Analysis and design of low noise transconductance amplifier for selective receiver front-end

Abstract

Analysis and design of a low-noise transconductance amplifier (LNTA) aimed at selective current-mode (SAW-less) wideband receiver front-end is presented. The proposed LNTA uses double cross-coupling technique to reduce noise figure (NF), complementary derivative superposition, and resistive feedback to achieve high linearity and enhance input matching. The analysis of both NF and IIP3 using Volterra series is described in detail and verified by SpectreRF ^® circuit simulation showing NF < 2 dB and IIP3 = 18 dBm at 3 GHz. The amplifier performance is demonstrated in a two-stage highly selective receiver front-end implemented in 65 nm CMOS technology. In measurements the front-end achieves blocker rejection competitive to SAW filters with noise figure 3.2–5.2 dB, out of band IIP3 >+17 dBm and blocker P_1dB >+5 dBm over frequency range of 0.5–3 GHz.

Appendices

Appendix 1: Derivation of the Volterra operators for the proposed LNTA: G1, G2 and G3

For the circuit shown in Fig. 9 the respective currents and voltages can be expressed as

$$v_{sgp} = \frac{{ - n_{p} v_{in} - R_{sp} (i_{dp} - \frac{{v_{outn} }}{{r_{dp} }})}}{{m_{p} }}$$

(43)

$$v_{gsn} = \frac{{n_{n} v_{in} - R_{sn} (i_{dn} + \frac{{v_{outn} }}{{r_{dn} }})}}{{m_{n} }}$$

(44)

$$v_{sp} = v_{sgp} + \frac{{v_{in} }}{2},v_{sn} = \frac{{v_{in} }}{2} - v_{gsn}$$

(45)

$$v_{outn} = k_{loadn} (i_{dp} - i_{dn} + \frac{{v_{sp} }}{{r_{dp} }} + \frac{{v_{sn} }}{{r_{dn} }})$$

(46)

where

$$m_{p} = 1 + R_{sp} (C_{gsp} s + \frac{1}{{r_{dp} }}),m_{n} = 1 + R_{sn} (C_{gsn} s + \frac{1}{{r_{dn} }})$$

(47)

$$n_{p} = 1 + \frac{{R_{sp} }}{{2r_{dp} }},n_{n} = 1 + \frac{{R_{sn} }}{{2r_{dn} }}$$

(48)

$$k_{p} = \frac{{g_{1p} R_{sp} }}{{m_{p} }},k_{n} = \frac{{g_{1n} R_{sn} }}{{m_{n} }}$$

(49)

$$\hat{Z}_{L} = \frac{{Z_{L} }}{{1 + Z_{L} (\frac{1}{{r_{dp} }} + \frac{1}{{r_{dn} }})}}$$

(50)

Substituting (22, 23) into (43, 44) with maximum 3rd order of v _in , we have

$$v_{sgp}^{3} = \frac{{B_{p}^{3} }}{{m_{p}^{3} (1 + k_{p} )^{3} }},v_{gsn}^{3} = \frac{{B_{n}^{3} }}{{m_{n}^{3} (1 + k_{n} )^{3} }}$$

(51)

$$v_{sgp}^{2} = \frac{{B_{p}^{2} }}{{m_{p}^{2} (1 + k_{p} )^{2} }}\left( {1 - \frac{{2R_{sp} g_{2p} B_{p} }}{{m_{p}^{2} (1 + k_{p} )^{2} }}} \right)$$

(52)

$$v_{gsn}^{2} = \frac{{B_{n}^{2} }}{{m_{n}^{2} (1 + k_{n} )^{2} }}\left( {1 - \frac{{2R_{sn} g_{2n} B_{n} }}{{m_{p}^{2} (1 + k_{p} )^{2} }}} \right)$$

(53)

$$v_{sgp}^{{}} = \frac{1}{{m_{p}^{{}} (1 + k_{p} )^{{}} }}\left\{ {B_{p}^{{}} - \frac{{R_{sp} g_{2p} B_{p}^{2} }}{{m_{p}^{2} (1 + k_{p} )^{2} }}\left( {1 - \frac{{2R_{sp} g_{2p} B_{p} }}{{m_{p}^{2} (1 + k_{p} )^{2} }}} \right) - \frac{{R_{sp} g_{3p} B_{p}^{3} }}{{m_{p}^{3} (1 + k_{p} )^{3} }}} \right\}$$

