JPH06310961A - Signal gain control circuit - Google Patents

Abstract

PURPOSE:To adopt lower drive voltage for the signal gain control circuit by applying an input signal to a voltage-current conversion circuit, applying a control signal between bases of transistors(TRs), and obtaining an output signal from collectors of the TR. CONSTITUTION:A differential amplifier 4 is made up of 1st and 2nd transistors(TRs) 1, 2 whose emitters are connected in common. A constant current circuit 5 connecting to a common emitter of the amplifier 4 and supplying a bias current to the TRs 1, 2, and a load circuit 3 connecting to collectors of the TRs 1, 2 and forming a load of the amplifier 4 are provided. Furthermore, a control means applying a control signal Vc between bases of the TRs 1, 2 and a voltage-current conversion circuit 6 whose output terminal connects to a common emitter of the amplifier 4 and changing the current of the output terminal in response to an input signal ej applied to the input terminal are provided. Then the input signal ej is obtained and an output signal eo whose voltage gain is changed in response to the control signal Vc by the load circuit 3. Thus, the signal gain control circuit is obtained, which is operated at a lower power supply voltage.

Description

Detailed Description of the Invention

[0001]

BACKGROUND OF THE INVENTION 1. Field of the Invention The present invention relates to an improvement of a signal gain control circuit for reducing the required power supply voltage.

[0002]

2. Description of the Related Art The circuit shown in FIG. 2 is a basic circuit of an analog multiplier which has been widely used as a signal gain control circuit. This circuit is composed of transistors 1 and 2 and a load resistance 3 (resistance value R _L ) of the differential amplifier 4, and a constant current circuit 5 for supplying a DC bias current to the differential amplifier 4.
And a transistor 17 cascade-connected between the constant current circuit 5 and the common emitter of the differential amplifier 4, and a resistor 18 (resistance value R _E ) connected between the emitter of the transistor 17 and ground. Therefore, the input signal e _i is applied to the base of the transistor 17, the control signal Vc is applied between the bases of the transistors 1 and 2, and the output signal e _o is obtained from the collector of the transistor 2. In this circuit, when the input signal e _i is applied to the base of the transistor 17, a signal current i = e _i / R _E almost flows through the collector of the transistor 17. This signal current i is shunted to the transistors 1 and 2, and the signal currents i _L1 and i _L2 (i _L1
+ I _L2 = i) flows. As a result, the output signal e _o = −i _L2 · R _L is obtained from the collector of the transistor 2.
On the other hand, the ratio (i _L1 : i _L2 ) of the signal currents flowing through the transistors 1 and 2 changes depending on the potential difference between the bases of the transistors 1 and 2, that is, the value of the control signal Vc.
That is, the value of the signal current i _L2 is changed by changing the value of the control signal Vc, and thus the gain of the output signal e _o can be controlled. This output signal e _o is

[0003]

[Equation 1] K: Boltzmann's constant T: absolute temperature q: unit charge amount Here, the voltage of each part necessary for operating the circuit shown in FIG.
Find the required value of cc). The constant current circuit 5 is usually
Since the current mirror circuit 12 shown in the figure is used, it will be replaced with this here. Figure 2
All of the transistors in the circuit of FIG. 2 operate in class A, but assuming that the bias voltage V _CE (collector-emitter voltage) required for this is V _CE ≧ V _BE (base-emitter voltage), Since the transistors 1, 2, 17 and 10 form a three-stage cascade circuit, a total voltage of V _TR ≧ 3 · V _BE ··· (2) is required for these operations. Become.

The DC bias current I supplied by the current mirror circuit 12 is equally divided into two transistors 1 and 2 when the control signal Vc = 0V, and I ₁ = I ₂ is applied to the collectors of the respective transistors. A DC bias current of = I / 2 flows. In the current mirror circuit 12, the voltage drop of each part due to these DC bias currents is
R _b · I, in the load resistance 3 of the differential amplifier 4, R _L · I
₂ = _RL · I / 2, and the total is _Vb = I · ( _Rb + _RL / 2) (3). Assuming that the maximum amplitude of the input signal e _i is e _{i MAX} and the control signal Vc = 0V, the maximum amplitude e of the output signal e ₀ is e.
_{o MAX} is _calculated from the equation (1), e _{O MAX} = e _{i MAX} · R _L / 2R _E
Becomes Therefore, in order to secure the required dynamic range of input / output signals in the circuit of FIG. 2, a voltage of Vs ≧ e _{i MAX} · (1 + R _L / 2R _E ) (4) is required. As a result, the required value of the power supply voltage Vcc of the circuit of FIG. 2 is Vc from the above equations (2), (3) and (4).
c ≧ (V _TR + V _b + V _S ) / 2, that is, Vcc ≧ {3V _BE + I · (R _b + R _L / 2) + e _{i MAX} · (1 + R _L / 2R _E )} / 2・ ・ (5)

