MOS linear transconductance amplifier

[0002] 1. Field of the Invention

[0003] The present invention relates to a MOS differential amplifier circuit having linear transconductance formed on a semiconductor integrated circuit, and in particular, to a MOS linear transconductance amplifier that operates in a low voltage and is excellent in a frequency characteristic.

[0004] 2. Description of the Prior Art

[0005] FIG. 8 is a circuit diagram showing a conventional MOS linear transconductance amplifier disclosed in Japanese Patent Laid-Open No. 11-251848.

[0006] This MOS linear transconductance amplifier comprises a MOS differential pair 1 formed with two n-channel MOS transistors M1 and M2 whose sources are connected with each other, and two n-channel MOS transistors M3 and M4 which operate as their load.

[0007] Sources of the MOS transistors M1 and M2 that form the MOS differential pair 1 are grounded through a constant current source 2 (current value: I_ss). This MOS differential pair 1 is driven by the constant current I_ssgenerated by the constant current source 2. Gates of the MOS transistors M1 and M2 form an input terminal pair of the amplifier concerned, and an input voltage V_iis applied between those gates. Each ratio (W/L) of gate width (W) and gate length (L) of the MOS transistors M1 and M2 is K₁as large as that of a unit MOS transistor (K₁is a constant, but K₁≧1).

[0008] The MOS transistor M3 operates as the load of the MOS transistor M1. A source of the MOS transistor M3 is connected to a drain of the MOS transistor M1, and the drain of the MOS transistor M3 is connected to a supply voltage line from which supply voltage V_DDis applied, and a bias voltage (DC constant voltage) V_Bis applied to a gate of the MOS transistor M3.

[0009] The MOS transistor M4 operates as the load of the MOS transistor M2. A source of the MOS transistor M4 is connected to a drain of the MOS transistor M2, a drain of the MOS transistor M4 is connected to the supply voltage line from which the supply voltage V_DDis applied, and the same bias voltage V_Bas that applied to the MOS transistor M3 is applied to a gate of the MOS transistor M4. Each ratio (W/L) of gate width (W) and gate length (L) of the MOS transistors M3 and M4 is K₂as large as that of the unit MOS transistor (K₂is a constant, however K₂≧1).

[0010] Next, the operation principle of a MOS linear transconductance amplifier shown in FIG. 8 will be described.

[0011] It is assumed that a body effect and a channel length modulation are disregarded and the relation between a drain current I_Dand a voltage between the gate and source of a MOS transistor which is operating in a saturation region follows a square-law. Then, the drain current I_Dis expressed as shown in a following formulas (1a) and (1b): $\{\begin{array}{cc}{I}_D=K{\beta \left(V_{\mathrm{GS}}-V_{\mathrm{TH}}\right)}^2& \left(V_{\mathrm{GS}}\ge V_{\mathrm{TH}}\right)\\ I_D=0& \left(V_{\mathrm{GS}}\le V_{\mathrm{TH}}\right)\end{array}\begin{array}{c}\text{(1a)}\\ \text{(1a)}\end{array}$

[0012] In these formulas (1a) and (1b), a symbol K is a ratio of the ratio (W/L) of the gate width (W) and gate length (L) of the MOS transistor to that of the unit MOS transistor. In addition, a symbol β is a transconductance parameter and a symbol V_THis threshold voltage. Assuming that an effective mobility of a carrier is μ and a gate oxide film capacity per unit area is C_OX, the transconductance parameter β will be defined by β=μ (C_OX/2) (W/L).

[0013] If it is assumed that characteristics of elements are almost consistent with each other, the two output currents I_D1and I_D2of the MOS differential pair 1, i.e., the drain currents of these MOS transistors M1 and M2, are expressed as shown in the following formulas (2a) and (2b), respectively. $\begin{array}{cc}{I}_{\mathrm{D1}}=\frac{1}{2}\left\{I_0+K_1\beta V_i\sqrt{\frac{2I_{\mathrm{ss}}}{K_1\beta }-V_i^2}\right\}\left(\left|V_i\right|\le \sqrt{\frac{I_{\mathrm{ss}}}{K_1\beta }}\right)& \text{(2a)}\\ I_{\mathrm{D2}}=\frac{1}{2}\left\{I_0-K_1\beta V_i\sqrt{\frac{2I_{\mathrm{ss}}}{K_1\beta }-V_i^2}\right\}\left(\left|V_i\right|\le \sqrt{\frac{I_{\mathrm{ss}}}{K_1\beta }}\right)& \text{(2b)}\end{array}$

[0014] As shown in formulas (2a) and (2b), an operational input voltage range of the MOS differential pair 1 is |V_i|≦{I_ss/(K₁β)}.

[0015] The drain currents I_D1and I_D2of the MOS transistors M1 and M2 expressed in formulas (2a) and (2b) are converted into voltages respectively by square root (root) compression performed by the MOS transistors M3 and M4 serving as their loads. Therefore, two output voltages V₀₁and V₀₂of the MOS differential pair 1 having the MOS transistors M3 and M4 as loads are generated in the drains of the MOS transistors M1 and M2 respectively, and are expressed in the following formulas (3a) and (3b). $\begin{array}{cc}{V}_{\mathrm{D1}}=V_B-V_{\mathrm{TH}}-\sqrt{\frac{I_{\mathrm{D1}}}{K_2\beta }}\left(\left|V_i\right|\le \sqrt{\frac{I_{\mathrm{ss}}}{K_1\beta }}\right)& \text{(3a)}\\ V_{\mathrm{D2}}=V_B-V_{\mathrm{TH}}-\sqrt{\frac{I_{\mathrm{D2}}}{K_2\beta }}\left(\left|V_i\right|\le \sqrt{\frac{I_{\mathrm{ss}}}{K_1\beta }}\right)& \text{(3b)}\end{array}$

[0016] That is, when a differential output voltage of the MOS differential pair 1 is defined as ΔV, the ΔV is expressed as follows. $\begin{array}{cc}\Delta V=V_{\mathrm{O1}}-V_{\mathrm{O2}}=-\frac{1}{\sqrt{K_2\beta }}\left(\sqrt{I_{\mathrm{D1}}}-\sqrt{I_{\mathrm{D2}}}\right)& \left(4\right)\end{array}$

[0017] Here, a following formula (5) is introduced. In the formula (5), symbols a and b are constants and symbol x is a variable. $\begin{array}{cc}b\left(\sqrt{a+\sqrt{2}x\sqrt{1-\frac{x^2}{2}}}-\sqrt{a-\sqrt{2}x\sqrt{1-\frac{x^2}{2}}}\right)=b\sqrt{2}x& \left(5\right)\end{array}$

[0018] In addition, in the formula (5), these symbols a, b, and x are set up as follows. $\begin{array}{cc}a=1,b=\sqrt{I_{\mathrm{ss}}/2},x=V_i/\sqrt{\frac{I_{\mathrm{SS}}}{K_1\beta }}& \left(6\right)\end{array}$

[0019] Then, the left side of the formula (5) becomes equal to what is obtained by substituting the formulas (2a) and (2b) for the formula (4). At this time, the right-hand side of the formula (5) becomes (K₁β)·V_I. Therefore, the following formula (7) is obtained. $\begin{array}{cc}\Delta V=\frac{1}{\sqrt{K_2\beta }}\left(\sqrt{I_{\mathrm{D1}}}-\sqrt{I_{\mathrm{D2}}}\right)=\frac{1}{\sqrt{K_2\beta }}\sqrt{K_1\beta }{V}_i\left(\left|V_i\right|\le \sqrt{\frac{I_{\mathrm{SS}}}{K_1\beta }}\right)& \left(7\right)\end{array}$

[0020] It is obvious from FIG. 7, the differential output voltage ΔV of the MOS differential pair 1, i.e., a difference between the square roots of the drain current I_D1and drain current I_D2expressed by the formulas (2a) and (2b), respectively, is proportional to an input voltage V_ias apparent from this formula (7).

[0021] In addition, when a differential output current of the MOS differential pair 1 is defined as ΔI_D, ΔI_Dis expressed like a following formula (8) with using drain currents I_D1and I_D2. $\begin{array}{cc}\begin{array}{c}\Delta I_D=I_{\mathrm{D1}}-I_{\mathrm{D2}}=\left(\sqrt{I_{\mathrm{D1}}}-\sqrt{I_{\mathrm{D2}}}\right)\left(\sqrt{I_{\mathrm{D1}}}+\sqrt{I_{\mathrm{D2}}}\right)\\ =K_1\beta V_i\sqrt{\frac{2I_{\mathrm{SS}}}{K_1\beta }-V_i^2}\left(\left|V_i\right|\le \sqrt{\frac{I_{\mathrm{SS}}}{K_1\beta }}\right)\end{array}& \left(8\right)\end{array}$

[0022] Therefore, it can be seen that the differential output current ΔI_Dof the MOS differential pair 1 includes a linear term shown in formula (9) and a nonlinear term shown in formula (10) as follows: $\begin{array}{cc}\sqrt{I_{\mathrm{D1}}}-\sqrt{I_{\mathrm{D2}}}=\sqrt{K_1\beta }{V}_i\left(\left|V_i\right|\le \sqrt{\frac{I_{\mathrm{SS}}}{K_1\beta }}\right)& \left(9\right)\\ \sqrt{I_{\mathrm{D1}}}+\sqrt{I_{\mathrm{D2}}}=\sqrt{K_1\beta }\sqrt{\frac{2I_{\mathrm{SS}}}{K_1\beta }-V_i^2}\left(\left|V_i\right|\le \sqrt{\frac{I_{\mathrm{SS}}}{K_1\beta }}\right)& \left(10\right)\end{array}$

[0023] Letting the source voltage of the MOS transistors M1 and M2, which form the MOS differential pair 1 and whose sources are mutually connected, be a common source voltage V_S1, the common source voltage V_S1is expressed in the following formula (11). $\begin{array}{cc}{V}_{\mathrm{S1}}=V_{\mathrm{CM1}}-V_{\mathrm{TH}}-\frac{1}{2}\sqrt{\frac{2I_{\mathrm{SS}}}{K_1\beta }-V_i^2}& \left(11\right)\end{array}$

[0024] In the formula (11), V_CM1is a common mode voltage of the input voltage V_ithat is differentially inputted. As shown in the formula (11), since the common source voltage V_S1is a function of the input voltage V_i, the common source voltage V_S1is changed with an input voltage V_i. Moreover, a 3rd term (square root term) in the formula (11) is equal to a second square root portion in the nonlinear term (10) with disregarding a value of a coefficient that is a multiplier to it. Therefore, it is understood that the nonlinear term (9) of the differential output current ΔI_Din the MOS differential pair 1 originates in the change of the common source voltage V_S1.