(54)

$$v_{gsn}^{{}} = \frac{1}{{m_{n}^{{}} (1 + k_{n} )^{{}} }}\left\{ {B_{n}^{{}} - \frac{{R_{sn} g_{2n} B_{n}^{2} }}{{m_{n}^{2} (1 + k_{n} )^{2} }}\left( {1 - \frac{{2R_{sn} g_{2n} B_{n} }}{{m_{n}^{2} (1 + k_{n} )^{2} }}} \right) - \frac{{R_{sn} g_{3n} B_{n}^{3} }}{{m_{n}^{3} (1 + k_{n} )^{3} }}} \right\}$$

(55)

where

$$B_{p} = - n_{p} v_{in} + \frac{{R_{sp} }}{{r_{dp} }}v_{outn} ,B_{n} = n_{n} v_{in} - \frac{{R_{sn} }}{{r_{dn} }}v_{outn}$$

(56)

Substituting (22, 23), (45, 51–55) into (46), we have

$$\begin{aligned} v_{outn} &= \frac{{\hat{Z}_{L} B_{p} (g_{1p} + \frac{1}{{r_{dp} }})}}{{(1 + k_{p} )m_{p} }} + \frac{{\hat{Z}_{L} B_{p}^{2} }}{{m_{p}^{2} (1 + k_{p} )^{3} }}\left\{ {g_{2p} \left( {1 - \frac{{2R_{sp} g_{2p} B_{p} }}{{m_{p}^{2} (1 + k_{p} )^{2} }}} \right) + \frac{{g_{3p} B_{p}^{{}} }}{{m_{p}^{{}} (1 + k_{p} )^{{}} }}} \right\} \\ &\quad- \,\frac{{\hat{Z}_{L} B_{n} (g_{1n} + \frac{1}{{r_{dn} }})}}{{(1 + k_{n} )m_{n} }} - \frac{{\hat{Z}_{L} B_{n}^{2} }}{{m_{n}^{2} (1 + k_{n} )^{3} }}\left\{ {g_{2n} \left( {1 - \frac{{2R_{sn} g_{2n} B_{n} }}{{m_{n}^{2} (1 + k_{n} )^{2} }}} \right) + \frac{{g_{3n} B_{n}^{{}} }}{{m_{n}^{{}} (1 + k_{n} )^{{}} }}} \right\} \hfill \\ \end{aligned}$$

(57)

Appendix 2: Derivation of the Volterra operators for the proposed LNTA: A1, A2, A3, H1, H2 and H3

For the circuit of Fig. 9 the current and voltage equations follow

$$v_{in} = v_{so} - 2R_{so} i_{so}$$

(58)

$$\begin{aligned} i_{so} &= - sC_{gsp} v_{sgp1} + sC_{gsn} v_{gsn1} + v_{inp} (\frac{1}{{sL_{p} }} + \frac{1}{{sL_{n} }}) + i_{dp2} - i_{dn2} \\ & \quad +\, sC_{gsp} v_{sgp2} - sC_{gsn} v_{gsn2} + \frac{{v_{sp2} - v_{outp} }}{{r_{dp} }} - \frac{{v_{outp} - v_{sn2} }}{{r_{dn} }} \hfill \\ \end{aligned}$$

(59)

where

$$v_{{sgp_{1} }} = v_{{sgp}} \left( {v_{{in}} ,v_{{out}} } \right),v_{{sgp_{2} }} = v_{{sgp}} \left( { - v_{{in}} , - v_{{out}} } \right)$$

(60)

$$v_{{gsn_{1} }} = v_{{gsn}} \left( {v_{{in}} ,v_{{out}} } \right),v_{{gsn_{2} }} = v_{{gsn}} \left( { - v_{{in}} , - v_{{out}} } \right)$$

(61)

$$v_{{sp_{2} }} = v_{{sp}} \left( { - v_{{in}} , - v_{{out}} } \right),v_{{sn_{2} }} = v_{{sn}} \left( { - v_{{in}} , - v_{{out}} } \right)$$

(62)

$$i_{{dp_{2} }} = i_{{dp}} \left( { - v_{{in}} , - v_{{out}} } \right),i_{{dn_{2} }} = i_{{dn}} \left( { - v_{{in}} , - v_{{out}} } \right)$$

(63)

For differential mode, the single voltages should be

$$v_{inp} = - v_{inn} = \frac{{v_{in} }}{2},v_{outp} = - v_{outn} = \frac{{v_{out} }}{2}$$

(64)