JPH048963B2 - - Google Patents

Description

DETAILED DESCRIPTION OF THE INVENTION Field of Industrial Application The present invention relates to an amplifier circuit used in radio receivers, stereo equipment, and other equipment.

Configuration of conventional example and its problems FIG. 1 is a schematic electrical connection diagram of a conventional amplifier circuit. In Fig. 1, the amplifier 3 has a positive input terminal (+), a negative input terminal (-), and an output terminal 6, and generally the DC voltage of the output terminal 6 is connected to one power supply (Vcc) and the other power supply (ground). ), it is constructed so that a large vibration signal can be extracted. So the resistance R ₁ and the resistance
Set R ₂ to almost the same value and set the bias voltage of the positive input terminal (+) to approximately the middle voltage between one power supply (Vcc) and the other power supply (earth).
Negative feedback is applied to the negative input terminal (-) from the output terminal 6 through the resistor _R3 , so that the voltage at the output terminal 6 operates as a voltage approximately halfway between one power supply (Vcc) and the other power supply (earth). It is configured as follows.

FIG. 2 shows more specifically the amplifier 3 shown in FIG. 1, and as the amplifier 3, a differential amplifier consisting of two transistors 9 and 10 is often used. In FIG. 3, the base of transistor 9 is a positive input terminal, and the base of transistor 10 is a negative input terminal.

Then, resistors R ₁ and R ₂ are connected so that the base of transistor 9, that is, the positive input terminal, has a voltage approximately halfway between one power supply (Vcc) and the other power supply (earth).
The values of are almost the same. From this, since the contact potential difference V _BE between the base and emitter of transistors 9 and 10 is approximately 0.7V, transistors 9 and 1
The emitter voltage of 0 is approximately 0.7V lower than the base of transistor 9. Therefore, when one power supply (Vcc) becomes 1.4V or less, the base voltage of transistor 9 becomes 0.7V or less, and the voltage between the emitters and bases of transistors 9 and 10 becomes 0.7V or less.
Current will no longer flow and the device will no longer operate.

That is, the conventional amplifier circuit has a drawback that the amplifier circuit no longer operates when the Vcc voltage becomes 1.4V or less.

FIG. 3 shows another conventional example, in which a signal is applied to the base of the transistor 15,
This is amplified by a differential amplifier consisting of transistors 17 and 18, and applied from the collector of transistor 17 to the base of transistor 27.
The output signal is taken out from the collector of No. 7. In this case, the positive input terminal is transistor 1
The base of transistor 15 and the negative input terminal are the base of transistor 18, and the base of transistor 15 is almost the same as the ground voltage, so the emitter of transistor 15 and the base of transistor 17 are approximately 0.7V.
It is. Therefore, if the negative input terminal, which is the base of the transistor 18, is also set to approximately 0.7V, the voltage at the output terminal 6 will be approximately half the voltage of one power supply (Vcc) and the other power supply (earth), and a large output signal can be extracted. I can do it.

In this case, the current flowing through the negative feedback resistor 26 ranges from half the power supply voltage Vcc (voltage of output 6) to 0.7V (base voltage of transistor 18).
The voltage obtained by subtracting the voltage is divided by the resistor 26 (R26), but in order to always keep the base voltage of the transistor 18 at approximately 0.7V even if the power supply voltage Vcc changes, the current flowing through the resistor 26 is transferred to the transistor 22. I try to make it flow. That is, the value of the resistor 24 is set to twice the value of the resistor 26, and this is set to the value of the diode 2.
By connecting 5 and 23, the current flowing through the resistor 24 and the current flowing through the resistor 26 are made the same, the transistor 22 and the diode 23 operate as a current mirror circuit, the power supply voltage Vcc changes, and the voltage at the output terminal 26 changes. If the Vcc voltage is about to change by about 1/2, the negative feedback effect of the resistor 26 is used to correct this, and the voltage at the output terminal 6 is adjusted.
Use it so that the voltage is 1/2 of the Vcc voltage. Note that the resistor 4 is a resistor for setting the amount of negative feedback of the AC signal, and the capacitor 5 is a bypass capacitor.

