base current by N to feed back it. CONSTITUTION:A 1/N current of an output current IO of a constant current source 9 is generated by a constant current source 11, and a base current IB/N of 1/N of a base current IB of a transistor TR 8 is obtained as the base current of a TR 12. This current IB/N is multiplied by N in a current mirror circuit consisting of TRs 13 and 14, and the multiplied current is fed back to the collector of a TR 3. Thus, the unbalance between collector currents of TRs 2 and 3 due to the base current IB of the TR 8 is compensated to eliminate the offset voltage. Since an offset eliminating circuit of a TR 14 is not connected to an output terminal 10 from the constant current source 11, the dynamic range of the output voltage is not affected.

Description

DETAILED DESCRIPTION OF THE INVENTION [Field of Industrial Application] The present invention relates to a buffer amplifier for extracting the output voltage of a voltage stabilizing circuit in an IC, for example.

[Summary of the invention]

The present invention relates to a buffer amplifier, in which the output current is detected as -, a base current is formed based on this detected current, and this base current is multiplied by N and fed back, thereby increasing the input/output voltage of the buffer amplifier. This allows the offset between the two to be canceled.

[Conventional technology]

FIG. 4 shows the basic buffer amplifier connection.

In the figure, input terminal +11 is connected to the base of transistor (2), the emitter of this transistor (2) and the emitter of transistor (3) are connected to each other, and the midpoint of the connection of these emitters is connected to the constant current source (4). grounded through. The collector of this transistor (2) is connected to the Vcc power supply terminal (6) through the transistor (5) which constitutes the diode side of the current mirror, and the collector of the transistor (3) constitutes the transistor side of the current mirror. It is connected to the power supply terminal (6) through the transistor (7). More transistors (3)
The collector of the transistor (8) is connected to the base of the transistor (8). The collector of the transistor (8) is connected to the power supply terminal (6
), its emitter is grounded through a constant current source (9), this emitter is connected to the base of the transistor (3), and the output terminal (1
0) is derived.

In this circuit, a differential amplifier is formed by the transistor (21) (31), and its output is taken out through the emitter follower of the transistor (8) and is fed back to the base of the transistor (3) to form a bumper amplifier.

However, in this circuit, the main body current 2■1 flowing through the constant current source (4) and the output current I flowing through the constant current source (9)
If there is a difference of one fold or more between the input terminal VIN and the output voltage Vo, the offset voltage ΔV (= l
VIN VOl) becomes large.

In other words, in the above circuit, if the currents flowing through the two transistors T5+ (7) of the current mirror are both I, the collector current flowing through transistor (2) is 11, while the collector current flowing through transistor (3) is As mentioned above! The value obtained by subtracting the base current IB4c flowing through the transistor (8) from 1 is (11-18). Therefore, an offset of 1 occurs, and here 211-80 (μA), I
o = 2 (mA), h FE of transistor (8)
= 100, 11 = 40 (μA), and the offset voltage Δ■ is ΔV = 26 in −= 18 (m
V).

Therefore, various methods have been proposed to eliminate such offset voltage. First, it is conceivable to increase the current value of the constant current source (4) to the same level as Io in the above-mentioned circuit. However, with this method, the power consumption increases significantly, and since the transistor (51(?)) constituting the current mirror is a PNP type, the current capacity is small, and in order to flow a large current through it, it is necessary to
There was a problem that the area of transistors (51 (71) in C was extremely large.

Further, as shown in FIG. 5, a dummy emitter follower circuit consisting of a transistor (51) and a constant current source (52) is also provided at the collector of the transistor (2), and this base current [B is defined as the collector current of the transistor (2). A circuit has also been proposed that equalizes the collector current of the transistor (21 (31) and cancels the offset voltage Δ■ by subtracting it from the constant current source (52
is completely wasted, and the current consumption becomes extremely large.

On the other hand, in the circuit of FIG. 6, a transistor (61) is connected in series with the transistor (8), and the output current IO is caused to flow through this transistor (61).
The base of this transistor (61) is a transistor (
A base current IB equal to 8) is formed, and this base current IB is turned back by the current mirror of the transistors (62) and (63) to supply the collector current of the transistor (3). According to this, the base current Is drawn to the transistor (8) is supplied from the transistor (63), so that the collector currents of the transistors (2) (31) become equal.

However, in this circuit, transistors (61) to (
In order to make the circuit of 63) more convenient, a transistor (63)
1) Base-emitter voltage drop VBE such as (62)
Therefore, the dynamic range of the output voltage VO is V, ) < Vcc -2VBE Vci(sl
However, VctfS) is the saturation voltage between the collector and emitter of the transistor (8). On the other hand, the original dynamic range of the buffer amplifier is from VCl2 (Il+ of transistor (7) and VBIE of transistor (8) to Vo<Vcc
-Vcs (sl -Vsi), which lowered the dynamic range by 78 sections (approximately 0.7V).

[Problem that the invention seeks to solve]

As described above, the conventional technology has problems such as an increase in current consumption and occupied area or a decrease in dynamic range in order to eliminate the offset voltage.

[Means for solving problems]

The present invention provides differentially connected first and second active elements (two
+ (31 is provided, and an input terminal is applied to the base of this first active element (input terminal (1)) L, from the collector of the second active element to the third active element (8) of the emitter follower. Take out the output voltage (output terminal (10))
In addition, in the buffer amplifier configured to feed back this output voltage to the base of the second active element, the third
Detection by setting the output current of the active element to □ (constant current source (91
(11) ) L, this detection current is connected to the fourth active element (1
2) to form a base current, and multiply this base current by N times (transistor (13) (
14) ') A buffer amplifier configured to feed back to either of the first and second active elements.

