JPH01316014A - Active filter circuit - Google Patents

Abstract

PURPOSE:To obtain a time constant in matching with a prescribed resonance frequency at a low frequency band, to improve S/N and to realize the complete circuit integration of a low frequency band filter through small capacitance by using plural resistors forming the filter circuit and increasing the capacitance of the capacitor. CONSTITUTION:The following circuits are provided to the output size of transconductance amplifiers 3, 4 being components of the active filter circuit; a resistor RA corresponding to a resistor R1(R2) and a resistor RB corresponding to a resistor R2(R4) are provided on a signal current input terminal Pin of the amplifier 3(4), an operational amplifier 12 of an amplifier A is made correspondent with an operational amplifier OP1(OP2) and a capacitor CA is made correspondent to the capacitor C1(C2). Through the constitution above, an amplifier 12' with the amplification factor of unity which is equivalent to the amplifier 12 when viewing an output terminal Vo from the input terminal Pin, is as if to be the serial connection of the parallel connection of the RA, RB and of an equivalent capacitance of (RA+RB).CA/RA. Thus, a large capacitance is obtained while employing a small capacitance of the capacitor.

Description

[Detailed Description of the Invention] [Object of the Invention] (Industrial Application Field) The present invention relates to an accu-dip filter circuit suitable for a low frequency band, and relates to an internal configuration of a capacitor and other active elements in an integrated circuit. It is characterized by its characteristics.

(Conventional technology) In recent years, in order to improve the packaging density of electronic circuits and reduce costs,
2. Description of the Related Art Attempts are being made to incorporate electronic components such as capacitors, which have traditionally been provided around integrated circuits (ICs), inside the ICs. Among them, the built-in filter circuit has great advantages.

The most common method of completely incorporating a highly accurate filter circuit into an IC is to use a no-active filter circuit whose unit component is an integrator consisting of a transconductance snoop and a capacitor.

The problem when configuring a filter in an IC is
Due to variations in resistance, capacitance, etc., the characteristics may differ from the set a.
It means to deviate more than one characteristic and create a large gap. Therefore, in the active filter using the 1-lance conductance amplifier described above, each 1-lance conductance (hereinafter referred to as gm or Gm) of the multi-stage transconductance amplifier is made to have a proportional relationship. As a result, absolute variations in resistance and capacitance values can be corrected by -C, and characteristic accuracy can be suppressed by ratio accuracy of resistance and capacitance. In an IC, high ratio accuracy of resistance and capacitance can be achieved, so that variations in filter characteristics obtained by proportional adjustment of gm can be prevented from becoming a problem in practice.

Also, the g of this transconductance amplifier is
The method of adjusting m has the advantage of having a wide adjustment range and less distortion.

FIG. 3 shows a second-order low-pass filter as an example of the above active filter.

The 7I-1 corpus filter shown in Fig. 3 consists of operational amplifiers (71 amplifiers) connected in series.
Capacitors C3 and C4 are respectively connected between the output terminals of the inductance amplifiers 7 and 8 and the standard potential point, and the signals appearing at the output terminals of the subsequent stage conductance amplifiers 1 to 8 are connected to the respective preceding stages. Negative feedback is provided to each inverting input terminal of the transformer 21 inductance amplifier 7 and subsequent stage amplifier 8. The input signal is input to the non-inverting input terminal of the front-stage transconductance amplifier 7 via the input terminal 5, and the output signal is the negative feedback signal of the rear-stage transconductance amplifier 8, which is input to the output terminal 6. It is derived.

The transfer function of this circuit is expressed as follows.

here, It is.

By using a circuit with such a configuration, for example, the band can be increased to 20[K
Consider implementing an audio band D-pass filter limited to 1-17] in an IC. The time constant of the audio band (the platform realized in the IC is quite small. In other words, accepting C0 and realizing a small a>O in equation (2), ■C3, Larger C/1, ■g
m3. There are two possible ways to make gm4 smaller. Of these, (2) increases the chip area within the IC, resulting in high costs and is uneconomical.

Therefore, without increasing the capacity fi'L too much, we decided to improve the characteristics of the low frequency band. In order to satisfy 1, consider combining method ①. Therefore, we will examine method ■.

In order to examine the method (2), I will give a specific example of a typical 1-Herance conductance amplifier and further explain it to the audience.

FIG. 4 shows the most general concrete circuit example of a transformer conductance amplifier consisting of an input differential stage, a second differential stage, and a current folding stage. A differential input signal voltage vin is applied to terminals 15 and 16, and this signal voltage is applied.

Vin connects 7111 to the bases of input differential transistors Q1 and G2. In the transistors Q1 and 02, a series connection of resistors RE and RE is connected between the limiters, and a bias current source 211 is connected between the intersection of this series connection and a reference potential point.

