JPS61244107A - Negative feedback amplifier - Google Patents

Abstract

PURPOSE:To remove the current amplification factor of a PNP transistor (TR) from determination factors of an operation point, to prevent the operation point from deviating from a set value even if the PNP TR has variance in current amplification factor, and to prevent the dynamic range of an output waveform from narrowing by adding the PNP TR to a negative feedback amplifier. CONSTITUTION:A PNP TR 20 is provided; and its base is connected to the emitter of a PNP TR 1, the emitter is connected to the collector of a PNP TR 6, and the collector is connected to the collector of an NPN TR 4. The operation point of an output terminal (b) is determined only by the voltage at a negative feedback input terminal. The determination factors of the operation point V0 do not include the current amplification factor hFE2 of the PNP TR 2. The operation point V0 of the output terminal (b) is set to a proper value by letting currents from the collectors of the PNP TRs 5 and 6 to a feedback circuit so that their constant current values are different, and the determination factors of the operation point V0 do not include the current amplification factor hFE2 of the PNP TR 20.

Description

DETAILED DESCRIPTION OF THE INVENTION [Field of Industrial Application] The present invention relates to a negative feedback amplifier mainly composed of semiconductor integrated circuits.

[Conventional technology]

A conventional example of a negative feedback amplifier is shown in FIG. a is the input terminal,
b is the output terminal, C is the negative feedback input terminal, d is the power supply terminal, e
is a ground terminal, 1, 2, 5, 6.7 are PNP) transistors, 3.4 is an NPN) transistor, 8 is a constant current source, 9
is an inverting amplifier. The output amplified by the PNP transistor 1.2 is generated in a load composed of a transistor 3.4 and amplified by an inverting amplifier 9. The output of the inverting amplifier 9 is fed back to the emitter of the transistor 2 via a negative feedback circuit including resistors 16 and 17 and a capacitor f'18.

Transistors 5, 6.7 constitute a constant current source that provides an operating current. The resistor 15 is a load resistor, and the capacitor 19 is a pulling capacitance.

PNP) Run raster 5, 6. F, resistance 12, 13゜1
4 and 11 are constant current circuits having a current mirror circuit configuration. Supply current Io, P of constant current source 8
NP transistor 5, 6.7 base current from 5s to 6 + from 7+ ””P transistor 5,
6 collector currents respectively. 5 + engineering C6.

PNP) All transistors 5, 6.7 have the same characteristics,
resistance! 2, 13. When the resistance values of 14 are all equal, I
bs''Ib5=Ib7, and if this is taken as a formula, 5+''C6 becomes Ics=IC6=IO3Ib, and the collectors of transistors 5 and 6 supply the same constant current. This constant current is set to 3.

NPN) transistors 3 and 4 and resistors 10 and 11 are N)'
Construct a current mirror circuit in which the collector of N) transistor 4 is input and the collector of NPN transistor 3 is output, and when NPN) transistor 3 and 4 have the same characteristics and the resistance values of resistors 10 and 11 are equal, An output current approximately equal to the input current of the current mirror circuit is obtained from the collector of the NPN transistor 3. Here, the input current of the current mirror circuit is a constant current generator supplied from the collector of the PNP transistor 6. Since it is the output current of the current mirror circuit (NPN), the collector current of the transistor 3 becomes I'o.

When the input impedance of the inverting amplifier 9 whose input is the collector of the PNP transistor 2 is very large, the input current of the inverting amplifier 9 becomes almost @0'', and the collector current of the PNP transistor 2 is equal to the collector current of the NPN transistor 3. be equal.

In other words, the collector current of transistor 2 is 0
becomes. Therefore, if the base current of PNP transistor 2 is Ib2, the emitter current of PNP transistor 2 is IQ + I
It becomes b2.

Therefore, from the collector of PNP transistor 5 to PN
P) The current supplied to the emitter of transistor 2 is P.

Therefore, 2 flows from the negative feedback input terminal C').

The operating point of the output terminal is determined by the negative feedback input terminal voltage and the voltage generated in the negative feedback circuit.

The negative feedback input terminal voltage is determined by the voltage between the bases and emitters of the PNP transistors 1 and 2. If the PNP transistors 1 and 2 have the same characteristics, the voltage between the bases and emitters of each is independent of the collector current. The value is almost constant and the same.

If this is 't VBB, the negative feedback input terminal voltage is 2V
It becomes BB.

Since the current flowing through the resistor 16 of the negative feedback circuit is Ibz, the voltage generated in the negative feedback circuit is R16·Ibz, assuming the resistance value tR16 of the resistor 16. Furthermore, if the current amplification factor t'hPB2 of the PNP transistor 2 is Ibz = I
o/ fipgz, so the voltage generated in the negative feedback circuit is R16. /hpgz.

■ Determine the operating point of the output terminal. Then Vo =
2VBB + Rlg IO/11FB2.

[Problem that the invention seeks to solve]

The conventional negative feedback amplifier described above has P as a determining factor of the operating point.
It includes the current amplification factor of the NP transistor, and this P
Since the current amplification factor of NP) transistors varies widely depending on the product, there is a drawback that the operating point deviates greatly from the set value and the dynamic range of the output waveform becomes narrow.

[Means for solving problems]

The uninvented negative feedback amplifier includes a first transistor whose base is an input terminal and whose collector is grounded, and a second transistor whose base is connected to the emitter of the first transistor and whose collectors are connected by a current mirror circuit. ．．
a third transistor; The first . The device includes a second constant current source, an amplifier to which an input is applied to the collector of the second transistor, and a feedback circuit that feeds back the output of the amplifier to the emitter of the second transistor.