[Effect]

According to this, the offset between the input and output voltages is canceled by feeding back the base current, the current consumption of the cancellation circuit is small, the area occupied in the IC is small, and the dynamic range does not deteriorate. .

〔Example〕

In FIG. 1, a constant current source (
11) is provided, and the constant current source (11) is configured so that a current of □ flows with respect to the current value 1o of the constant n' current source (9). Furthermore, the emitter of a transistor (12) whose collector is connected to the power supply terminal (6) is connected to a constant current source (11), and the base of this transistor (12) is connected through a transistor (13) constituting the diode side of the current mirror. Connected to the power supply terminal (6). The transistor (14) constituting the transistor side of this current mirror is configured to flow N times as much current as the transistor (I3), and this transistor (
The emitter of 14) is connected to the collector of transistor (3).

Therefore, in this circuit, the constant current source (9) has an output current of 1
When o flows, a current of □ flows through the constant current source (■1), and each time this current flows through the transistor (12), it is multiplied by N by the transistor (14),
By feeding back the current of the formed ra to the collector of the transistor (3), the unbalance of the collector current of the transistor (21 (31) due to the base current 1B of the transistor (8) is corrected, and the voltage between the input and output voltages is Offset voltage Δ■ is canceled.

In this circuit, since the cancellation circuit of the constant current source (11) to the transistor (14) is not connected to the output terminal (10), the dynamic range of the output voltage Vo is not affected by this circuit, and the original buffer The dynamic range of the amplifier can be used effectively.

O Also, the increase in current consumption is only -, which is significantly smaller than in the conventional technology.

In this way, the offset between the input and output voltages is canceled, but according to the circuit described above, the dynamic range is not reduced by this, and the increase in current consumption is also small.

Furthermore, FIG. 2 shows a case in which the area occupied by the cancellation circuit within IC is also considered. That is, in the above example, since the current is multiplied by N times by the P-type transistors (13) and (14), there is a possibility that the area occupied by the transistor (14) in particular becomes extremely large. Therefore, in this example, the P-type transistors (13) and (14) are set at a ratio of 1:1, and this current is further supplied to the N-type transistors (21) and (22).
Here, it is multiplied by N.

According to this, since an N-type transistor originally occupies a small area, even if this is multiplied by N, the area occupied by the entire circuit does not increase much.

In this example, since the operation is performed in the direction in which the current IB is taken out by the transistor (21), this transistor (21)
21) is connected to the collector of the transistor (2), and the base current 1B is subtracted to eliminate the unbalance.

In addition, in this example, in order to correct the pair characteristics of the transistors (13) (14), the base-collector of the transistor (13) is connected between the base emitter of the transistor (21), and each transistor (13) (14) is connected between the base and collector of the transistor (21).
4) Connect resistors (24) to (26) to the emitter of (23).
) has been established.

In addition, a circuit that increases the current by N times using transistors (21) and (22) does not require increasing the current by N times using transistors (22).
You can also set it to □.

Further, FIG. 3 shows a specific circuit when the circuit is applied to an output buffer amplifier of a voltage stabilizing circuit. In this figure, the output of the voltage stabilizing circuit (31) is passed through a voltage divider circuit consisting of a resistor (32), four diodes (33) to (36), and a resistor (37). ) and diode (3
3) is supplied to the base of transistor (2), and the output from transistor (8) is supplied to the diode side of the current mirror consisting of transistors (9b) and (9c) via resistor (9a). The transistor side of this current mirror is connected to current sources (4) and (
11) transistor (4a) (lla
).

Therefore, in this circuit, by making the emitter resistance (4b) of the transistor (4a) larger than the emitter resistance (9d) of the transistor (9c), 21+<<
The operating current of Io can be supplied to the transistor (21F31 etc.) through the transistor (4a), and by making the emitter resistance (llb) of the 1-transistor (lla) N times the O resistance (9d), the current of □ can be supplied to the transistor (21F31 etc.). (12).

In addition, in this circuit, it is possible to divide the output of the buffer amplifier and supply it to various parts of the IC, and in that case, if it is divided into N, the power of one current source is 1.

The current value can be set to □. So this place I.

In this case, the current from the current source in the box □ may be reversed at a ratio of 1:1 and supplied to the transistor (12).

〔Effect of the invention〕

According to the invention, the base current is fed back to cancel the offset between the output voltages, the current consumption of the cancellation circuit is small, the area occupied in the IC is small, and the dynamic range is also reduced. You can prevent this from happening.

BRIEF DESCRIPTION OF THE DRAWINGS FIG. 1 is a configuration diagram of an example of the present invention, FIGS. 2 and 3 are diagrams for explaining the same, and FIGS. 4 to 6 are diagrams for explaining the conventional technology. be.

(11 is the input terminal, (2) (31(5) (71f8
1 (12) to (14) are transistors, (4) (91
(11) is a constant current source, (6) is a power supply terminal, and (10) is an output terminal.

Claims (1)

[Claims] First and second active elements are provided which are differentially connected, and an input voltage is applied to the base of the first active element, and an input voltage is applied from the collector of the second active element to the emitter follower of the emitter follower. In the buffer amplifier, the output voltage is taken out through the active element No. 3, and this output voltage is fed back to the base of the second active element, and the output current of the third active element is set to 1/N. detect the detection current, supply this detection current to the collector-emitter of the fourth active element to form a base current, multiply this base current by N, and feed it back to either the first or second active element. buffer amplifier.