The second differential stage is illustrated by transistors Q3 to Q6. These transistors Q3 to Q6 form the so-called gain cells of Gilbar 1, in which a transistor Q3°G4, to which a fixed voltage source VB is applied to the base, is cascade-connected with transistors Q1 and G2, and a bias current source 21
Transistors Q5 and G6, which are driven by
The signal from the collector of transistor Q1~G2 is input to the base, and from the collector, an output signal current 1out is derived to terminal 17 via a current folding stage composed of current sources ■3 to ■8. There is. Note that the control voltage ■a from the terminal 18 can adjust the value of the bias current source 212, thereby making it possible to vary the G11l value.

In such a circuit, the human input signal voltage Vin is connected to the series resistance R
E, RE is converted into a current by the resistance value of 2RE, and this current flows through the collector-emitter paths of transistors Q1 and G2 and the transistors Q5 and G6, and the bias currents 11 and 11. controlled by I2, to the collectors of transistors Q5, G6, Vi
A differential current proportional to n appears. This is converted into a single current by a current folding stage and used as an output signal.

The 1-lance-1 inductance Qm of this circuit is Gm
=-! -・I2...(3
)E 11 .

Further, noise in this circuit is dominated by thermal noise of the base resistance of the transistors Q3 to Q6 and shot current of the collector current. Among these, the thermal noise of the base resistor can be reduced by selecting a low-noise process in the IC manufacturing process and increasing the size of the transistor by E1m.

Regarding the collector current ((,1, output current 1o+, +1
The S/N of the signal is degraded. First, 1 to transistor Q3,
The rector shot current 10 of Q4, -2 is +n = 2Q11Δfx < -L? -) 2・(4
)・-2 Collector shock 1 to current of transistors Q5 and Q6,
-2 10 is i″″″″n2-2qI2Δf ・
(5) where Δf represents the noise bandwidth and q represents the charge of the electron. From this, the 1-tal noise current appearing at terminal 17 is -, if noise other than the above noise is ignored, -.
The signal current appearing at terminal 7 is determined by using equation (3) and from equations (6) and (7), the S/N of the current at terminal 17 is
The S/N of the output current Iout in the amplifier is determined.

Now, ci m3. If we try to satisfy the characteristics by reducing g m4, from equation (3), ■R
There are two possible methods: increasing E11 and decreasing ■■2.

However, as a result of deriving the S/N of the output horn current using the formula (8), the means of (1) and (2) above show that no matter whether R[11 is increased or I2 is decreased, S/N h< worsens. It turns out that this is fatal as a filter for the audio A signal band.

In fact, the S/N of a filter circuit realized by combining such a transconductance amplifier with a -1:1 shibashita varies depending on the configuration method and various constants of the filter, and is expressed by equation (8). Although it is not the same as the S/N determined by
changes depending on the S/N of the transformer-1 inductance amplifier alone, so in this specification, the S/N is (
8) The explanation will be based on the assumption that it is determined only by the formula.

(Problem to be solved by the invention) An active filter circuit configured by connecting a capacitor to the output end of a transconductance amplifier has a built-in capacitor, and when forming a low frequency band filter circuit, it is practically difficult to use a capacitor. Since the large serration is limited, characteristics satisfying a filter in a low frequency band can be achieved by combining this with a means for reducing the l-lance conductance. However, reducing the transconductance leads to deterioration of the S/N ratio, and there is a problem in that the capacity cannot be made sufficiently small.

This invention eliminates the above problems, eliminates the deterioration of the S/N ratio and the narrowing of the capacity as design constraint factors, and satisfies sufficient characteristics as a filter in the low frequency band. The purpose is to provide circuits.

[Structure of the Invention] (Means for Solving the Problems) The present invention provides a method for connecting the non-inverting input terminal and the output terminal of an operational amplifier in which the inverting input terminal and the output terminal are connected via a short circuit or a resistor. , a capacitance multiplier circuit comprising a first and a second resistor connected in series, and one end of a capacitor connected to the non-inverting input end, and a differential input voltage between the input signal and a predetermined feedback signal. A proportional current is outputted to the output end, and this output end is connected to the first. This is an ac-dip filter circuit configured in the same integrated circuit by combining a transconductance amplifier connected to the connection intersection of the second resistor.

(Function) According to such a configuration, the capacitance multiplier circuit has a capacitance multiplier whose capacitance h1 is the first. This is equivalent to a capacitive element multiplied by a ratio determined by the ratio of the second resistance. Therefore, if such a capacitance multiplier circuit is
~Transconductance] By connecting to the output end of the inductance amplifier, a filtering effect due to a capacitance larger than that of the actually used capacitor can be obtained, and the transconductance can be adjusted to the specified resonant frequency in the low frequency band without reducing the transconductance. This will improve the S/N ratio, reduce the size of the filter, and make the low-frequency filter completely integrated into an IC.

