A Highly Linear Wideband CMOS LNTA Employing Noise/Distortion Cancellation and Gain Compensation

Abstract

A wideband low-noise transconductance amplifier (LNTA) with high linearity is proposed. The differential LNTA adopts complementary common-gate (CG), current mirror (CM) and common-source (CS) schemes to obtain noise and distortion cancellation, and $g_{m}$-enhancement. A high third-order input intercept point (IIP3) is obtained due to the distortion cancellation and the complementary characteristics of NMOS and PMOS transistors. The gain expansion of the class-AB CS stage compensates the gain compression of the CG–CM stage, which leads to a high large-signal linearity. A wide input matching bandwidth is achieved by utilizing a $\uppi$-type input matching network without the use of on-chip bulky inductors, and the passive voltage gain of the $\uppi$-network further enhances the effective transconductance. Designed in a 0.18-$\upmu \hbox {m}$ CMOS process, the simulation results show that it provides a minimum noise figure (NF) of 2.95 dB and a maximum transconductance of 79 mS from 0.1 to 3.6 GHz. An input 1-dB compression/desensitization point and an IIP3 of 8.1/5.59 and 18.14 dBm are obtained, respectively. The NF is degraded by 0.3 dB with a 0-dBm blocker. The circuit draws 10.7 mA from a 2.5-V supply, and the core area is only $0.5 \times 0.12\,\hbox {mm}^{2}$.

Appendix Volterra Series Analysis

To capture high-frequency effect, Volterra series analysis is performed as below. Referring to Fig. 8 and repeating (9)–(10) for $V_{x}$ and $V_{1}$,

$$\begin{aligned} V_{x}= & {} A_{1} (s_{1}) \circ V_{a} +A_2 (s_1 ,s_2)\circ V_{a}^{2} +A_{3} (s_1 ,s_2, s_3) \circ V_{a}^{3}\end{aligned}$$

(12)

$$\begin{aligned} V_1= & {} B_1 (s_1 )\circ V_a {+}B_2 (s_1 ,s_2 )\circ V_a^2{ +}B_3 (s_1 ,s_2 ,s_3 )\circ V_a^3 \end{aligned}$$

(13)

Applying Kirchhoff’s current law at each node of the circuit in Fig. 8, the following equations can be obtained:

$$\begin{aligned} \frac{V_x }{Z_x }+\frac{V_x -V_a }{Z_a }+\frac{V_x -V_1 }{r_{o1} }= & {} g_{m1} (-V_x )+\frac{{g}'_{m1} }{2}(-V_x )^{2}+\frac{{g}''_{m1} }{6}(-V_x )^{3} \end{aligned}$$

(14)

$$\begin{aligned} \frac{V_1 }{Z_1 }{+}\frac{V_1 -V_x }{r_{o1} }= & {} -\left[ {g_{m1} (-V_x ){+}\frac{{g}'_{m1} }{2}(-V_x )^{2}{+}\frac{{g}''_{m1} }{6}(-V_x )^{3}} \right] \end{aligned}$$

(15)

where $Z_x (s)=\frac{1}{sC_x }$, $Z_1 (s)=R_1 \left\| {\frac{1}{sC_1 }} \right.$, and $Z_{a}$ is given in (8). $g_{m1}$, ${g}'_{m1}$ and ${g}''_{m1}$ are given in (1).

To find the linear transfer functions $A_1(s)$ and $B_1 (s)$, we excite the circuit with a single tone $v=e^{st}$. Substituting (12) and (13) in (14) and (15), equating the coefficients of $e^{st}$ on both sides of (14) and (15), and solving for $A_1 (s)$ and $B_1 (s)$, we get

$$\begin{aligned} A_1 (s)= & {} \frac{Z_1 (s)+r_{o1} }{H(s)} \end{aligned}$$

(16)

$$\begin{aligned} B_1 (s)= & {} \frac{Z_1 (s)(1+g_{m1} r_{o1} )}{Z_1 (s)+r_{o1} }A_1 (s) \end{aligned}$$

(17)

where

$$\begin{aligned} H(s)=Z_a (s)(1+g_{m1} r_{o1} )+\left( {Z_1 (s)+r_{o1} } \right) \left( {1+\frac{Z_a (s)}{Z_x (s)}} \right) \end{aligned}$$

(18)