In the conventional example shown in FIG.
The emitter voltage of 7 and 18 is operating at 1.4V,
When the Vcc voltage becomes 1.4V or less, the transistor 17,
No current flows through 18 and it stops operating.

As explained above, in the conventional amplifier circuit, 1.4V
The problem was that it did not work at all at low voltages.

OBJECTS OF THE INVENTION The present invention eliminates the above-mentioned conventional drawbacks and provides an excellent amplifier circuit that operates satisfactorily at a lower voltage than conventional amplifier circuits.

Structure of the Invention The present invention operates with the bias voltage of the positive input terminal and negative input terminal of the amplifier being less than the contact potential difference V _BE between the base and emitter of the transistor, thereby significantly reducing the power supply voltage. It is also designed to operate satisfactorily.

DESCRIPTION OF EMBODIMENTS FIG. 4 is an electrical wiring diagram of an embodiment of the amplifier circuit of the present invention.

In Figure 4, for example, the value of resistor R ₁ is 2KΩ,
Assuming that the value of resistor _R2 is 8KΩ and the power supply voltage Vcc is 1V, there will be a voltage drop of approximately 0.2V across resistor _R1 . From this, the base voltage of transistor 9 as a positive input terminal is 0.8V, so the base voltage of transistor 10 as a negative input terminal is also set to about 0.8V. Then, the voltage at output terminal 6 is set to the power supply voltage.
A large output voltage can be obtained by setting the voltage to 1/2 of Vcc. For this purpose, the resistors R ₃ , R ₄ , 18,
_R5 transistors 17, 19, 21 and a diode 20 are provided. For example, if the value of resistor R ₃ is 8KΩ,
If the value of resistor _R4 is 2KΩ, the voltage drop due to resistor _R4 is approximately 0.2V, which is the voltage drop due to resistor _R1 and the resistance.
The voltage drop due to R ₄ will be the same. That is, the connection point between resistor _R3 and resistor _R4 is 0.2V, and the base of emitter follower transistor 17 is connected to this point, so the emitter of transistor 17 is approximately
It becomes 0.9V. (The contact potential difference V _BE between the base and emitter of transistor 17 is 0.7V) Then, the base of transistor 19 is connected to the emitter of transistor 17, and the resistor R ₅ is connected between the emitter of transistor 19 and ground. A current mirror circuit consisting of a diode 20 and a transistor 21 is connected to the collector of the transistor 19, and the collector of the transistor 21 is connected to the base of the transistor 10.

Here, if the value of resistor R ₅ is set to 2KΩ, the resistor R ₅ will be
A current of 100μA flows. (Because the voltage drop across resistor _R5 is approximately 0.2V and the base-emitter voltage V _BE of transistor 19 is 0.7V). A current of 100μA also flows through the collector of transistor 19, so if the value of resistor _R6 is set to 3KΩ, output terminal 6
Since the voltage is 0.5V, there is a voltage drop of 0.3V across the resistor _R6 , and the base of the transistor 10 can be set to 0.8V. Therefore, in this state, output terminal 6
A large output signal can be extracted.

That is, the values of resistors R ₁ , R ₂ , R ₃ , R ₄ , R ₅ , and R ₆ are:
The value of R ₁ /R ₁ +R ₂ and the value of R ₄ /R ₃ +R ₄ are almost the same, and the value of R ₁ /R ₁ +R ₂ and the value of R ₅ /R ₅ +R ₆ × 1/2 If Vcc is almost the same, even if Vcc changes, transistor 9,
10 base voltages are almost the same, and the output terminal 6
The voltage will also be approximately in the middle between the power supply voltage Vcc on one side and ground, making it possible to obtain a large output.

For example, the value of resistor _R5 is 1KΩ, and the value of resistor _R6 is 1.5K.
Even when considering the case of Ω, R ₁ /R ₁ +R ₂ = 2/2 + 8 = 2/10 = 1/5 R ₅ /R ₅ +R ₆ × 1/2 = 1/1 + 0.5 × 1/2 = It becomes 1/5 and becomes the same. Therefore, in this case too, the resistance
A current of 200 μA flows through R ₆ and R ₅ , the voltage drop across the resistor R ₆ becomes 0.3V, and the voltage at the base of the transistor 10 becomes 0.8V. Since the collector impedance of the transistor 21 is high, the amount of negative feedback from the output terminal ₆ can be increased by the resistor R6, and the voltage at the output terminal 6 can be accurately controlled.