(Example) The present invention will be explained below with reference to illustrated embodiments.

FIG. 1 is a circuit diagram showing an example of an actual flow of an active filter circuit according to the present invention.

The circuit shown in FIG. 1 is a second-order low-pass filter circuit, and an input terminal 1 receives a low frequency signal of the A-day A band. The signal from this terminal 1 enters the non-inverting input of a transconductance amplifier 3 having a constant gml. The signal appearing at the current output terminal of the transconductance amplifier 3 is introduced to the non-inverting input terminal of the -C next stage transconductance amplifier 4 via the resistor R1, and is also introduced to the non-inverting input terminal of the operational amplifier OP1 via the resistor R2. lead to. A capacitor C1 is connected between the non-inverting input terminal of the operational amplifier OP1 and a reference potential point.

Further, the inverting input terminal and output terminal of the operational amplifier OP1 are short-circuited to each other and connected to the non-inverting input terminal of the next-stage transconductance amplifier 4.

Also on the current output side of the transformer conductance amplifier 4, 18 circuits having the same configuration as the circuit including the resistors R1 and R2, the capacitor C1, and the operational amplifier OPi are connected. That is, the output 9''1:; of the transconductance amplifier 4 is connected to the C output terminal 2 via a resistor R3, and also to the reference potential point via a resistor R4 and a capacitor C2. The connection intersection with the resistor R4 is connected to the non-inverting input terminal of the operational amplifier OP2, and the inverting input terminal and the output terminal of the operational amplifier OP2 are connected to each other and connected to the output terminal 2. The output signal appearing at 2 is supplied to each inverting input terminal of transconductance amplifiers 3 and 4 as a negative feedback signal.

The second-order [-1-pass filter according to this embodiment is constructed as described above, and its operation will be explained next.

Each transconductance amplifier in the circuit of FIG. The circuits connected to the output end of /l are all connected to the second
It has a configuration as shown in Figure (A).

In FIG. 2 (8), Pln indicates the input terminal of the signal current from the transconductance amplifier 3 (4), the resistor R8 is connected to the resistor R1 (R2), and the resistor RB is connected to the resistor R2 (
R4) respectively.

Further, the operational amplifier 12 with the amplification factor △ is the operational amplifier OP1
(OR3), C8 is the capacitor CI (C2)
It corresponds to

Now, let the output impedance of the current output terminal of the transconductance amplifier 3 (4) be infinite, and the signal voltage generated at the connection intersection of resistors RA and RB < p in) by the signal @ current Ii from the output terminal is Vl, The voltage appearing across the A7 pacita C^ is output via the VC1 resistor RA.
If we set the formula for the circuit shown in Figure 2 (8) with the signal voltage visible at the end as VO, then VO=A (Vc - Vo )
... (10) (9) , (10'),
When formula (11) is combined and solved for Vi/Ii and VO, Vo=-A-Vc...(13)
1−+− to −1/I − A−+■ and −C approximation, equations (12) and (13) become vo = vc ・・
・It becomes (15).

(14), (From formula 151, the circuit in Figure 2 (A) is
This can be represented by the circuit shown in FIG. 2(B). Bll'
), looking from the input terminal Pin to the output terminal (■0), an amplifier 12' with an amplification factor of 1 which is equivalent to the operational amplifier 12' i, l
:, RA and the capacitance are connected in series, and the voltage VC appearing at the connection intersection of the circuit is amplified and delivered to the output terminal. In other words, they will have a baby. Therefore, if the magnification of the resistors RA and RB is chosen to be large, it means that a small 4-power capacitor can achieve a large capacity on the same surface.

In this way, the circuit shown in FIG. 1 is a circuit that multiplies the capacity on the same page as the conventional -1:Yapashita 03. 4!J, J,
It's something that makes me happy.

However, in order for the capacitance multiplier circuit to not affect the filter characteristics, the value of the parallel resistor RA//RB connected in series with the signal path is required to be small.

In this invention, it is proposed to connect the capacitance multiplier circuit to the current output terminal of the transformer conductance amplifier, and considering that the impedance of this current output terminal is infinite, "two parallel It can be inferred that even if the resistance value is slightly different, it will not be a factor that changes the frequency characteristics. However, in reality, there may be some influence due to the fact that the output impedance of the transconductance amplifier is finite and the dynamic range of output J is finite.

In order to eliminate the above-mentioned influence, it is sufficient to reduce RA and RB without changing their ratio. In other words, the isometric internal time constant of the capacitance multiplier circuit is = 16- (r<A//11) ・RA-1-R8C8-R
If the value of BCAR8 (16) is a sufficiently small value (for example, 1 μsec or less) than the constant that corresponds to the predetermined resonance frequency of low frequency (for example, A-dio band), then 1) The influence of column resistance of RA and R8 can be made negligibly small.