[0025] This means that, if it is possible to fix the common source voltage V_S1of the MOS differential pair 1 to a constant voltage, a linear operation of the MOS differential pair 1 can be performed. Therefore, the differential output voltage ΔV of the MOS differential pair 1 is expressed like the following formula (12). $\begin{array}{cc}\Delta V=V_{\mathrm{O1}}-V_{\mathrm{O2}}=-\sqrt{\frac{K_1}{K_2}}{V}_i(\left|V_i\right|\le \sqrt{\frac{I_{\mathrm{SS}}}{K_1\beta }})& \left(12\right)\end{array}$

[0026] In the formula (12), if the ratio K₂of the gate width (W) and gate length (L) of the MOS transistors M3 and M4 that serve as loads is larger than the ratio K₁of the gate width (W) and gate length (L) of the MOS transistors M1 and M2 which form the MOS differential pair 1, this MOS differential pair 1 serves as a antiphase linear attenuator. On the other hand, if K₂is equal to or smaller than K₁, this MOS differential pair 1 serves as a antiphase linear amplifier. Here, its good linearity is realized over the entire operation input voltage range |V_i|≦{I_ss/(K₁β)}.

[0027] As obvious from the formula (12), the differential output voltage ΔV of the MOS differential pair 1 having the MOS transistors M3 and M4 as loads is proportional to the input voltages V_i. In other words, the MOS differential pair 1 having the MOS transistors M3 and M4 as loads operates as the linear attenuator or the linear amplifier to the input voltage V_i. Then, if (K₂/K₁) is set as a small value, a high gain can be realized.

[0028] In addition, when a common mode voltage of output voltages V₀₁and V₀₂is defined as V_CM2, the common mode voltage V_CM2is expressed with the following formula (13). $\begin{array}{cc}\begin{array}{c}{V}_{\mathrm{CM2}}=\frac{V_{\mathrm{o1}}+V_{\mathrm{o2}}}{2}=V_B-V_{\mathrm{TH}}-\sqrt{\frac{I_{D_1}}{K_2\beta }}-\sqrt{\frac{I_{\mathrm{D2}}}{K_2\beta }}\\ =V_B-V_{\mathrm{TH}}-\frac{1}{2}\sqrt{\frac{K_1}{K_2}}\sqrt{\frac{2I_{\mathrm{SS}}}{K_1\beta }-V_i^2}\\ =V_B-V_{\mathrm{TH}}-\sqrt{\frac{K_1}{K_2}}\left(V_{\mathrm{CM1}}-V_{\mathrm{TH}}-V_{\mathrm{S1}}\right)\left(\left|V_i\right|\le \sqrt{\frac{I_{\mathrm{SS}}}{K_1\beta }}\right)\end{array}& \left(13\right)\end{array}$

[0029] The formula (13) shows that the common mode voltage V_CM2of the output voltages V₀₁and V₀₂of the MOS differential pair 1 which have the MOS transistors M3 and M4 as loads are expressed with using the common source voltage V_S1(refer to the above-described formula (11)).

[0030] FIG. 9 is a graph showing a calculated values of the output voltage characteristic of the MOS differential pair 1 shown in FIG. 8.

[0031] In FIG. 9, two curves 31 and 32 show the output voltages V₀₁and V₀₂of the MOS differential pair 1 respectively, and a curve 33 shows the common mode voltage V_CM2of the input voltage V_i. A curve 34 shows a voltage [−V₀₁+2(V_B−V_TH)], and a straight line 35 shows a voltage [V₀₂−V₀₁+V_B−V_TH]. It is obvious from straight line 35, the differential output voltage ΔV of the MOS differential pair 1 is proportional to the input voltage V_i.

[0032] FIG. 10 is a circuit diagram showing a MOS triple-tail cell 3 used for a MOS linear transconductance amplifier (refer to FIG. 11) according to an embodiment of the present invention.

[0033] This MOS linear transconductance amplifier has a configuration of combining the MOS triple-tail cell shown in FIG. 10, and the MOS linear transconductance amplifier shown in FIG. 8. As shown in FIG. 10, this MOS triple-tail cell 3 is formed with three n-channel MOS transistors M5, M6 and M7 whose sources are connected each other. The sources of the MOS transistors M5, M6 and M7 are grounded in common through a constant current source4 (current value: I₀). This MOS triple-tail cell 3 is driven by the constant current I₀generated by the constant current source 4.

[0034] Each ratio (W/L) of the gate width (W) and gate length (L) of the MOS transistors M5 and M6 are equal to that of the unit MOS transistor, respectively. A ratio (W/L) of the gate width (W) and gate length (L) of the MOS transistor M7 is K₃times as large as that of the unit MOS transistor (K₃is a constant, however K₃≧1).

[0035] The first output voltage V₀₁of the MOS differential pair 1 generated in the drain of the MOS transistor M1 forming the MOS differential pair 1 (namely, the MOS linear transconductance amplifier in FIG. 8) is applied to the gate of the MOS transistor M5. At the same time, the second output voltage V₀₂of the MOS differential pair 1 generated in the drain of MOS transistor M2 forming the MOS differential pair 1 is applied to the gate of MOS transistor M6. The Difference (namely, differential output voltage of the MOS differential pair) ΔV between these two output voltages V₀₁and V₀₂becomes the input voltage of the triple-tail cell 3.

[0036] A drain of the MOS transistor M7 is connected to a supply voltage line (V_DD), and a control voltage (DC constant voltage) V_cis applied to its gate. Two drains of the MOS transistors M5 and M6 forms an output terminal pair of this triple-tail cell 3, and output currents I⁺and I⁻are taken out from the output terminal pair, respectively. An n-channel MOS transistor M8 and a constant current source 5 (current value: I_ss/2) constitute a control voltage generating circuit for generating the control voltage V_cfor the MOS transistor M7. A source of the MOS transistor M8 is grounded through the constant current source 5, a drain of the MOS transistor M8 is connected to a supply voltage line (V_DD), and a bias voltage V_Bis applied to its gate.

[0037] Next, an operation principle of this MOS triple-tail cell 3 will be described.

[0038] A MOS quadritail cell consists of the unit transistors M5 and M6 and the transistors M7 and M8. Here, transistor size ratios of the transistors M7 and M8 to the unit transistor are K₃, and the sources of the transistors M5 to M7 are commonly connected and are driven by the constant current I₀. When a voltage ΔV is applied between the gates of the transistors M5 and M6, and a voltage V_cis applied to the gate of the transistor M7, and V_CM3is input common mode voltage, the drain currents of respective transistors are expressed as follows:

[0039] I_D5=β(V_CM3+ΔV/2−V_S2−V_TH)²  (14)

[0040] I_D6−β(V_CM3−ΔV/2−V_S2−V_TH)²  (15)

[0041] I_D7=I_D8=K₃β(V_CM3+V_C−V_S2−V_TH)²  (16)

[0042] where an input common mode voltage is made to be V_CM3. Here, a symbol V_S2is a common source voltage of the MOS quadritail cell.

[0043] Moreover, the following formula is obtained from a condition of the tail current:

[0044] I_D5+I_D6+I_D7+I_D8=I₀  (17)

[0045] With solving formula (V_CM3−V_S2−V_TH) by substituting the formulas (14) to (16) for the formula (17), the solution is as follows: $\begin{array}{cc}{V}_{\mathrm{CM3}}-V_{\mathrm{S2}}-V_{\mathrm{TH}}=\frac{-K_3{V}_C+\sqrt{\left(K_3+1\right)\frac{I_0}{2\beta }-\frac{K_3+1}{4}{\left(\Delta V\right)}^2-K_3{V}_C^2}}{K_3+1}& \left(18\right)\end{array}$

[0046] When the drain currents of MOS transistors M5 and M6 are defied as I_D5and I_D6respectively, a differential output current ΔI (=I⁺−I⁻) of this triple-tail cell 3 is expressed in the following formula (19). $\begin{array}{cc}\begin{array}{c}\Delta I=U_{\mathrm{D5}}-I_{\mathrm{D6}}=2\beta \left(\Delta V\right)\left(V_{\mathrm{CM3}}-V_{\mathrm{S2}}-V_{\mathrm{TH}}\right)\\ =\frac{\begin{array}{c}-2K_3\beta \left(\Delta V\right)V_C+\\ 2\beta \left(\Delta V\right)\sqrt{\left(K_3+2\right)\frac{I_0}{\beta }-\frac{K_3+2}{2}{\left(\Delta V\right)}^2-2K_3{V}_C^2}\end{array}}{K_3+2}\\ \left(\left|\Delta V\right|\le \min \left\{\sqrt{\frac{2I_0}{\beta }-4V_C^2},\frac{\begin{array}{c}-2K_3{V}_C+\\ 2\sqrt{\left(K_3+4\right)\frac{I_0}{\beta }-4K_3{V}_C^2}\end{array}}{K_3+4}\right\}\right)\end{array}& \left(19\right)\end{array}$

[0047] Here, the differential output voltage ΔV of the MOS differential pair 1 that is inputted between the gates of the MOS transistors M5 and M6 is linear to the input voltage V_iinputted to the MOS differential pair 1 (that is, this amplifier). Moreover, the drain currents I_D5and I_D6of the MOS transistors M5 and M6 which form the triple-tail cell 3 have a square-law characteristics to the input voltage ΔV inputted to the triple-tail cell 3, respectively. In consideration of these, in order that the amplifier shown in FIG. 8 outputs a current having the square-law characteristics, it is necessary that the differential output current ΔI of this triple-tail cell 3 that is expressed in the formula (19) becomes linear to the input voltage ΔV, i.e., that the differential output current ΔI is proportional to the input voltage ΔV.

