JPS6053311A - Power amplifier - Google Patents

Abstract

PURPOSE:To eliminate the unbalance of a distortion factor between output siggnals by connecting respectively a power amplifier to two output terminals of a parallel feedback amplifier and connecting a load between output terminals of the power amplifier. CONSTITUTION:A base of transistors (TR) Q1, Q3, a base of TRs Q2, Q4 and an emitter of TRs Q1, Q2, Q3 and Q4 are connected respectively in common, an input signal is fed to a base of the TRs Q1, Q3 via a TRQ7, feedback is applied from the collector of the TRQ1 to the base of the TRs Q2, Q4 so as to constitute a parallel feedback type amplifier. A collector output of the TRs Q3, Q4 is fed respectively to power amplifiers A4 and A1 and a load RL is connected between output terminals of the amplifiers A4, A1. Thus, the unbalance of the distortion factor between the output signals is eliminated and also the decrease in the input impedance is avoided.

Description

DETAILED DESCRIPTION OF THE INVENTION The present invention relates to a power amplifier device that supplies the entire output from a pair of power amplifiers to a load such as a speaker.

Generally, the circuit configuration of a power amplifier in which a load such as a speaker is connected between the outputs of a pair of power amplifiers is BTL (Bala
nced Transformer-Less) circuit used in audio amplification circuits such as toy-stereos. An example of a conventional BTL circuit? It is shown in FIGS. 1 and 2.

In FIG. 1, Ax positive phase negative feedback power amplifier.

A2 anti-phase negative feedback power amplifier, H+t −H, ur
i resistor, and the load (speaker L) is the C1-C3H capacitor, and these constitute the Also, voltage source directly to Voo, 80
indicates the signal generator 0 amplifier AIC to signal generator SG
The output signal #ri of this amplifier AI becomes a composite signal of the fundamental signal from the signal generator 8G and the distortion component and noise component generated in the amplifier 1. Ten thousand,
Is the output signal of amplifier AI the input signal of amplifier A2? Therefore, the output signal of amplifier A2 is a composite signal of the basic signal from signal generator SG, the distortion removed and noise components generated by amplifiers A, I, and A2, respectively. Therefore, the distortion and noise components of amplifier A1 are added, and therefore, the distortion and noise components of amplifier A1. An unbalanced sound of distortion and noise components occurs between each output signal of A2 ○ Is this more specific rii5? , reveal. In FIG. 1, the following equation stands for Q. Here, the distortion factor of the signal generator S'() is ignored.

vo=Avvl+v, +++ ・+・・−・(t)V
l:':el-βvo---(2) L (7, self, vl, vo, vn, Ay, and β are respectively input cuff f℃ pressure of amplifier A1, inverting input voltage, skewer voltage, ( Distortion + noise) voltage, open-loop gain, and feedback circuit gain O Here, if we eliminate to V+ from these equations, the output 'voltage of amplifier AI becomes V. The following shows vc. ○Here, amplification RyA2
On the input surface of , the pressure e1 is the output voltage V of the amplifier A1. fp,
5. Since the input is made with a resistor division ratio of Re, 5 1 (, s + HτvoVC is compressed. At this time, the input voltage CI is equal to or less than 1 ○ Here, the amplifier A2 is a negative-phase negative feedback power amplifier. So,
Assuming that the gain is represented by a symbol similar to amplifier AI, and that the current flowing through resistor 3 is 1, the following equations hold true.

One = = Av'vl'tenvn' ・・・・・・・・・
・(5)v(=e(-kLs■+βv,/ , , ,
, , , , (6) Where is the input impedance of amplifier A2? If it is set to infinity, the following formula holds true.

By eliminating vl' and ITh from these equations (5) to (7), the following equation is obtained from β'=l(,3 ball T.

Roughly speaking, since amplifiers A, I, and A2 are the same custom-made amplifiers, vn=vn', Av=Av', and β=β'.0 Also, here, 4IPgA1. In order to make the respective output voltages of amplifiers AI and A2 equal, the input voltage of amplifier A2 is set to the input voltage of amplifier A1. That is, the output voltage V2 of the amplifier A1 is equal to the output voltage of the amplifier AI. ′ becomes as follows.