To find the second-order term $A_2 (s_1 ,s_2 )$ and $B_2 (s_1 ,s_2 )$, we excite the circuit with two tones $v=e^{s_1 t}+e^{s_2 t}$. Substituting (12) and (13) in (14) and (15), equating the coefficients of $e^{(s_1 +s_2 )t}$ on both sides of (14) and (15), and solving for $A_2 (s_1 ,s_2 )$ and $B_2 (s_1 ,s_2 )$, we get

$$\begin{aligned} A_2 (s_1 ,s_2 )= & {} \frac{\frac{1}{2}{g}'_{m1} r_{o1} Z_a (s_1 +s_2 )A_1 (s_1 )A_1 (s_2 )}{H(s_1 +s_2 )} \end{aligned}$$

(19)

$$\begin{aligned} B_2 (s_1 ,s_2 )= & {} \frac{-Z_1 (s_1 +s_2 )}{Z_x (s_1 +s_2 )\left\| {Z_a (s_1 +s_2 )} \right. }A_2 (s_1 ,s_2 ) \end{aligned}$$

(20)

Likewise, $A_3 (s_1 ,s_2 ,s_3 )$ and $B_3 (s_1 ,s_2 ,s_3 )$ are given by

$$\begin{aligned}&A_3 (s_1 ,s_2 ,s_3 )\nonumber \\&\quad =\frac{-Z_a (s_1 +s_2 +s_3 )r_{o1} \left( {-{g}'_{m1} \overline{A_1 (s_1 )A_2 (s_2 ,s_3 )} +\frac{1}{6}{g}''_{m1} A_1 (s_1 )A_1 (s_2 )A_1 (s_3 )} \right) }{H(s_1 +s_2 +s_3 )} \nonumber \\\end{aligned}$$

(21)

$$\begin{aligned}&B_3 (s_1 ,s_2 ,s_3 )=\frac{-Z_1 (s_1 +s_2 +s_3 )}{Z_x (s_1 +s_2 +s_3 )\left\| {Z_a (s_1 +s_2 +s_3 )} \right. }A_3 (s_1 ,s_2 ,s_3 ) \end{aligned}$$

(22)

where

$$\begin{aligned} \overline{A_1 (s_1 )A_2 (s_2 ,s_3 )} =\frac{1}{3}\left[ {A_1 (s_1 )A_2 (s_2 ,s_3 )+A_1 (s_2 )A_2 (s_1 ,s_3 )+A_1 (s_3 )A_2 (s_1 ,s_2 )} \right] \end{aligned}$$

(23)

For the common-source stage and current mirror output stage, the parasitic capacitances and output impedances of $M_{2}$ and $M_{4}$ can be ignored since the load resistance $R_\mathrm{L}$ is small enough, and the memoryless Taylor series is applied to reduce the analysis complexity. The output current can be expressed as

$$\begin{aligned} i_o =-i_{m2} -i_{m4}= & {} -\left( {g_{m2} V_x +\frac{{g}'_{m2} }{2}V_x^2 +\frac{{g}''_{m2} }{6}V_x^3 } \right) \nonumber \\&-\left( {g_{m4} V_1 +\frac{{g}'_{m4} }{2}V_1^2 +\frac{{g}''_{m4} }{6}V_1^3 } \right) \end{aligned}$$

(24)

where $g_{m2}$, $g_{m4}$, ${g}'_{m2}$, ${g}'_{m4}$, ${g}''_{m2}$ and ${g}''_{m4}$ are given in (1).

Plugging in the above Volterra kernel expressions, the fundamental and third-order output current $i_{o}$ expressions are found to be

$$\begin{aligned} -i_{o,\mathrm{fund}}= & {} \left( {A_1 (s)\circ V_a } \right) \times g_{m2} +\left( {B_1 (s)\circ V_a } \right) \times g_{m4} \end{aligned}$$

(25)

$$\begin{aligned} -i_{o,\mathrm{3rd}}= & {} \left( {A_3 (s_1 ,s_2 ,s_3 )\circ V_a^3 } \right) \times g_{m2} +\left( {B_3 (s_1 ,s_2 ,s_3 )\circ V_a^3 } \right) \times g_{m4}\nonumber \\&+\left( {\overline{A_1 (s_1 )A_2 (s_2 ,s_3 )} \circ V_a^3 } \right) \times {g}'_{m2} +\left( {\overline{B_1 (s_1 )B_2 (s_2 ,s_3 )} \circ V_a^3 } \right) \times {g}'_{m4} \nonumber \\&+\left( {A_1 (s)\circ V_a } \right) ^{3}\times \frac{{g}''_{m2} }{6}+\left( {B_1 (s)\circ V_a } \right) ^{3}\times \frac{{g}''_{m4} }{6} \end{aligned}$$

(26)