[0048] That is, when c is constant, it is required that satisfying a following formula:

[0049] ΔI=cΔV  (20)

[0050] Therefore, a coefficient multiplied to ΔV of numerator of the above-described formula (19) must be equal to a constant c. In other words, it is necessary to satisfy a following formula (21). $\begin{array}{cc}-K_3{V}_C+\sqrt{\left(K_3+2\right)\frac{I_0}{2\beta }-\frac{K_3+2}{4}{\left(\Delta V\right)}^2-2K_3{V}_C^2}=c\left(\mathrm{constant}\right)& \left(21\right)\end{array}$

[0051] Here, the differential output current ΔI of the triple-tail cell 3 is as follows. $\begin{array}{cc}\Delta I=\frac{2\mathrm{c\beta }}{K_3+2}\left(\Delta V\right)& \left(22\right)\end{array}$

[0052] Moreover, with determining the control voltage V_cat this time based on the formula (21), the control voltage V_cis expressed as a following formula (23). $\begin{array}{cc}{V}_C=\frac{-K_{3}c+\sqrt{{K_3\left(K_3+2\right)}^2\frac{I_0}{\beta }-\frac{{K_3\left(K_3+2\right)}^2}{2}{\left(\Delta V\right)}^2-2K_3{c}^2}}{K_3\left(K_3+2\right)}& \left(23\right)\end{array}$

[0053] Therefore, the differential output current ΔI of this triple-tail cell 3 expressed in the above-described formula (19) is linear to the input voltage ΔV. That is, in order that the MOS linear transconductance amplifier shown in FIG. 8 outputs a current having a linear characteristics, the control voltage Vc must be set up so as to satisfy the formula (21). Then, the differential output current ΔI of the triple-tail cell 3 at that time is expressed in the above-described formula (22).

[0054] For example, when defining C²as follows: $\begin{array}{cc}{c}^2={\left(K_3+2\right)}^2\frac{I_0}{4\beta }& \left(24\right)\end{array}$

[0055] the control voltage V_cis required to be set as follows: $\begin{array}{cc}{V}_C=-\frac{1}{2}\sqrt{\frac{I_0}{\beta }}+\sqrt{\frac{I_0}{2K_3\beta }-\frac{1}{2K_3}{\left(\Delta V\right)}^2}& \left(25\right)\end{array}$

[0056] As described above, if the control voltage V_cto the MOS transistor M7 of the triple-tail cell 3 is set up so as to satisfy the above-described formula (23), the differential output current ΔI of this triple-tail cell 3 expressed in the above-described formula (19) becomes linear to the input voltage ΔV. Thus, the differential output current ΔI is expressed in the above-described formula (22).

[0057] FIG. 11 is a circuit diagram showing a MOS linear transconductance amplifier obtained in this way.

[0058] In this amplifier, a cascade connection of the MOS triple-tail cell 3 shown in FIG. 10 is performed to an output terminal of the MOS differential pair 1 (namely, MOS linear transconductance amplifier shown in FIG. 8) which consists of the MOS transistors M1 and M2, and MOS transistors M3 and M4 used as loads thereof. For this reason, respective gate voltages of the MOS transistors M5, M6 and M7 which form the triple-tail cell 3 become V₀₁, V₀₂and (V_CM2+V_c). If the gate voltage (V_CM2+V_c)=V_G7of the MOS transistor M7 becomes a constant value, a gate bias circuit for generating the control voltage V_ccan be simplified remarkably. Thereat, next, conditions necessary for it will be obtained.

[0059] Since the common mode voltage V_CM2of the output voltages V₀₁and V₀₂is expressed in the above-described formula (13) and the control voltage V_cis expressed as the above-described formula (23), the gate voltage V_G7=(V_CM2+V_c) of the MOS transistor M7 is expressed in the following formula (26). Here, symbol d is a constant. $\begin{array}{cc}\begin{array}{c}{V}_{\mathrm{G7}}=V_{\mathrm{CM2}}+V_C\\ =V_B-V_{\mathrm{TH}}-\frac{1}{2}\sqrt{\frac{K_1}{K_2}}\sqrt{\frac{2I_{\mathrm{SS}}}{K_1\beta }-V_i^2}+\\ \frac{-K_{3}c+\sqrt{{K_3\left(K_3+2\right)}^2\frac{I_0}{\beta }-\frac{K_1{K_3\left(K_3+2\right)}^2}{2K_2}{\left(\Delta V\right)}^2-2K_3{c}^2}}{K_3\left(K_3+2\right)}\\ =d\left(\mathrm{constant}\right)\end{array}& \left(26\right)\end{array}$

[0060] As mentioned above, it is required for the differential output current ΔI of the triple-tail cell 3 to be proportional to the input voltage ΔV in order that the MOS linear transconductance amplifier of FIG. 11 outputs the current having linear characteristics. Therefore, in the formula (26), all coefficients of the term containing the input voltage ΔV must be zero. That is, the formula (26) must be simplified like the following formula (27). $\begin{array}{cc}{V}_{\mathrm{CM2}}+V_C=V_B-V_{\mathrm{TH}}-\frac{c}{K_3+2}=d\left(\mathrm{constant}\right)& \left(27\right)\end{array}$

[0061] Necessary conditions for satisfying the formula (27) are that satisfying two relational expressions (28a) and (28b) in the formula (26).

[0062] K₃=2  (28a) $\begin{array}{cc}\frac{I_0}{\beta }=\frac{I_{\mathrm{SS}}}{K_2\beta }+\frac{c^2}{8}& \text{(28b)}\end{array}$

[0063] Therefore, when values such as current values I₀and I_ssare set up so as to satisfy such relational expressions (28a) and (28b), satisfying the formula (27) and the gate voltage V_G7=(V_CM2+V_c) of the MOS transistor M7 becomes a constant value. Consequently, the bias circuit for generating the control voltage V_capplied to the MOS transistor M7 is remarkably simplified as shown in FIG. 11. Then, in that case, since the control voltage V_cfollows the above-described formula (23) in the circuit configuration shown in FIG. 11, the differential output current ΔI of this triple-tail cell 3 becomes linear to the input voltage ΔV as expressed in the above-described formula (22).

[0064] Moreover, as mentioned above, the input voltage ΔV to the MOS triple-tail cell 3 is the differential output voltage ΔV of the MOS differential pair 1 having the MOS transistors M3 and M4 as loads, and is proportional to the input voltage V_ito this amplifier concerned. In this way, it is assured that the amplifier in FIG. 11 outputs the output current ΔI having linear characteristics to the input voltage V_ias the differential output current ΔI of the MOS triple-tail cell 3.

[0065] Next, an operational input voltage range of the MOS linear transconductance amplifier shown in FIG. 11 will be described.

[0066] The drain currents I_D5, I_D6and I_D7of the MOS transistors M5, M6 and M7 which form the triple-tail cell 3 are expressed in regard to the input voltage ΔV in the following formulas (29a), (29b) and (29c), respectively. $\begin{array}{cc}{I}_{\mathrm{D5}}=\frac{\beta }{4}{\left(\Delta V-\sqrt{\frac{I_0}{\beta }}\right)}^2\left(\left|\Delta V\right|\le \sqrt{\frac{I_0}{2\beta }}\right)& \text{(29a)}\\ I_{\mathrm{D6}}=\frac{\beta }{4}{\left(\Delta V+\sqrt{\frac{I_0}{\beta }}\right)}^2\left(\left|\Delta V\right|\le \sqrt{\frac{I_0}{2\beta }}\right)& \text{(29b)}\\ I_{\mathrm{D7}}=\frac{1}{2}\left\{I_0-{\beta \left(\Delta V\right)}^2\right\}\left(\left|\Delta V\right|\le \sqrt{\frac{I_0}{2\beta }}\right)& \text{(29c)}\end{array}$

[0067] Therefore, an effective tail current of two transistors that constitute a differential pair of a quadri-tail cell is shown as a following formula (30). $\begin{array}{cc}{I}_{\mathrm{D5}}+I_{\mathrm{D6}}=\frac{1}{2}\left\{I_0+{\beta \left(\Delta V\right)}^2\right\}\left(\left|\Delta V\right|\le \sqrt{\frac{I_0}{2\beta }}\right)& \left(30\right)\end{array}$

[0068] It is obvious from the formulas (29c) and (30), a sum current of two output currents I⁺and I⁻of the triple-tail cell 3 and I_D7are proportional to the square value of the input voltage ΔV respectively, the sum current of those output currents I⁺and I⁻are equiphase square currents, and I_D7shows the square current in an antiphase.

[0069] Next, the conditions that the linear input voltage range of the MOS triple-tail cell 3 and the operational input voltage range of the MOS differential pair 1 become equal will be obtained.