...(8)' Here, (31, (ff) Comparing all the formulas, the distortion and noise voltage of the output voltage of amplifier A1 are also reduced to the distortion and noise voltage of the output voltage of amplifier A2 (1-7 , ) is unbalanced by v.

As a specific example of this, AV=1000 (60d/),
In the case of β = virtual vn-10nv, the increase in (distortion and noise) and pressure at the output of amplifier A2 △vn'i△vn-
()) (l p,)" 1 mv in o, and the distortion and noise voltage of amplifier A2 is 1 mv from the distortion and noise voltage of amplifier A1.
(It increases by 1% and causes an unbalanced sound by that amount.)Next, the input impedance Z1 at the input terminal 11c is
Assuming that the input impedance of the amplifier A1 is 1Zio, it becomes 0 as shown in the following equation.In the qo() equation, since US is a capacitor for bypassing the AC signal, it becomes l(,R=)LIO since it is sufficiently small compared to torc. Also, since Z+o rr is the input impedance of the positive phase negative feedback power amplifier Al, Rrr
Since the amount of light is larger in comparison, it can be set as 1L+o during ZI.

For example, Rs=R9=10Ω, 6=47μF, J=
, 1. [Se.

R+o =: 1(Il+ = 50KQ, Zio =
If it is 5MΩ, then Zi=49.5X10 KiRIO.

Furthermore, the output of amplifier A1 is connected to the input of amplifier A2 through resistor R1.
5. Since the IC is made of 6-piece split IC Elly material, resistance loss, R
8, so that the DC operating point 2 has a variation of 1.7, the DC cut capacitance (: 4 is required). Figure 2 shows the circuit diagram of another conventional BTL circuit. Input terminal IVC of phase negative feedback power amplifier AI
, a resistor 3. from the negative phase input terminal 3 of the negative phase negative feedback amplifier A2.
The resistor I(, s,
It is transparent to the intersection of R9, H and IO. In this figure, the input signal rX of the amplifier AI, A2 is input from the direct signal generator S(), so the amplifier /l+, A
Each output signal of 2 is output as a basic signal from the signal generator SG and a current signal of the distortion and noise components generated by the amplifiers AI and A2, respectively. Here, when amplifiers AI and A2 are configured with the same circuit, the distortion and noise components generated in the amplifiers are almost the same, so the distortion and noise between the respective output signals of amplifiers AI and A2 are Unbalance? It doesn't happen.

Next, the input impedance Ziho of the input terminal l is expressed by the following equation as in the case of FIG.

Here, from input terminal l to amplifier A2? This amplifier is a reverse-phase negative feedback amplifier.
Since it is sufficiently smaller than c, it becomes R3, and as in FIG. 1, it becomes halo in rcR. (Zj in formula 10.

Since it is the input impedance of the in-phase negative feedback power amplifier AI, I(, (kiRto), )(,31c is larger by an optical amount than
It is set with the parallel resistance value of R3, but the general resistance value of R3 is v
c The resistance is from several hundred Ω to several Ω, and the threshold voltage gain A, v of the negative-phase negative feedback power amplifier A2 is
If the open circuit voltage gain AVO increases by an optical amount h, Av = "/R4. Therefore, in order to increase the input impedance Z+ seen from the input terminal l, the resistance value R3 is increased taX.
Therefore, the resistance value R4 inevitably becomes large, and the resistance value R7 also needs to be made large. This is because the resistor 7 is set to the same value as the resistor I (, 4) in order to keep the current bias voltage at the input terminal of the amplifier constant.In this case, the resistor connected to the amplifier AHC R
4,') The resistance value of R7 is thick (2#.

and the unbalance 1' between the resistance values of the resistors R2 and 10 connected to the amplifier A1.
This will cause a difference in the DC bias voltage at the common-mode input terminal of A2, and the DC current will flow into the output load, which may burn out. Input impedance ZiIHR (R+o in lt
) and R3, the resistance value of the input impedance Zi + uR3 seen from the input terminal l becomes dominant,
Compared to the input impedance Zi (e.g., tJf, 50Ω) in FIG.
Is it a problem of a decrease in T (Ω)? arise. To give a specific example, 几s = R9 = l (lJ(rλ R10 = 5(
IKΩ, e5==47 try, f=IJ<1llz.

■If 3=2 and Ω+”0=5MΩ, then 1(・==
Rtθ, ZI′=1.92×10 (,3).