[0005]

By the way, today, there is an increasing demand for low power consumption of electronic devices, especially in portable devices that require battery drive.
As a method for reducing the power consumption of this electronic device, lowering the voltage of the circuit power supply is extremely effective. However, the multiplier shown in FIG. 2 has a drawback in that it is difficult to lower the voltage of the power source because of the circuit configuration in which a large number of transistors are cascade-connected between the positive and negative power sources. That is, according to the above equation (5), the first term on the right side of the equation is the sum of the minimum bias voltages necessary for the operation of the transistors in the circuit of FIG. Is a fixed part that has nothing to do with it. Therefore, the power supply voltage V on the left side of the equation
Even if we try to reduce cc, the first term on the right-hand side of the equation naturally determines the lower limit. The actual value of the first term on the right side is usually V _BE = 0.6V to 0.8V, so 3 ×
V _BE = 1.8V to 2.4V. This value can be said to be extremely large in recent years when a circuit that operates under a low voltage power supply of only a few volts is required. For the above reasons, it is desired to realize a circuit having the same function as that of the circuit of FIG. 2 and capable of operating with a lower voltage power source, and this is also an object of the present invention.

[0006]

In order to achieve the above object, the present invention provides a differential amplifier 4 comprising transistors 1 and 2 and a load resistor 3 thereof, as shown in FIG. A constant current circuit 5 for supplying a DC bias current and a voltage-current conversion circuit 6 whose output end is connected to the common emitter of the differential amplifier 4 are provided. The voltage-current conversion circuit 6 receives an input signal e _i. Is applied, and transistors 1 and 2
The control signal Vc is applied between the bases of the two, and the output signal e ₀ is obtained from the collector of the transistor 2.

[0007]

In the circuit of FIG. 1, when the conversion coefficient of the voltage-current conversion circuit 6 is set to 1 / _RE and the input signal e _i is applied to the voltage-current conversion circuit 6, the common emitters of the transistors 1 and 2 are connected. The signal current i = e _i / R _E flows. This signal current i is shunted to the transistors 1 and 2, and the signal currents i _L1 and i _L2 (i _L1
+ I _L2 = i) flows. As a result, the output signal e _o = −i _L2 · R _L is obtained from the collector of the transistor 2. On the other hand, the ratio (i _L1 : i _L2 ) of the signal currents flowing through the transistors 1 and 2 changes depending on the potential difference between the bases of the transistors 1 and 2, that is, the value of the control signal Vc.
That is, the value of the signal current i _L2 is changed by changing the value of the control signal Vc, and thus the gain of the output signal e _o can be controlled. This output signal e _o is similar to the circuit of FIG.

[0008]

[Equation 2] K: Boltzmann's constant T: absolute temperature q: unit charge amount Here, similarly to the case of the circuit of FIG. 2 described above, the voltage of each part necessary for operating the circuit of FIG. 1 is obtained, and the required value of the power supply voltage is obtained. The constant current circuit 5 will be replaced with the current mirror circuit 12 as in the circuit of FIG. First, all the transistors in the circuit of FIG. 1 perform class A operation, but if the bias voltage V _CE required for this is assumed to be V _CE ≧ V _BE , the transistors 1, 2, 10 of the circuit of FIG. Since a cascade circuit of stages is configured, a total of V _TR ≧ 2 · V _BE (7) is required for these operations.

Next, the DC bias current I supplied by the current mirror circuit 12 is equally divided into two transistors 1 and 2, assuming that the control signal Vc = 0V, and the collector of each of these transistors has I ₁ A DC bias current of = I ₂ = I / 2 flows. The voltage drop in each part due to the DC bias current is R _b · I in the current mirror circuit 12 and R _L in the load resistance 3 of the differential amplifier 4.
_{· I 2 = R L · I} / 2 , and the sum in the same manner as the circuit of aforementioned Figure _{2, V b = I · (} R b + R L / 2) becomes ... (8). In addition, the maximum value of the amplitude of the input signal e _i is represented by e _{i MAX} ,
Assuming that the control signal Vc = 0V, the maximum value e _{o MAX} of the amplitude of the output signal e _o can be calculated from equation (6) as e _{o max} = e _{i max RL}
/ 2R _E. Therefore, in order to secure the required dynamic range of the output signal e _o , a voltage of Vs ≧ e _{i MAX} · R _L / 2R _E (9) is required.