[0070] First, if V_iis zero in the formula (28b), the formula (27) is satisfied also in the case of V_o=0. At this time, formulas (31) and (32) are obtained. $\begin{array}{cc}d=V_B-V_{\mathrm{TH}}-\sqrt{\frac{I_{\mathrm{SS}}}{2K_2\beta }}& \left(31\right)\\ c=4\sqrt{\frac{I_{\mathrm{SS}}}{2K_2\beta }}=2\sqrt{\frac{2I_{\mathrm{SS}}}{K_2\beta }}& \left(32\right)\end{array}$

[0071] Moreover, a following formula (33) is obtained by substituting the formula (32) for the formula (28b) and solving this. $\begin{array}{cc}{I}_0=2\frac{I_{\mathrm{SS}}}{K_2}& \left(33\right)\end{array}$

[0072] In the case of K₁=K₂=1, K₃=2 and I_ss=I₀/2, the circuit configuration shown in FIG. 11 can be most simplified. At this time, a value of the constant c can be calculated from formula (34). $\begin{array}{cc}c=2\sqrt{\frac{I_0}{\beta }}& \left(34\right)\end{array}$

[0073] Moreover, the constant d and the control voltage V_cat this time become as shown in a following formulas (35) and (36), respectively. $\begin{array}{cc}d=V_{\mathrm{CM2}}+V_C=V_B-V_{\mathrm{TH}}-2\sqrt{\frac{I_0}{\beta }}& \left(35\right)\\ V_C=\frac{1}{2}\left\{-\sqrt{\frac{I_0}{\beta }}+\sqrt{\frac{I_0}{\beta }-{\left(\Delta V\right)}^2}\right\}& \left(36\right)\end{array}$

[0074] At this time, the differential output current ΔI of the linear transconductance amplifier shown in FIG. 11 is expressed in formula (37). $\begin{array}{cc}\Delta I=I_{\mathrm{D5}}-I_{\mathrm{D6}}=-\sqrt{\beta I_0}{V}_i\left(\left|V_i\right|\le \sqrt{\frac{I_0}{2\beta }}\right)& \left(37\right)\end{array}$

[0075] At this time, an operation range |V_i|≦(I₀/β) becomes equal to the operation range of the MOS differential pair having transistors as loads. A transconductance is expressed in formula (38). $\begin{array}{cc}\frac{d\left(\Delta V\right)}{d{V}_i}=-\sqrt{\beta I_0}\left(\left|V_i\right|\le \sqrt{\frac{I_0}{2\beta }}\right)& \left(38\right)\end{array}$

[0076] The MOS transistor M7 constituting the triple-tail cell 3 shown in FIG. 10 has a ratio (W/L) between the gate width (W) and gate length (L) that is two times of the ratio of the unit MOS transistor (K₃=2). Therefore, the MOS transistor M7 can be divided into two unit MOS transistors M7A and M7B all of whose sources, drains and gates are connected each other. Thus, the triple-tail cell 3 can be modified into a quadri-tail cell 3′ shown in FIG. 12. Here, sources of the MOS transistors M7A and M7B of the quadritail cell 3′ are connected to each source of the MOS transistors M5 and M6, respectively. In addition, two resistors (not shown) having resistances each are R_Lare added as loads of the MOS transistors M5 and M6. Thus, the amplifier performs class A operation. This will be described with using FIG. 11.

[0077] FIG. 13 is a graph showing a characteristics of drain currents I_D5, I_D6, I_D7Aand I_D7Bof the MOS transistors M5, M6, M7A and M7B constiting the quadri-tail cell 3′.

[0078] In FIG. 13, curves 51, 52 and 53 show that each of the drain currents I_D5, I_D6, I_D7Aand I_D7Bhas square-law characteristics. Moreover, a curve 54 shows that a sum of the drain currents I_D7Aand I_D7Balso has the square-law characteristics. Furthermore, it can be also seen from a curve 55 shows that a sum of the drain currents I_D5and I_D7Ahas linear characteristics, and also from a curve 56 that a sum of drain currents I_D6and I_D7Bhas linear characteristics. In this way, the differential output current ΔI in the linear transconductance amplifier is converted into a voltage, which is taken out as the differential output voltage ΔV_o.

[0079] It is obvious from FIG. 13, it can be seen from the curve 55 that the sum of the drain currents I_D5and I_D7Ahas linear characteristics, and from the curve 56 that the sum of the drain currents I_D6and I_D7Bhas linear characteristics. Hence, in the amplifier according to this fifth embodiment, the differential output current ΔI can be expressed as ΔI=(I_D5+I_D7A)−(I_D6+I_D7B) and this also becomes linear in a wide input voltage range |V_i|≦(I₀/2β). Therefore, the differential output voltage V_oobtained by performing the voltage conversion of the differential output current ΔI by the resistors also has linear characteristic in a wide input voltage range |V_i|≦(I₀/2β).

[0080] FIG. 14 is a graph showing the calculated values of transfer characteristics of the MOS linear transconductance amplifier shown in FIG. 12 are shown.

[0081] In FIG. 14, two curves 81 and 82 show changes of the output currents (I_D6+I_D7B) and (I_D5+I_D7A) in the MOS linear transconductance amplifier shown in FIG. 12. FIG. 14 shows that the output currents (I_D6+I_D7B) and (I_D5+I_D7A) in the MOS linear transconductance amplifier have the linear characteristics.

[0082] Otherwise, this kind of MOS linear transconductance amplifier is described in Japanese Patent Laid-Open No. 10-209781. This circuit is shown in FIG. 15. Although an operation of this circuit is analyzed in preferred embodiments in the above-described Japanese Patent Laid-Open No. 10-209781 using formulas. However, each formula does not have a middle voltage V_CM. Therefore, it is difficult to understand the operation. Moreover, since an operational input voltage range and a linear input voltage range are not shown and it is also not shown why the linear input voltage range is wide or how the linear input voltage range can be extended, technical disclosure is not sufficient.

[0083] Then, the present inventor will simply show these by performing a circuit analysis. In this circuit, since all transistors M1 to M9 are the same transistors, the circuit analysis becomes simpler by several steps than that in the case of being Japanese Patent Laid-Open No. 11-251848. In FIG. 15, the transistors M5 to M9 are mutually connected at their sources, and are driven by a constant current source 5I_ss.

[0084] I_D5+I_D6+I_D7+I_D8+I_D9=5I_SS  (39)

[0085] Moreover, a current flowing in the transistor M5 drives four transistors M5 to M8 whose drains are mutually connected through a 1:4 current mirror circuit consisting of transistors M14 to M18.

[0086] Thus, the following formula is obtained:

[0087] I_D5+I_D6+I_D7+I_D8=4I_D9  (40)

[0088] Hence, solving the formulas (39) and (40), the following formulas can be obtained:

[0089] I_D5=β(V_CM−V_n2−V_TH)²=I_SS  (41)

[0090] I_D5+I_D6+I_D7+I_D8=4I_SS  (42)

[0091] However, V_n2denotes a common source voltage.

[0092] From the formula (41), the following formula can be obtained: $\begin{array}{cc}{V}_{\mathrm{CM}}-V_{\mathrm{n2}}-V_{\mathrm{TH}}=\sqrt{\frac{I_{\mathrm{SS}}}{\beta }}& \left(43\right)\end{array}$

[0093] Moreover, satisfy the following formulas: $\begin{array}{cc}{I}_{\mathrm{D5}}={\beta \left(V_{\mathrm{CM}}+\frac{V_i}{2}-V_{\mathrm{n2}}-V_{\mathrm{TH}}\right)}^2& \left(44\right)\\ I_{\mathrm{D6}}={\beta \left(V_{\mathrm{CM}}-\frac{V_i}{2}-V_{\mathrm{n2}}-V_{\mathrm{TH}}\right)}^2& \left(45\right)\end{array}$  I_D7=I_D8=β(V_CONT−V_n2−V_TH)²  (46)

[0094] Hence, with substituting the formulas (44) to (46) for the formula (42) and rewriting the formula (42) with using the formula (43), the following formula can be obtained: $\begin{array}{cc}{V}_{\mathrm{CONT}}-V_{\mathrm{CM}}=-\sqrt{\frac{I_{\mathrm{SS}}}{\beta }}+\sqrt{\frac{I_{\mathrm{SS}}}{\beta }-\frac{V_i^2}{4}}\left(\left|V_i\right|\le \sqrt{\frac{2I_{\mathrm{SS}}}{\beta }}\right)& \left(47\right)\end{array}$

[0095] The input operation range where the formula (47) can be obtained is given from the following condition in which no transistors M5 to M9 become pinch-off: $\left(\sqrt{\frac{I_{\mathrm{SS}}}{2\beta }}\le \sqrt{\frac{I_{\mathrm{SS}}}{\beta }-\frac{V_i^2}{4}}\left(\le \sqrt{\frac{I_{\mathrm{SS}}}{\beta }}\right)\right)$

[0096] With substituting the formulas (43) and (47) for the formulas (44) to (46), the followings can be obtained: $\begin{array}{cc}{I}_{\mathrm{D5}}=\frac{\beta }{4}{V}_i^2+\sqrt{\beta I_{\mathrm{SS}}}{V}_i+I_{\mathrm{SS}}\left(\left|V_i\right|\le \sqrt{\frac{2I_{\mathrm{SS}}}{\beta }}\right)& \left(48\right)\\ I_{\mathrm{D6}}=\frac{\beta }{4}{V}_i^2-\sqrt{\beta I_{\mathrm{SS}}}{V}_i+I_{\mathrm{SS}}\left(\left|V_i\right|\le \sqrt{\frac{2I_{\mathrm{SS}}}{\beta }}\right)& \left(49\right)\\ I_{\mathrm{D7}}=I_{\mathrm{D8}}=I_{\mathrm{SS}}-\frac{\beta }{4}{V}_i^2\left(\left|V_i\right|\le \sqrt{\frac{2I_{\mathrm{SS}}}{\beta }}\right)& \left(50\right)\end{array}$

[0097] Hence, the following formulas can be obtained: $\begin{array}{cc}{I}_{\mathrm{D5}}+I_{\mathrm{D7}}=2I_{\mathrm{SS}}+\sqrt{\beta I_{\mathrm{SS}}}{V}_i\left(\left|V_i\right|\le \sqrt{\frac{2I_{\mathrm{SS}}}{\beta }}\right)& \left(51\right)\\ I_{\mathrm{D6}}+I_{\mathrm{D8}}=2I_{\mathrm{SS}}-\sqrt{\beta I_{\mathrm{SS}}}{V}_i\left(\left|V_i\right|\le \sqrt{\frac{2I_{\mathrm{SS}}}{\beta }}\right)& \left(52\right)\end{array}$

[0098] Thus, both I_D5+I_D7and I_D6+I_D5are linear differential currents.

[0099] Here, a control voltage generating circuit consisting of the transistors M5 to M9 will be compared to an output circuit consisting of transistors M1 to M4. Supposing that the constant current I_ssis removed since the current always flows in the transistor M5, it can be seen that the transistors M1 to M4 and transistors M5 to M8 have mutually equal bias conditions.