Next, we will explain how to make the capacitance multiplier circuit too small.

Consider obtaining the same frequency characteristics as in FIG. 3 using the circuit in FIG. First, gml and gm2 of 1 to lance conductance amplifiers 3 and 4 of both circuits are respectively gm3. gm
It is equal to 4. The capacitance multiplier circuit in Figure 1 is shown in Figure 2 (B
) and change d1, in order to have the same frequency characteristics in both circuits, refer to equation (2) and calculate c 4 = R3-t R4c 2-17 = (However, R1//R2 and R37/R4 are selected to have relatively small resistance values that do not affect the frequency characteristics). This means that to obtain the same characteristics, CI, C
The capacity of 2 is the same as that of the conventional configuration.
R3- will suffice.

R1+R2R3+R4 For example, R1:R2=1:9, R
If the difference is 3:R4=1:9, the capacitance of C1:C2 is 1
This means that the cost will be reduced by 1/0'c. A reduction in the capacity &'L value directly results in a reduction in the area on the IC circuit, and it is possible to eliminate the low=1-stroke acceleration filter IG.

Further, the multiplication of the capacitance described above (J) directly increases the transconductance.

This is a condition in which, contrary to the above, the capacitance values of the circuit in FIG. 1 and the circuit in FIG. 3 are poor (C3=C1, C4=C2). Therefore, in this case, For example R4:
R2=i:9. If R3:R4=1:9, then gm
l and gm2 are each increased by 10 times.

Here, consider the specific circuit shown in FIG. Molecule 1 of formula (3)
(Ell requires the input dynamic range to be at least V1nHAX, the maximum value of 7 songs, and no distortion conditions), and because of the contradictory relationship of wanting to obtain S/N that can be achieved under these conditions, I decided to change to V inM 8X. J: Select a slightly larger value. In addition, REll does not need to be changed if an optimal design is made, so it is a fixed value.
In order to obtain S/N by increasing 2, 11, 12
Multiply by 10 and 11 E becomes 1/10. this is,
Instead of saving RE 11, Qm will be multiplied by 10.

In this case, from equation (8), the S/N of the transconductance amplifier is m- times, that is, 10 [dB] up J-
That will happen.

In addition to this, the active filter circuit of this embodiment is
By setting circuit constants that reduce the capacitance value, the S/N
There are some things that can be made higher.

Although the above embodiment has been explained in the case of a low-pass filter, for example, a chitopacitor C4 connected to the reference potential point side
, one end of C2 is a return signal path (low impedance terminal)
By connecting it to, you can realize a nop filter. Further, the terminals of the capacitors CI and C2 are
It may also be connected to the voltage source terminal. In short, capacitor CI
, C2, filters with various characteristics can be realized by selecting the connection points of the terminals of C2.

In addition, in the example, the output terminal and the inverting input terminal are short-circuited, and a capacitance multiplier circuit is realized using an operational amplifier connected by a porticebo [1W], but a resistor of a predetermined value is connected between the same terminals. It's good. This allows the gain to be set to a desired value.

Furthermore, the capacitance multiplier circuit according to the present invention has an operational amplifier OP
Since I (O12) serves as a buffer for the capacitance end voltage VC, it has the advantage that it can be easily connected to other elements constituting the accdip filter.

``Effects of the Invention'' As explained above, this invention makes it possible to design a design that improves S/N and reduces capacity at the same time, making it a very effective means for IC filters, especially in the audio band. .

[BRIEF DESCRIPTION OF THE DRAWINGS] FIG. 1 is a circuit diagram showing one embodiment of the active filter circuit according to the present invention, FIG. 2 is an explanatory circuit diagram for explaining the circuit of FIG. 1 in detail, and FIG. The figure shows a conventional active filter circuit using a conventional low-pass filter as an example.
Circuit Diagram FIG. 4 is a circuit diagram showing an example of a specific circuit of a conventional transconductance amplifier. 1 input terminal, 2...output terminal, 3. /I...Transconductance amplifier, R1, R2, R3, R4-
411; anti, CI, C2-*vpacita, OPl,
O12...Operation amplifier, gml, 0m2...Transconductance.

Claims (1)

[Claims] Between the non-inverting input terminal and the output terminal of an operational amplifier, the inverting input terminal and the output terminal being connected directly or through a resistor,
a capacitance multiplier circuit comprising first and second resistors connected in series and one end of a capacitor connected to the non-inverting input terminal; The current is led out as an output to an output end, and this output end is connected to the first
, a transconductance amplifier connected to the connection intersection of the second resistor, and the other end of the capacitor is connected to a reference potential point or a predetermined signal path. 1. An active filter circuit characterized in that an active filter circuit is constructed by mounting one or several components on the same integrated circuit.