As a result of the above, the transistors 1, 2,
The required value of the power supply voltage Vcc in the cascade circuit composed of 10 is Vcc from the equations (7), (8) and (9).
≧ (V _TR + V _b + Vs) / 2, that is, Vcc ≧ {2 · V _BE + I · {R _b + R _L / 2) + e _{i MAX} · R _L / 2R _E )} / 2 ( 10) becomes. The voltage-current conversion circuit 6 shown in FIG. 1 can be configured by a simple circuit such as the base-grounded amplifier 15 shown in FIG. This circuit can sufficiently operate within the required value of the power supply voltage (Vcc) of the cascade circuit of the transistors 1, 2 and 10, and the dynamic range of the input signal e _i can be easily secured. Therefore, it may be considered that the required value of the power supply voltage of the circuit of FIG. 1 is given only by the above equation (10). Now, the required value of the power supply voltage between the conventional signal gain control circuit shown in FIG. 2 and the circuit of the present invention shown in FIG. 1 will be compared. That is, comparing the values of Vcc in the equation (5) and the equation (10), it is apparent that the equation (10) is smaller. From the two equations, the difference is (V _BE + e _{i MAX} ) / 2, which is several V depending on the value of e _{i MAX} . That is, the circuit of the present invention can be driven by a power supply of a lower voltage than that of the conventional circuit. That is, it can be said that it is advantageous for lowering the voltage of the circuit power supply, which has been a technical problem in recent years.

[0011]

EXAMPLES An example of the present invention will be described below.
The circuit shown in FIG. 3 is a signal gain control circuit to which the circuit of the present invention shown in FIG. 1 is applied. That is, the circuit shown in FIG. 3 has the same structure as the first multiplier constituted by the transistors 1, 2, 10 and 13 and the transistors 1 ′, 2 ′ and 1 similarly.
0 ', 13', and a second multiplier that is completely symmetrical to the first multiplier,
A differential amplifier 4 composed of 2 and a base and collector of a differential amplifier 4'composed of transistors 1'and 2'are connected in common, and a resistor 3 ( _RL ) is used as a common load. An input signal e _i is applied to the input terminal of 15 (15 '), a control signal Vc is applied between the common bases of the differential amplifiers 4 and 4', and an output signal e _{o is also} from the common collector.
Is to get. The first and second multipliers are basically the same as the circuit of the present invention shown in FIG. In the multiplier of FIG. 1, when the control signal Vc changes, the DC bias current I ₂ flowing through the load resistor 3 ( _RL ) changes, and the DC potential of the output signal e _o changes.

However, in the circuit of FIG. 3, DC bias currents I ₁ and I ₂ are supplied from the constant current circuit (current mirror circuit 12 ') formed by the transistors 10 and 10' to the differential amplifiers 4 and 4 ', respectively. Is equal to I ₁ = I ₂ = I, even if the control signal Vc changes, the transistors 2, 2 '
DC bias currents I ₁₂ , I 3 supplied to the load resistor 3 from
_Since the sum of ₂₂ is always constant as I ₁₂ + I ₂₂ = I, there is an advantage that the DC potential of the output signal e _o does not change. The circuit configuration method as shown in FIG. 3 is a known technique, and is widely used in the case of the conventional multiplier shown in FIG. Figure 3
In the circuit (1), only the first multiplier composed of the transistors 1, 2, 10, 13 is involved in the actual control of the input signal e _i . Therefore, the output signal e _o is similar to the circuit of FIG.

[0013]

[Equation 3] K: Boltzmann's constant T: absolute temperature q: unit charge amount That is, the circuit of FIG. 3 has a function as a gain control circuit that controls the gain of the output signal e _o according to the control signal Vc.

[0014]

According to the present invention, it is possible to realize a signal gain control circuit which operates with a lower voltage power source, and in particular, to reduce the power consumption of electronic equipment centering on portable equipment which requires battery drive. It will help you in your efforts.

[Brief description of drawings]

FIG. 1 is a basic circuit of a signal gain control circuit of the present invention.

FIG. 2 is a basic circuit of a conventional signal gain control circuit.

FIG. 3 shows an embodiment of a signal gain control circuit to which the present invention is applied.

[Explanation of symbols]

 4,4 'Differential amplifier 5 Constant current circuit 6 Voltage-current conversion circuit 12, 12' Current mirror circuit 15, 15 'Grounded base amplifier

Claims (1)

[Claims] 1. A differential amplifier composed of first and second transistors whose emitters are commonly connected, and a bias current which is connected to the common emitter of the differential amplifier and is supplied to the first and second transistors. A constant current circuit for flowing the current, a load circuit connected to the collectors of the first and second transistors and forming a load of the differential amplifier, and the first and second
Control means for applying a control signal between the bases of the second transistors and the output end are connected to the common emitter of the differential amplifier, and the current at the output end is changed according to the input signal applied to the input end. A signal gain control circuit comprising: a voltage-current conversion circuit, wherein the input signal is obtained and an output signal whose voltage gain is changed according to the control signal is obtained from the load circuit.