JPS6130447B2 - - Google Patents

Description

DETAILED DESCRIPTION OF THE INVENTION The present invention relates to an amplifier circuit, and particularly to an amplifier circuit using bipolar transistors.

In amplifiers that use transistors as amplifying elements, the input/output characteristics between the base and emitter of the transistor exhibit nonlinearity, so negative feedback is generally applied to suppress distortion due to this nonlinearity. was. However, if negative feedback is applied, a reduction in amplification cannot be avoided, and thus not only will many transistors be required to obtain the desired amplification, but there is also a risk that the stability of the circuit will deteriorate and oscillation may occur. Gender also exists.

To this end, amplifiers have been devised that can improve distortion due to transistor nonlinearity without applying negative feedback. An example of such an amplifier is shown in FIG.
_IN is a PNP transistor with emitter follower configuration
This becomes the base input of _Q1 , and the emitter follower output of this transistor _Q1 becomes the base input of the next stage amplifying NPN type transistor _Q2 . An emitter resistor _R1 is connected between the emitter of the transistor _Q2 and a reference potential (ground) point. A constant ratio of currents I ₁ and I ₂ flows through both transistors Q ₁ and Q ₂ .
(I ₁ /I ₂ = 1/α, α is constant), for example, is provided with a current mirror circuit 1, and as shown in the figure, this current mirror circuit 1 consists of PNP transistors Q whose bases are commonly connected to each other. ₃ and _Q4 . The transistor Q ₄ has a diode configuration in which the base and collector are commonly connected, and the transistors Q ₃ and Q ₄ are so-called mirror coupled to each other, and each emitter resistance R ₂ of the transistors Q ₃ and Q ₄
By appropriately selecting and R ₃ , the ratio of currents supplied to transistors Q ₁ and Q ₂ (1/α) can be set. Furthermore, the current mirror circuit 1 includes a transistor.
A PNP transistor _Q5 whose bases are commonly connected to _Q3 and _Q4 is provided, and this transistor
The collector of Q ₅ is connected to a reference potential (ground) point via a resistor R _L . And this resistance R _L
The configuration is such that the amplified output V _OUT is derived from both ends of the . Here, the ratio of the currents I ₂ and I ₃ supplied to the transistor Q ₂ and the resistor R _L is determined to be 1/α' by selecting the value of the emitter resistor R ₄ of the transistor Q ₅ .

In such a configuration, transistors Q ₁ and Q ₂
Let the base-emitter voltages of the transistors be V _BE1 and V _BE2 respectively, the emitter line voltage of the transistor Q ₁ be V _A , and the transistors Q ₁ and Q ₂ and the resistor R
If the currents flowing through _L are respectively I ₁ , I ₂ and I ₃ , the following equation holds true.

V _A = V _IN + V _BE1 …(1) I ₂ = (V _A − V _BE2 )/R ₁ …(2) V _OUT = I ₃・R _L …(3) Here, I ₃ = α Since 'I ₂ , the following equation is obtained from equations (1) to (3).

V _OUT = (V _IN +V _BE1 −V _BE2 )・R _L /R ₁・α′
...(4) Generally, the relationship between the collector current Ic of a transistor and V _BE is expressed by the following equation.

Ic=Is( _espqVBE /kT-1)...(5) Here, q is the electronic charge, k is Boltzmann's constant,
T is the absolute temperature, and Is is the base-emitter reverse saturation current. Transforming equation (5), we obtain the following equation.

V _BE =kT/qln (Ic/Is+1) (6) Therefore, (V _BE1 - V _BE2 ) in equation (4) is expressed by the following equation.

V _BE1 −V _BE2 =k/q{T ₁ ln (I ₁ /Is ₁ +1) − T ₂ ln (αI ₁ /Is ₂ +1)}...(7) Here, T ₁ and T ₂ are transistors Q ₁ , Q ₂ is the base-emitter junction temperature.

Also, since Is is a constant specific to the transistor,
It can be set as Is ₂ =βIs ₁ (β is constant), and since Is is an extremely small value and if a sufficient collector current is allowed to flow, Ic/Is≫1 holds true, so the following equation can be obtained.