[0100] Hence, the following formulas can be obtained: $\begin{array}{cc}{I}_{\mathrm{D1}}+I_{\mathrm{D3}}=2I_{\mathrm{SS}}+\sqrt{\beta I_{\mathrm{SS}}}{V}_i\left(\left|V_i\right|\le \sqrt{\frac{2I_{\mathrm{SS}}}{\beta }}\right)& \left(53\right)\\ I_{\mathrm{D2}}+I_{\mathrm{D4}}=2I_{\mathrm{SS}}-\sqrt{\beta I_{\mathrm{SS}}}{V}_i\left(\left|V_i\right|\le \sqrt{\frac{2I_{\mathrm{SS}}}{\beta }}\right)& \left(54\right)\end{array}$

[0101] Both I_D1+I_D3and I_D2+I_D4are linear differential currents. That is, two output currents are expressed as follows: $\begin{array}{cc}{I}_{\mathrm{On}}=-\sqrt{\beta I_{\mathrm{SS}}}{V}_i\left(\left|V_i\right|\le \sqrt{\frac{2I_{\mathrm{SS}}}{\beta }}\right)& \left(55\right)\\ I_{\mathrm{Op}}=+\sqrt{\beta I_{\mathrm{SS}}}{V}_i\left(\left|V_i\right|\le \sqrt{\frac{2I_{\mathrm{SS}}}{\beta }}\right)& \left(56\right)\end{array}$

[0102] Hence, a linear OTA is realizable.

[0103] However, since the conventional MOS linear transconductance amplifier described above reverses a voltage with the antiphase amplifier, it is not able to lower a power supply voltage in order to maintain a linear operating input voltage. Alternatively, there is a disadvantage that, even if it is possible to lower the voltage, a circuit current in the control voltage generating circuit becomes larger than a circuit current in the main output circuit.

[0104] In analog signal processing, a differential amplifier circuit is an essential function block. In particular, a demand of a linear MOS differential amplifier circuit has increased further. Therefore, an object of the present invention is to provide a MOS linear transconductance amplifier that can be realized with an LSI and has the linear transconductance excellent in the frequency characteristic.

[0105] A MOS linear transconductance amplifier according to claim 1 is characterized in: that gates of first and second transistors whose sources are grounded form an input pair, to which a differential voltage is inputted; that a gate and drain of a third transistor are mutually connected; that drains of the above first, second and third transistors are mutually connected and are driven by a constant current; wherein the MOS linear transconductance amplifier comprises means for adding a current flowing in the above-described first transistor to a current that is a half of a current flowing in the above-described third transistor, and means for adding a current flowing in the above-described second transistor, and a current that is a half of a current flowing in the above-described third transistor; and that these two sum currents are made to be a differential output current.

[0106] A MOS linear transconductance amplifier according to claim 2 is characterized in: that gates of first and second transistors whose sources are grounded form an input pair, to which a differential voltage is inputted; that the gate and drain of a third transistor are mutually connected; that drains of the above-described first, second and third transistors are mutually connected and are driven by a constant current; that a fourth transistor whose gate is mutually connected to the gate of the above-described first transistor, a fifth transistor whose gate is mutually connected to the gate of the above-described second transistor, and sixth and seventh transistors whose gates are mutually connected to the gate of the third transistor are mutually connected at sources and are driven by a constant current; that drains of the above-described fourth transistor and above-described sixth transistor are mutually connected and form one of a differential output pair; that drains of the above-described fifth transistor and above-described seventh transistor are mutually connected, and form the other of the differential output pair; and that each gate ratio (W/L) of the above-described sixth and seventh transistors is a half of a gate ratio (W/L) of the above-described third transistor.

[0107] A MOS linear transconductance amplifier according to claim 3 is characterized in that, in the MOS linear transconductance amplifier according to claim 2, sources of the above-described first, second and third transistors are mutually connected and are grounded through a resister or a diode that is formed by mutually connecting a gate and a source.

[0108] A MOS linear transconductance amplifier according to claim 4 is characterized in: that the MOS linear transconductance amplifier comprises first, second, third and forth transistors whose sources are mutually connected, and which are driven by a constant current; that a common voltage V_CMof a differential input voltage is applied to the gate of the above-described first transistor; that gates of the above-described second and third transistors constitute an input pair, to which a differential voltage is inputted; that the gate and drain of the above-described fourth transistor are mutually connected; that the above-described second, third and fourth transistors are mutually connected at their drains and are driven by a constant current; that a fifth transistor whose gate is mutually connected to the gate of the above-described second transistor, a sixth transistor whose gate is mutually connected to the gate of the above-described third transistor, and seventh and eighth transistors whose gates are mutually connected to the gate of the fourth transistor are mutually connected at their sources and are driven by a constant current; that drains of the above-described fifth transistor and above-described seventh transistor are mutually connected and form one of a differential output pair; that drains of the above-described sixth transistor and above-described eighth transistor are mutually connected, and form the other of the differential output pair; and that each gate ratio (W/L) of the above-described seventh and eighth transistors is a half of a gate ratio (W/L) of the above-described fourth transistor.

[0109] A MOS linear transconductance amplifier according to claim 5 is characterized in that, in the MOS linear transconductance amplifier according to claim 4, wherein a current flowing in the above-described first transistor is a reference current, currents having predetermined mirror ratios are supplied to the common drain of the above-described second, third and fourth transistors, and one and the other of the above-described differential output pair respectively.

[0110] A MOS linear transconductance amplifier according to claim 6 or 7 is characterized in that, in the MOS linear transconductance amplifier according to claim 4 or 5, a common source voltage of the above-described first, second, third and fourth transistors is different from a common source voltage of the above-described fifth, sixth, seventh and eighth transistors.

[0111] A MOS linear transconductance amplifier according to claim 8 or 9 is characterized in that, in the MOS linear transconductance amplifier according to claim 4 or 5, sources of the above-described first, second, third, fourth, fifth, sixth, seventh and eighth transistors are mutually connected.

[0112] Next, in other words, the configuration of the MOS linear transconductance amplifier according to the present invention will be described once again. In the MOS linear transconductance amplifier according to the present invention, that gates of the first and second transistors whose sources are grounded constitute the input pair, the differential voltage is inputted to the input pair, and the gate and drain of the third transistor is mutually connected, the above-described first, second and third transistors are mutually connected at their drains, and are driven by the constant current. Furthermore, the MOS linear transconductance amplifier comprises means for adding a current flowing in the above-described first transistor to a current that is a half of a current flowing in the above-described third transistor, and means for adding the current flowing in the above-described second transistor to the current that is a half of the current flowing in the above-described third transistor, and these two sum currents are made to be differential output currents.

[0113] Alternatively, sources of the above-described first, second and third transistors are mutually connected and are grounded directly, or of through a resister or a diode that is formed by mutually connecting a gate and a source. Gates of the above-described first and second transistors constitute an input pair, to which the differential voltage is inputted. The gate and drain of the above-described third transistor are mutually connected. Furthermore, the above-described first, second and third transistors are mutually connected at their drains, and are driven by a constant current. The fourth transistor whose gate is mutually connected to the gate of the above-described first transistor, a fifth transistor whose gate is mutually connected to the gate of the above-described second transistor, and six and seventh transistors whose gates are mutually connected to the gate of the above-described third transistor are mutually connected at their sources and are driven by a constant current. Drains of the above-described fourth transistor and above-described sixth transistor are mutually connected and form one of a differential output pair. Drains of the above-described fifth transistor and above-described seventh transistor are mutually connected, and form the other of the differential output pair. Then, each gate ratio (W/L) of the above-described sixth and seventh transistors is a half of a gate ratio (W/L) of the above-described third transistor.

[0114] Alternatively, the MOS linear transconductance amplifier comprises first, second, third and forth transistors whose sources are mutually connected, and which are driven by the constant current. The common voltage of the differential input voltage is applied to the gate of the above-described first transistor. Gates of the above-described second and third transistors constitute an input pair, to which the differential voltage is inputted. The gate and drain of the above-described fourth transistor are mutually connected. Furthermore, the above-described second, third and fourth transistors are mutually connected at their drains, and are driven by the constant current. A fifth transistor whose gate is mutually connected to the gate of the above-described second transistor, a sixth transistor whose gate is mutually connected to the gate of the above-described third transistor, seventh and eighth transistors whose gates are mutually connected to the gate of the fourth transistor are mutually connected at their sources and are driven by a constant current. Drains of the above-described fifth transistor and above-described seventh transistor are mutually connected and form one of the differential output pair. Drains of the above-described sixth transistor and above-described eighth transistor are mutually connected, and form the other of the differential output pair. In addition, each gate ratio (W/L) of the above-described seventh and eighth transistors is a half of the gate ratio (W/L) of the above-described fourth transistor.

[0115] The MOS linear transconductance amplifier including a current mirror circuit, wherein the current flowing in the above-described first transistor is the reference current, currents having predetermined mirror ratios are supplied to the common drain of the above-described second, third and fourth transistors, and one and the other of the above-described differential output pair, respectively.

[0116] Next, with changing words, an operation of the MOS linear transconductance amplifier according to the present invention will be described once again. As a method of obtaining a straight line from a curve with the square-law characteristics is generally called a parabolic characteristics, there is a method of using a class A operation obtained by adding square characteristics to the parabolic characteristics besides a class AB operation obtained by reducing the parabolic characteristics from the parabolic characteristics. Since it is possible to configure a circuit without including a p-channel transistor on a signal path in case of current addition procedure, improvement in its frequency characteristic can be estimated.