Further, as in FIG. 1, a capacitor 4 for DC cut is required so that the DC operating point 3 does not vary with respect to the input of the negative phase negative feedback amplifier amplifier 82.

The purpose of the present invention is to solve all of these problems - to eliminate the unbalance of distortion between output signals in the hiL circuit,
Power increase @ equipment f*:! that eliminates the drop in input impedance and eliminates the need for the specified DC cut coupling capacitance of the feedback input. r
YC has something to offer.

In the first fnry of the power amplifier of the present invention, each pace and emitter are warped, and it is connected to a common ground.
and a third transistor whose bases and emitters are connected in common. A F differential amplifier is configured with the first transistor and the fourth transistor. Input from the base of the second transistor? supplying said third. The base of the fourth transistor is used as a feedback input, and the first... a parallel feedback amplifier with nine output terminals for each collector of the third transistor;
Did you connect each output terminal of this parallel feedback amplifier with seven connections and connect the input terminals with each other? Features.

FIG. 3 is a circuit diagram of a second embodiment of the present invention, and FIG. 4 is a detailed circuit diagram of the amplifier Al (A4) in FIG. 3.

In the figure, A1. A41”I positive phase negative feedback power amplifier, A3 is a parallel feedback amplifier, D1 to D4 are diodes,
QI~Qyrr transistor, ■ is also +4. IJ') resistor, 11M13i constant current source, with the same symbol as in Figure 1 and the same -S component? show. Amplifier A is installed in 4 and 5 respectively.
3 output terminal, 6. 71'; J transistors Q1, Q3 + Q2 and Q4 σ birth terminal, 8 stage amplifier A3
input terminal.

10.11, 12 are in phase with amplifier AI, anti-phase input terminals I and output terminals, 13, 14, 151-r are in phase with amplifier A4.

These are a negative phase input terminal and an output terminal. 1, parallel negative feedback power amplifier A3, transistor Q1. The base of Q3 and transistor Q2. The bases of Q4 are commonly connected and the transistors Ql, Q, 2 . Q3. Q4
The emitters of the transistors Ql and Q2 are also connected in common.
Work Q3. It constitutes a Q4 force amplifier. A constant wL current source 12 is connected to this common emitter, transistors Q, 1. The collector of Ql+ is connected to the base of transistor Q6, the base of transistor Q5 is connected to the collector of transistor Q2, and a diode D5 is connected to the collector of transistor Q3.
and the common-mode input terminal 13 of amplifier A4 [connected to the collector of transistor Q4, the collector of transistor Q4, the common-mode input terminal of amplifier A1 connected to the diode D4, and the common paste of transistors Ql, Q3 and the castor current source 11. The connection point with the diode D1 is connected to the emitter Vc of the input transistor Q7. Connection 9, transistor Q
21Q4 common pace constant current source ■3 and transistor Q
In this figure, the transistor QI, Q = base and emitter are common, so it is a single transistor QA with two collectors.
In addition, the transistors Q2 and Q4 are considered to be one transistor Q and B having two collectors, and the emitters of the transistors QA and QB are considered to be a common difference amplifier.

The base IC (2) of transistor Q6 is connected to the collector of this transistor QA, and this transistor Q,
Since the collector of the transistor Q6 is connected to the constant current source 3 and the base V'C connection point vc of the transistor QB, this differential amplifier serves as a total feedback amplifier. Therefore, since the pace potential of the transistor QA (the potential at the connection point 6) and the base potential of the transistor Qn (the potential at the connection point 7) are equal, there is one collector.

(The collector of Q3 and one collector of transistor QB (collector of Q4)
, 5. For this reason, additional reheating device A],
There is no difference in the input voltage of A, but there is a difference in output onset voltage between the outputs of amplifiers AI and A4. Furthermore, the output signal of output 9 @ 5 becomes a negative phase signal with respect to the input signal, and it becomes a positive phase signal at output No. 3 of output terminal 4, so that it is connected to the primary stage amplifier (7 is positive phase negative feedback 1). Force amplifier (AI, A4
) is required, and an anti-phase negative feedback power amplifier is not required. Tfl, that is, since amplifiers with the same configuration are used, it is considered that the distortion and noise voltage generated from the amplifiers are equal.

Next, the present invention will be explained in detail.