Ic/Is+1≈Ic/Is (8) Therefore, by using equation (8), equation (7) becomes as follows.

V _BE1 −V _BE2 ≒k/q{T ₁ ln (I ₁ /Is ₁ ) −T ₂ ln (αI ₁ /Is ₁ )} …(9) In equation (9), assuming that the junction temperature T of the transistor is constant, For example, V _BE1 - V _BE2 = kT/qln (β/α)...(10) From equation (10), (V _BE1 - V _BE2 ) is a constant value, so
If this value is set as γ, equation (4) becomes the following equation.

V _OUT = (V _IN + γ)・R _L /R ₁・α′...(11) As can be seen from equation (11), the output V _OUT is independent of the V _BE of each transistor and is caused by V _BE . Distortion will no longer occur. In this way, it is possible to improve distortion due to nonlinearity of the transistor without applying negative feedback.

Here, if R ₁ = R ₂ = R ₃ = R ₄ is selected, γ = 0, α' = 1 in equation (11), and the gain G of the circuit is R
_L /R becomes ₄ . Now, if G = R _L /R ₄ ≒ 2, each voltage distribution of the series circuit of the output resistor R _L , transistor Q ₅ and resistor R ₄ will be the second
It will look like the figure. In the figure, V _R4 represents the voltage across the resistor R ₄ , V _CE represents the emitter-collector voltage of the transistor Q ₅ , and V _RL represents the voltage across the output resistor RL (V _OUT ). The amplitude at each point of the emitters and bases of transistors _Q1 and _Q2 in the input stage is approximately equal to V _R4 , and if the gain G is selected to be large, the amplitude at the output terminal will decrease as shown in Figure 2. As V _RL (=V _OUT ) becomes extremely large, the amplitude at the emitter and base of each transistor Q ₁ and Q ₂ becomes small. Therefore, if the circuit power supply +B ₁ is set to a large value in order to obtain the necessary gain G, not only will unnecessary voltage be generated in the transistors Q ₁ and Q ₂ , resulting in large power wastage, but also PC
This is inconvenient since it is necessary to use a large transistor.

Therefore, an object of the present invention is to provide a distortion-free amplifier circuit that consumes little power while ensuring sufficient circuit gain.

The amplifier circuit of the present invention is provided with a second transistor having a conductivity type opposite to that of the first transistor, with the output of the first transistor having the base inputted with an input signal applied to the base, and a current of a fixed ratio is applied to both transistors. A current output means is provided to generate and output a current having a constant ratio with the current flowing through either of these transistors. The present invention is characterized in that an operating voltage source for the current supply means and an operating voltage source for the current output means are provided separately so that the absolute value of the latter voltage is larger than that of the former.

The present invention will be explained below using the drawings.

FIG. 3 is a circuit diagram of one embodiment of the present invention, and parts equivalent to those in FIG. 1 are designated by the same symbols.
Transistor of current mirror circuit 1 in Fig. 1
Q ₅ is removed and a current mirror circuit 1' consisting of transistors Q ₃ and Q ₄ is provided to supply constant ratio currents I ₁ and I ₂ to both transistors Q ₁ and Q ₂ , respectively. A current mirror circuit 2 is provided which inputs the current I ₁ flowing through the transistor Q ₁ and outputs a current I ₄ having a constant ratio thereto. This circuit 2 is a diode-connected NPN transistor
Q ₆ and an NPN transistor Q ₇ which has a common base with the base of this transistor and an emitter resistor
Consists of R ₅ and R ₆ . The output current I ₄ from the transistor Q ₇ is supplied to the load resistor _RL , and the output V _OUT is obtained from both ends thereof. As an operating voltage source for the current mirror circuit 2, a voltage source _-B2 different from the operating voltage source + _B1 of the current mirror circuit 1' is provided. Here, by appropriately setting each emitter resistance of the current mirror circuit, the currents I ₁ and I ₄
The ratio of

Even in this configuration, equations (1) and (2) hold true, and the output V _OUT is expressed by the following equation.

V _OUT =-I ₄ ·R _L (12) Since I ₄ = βI ₁ and I ₂ = αI ₁ , the following equation is obtained from equations (1), (2), and (12).