[0117] FIG. 1 is a circuit diagram showing an embodiment of the linear transconductance amplifier according to claim 1;

[0118] FIG. 2 is a circuit diagram showing an embodiment of the linear transconductance amplifier according to claim 2;

[0119] FIG. 3 is a circuit diagram showing an embodiment of the linear transconductance amplifier according to claim 3;

[0120] FIG. 4 is a circuit diagram showing an embodiment of the linear transconductance amplifier according to claim 4;

[0121] FIG. 5 is a circuit diagram showing an embodiment of the linear transconductance amplifier according to claim 5;

[0122] FIG. 6 is a circuit diagram showing an embodiment of the linear transconductance amplifier according to claim 7;

[0123] FIG. 7 is a circuit diagram showing another embodiment of the linear transconductance amplifier according to claim 7;

[0124] FIG. 8 is a circuit diagram showing a conventional linear transconductance amplifier

[0125] FIG. 9 is an explanatory diagram showing the linear operation of the conventional linear transconductance amplifier;

[0126] FIG. 10 is a circuit diagram showing a conventional linear transconductance amplifier (quadritail cell);

[0127] FIG. 11 is a circuit diagram showing another conventional linear transconductance amplifier;

[0128] FIG. 12 is a circuit diagram showing a conventional class A linear transconductance amplifier;

[0129] FIG. 13 is an explanatory diagram showing class A operation in the conventional linear transconductance amplifier;

[0130] FIG. 14 is a characteristic diagram showing the output current of the conventional linear transconductance amplifier; and

[0131] FIG. 15 is a circuit diagram showing another example of a conventional linear transconductance amplifier (quadritail cell).

[0132] FIG. 1 is a circuit diagram showing an embodiment of a MOS linear transconductance amplifier according to claim 1.

[0133] Transistors M6 and M7 whose sources are grounded constitutes a current mirror circuit for setting a bias current. Here, a current flowing in the unit transistor M7 is a reference current of another current mirror circuit (mirror ratio is 1:K₁+2) when the input common mode voltage V_CMis applied. In transistors M1 to M3 whose sources are grounded, the transistors M1 and M2 are unit transistors. It is assumed that a gate ratio (W/L) of the transistor M3 is K₁time as large as that of the unit transistor. Two gates of transistors M1 and M2 constitute a differencial input pair, and a gate and drain of the transistor M3 is connected each other. Furthermore, the transistors M1 to M3 are mutually connected at their drains, and are driven by a constant current I_B.

[0134] Drain currents of respective transistors M1, M2 and M3 at this time are shown as follows: $\begin{array}{cc}{I}_{\mathrm{D1}}={\beta \left(V_{\mathrm{CM}}+\frac{V_i}{2}-V_{\mathrm{TH}}\right)}^2& \left(57\right)\\ I_{\mathrm{D2}}={\beta \left(V_{\mathrm{CM}}-\frac{V_i}{2}-V_{\mathrm{TH}}\right)}^2& \left(58\right)\end{array}$  I_D3=K₁β(V_CM+V_C−V_TH)²  (59)

[0135] Here, since these transistors are driven by the constant current I_B, the following formula is satisfied:

[0136] I_D1+I_D2+I_D3=I_B  (60)

[0137] Here, if the bias current I_Bis set so as to be V_C=0 when V_i=0, a following formula is obtained: $\begin{array}{cc}{\beta \left(V_{\mathrm{CM}}-V_{\mathrm{TH}}\right)}^2=\frac{I_B}{K_1+2}& \left(62\right)\end{array}$

[0138] With solving formula (V_CM+V_C−V_TH) through substituting the formulas (57) to (59) for the formula (60) and replacing it with the formula (62), the following formula can be obtained: $\begin{array}{cc}{V}_{\mathrm{CM}}+V_C-V_{\mathrm{TH}}=\frac{1}{\sqrt{K_1}}\sqrt{\frac{K_1}{K_1+2}\frac{I_B}{\beta }-\frac{1}{2}{V}_i^2}\left(\left|V_i\right|\le \sqrt{\frac{K_1}{K_1+2}\frac{I_B}{\beta }}\right)& \left(63\right)\end{array}$

[0139] Hence, the voltage V_Ccan be obtained as follows: $\begin{array}{cc}{V}_C=-\sqrt{\frac{1}{K_1+2}\frac{I_B}{\beta }}+\sqrt{\frac{1}{K_1+2}\frac{I_B}{\beta }-\frac{V_i^2}{2K_1}}\left(\left|V_i\right|\le \sqrt{\frac{K_1}{K_1+2}\frac{I_B}{\beta }}\right)& \left(64\right)\end{array}$

[0140] Here, V_c≦V_CM

[0141] Furthermore, if K₁can be equally divided into two K₁/2, the linear differential currents can be obtained as follows: $\begin{array}{cc}{I}_{\mathrm{D1}}+\frac{I_{\mathrm{D3}}}{2}=\frac{I_B}{2}+\sqrt{\frac{\beta I_B}{K_1+2}}{V}_i\left(\left|V_i\right|\le \sqrt{\frac{K_1}{K_1+2}\frac{I_B}{\beta }}\right)& \left(65\right)\\ I_{\mathrm{D2}}+\frac{I_{\mathrm{D3}}}{2}=\frac{I_B}{2}-\sqrt{\frac{\beta I_B}{K_1+2}}{V}_i\left(\left|V_i\right|\le \sqrt{\frac{K_1}{K_1+2}\frac{I_B}{\beta }}\right)& \left(66\right)\end{array}$

[0142] Therefore, the MOS linear transconductance amplifier is realizable using MOS transistors M4 to M6A and M6B. That is, sources of transistors M4 to M6A and M6B is grounded. And MOS transistors M4 to M6A and M6B share gate voltages with the transistors M1 to M3, respectively. Additionally, the drains of the transistors M4 and M6A are mutually connected, the drains of the transistors M5 and M6B are mutually connected, and these drains constitutes a differential output pair. Thus, the MOS linear transconductance amplifier is realizable Here, an important thing is that, even if a transistor size of the transistor M3 is set to an arbitrary value, the MOS linear transconductance amplifier is realizable. Moreover, if the gate ratios of the transistors M4 and M5 is set to K₂times as large as that of a unit transistor and the gate ratios of the transistors M6A and M6B is set to (K₁/2)K₂times as large as that of the unit transistor, it is possible to make the output current be K₂times of the current value shown in the formulas (65) and (66) as follows: $\begin{array}{cc}{I}^{+}=I_{\mathrm{D4}}+I_{\mathrm{D6A}}=\frac{K_2{I}_H}{2}+K_2\sqrt{\frac{\beta I_B}{K_1+2}}{V}_i\left(\left|V_i\right|\le \sqrt{\frac{K_1}{K_1+2}\frac{I_B}{\beta }}\right)& \left(67\right)\\ I^{-}=I_{\mathrm{D5}}+I_{\mathrm{D6B}}=\frac{K_2{I}_H}{2}-K_2\sqrt{\frac{\beta I_B}{K_1+2}}{V}_i\left(\left|V_i\right|\le \sqrt{\frac{K_1}{K_1+2}\frac{I_B}{\beta }}\right)& \left(68\right)\end{array}$

[0143] Therefore, the MOS linear transconductance amplifier can be realized. Here, a noteworthy thing is that, even if the transistor size of the transistor M3 is set to an arbitrary value, the MOS linear transconductance amplifier is realizable. Moreover, a consumption current in a control voltage generating circuit can be made lower than a consumption current in an output circuit by setting K₂to be K₂>2/(K₁+2)+1. An operation of a circuit with such an output circuit is called a class A operation. In addition, it is possible to realize an OTA (operational transconductance amplifier) by driving the drains of the transistors M4 and M6A, which are mutually connected, and the drains of the transistors M5 and M6B, which are mutually connected with a constant current 2K₂I_B/(K₁+2) respectively.

[0144] Moreover, as for the value of K₁, K₁=2 from which the value of K₁/2 becomes a unit transistor seems to be reasonable on the basis of a conventional circuit analysis. However, actually, K₁is set in a range of 1≦K₁≦2 so as to become V_c=V_CMat V_i=0 by reducing a value of the bias current I_B. On the other hand, gate ratios of the transistors M6A and M6B were set to be K₂times as large as that of the unit transistor. It is understood that this case further increases linearity according to a simulation. It is considered that this is caused by a fact that a power value small than a value coming from the square-law, for example, the power of 1.5 generates an error smaller than the value coming from the square-law in the case of approximating an entire operational current range since a gradient gradually decreases as the drain current increase in the MOS transistor although this assumes the square-law of a MOS transistor in circuit analysis and the formula is solved. In this case, a unit transistor must be made to be a transistor larger than the minimum transistor in a design rule, and must enable to realize a transistor smaller than a unit transistor.

[0145] Next, it will be described that a MOS linear transconductance amplifier is also realizable by making transistors M1 to M3 and transistors M4 to M6A and M6B, which share gate voltages with transistors M1 to M3 respectively, floated through driving the transistors M1 to M3 and transistors M4 to M6A and M6B with a constant current, and making the transistors M4 and M6A, and transistors M5 and M6B be a differential output pair through mutually connecting drains of the transistors M4 and M6A, and mutually connecting drains of the transistors M5 and M6B.

[0146] FIG. 2 is an example of a circuit showing an embodiment of the MOS linear transconductance amplifier according to claim 2.

[0147] It is assumed that each gate ratio (W/L) of transistors M4 and M5 is K₂times as large as that of a unit transistor and each gate ratio (W/L) of the transistors M6A and M6B is K₂(K₁/2) times as large as that of the unit transistor. With letting K₂be K₂≧1, it is possible to arrange a constant current source between a common source of the transistors M4 to M6A and M6B, and a negative power supply −V_ss. A symbol I₀denotes a value of a constant current driving the transistors M4 to M6A and M6B whose sources are mutually connected.