Since the power supply voltage Vcc is 1V, the emitters of transistors 9 and 10 are 0.1V at this time.
, a sufficient current flows through the transistors 9 and 10, and the transistors 9 and 10 are ready for operation.

In the above embodiment, the base of transistor 9 is a positive input terminal, the base of transistor 10 is a negative input terminal, and an input signal that has been negatively fed back is applied from output terminal 6 to the negative input terminal via resistor _R6 . ing. The resistor 4 changes the amount of AC negative feedback by dividing it with the resistor _R6 , and the capacitor 5 is a bypass capacitor.

Furthermore, in the embodiment, the case where the voltage drop due to the resistors R ₁ , R ₄ , and R ₅ was approximately 0.2V was explained, but the voltage drop due to these resistors _is approximately
0.7V) or lower, and by doing so, it is possible to sufficiently operate the power supply voltage Vcc to a voltage lower than 1.4V.

FIG. 5 shows another embodiment, and in this case, the transistor 21 and diode 2 constituting the current mirror circuit in the embodiment shown in FIG.
0 is omitted.

The voltage between the emitter and collector of the transistor 19 is made to operate at a value close to zero.

Further, in these embodiments, the diodes 8, 20
However, it goes without saying that this may be replaced with a diode-connected transistor in which the base and collector of the transistor are connected together.

Effects of the Invention As explained above, the amplifier circuit of the present invention can operate satisfactorily even at low voltages, can increase the amount of DC feedback, and can be expected to operate stably, which is extremely advantageous in practice. It is something.

[Brief explanation of drawings]

Figures 1 to 3 are electrical wiring diagrams of a conventional amplifier circuit, Figure 4 is an electrical wiring diagram of one embodiment of the amplifier circuit of the present invention, and Figure 5 is an electrical wiring diagram of another embodiment. It is. 1... Signal input terminal, 2, 5... Capacitor,
4, 11, 13, 14, 16, 18, R ₁ ~ R ₆ ...
...Resistance, 8,20...Diode, 7,9,1
0, 12, 15, 17, 19, 21...transistor.

Claims (1)

[Claims] 1 has a positive input terminal, a negative input terminal, and an output terminal,
The base of the first transistor is connected to the positive input terminal, the base of the second transistor is connected to the negative input terminal, and the emitter of the first transistor and the emitter of the second transistor are connected to each other to make a difference. A circuit that constitutes a dynamic amplifier and supplies current to the first and second transistors is connected between the emitters of the first and second transistors and the other power supply, and A signal is taken out from between the collector of the transistor and one power source, and the signal obtained at at least one of the collectors of the first and second transistors is guided to the output terminal, and one power source and the other power source are connected to each other. A connection point between a first resistor and a second resistor connected in series between a power source and a second resistor is connected to the base of the first transistor, and a resistor connected in series between one power source and the other power source is connected to the base of the first transistor. The connection point between the third resistor and the fourth resistor is connected to the base of the third transistor, the emitter of the third transistor is connected to the base of the fourth transistor, and the emitter of the fourth transistor is connected to the base of the fourth transistor. A fifth resistor is connected, the collector of the fourth transistor is connected to the base of the second transistor directly or via a current mirror circuit, and a fifth resistor is connected between the output terminal and the base of the second transistor. 6 resistors are connected, the first resistor and the second
The voltage drop across the first resistor obtained by dividing by the resistor, the voltage drop across the fourth resistor obtained by dividing by the third resistor and the fourth resistor, and the fifth resistor.
The voltage drop of the above-mentioned second resistor shall be approximately the same value as the voltage drop of the second resistor.
and the voltage drop across the third resistor is the voltage drop across the emitter of each transistor and the base of the transistor.
The base of the first transistor and the base of the second transistor are An amplifier circuit characterized in that a sixth resistor is set so that voltages have approximately the same value.