V _OUT =-βR _L /αR ₁ (V _IN +V _BE1 -V _BE2 )...
(13) Also, since V _BE1 −V _BE2 is a constant value γ according to equation (10), V _OUT =−βR _L /αR ₁ (V _IN +γ) (14). As can be seen from equation (14), the output V _OUT is independent of the V _BE of the transistor, and no distortion occurs due to V _BE . Furthermore, α=1, β=1 (R ₁ = R ₂ =
It is clear that if R ₃ = R ₅ = R ₆ ), γ=0 and the output V _OUT becomes -R _L /R ₆ ·V _IN .

Under the above conditions, that is, α=1, β=1,
Assuming that the maximum amplitude of the output V _OUT is 28 volts and the circuit gain G is set to 20 dB, the amplitude at both ends of the resistor R _L is the same as V _OUT , and the resistance at that time is
The amplitude across R ₆ is 2.8 volts. Power supply
The voltage of B ₂ must be 30 volts or more.

However, since the amplitude across each end of resistors R ₁ , R ₂ , and R ₃ at this time is a maximum of 2.8 volts,
The minimum value of the circuit power supply +B ₁ can be calculated as follows.

+B ₁ (min) = 2.8 + V _CE +2.8 + V _BE4 Here, V _CE is the collector of transistor Q ₂ .
The emitter voltage, _VBE4 , is the base-emitter voltage of transistor _Q4 . Assuming that V _BE4 = V _CE ≈0.6 volts, +B (min) will be approximately 7 volts, and even if designed with a certain margin, it will be sufficient to use a circuit power supply of at most 10 volts. Therefore, the Pc and breakdown voltage of the transistors Q ₁ , Q ₂ and the like need only be set to low values, and power consumption can also be prevented.

FIG. 4 is a circuit diagram of another embodiment of the present invention,
Components equivalent to those in FIGS. 1 and 3 are designated by the same reference numerals. In the figure, the collector output current I ₃ of the transistor Q ₅ of the current mirror circuit 1 of FIG.
By supplying it to Q ₆ , the same effect as shown in Fig. 3 can be obtained.

In this example as well, if R ₁ = R ₂ = R ₃ = R ₄ = R ₅ = R ₆ , the circuit gain G becomes -R _L /R ₆ , and the third
As shown in the figure, it is possible to set the power supplies +B ₁ and -B ₂ to 10 volts and -30 volts, respectively, so that G=20 dB, resulting in low power consumption.

According to the present invention, there is an advantage that distortion-free output can be obtained and wasteful power consumption of the circuit can be suppressed to a minimum.

Although a current mirror circuit is used for both current supply and output current generation, it goes without saying that a circuit configuration having the same function as this can be used.

[Brief explanation of the drawing]

FIG. 1 is a diagram showing an example of a conventional amplifier circuit, FIG. 2 is a voltage distribution diagram of the circuit of FIG. 1, and FIGS. 3 and 4 are circuit diagrams of an embodiment of the present invention. Explanation of symbols of main parts: 1, 1', 2...Current mirror circuit, _Q1 to _Q7 ...Transistor.

Claims (1)

[Claims] 1 A first transistor to which an input signal is applied to a base, a base coupled to an emitter of the first transistor, and an emitter connected to a reference potential point via a resistor, and of a conductivity type opposite to that of the first transistor. and a second transistor driven by the first power supply via an emitter resistor to the first transistor.
and a first current mirror circuit consisting of a plurality of transistors mirror-coupled to each other that supplies a constant ratio of current between the emitter and collector of each of the second transistors, and an output that has a constant ratio to the current supplied by the first current mirror circuit. The second current mirror circuit includes a second current mirror circuit including a plurality of transistors mirror-coupled to each other that generates a current, and an output means that generates an output voltage that changes depending on the output current of the second current mirror circuit. Each transistor is driven through an emitter resistor by a second power source that produces a voltage of a different polarity and larger than that of the first power source, and the transistors forming its output stage are driven through a load resistor that produces the output voltage. An amplifier circuit characterized by having a collector connected to a reference potential point.