[0148] Drain currents of respective transistors at this time are expressed as follows: $\begin{array}{cc}{I}_{\mathrm{D4}}=K_2{\beta \left(V_{\mathrm{CM}}+\frac{V_{\mathrm{in}}}{2}-V_{\mathrm{S2}}-V_{\mathrm{TH}}\right)}^2& \left(69\right)\\ I_{\mathrm{D5}}=K_2{\beta \left(V_{\mathrm{CM}}-\frac{V_{\mathrm{in}}}{2}-V_{\mathrm{S2}}-V_{\mathrm{TH}}\right)}^2& \left(70\right)\\ I_{\mathrm{D6A}}=I_{\mathrm{D6B}}=K_2\frac{K_1}{2}{\beta \left(V_{\mathrm{CM}}+V_C-V_{\mathrm{S2}}-V_{\mathrm{TH}}\right)}^2& \left(71\right)\end{array}$

[0149] Since each transistor is driven by a constant current I₀, the following formula is satisfied:

[0150] I_D4+I_D5+I_D6A+I_D6B=I_O  (72)

[0151] A following formula can be obtained by solving (V_CM−V_S2−V_TH) through substituting the formulas (69) to (71) for the formula (72): $\begin{array}{cc}{V}_{\mathrm{CM}}-V_{\mathrm{S2}}-V_{\mathrm{TH}}=\frac{-K_1{V}_C+\sqrt{\frac{K_1+2}{K_2}\frac{I_0}{\beta }-\frac{K_1+2}{2}{V}_i^2-2K_1{V}_C^2}}{K_1+2}& \left(73\right)\end{array}$

[0152] Here, in order for this MOS lenear transconductance amplifier performs the linear operation, the common source voltage must be a constant voltage, and it necessary to satisfying a following formula: $\begin{array}{cc}-K_1{V}_C+\sqrt{\frac{K_1+2}{K_2}\frac{I_0}{\beta }-\frac{K_1+2}{2}{V}_i^2-2K_1{V}_C^2}=c\left(\mathrm{constant}\right)& \left(74\right)\end{array}$

[0153] A following formula can be obtained by solving V_cfrom the formula (74). $\begin{array}{cc}{V}_C=\frac{-K_{1}c+\sqrt{\frac{{K_1\left(K_1+2\right)}^2}{K_2}\frac{I_0}{\beta }-\frac{{K_1\left(K_1+2\right)}^2}{2}{V}_i^2-2K_1{c}^2}}{K_1\left(K_1+2\right)}& \left(75\right)\end{array}$

[0154] Here, since V_c=0 at V_i=0, in a case of a following formula is satisfied: $\begin{array}{cc}{c}^2=\frac{K_1+2}{K_2}\frac{I_0}{\beta }& \left(76\right)\end{array}$

[0155] Then, a following formula can be obtained. $\begin{array}{cc}{V}_C=-\sqrt{\frac{1}{K_2\left(K_1+2\right)}\frac{I_0}{\beta }}+\sqrt{\frac{1}{K_2\left(K_1+2\right)}\frac{I_0}{\beta }-\frac{1}{2K_1}{V}_i^2}\left(\left|V_i\right|\le \sqrt{\frac{K_1}{K_2\left(K_1+2\right)}\frac{I_0}{\beta }}\right)& \left(77\right)\end{array}$

[0156] Similarly to the case of obtaining the formula (47), V_cbecomes a control voltage for performing the linear operation.

[0157] Hence, this formula (77) should just become equal to the formula (64). Thus, in order to satisfy this condition, it is necessary to satisfy the formula (78). $\begin{array}{cc}\frac{1}{K_1+2}\frac{I_B}{\beta }=\frac{1}{K_2\left(K_1+2\right)}\frac{I_0}{\beta }& \left(78\right)\end{array}$

[0158] Therefore, a conditional formula for performing the linear operation can be obtained as follows:

[0159] I₀=K₂I_B  (79)

[0160] For example, if K₂is 2 then I₀=2I_B, and if K₂is 4 then I₀=4I_B. However, in this condition, a voltage difference between a common source and the ground is not produced. It is because the voltage between the gate and source of each transistor does not change if the drive current is also K₂times even if each transistor size of the control voltage generating circuit is made to be K₂times. Hence, as shown in FIG. 2, it is understood that the constant current source I₀should be arranged between a common source and negative power supply −V_ss. Alternatively, as shown in FIG. 3, a constant current flows in a resistor R by inserting resistor R between the common source of transistors M1 to M3, M6 and M7, and the ground. Hence, it is possible to set the voltage between the common source of transistors M1 to M3, M7 and M8 and the ground to be a constant value, and to arrange the constant current source between the common source and ground. Alternatively, it is also preferable to insert a transistor, having diode connection obtained by mutually connecting its gate and drain, instead of the resistor R.

[0161] An output current at this time is as follows: $\begin{array}{cc}{I}^{+}=I_{\mathrm{D4}}+I_{\mathrm{D6A}}=K_2{I}_B+K_2\sqrt{\frac{\beta I_B}{K_1+2}}{V}_i\left(\left|V_i\right|\le \sqrt{\frac{K_1{I}_B}{\left(K_1+2\right)\beta }}\right)& \left(80\right)\\ I^{-}=I_{\mathrm{D5}}+I_{\mathrm{D6B}}=K_2{I}_B-K_2\sqrt{\frac{\beta I_B}{K_1+2}}{V}_i\left(\left|V_i\right|\le \sqrt{\frac{K_1{I}_B}{\left(K_1+2\right)\beta }}\right)& \left(81\right)\end{array}$

[0162] Hence, the MOS linear transconductance amplifier is realizable. Here, a noteworthy thing is that, even if the transistor size of the transistor M3 is set to an arbitrary value, the MOS linear transconductance amplifier is realizable. Moreover, a current consumed in the control voltage generating circuit can be made lower than a current consumed in the output circuit by setting K₂as K₂>2/(K₁+2)+1. An operation of a circuit with such an output circuit form is called a class A operation. In addition, the OTA (operational transconductance amplifier) is realizable by driving the common drain of transistors M4 and M6A, and the common drain of transistors M5 and M6B with constant currents K₂I_B, respectively.

[0163] In each MOS linear transconductance amplifier described above, a linear operation input voltage range and a transconductance are determined with the drive currents of the control voltage generating circuit, and transconductance parameters of transistors. It is desired that the transconductance is variable in the MOS linear transconductance amplifier. For that purpose, it is necessary to make the transistors M1 to M3, which constitute a control voltage generating circuit and have the grounded sources, floating and driving them with the constant current source.

[0164] FIG. 4 is an example of a circuit showing an embodiment of a MOS linear transconductance amplifier according to claim 4.

[0165] Here, sources of the transistors M1 to M3 are mutually connected and their drains are also mutually connected. Hence, so as to normally operate, it is necessary to set a common source voltage value. As a method of setting a common source voltage value, it is common to add a transistor M0, to apply a gate voltage to the transistor M0, to perform a level shift with a voltage between its gate and source, and to make it be a common source voltage. What is necessary is to consider as the input common mode voltage V_CM, and just to consider transistor M0 as a constant current drive as volt age applied to the gate of transistor M0, since differential voltage is inputted into a control voltage generating circuit.

[0166] Here, supposing that a gate ratio (W/L) of the transistor M0 is K₀times as large as that of the unit transistor. In addition, a current I_ssindependent of the input common mode voltage V_CMis introduced. The transistors M1 to M3 whose drains are mutually connected and are driven by a constant current (K₁+2)I_ss. Supposing that a constant current K₀I_ssflows in the transistor M0, the transistors M0 to M3 are mutually connected at their sources, and are driven by a constant current (K₀+K₁+2)I_ss. Here, since the current flowing in the transistor M0 is a value obtained by subtracting the currents, flowing in the transistors M1 to M3 from the drive current (K₀+K₁+2)I_ss, it is not necessary to drive it by the constant current K₀I_ssfrom the drain side of transistor M₀.

[0167] At this time, I_D0is as follows:

[0168] I_D0=K₀I_SS=K₀β(V_CM−V_S1−V_TH)²  (82)

[0169] Hence, the common source voltage V_S1is obtained as follows: $\begin{array}{cc}{V}_{\mathrm{S1}}=V_{\mathrm{CM}}-\sqrt{\frac{I_{\mathrm{SS}}}{\beta }}-V_{\mathrm{TH}}& \left(83\right)\end{array}$

[0170] Drain currents of respective transistors at this time are shown as follows: $\begin{array}{cc}{I}_{\mathrm{D1}}={\beta \left(V_{\mathrm{CM}}+\frac{V_i}{2}-V_{\mathrm{S1}}-V_{\mathrm{TH}}\right)}^2& \left(84\right)\\ I_{\mathrm{D2}}={\beta \left(V_{\mathrm{CM}}-\frac{V_i}{2}-V_{\mathrm{S1}}-V_{\mathrm{TH}}\right)}^2& \left(85\right)\end{array}$  I_D3=K₁β(V_CM+V_C−V_S1−V_TH)²  (86)

[0171] Since these transistors are driven by the constant current (K₁+2) I_ss, the following formula is satisfied:

[0172] I_D1+I_D2+I_D3=(K₁+2)I_S  (87)

[0173] With solving (V_CM−V_S1−V_TH) through substituting the formulas (84) to (86) for the formula (87), the following formula can be obtained: $\begin{array}{cc}{V}_{\mathrm{CM}}-V_{\mathrm{S1}}-V_{\mathrm{TH}}=\frac{-K_1{V}_C+\sqrt{{\left(K_1+2\right)}^2\frac{I_{\mathrm{SS}}}{\beta }-\frac{K_1+2}{2}{V}_i^2-2K_1{V}_C^2}}{K_1+2}& \left(89\right)\end{array}$

[0174] Here, with substituting the formula (83) into the formula (89), the following formula can be obtained: $\begin{array}{cc}\sqrt{\frac{I_{\mathrm{SS}}}{\beta }}=\frac{-K_1{V}_C+\sqrt{{\left(K_1+2\right)}^2\frac{I_{\mathrm{SS}}}{\beta }-\frac{K_1+2}{2}{V}_i^2-2K_1{V}_C^2}}{K_1+2}& \left(90\right)\end{array}$

[0175] With solving this formula about V_c, the following formula can be obtained: $\begin{array}{cc}{V}_C=-\sqrt{\frac{I_{\mathrm{SS}}}{\beta }}+\sqrt{\frac{I_{\mathrm{SS}}}{\beta }-\frac{1}{2K_1}{V}_i^2}\left(\left|V_i\right|\le \sqrt{K_1\frac{I_{\mathrm{SS}}}{\beta }}\right)& \left(91\right)\end{array}$