〔Example〕

Next, the present invention will be explained with reference to the drawings.

FIG. 1 shows an embodiment of the present invention. For the sake of simplicity, parts corresponding to those in FIG. 2 showing the conventional example are given the same reference numerals.

In terms of configuration, a PNP transistor 20 is newly provided in addition to the conventional example, and the emitter of the base t-PNP transistor 1 and the collector of the emitter 1 PNP transistor 6 are connected to each other.
The collector of the NPN transistor 4 is connected to the collector of the NPN transistor 4.

As in the conventional example, PNP transistors 5 and 6.7 all have the same characteristics and the resistance values of resistors 12 and 13.14 are the same.PNP) The collectors of transistors 5 and 6 supply the same constant current, Worker. shall be. PNP) If the run raster 200 base current is Ib20, PNP
) The collector current of transistor 20 is I'o Ib2o
becomes.

Furthermore, as in the conventional example, the NPN transistors 3.4 have the same characteristics and the resistance values of the resistors 10 and 11 are equal, and when the input impedance of the inverting amplifier 9 is very large, the PNP
) The collector current of the transistor 2 becomes a value Io - Ibzo, which is equal to the collector current of the transistor 20.

(PNP) When transistors 2 and 20 have the same characteristics, if their collector currents are equal, their emitter currents are also equal. Therefore, the emitter current of the PNP transistor 2 is equal to 3, which is equal to the current supplied from the collector of the PNP transistor 5, so the negative feedback circuit V
The cJK style doesn't flow.

Therefore, the operating point of the output terminal is determined only by the negative feedback input terminal voltage. As in the conventional example, the negative feedback input terminal voltage is determined by the base-emitter voltage of the PNP transistor 1.2 and becomes 2VBB. Therefore, the operating point of the output terminal ■. ■. = 2 VBE, so in this case the operating point ■. The current amplification factor hFE2 of the PNP transistor 20 is not included in the determining factors.

Here, the operating point o of the output terminal is set to 1 of the normal power supply voltage VOO so that the dynamic range of the output waveform is maximized.
/2 voltage, but the negative feedback circuit of the present invention has a voltage of P
NP) By making a difference between the constant current values supplied from the respective collectors of transistors 5 and 6 and allowing current to flow through the feedback circuit, the υ operating point ■0 can be set to an appropriate value. Also, the current amplification factor hFB2 of the PNP transistor 20 is not included in the determining factor of the operating point Vo.

The collector current of each of the PNP transistors 5.6 of the present invention shown in FIG.
If the resistance value of transistor 3 is '12*'' 13, then the resistance value of transistor 20 can be set to 12.

When toss, the emitter current of transistor 20 is IC
Be 6, 9. PNP) The collector current of the transistor 20 becomes Ice - Ibzo. If the NPN transistors 3 and 4 have the same characteristics, the resistance value of the resistor 10.11 is the same, and the input impedance of the inverting amplifier 9 is very large, the collector current of the PNP transistor 2 will also be Ic.
6-Ib20. PNP) When transistors 2 and 20 have the same characteristics, if their collector currents are equal, their emitter currents are also equal.
) The emitter current of transistor 2 becomes IC6.

Since the current supplied from the collector of PNP) transistor 5 to the emitter of PNP) transistor 2 is ICl5, the current flowing from negative feedback input terminal C is 1. 6-I.

It becomes 5.

Negative input terminal voltage is IC5"IC6"l. As in the case of 2VBR, if the resistance value of the resistor 16 is 16, then the operating point of the output terminal is 2. is Vo = 2 ■BIH
+(I.a-Ics)Rts. Therefore, the current amplification factor hpmz of the transistor 20 (KPNP), which is the determining factor for the operating point (i)o, is not included.

〔Effect of the invention〕

As explained above, by adding a PNP transistor to a conventional negative feedback amplifier, the current amplification factor of the PNP transistor can be removed from the determining factor of the operating point. Even if the current amplification factor varies, the operating point will not deviate from the set value, and the dynamic range of the output waveform will not become narrow. Furthermore, by making a difference between the two constant currents, it is possible to set the operating point by flowing current through the feedback circuit, and there is also the effect that the dynamic range of the output waveform can be widened. In particular, when the present invention is used in a semiconductor integrated circuit, it is easy to make the characteristics of each transistor the same, so it is easy to obtain a negative feedback amplifier in which the dynamic range of the output waveform is not affected by variations in transistor characteristics among products. It has the advantage of being able to

[Brief explanation of drawings]

FIG. 1 is a circuit diagram showing an embodiment of a negative feedback amplifier according to the present invention, and FIG. 2 is a circuit diagram showing an example of a conventional negative feedback amplifier. 1.2,5,6,7.20...PNP) transistor, 3.4...NPN) transistor, 8
... Constant current source, 9 ... Inverting amplifier, 1
0,11,12,13,14゜15.16.17...
...Resistance, 18.19...Capacitor, a.
...Input terminal, b...Output terminal, C...
...Negative feedback input terminal, d...Power supply terminal, e
・・・・・・Ground terminal.

Claims (1)

[Claims] A first transistor whose base is an input terminal and whose collector is grounded; and second and third transistors whose bases are connected to the emitter of the first transistor and whose respective collectors are connected to a load of a current mirror circuit configuration. a transistor, first and second constant current sources that supply constant current to respective emitters of the second and third transistors, an amplifier to which an input is applied from the collector of the second transistor, and the amplifier a feedback circuit that feeds back the output of the second transistor to the emitter of the second transistor.