First, the input signal G input from the No. h generator SG is directly connected to the amplifier At via the parallel negative feedback amplifier A3.
It is input to each input terminal 10.13 of A4.

Therefore, the output signal of the amplifier AI is the composite signal of the basic signal from the signal generator SO, the distortion and noise components generated by the parallel negative feedback amplifier A3, and the distortion and miscellaneous components generated by the amplifier AI. be. Similarly, the output of the amplifier A4 is 1G, the basic signal from the pulse generator 8G, the distortion and noise components generated by the parallel negative feedback amplifier A3, and the distortion and noise components generated by the amplifier A4.
It becomes a composite signal with j1-55C. Here amplifier AI
, A4 have the same circuit configuration, then 1 to the amplifier.
．． The difference between the output signals of the A4 will result in an unbalanced sound due to the distortion and noise components.

Next, the input impedance Z+2 when viewed from the input terminal 8 within the circuit is expressed by the following equation.

TATASI, ZV:r is the input interference inside the circuit from the base of transistor Q7, and there is Ω, I(
+t is about several tens of Ω, so Z i zki1 (+14
Tonaotsu. Example, t, z==, 14Ω, R14=5
0, ΩtoTbtoZr2 = 3Ω 48.8XlO (KR・1).

-! In addition, in this embodiment, amplifier A2 is added as in the conventional circuit.
A DC cut capacitor (C4) k is not required to prevent the DC operating point from being fully submerged at the input of.

Furthermore, the output offset voltage FEt of amplifier AI, A4
There is no influence due to c. That is, even if the collector current of the transistor Q3 of the amplifier A3 is twice the collector current of the transistor Q4, the amplifier AI,
Only the potential difference i18mv between the input terminals 10 and 13 of A4 changes. In general, the permissible DC voltage across the load that does not cause load burnout is 150mV or more, so the output offset voltage is J:v.Load burnout? There is nothing that happens.

Fig. 419 shows a detailed circuit diagram of the A4 section of amplifier AI in Fig. 3. In the figures, the same numbers as in Figures 1 to 3,
Symbols indicate the same components, D6 and 7 are diodes, Q
2 to Q+7 transistors, 20 to 1L26 resistors, and 14 constant current sources. The positive phase negative feedback power amplifiers AI, A, and 4 have the same circuit configuration.

As explained above, by using the B 'I' L configuration circuit of the present invention, the unbalance of distortion between output signals, the reduction in input interference, and the DC operation of the output terminal in the conventional BTL circuit can be avoided. Problems such as the insertion of a current cutting capacitor at the feedback input terminal to prevent point fluctuations can be improved.

Note that transistor Q! in amplifier A3 in diagram N3!
and Q3. It is clear that the same effect can be obtained by constructing a differential amplifier consisting of transistors Q2 and Q4 by making each of these transistors into a multi-collector and constructing a differential amplifier.

[Brief explanation of drawings]

Fig. 1, Fig. 2 is an example of a power amplifier according to the company? FIGS. 3 and 4 are circuit diagrams showing first and second embodiments of the present invention. In the figure, A, l, A4...
... Positive phase negative feedback power amplifier, A2... Negative phase power feedback power amplifier, A3... Parallel negative feedback amplifier, +-C+
...Continued length, D++-D7...Diode, II-J3...Constant current source, Q1-Q
71 QIO~Q17...Transistor, 1 odor~It! 5. R2o~R26・・・・・・・・・
・Resistor, RL...Load such as speaker, 8G...Signal generator, l, l O, 13...Common mode input terminal of amplifier, 2...Resistance connection point, 3゜11.14... ...Reverse phase input terminal of amplifier, 4.5... Output terminal of amplifier can 3, 6,7... Common pace terminal of transistor, 8.
... Input terminal of amplifier A3, 12°15 ...... Output terminal of amplifier, 0

Claims (1)

[Claims] The bases and emitters of each base are connected in common. A third transistor whose base and emitter are commonly connected to the F2 transistor. A differential amplifier is configured by the fourth L transistor, and the first... An input is supplied from the base of the transistor F2, and the third. The base of the fourth transistor is a full feedback input; a parallel feedback amplifier with each collector of a third transistor as an output terminal; a first and second power increasing amplifier device in which each output terminal of the parallel feedback amplifier is connected to an input terminal; .