[0176] With substituting the formulas (83) and (91) into the formulas (84) to (86), the following formulas can be obtained: $\begin{array}{cc}{I}_{\mathrm{D1}}=\frac{\beta }{4}{V}_i^2+\sqrt{\frac{I_{\mathrm{SS}}}{\beta }}{V}_i+I_{\mathrm{SS}}\left(\left|V_i\right|\le \sqrt{K_1\frac{I_{\mathrm{SS}}}{\beta }}\right)& \left(92\right)\\ I_{\mathrm{D2}}=\frac{\beta }{4}{V}_i^2-\sqrt{\frac{I_{\mathrm{SS}}}{\beta }}{V}_i+I_{\mathrm{SS}}\left(\left|V_i\right|\le \sqrt{K_1\frac{I_{\mathrm{SS}}}{\beta }}\right)& \left(93\right)\\ I_{\mathrm{D3}}=K_1{I}_{\mathrm{SS}}-\frac{\beta }{2}{V}_i^2\left(\left|V_i\right|\le \sqrt{K_1\frac{I_{\mathrm{SS}}}{\beta }}\right)& \left(94\right)\end{array}$

[0177] Furthermore, if K₁can be equally divided into two K₁/2, the following formulas can be obtained: $\begin{array}{cc}{I}_{\mathrm{D1}}+\frac{I_{\mathrm{D3}}}{2}=\left(1+\frac{K_1}{2}\right)I_{\mathrm{SS}}+\sqrt{\frac{I_{\mathrm{SS}}}{\beta }}{V}_i\left(\left|V_i\right|\le \sqrt{K_1\frac{I_{\mathrm{SS}}}{\beta }}\right)& \left(95\right)\\ I_{\mathrm{D2}}+\frac{I_{\mathrm{D3}}}{2}=\left(1+\frac{K_1}{2}\right)I_{\mathrm{SS}}-\sqrt{\frac{I_{\mathrm{SS}}}{\beta }}{V}_i\left(\left|V_i\right|\le \sqrt{K_1\frac{I_{\mathrm{SS}}}{\beta }}\right)& \left(96\right)\end{array}$

[0178] Hence, a linear differential current can be obtained.

[0179] Here, a noteworthy thing is that, also in the case that a transistor size of the transistor M3 is set to an arbitrary value, the MOS linear transconductance amplifier is realizable. Hence, the MOS linear transconductance amplifier is realizable by making transistors M1 to M3 and transistors M4 to M6A and M6B, which share gate voltages with transistors M1 to M3 respectively, be a differential output pair through mutually connecting drains of the transistors M4 and M6A, and mutually connecting drains of the transistors M5 and M6B.

[0180] Here, in the case of that the gate rations of each transistors M4 and M5 set to K₂times as large as that of a unit transistor, gate ratios of the transistors M6A and M6B set to (K₁/2)K₂times as large as that of the unit transistor, and the drive current set to K₂times, it is possible to make a K₂times output current of a current value shown in the formulas (95) and (96) as follows: $\begin{array}{cc}{I}^{+}=I_{\mathrm{D4}}+I_{\mathrm{D6A}}=K_2\left(1+\frac{K_1}{2}\right)I_{\mathrm{SS}}+K_2\sqrt{\frac{I_{\mathrm{SS}}}{\beta }}{V}_i\left(\left|V_i\right|\le \sqrt{K_1\frac{I_{\mathrm{SS}}}{\beta }}\right)& \left(97\right)\\ I^{-}=I_{\mathrm{D5}}+I_{\mathrm{D6B}}=K_2\left(1+\frac{K_1}{2}\right)I_{\mathrm{SS}}-K_2\sqrt{\frac{I_{\mathrm{SS}}}{\beta }}{V}_i\left(\left|V_i\right|\le \sqrt{K_1\frac{I_{\mathrm{SS}}}{\beta }}\right)& \left(98\right)\end{array}$

[0181] Hence, the MOS linear transconductance amplifier is realizable. Here, a noteworthy thing is that it is not necessary to make all of four transistors M4 to M6A and M6B equal size. Moreover, a linear input voltage range can be also extended by setting the transistor size of the transistor M3 to be larger. However, it is necessary to set current per-unit-area (current density) flowing in the transistors M1 to M3 at the time of a non-signal (V_i=0) to be an equal constant value. Moreover, a current consumed in the control voltage generating circuit can be made lower than a current consumed in the output circuit by setting K₂as K₂>K₀/(K₁+2)+1.

[0182] An operation of a circuit with such an output circuit form is called class A operation. In addition, OTA (operational transconductance amplifier) is realizable by driving the common drain of transistors M4 and M6A, and the common drain of transistors M5 and M6B with constant currents K₂(1+K₁/2)I_ss, respectively.

[0183] Thus, desired currents are already generated in the control voltage generating circuit for supplying the gate voltages required in order to operate four transistors with the class A transistors. Hence, it is understood that, by changing the output of a transistor in the same circuit as this control voltage generating circuit, the MOS linear transconductance amplifier is realized.

[0184] FIG. 5 is a circuit diagram showing an embodiment of a MOS linear transconductance amplifier according to claim 5.

[0185] Thus, in the OTA where each common source of a control voltage generating circuit and an output circuit is made to be floated, a transconductance is proportional to the square root of a driving current. Therefore, it is generally that the transconductance is set at a desired value by adjusting the driving current. Hence, it is required that currents supplied to drains of transistors from a power supply line is equally interlocked to currents supplied to the ground from common sources change the transconductance.

[0186] Here, when supplying currents to the ground from the common sources, or when supplying currents to the drains of transistors from the power supply line, the currents are supplied through a current mirror circuit. Hence, it is possible to supply currents with desired current ratio even if two or more current supplies are used. However, it is troublesome to interlock the current mirror circuit in the ground side with that in the power supply side. Configuration of only one of two mirror circuits is simpler by several steps and can make the current ratio more accurate.

[0187] Hence, as shown in FIG. 5, by letting the current, flowing in the transistor M0, be a reference current of the current mirror circuit, supplying currents, flowing into respective common drains of the transistors M1 to M3, transistors M4 and M6A, and transistors M5 and M6B, in a current ratio from the current mirror circuit, setting the current ratio to be (K₁+2)/K₀, K₂(1+K₁/2)/K₀and K₂(1+K₁/2)/K₀respectively, and changing currents with setting the current ratio of two current sources in the ground side to be (K₀+K₁+2):K₂(K₁+2), simultaneously, desired currents whose current ratio is K₀:(K₁+2):K₂(1+K₁/2):K₂(1+K₁/2) are automatically supplied from the current mirror circuit in the power supply side.

[0188] Here, it is also possible to put a linear operating input voltage range into the desired voltage range by positively changing parameters so that the common source voltage V_s1of a first cell constituting the control voltage generating circuit differs from the common source voltage V_s2of a second cell constituting the output circuit without making the common source voltage V_s1of the first cell, constituting the control voltage generating circuit, and the common source voltage V_s2of the second cell, constituting the output circuit, as shown in FIGS. 4 and 5.

[0189] As described above, it is also considered that a power value smaller than a value coming from the square-law, for example, the power of 1.5 generates an error smaller than the value coming from the square-law in the case of approximating an entire operational current range since a gradient gradually decreases as the drain current increase in the MOS transistor although this assumes the square-rule of a MOS transistor in circuit analysis and the formula is solved.

[0190] Alternatively, a secondary effect disregarded in the above circuit analysis is a factor for degrading the linearity of a linear transconductance amplifier. For example, as the secondary effect, a channel length modulation and the like notably appear. Hence, with citing an example, it is also possible to set the output voltage V_cof the control voltage generating circuit so that the degradation of linearity may not become especially large even when an input voltage is comparatively high while allowing such a status that some extent of degradation of the linearity may arise when the input voltage is comparatively low.

[0191] In general, as expressed in the formulas (64), (77) and (91), since the output voltage V_cof the control voltage generating circuit is expressed in a form where the square value of the input voltage V_iis subtracted in the root (square root), the output voltage V_ccan be approximated as follows: $\begin{array}{cc}{V}_C\approx -b+b\sqrt{1-\frac{V_i^2}{a}}=-b\left(1-\frac{1}{\sqrt{a}}\right)V_i^2& \left(99\right)\end{array}$

[0192] This is a quadric curve (downward parabola) having a form like turning a bowl down. For example, suppose that it is better that the control voltage is as follows:

[0193] V_c=−cV₁^1.5  (100)

[0194] In approximating by the control voltage expressed with a formula (99), it shares zero point at the time of a non-signal (V_i=0). It is considerable to maintain the linearity over a comparatively high input voltage range by setting parameters so that the value of the formula (99) may be larger than the value of the formula (100) in a range where the input voltage is low, both values may become equal again when the input voltage becomes a considerably high value, and the value of the formula (100) may become larger than the value of the formula (99) in an input voltage range that is further high and is beyond this intersection.

[0195] Moreover, in the circuits shown in FIGS. 4 and 5, at the time of a non-signal (V_i=0), parameters of the transistors M0 to M5, M6A and M6B are set so that any current densities of transistors may become equal at the time of a non-signal (V_i=0). Hence, the common source voltage V_s1of the first cell constituting the control voltage generating circuit, and the common source voltage V_s2of the second cell constituting the output circuit become equal. Thus, even if one cell is made by mutually connecting the common source of the first cell and the common source of the second cell and driving currents of the two cells are added, the operation of this amplifier does not change at all. FIGS. 6 and 7 show examples of such circuits that all the sources of respective transistors are mutually connected in the circuits shown in FIGS. 4 and 5.

[0196] For example, an effect of the present invention is that a linear current can be obtained by adding the output currents. Thus, it is possible to realize an output circuit with N-channel transistors, and to improve frequency characteristics. This is because a class A operation can be realized and linearity can be provided by just adding a quadric curve, having a form of turning down a bowl, to either of a parabola rising to the right or a parabola rising to the left.

[0197] The invention may be embodied in other specific forms without departing from the spirit or essential characteristic thereof. The present embodiments are therefore to be considered in all respects as illustrative and not restrictive, the scope of the invention being indicated by the appended claims rather than by the foregoing description and all changes which come within the meaning and range of equivalency of the claims are therefore intended to be embraced therein.

[0198] The entire disclosure of Japanese Patent Application No. 2000-272597 (Filed on Sep. 8th, 2000) including specification, claims, drawings and summary are incorporated herein by reference in its entirety.