第六讲 Low-Noise Amplifier 游飞博导/副教授，电子科技大学 feiyou@uestc.edu.cn
第六讲 Low-Noise Amplifier 游飞 博导/副教授，电子科技大学 feiyou@uestc.edu.cn
General Considerations Noise figure Figure 5.1.(a)LNA with input-referred noise voltage,(b)simplified circuit. [View full size image] Ay LNA LNA Rs 4KTRs n.in Vout 4kTRs ,家男家等有用男家家家年果市等年 (a NF n.out A (b) 4kTRs =1+ 只n 4kTRs
General Considerations Noise figure
Example 5.1: A student lays out an LNA and connects its input to a pad through a metal line 200 um long.In order to minimize the input capacitance, the student chooses a width of 0.5 um for the line.Assuming a noise figure of 2 dB for the LNA and a sheet resistance of 40 m/for the metal line,determine the overall noise figure.Neglect the input- referred noise current of the LNA. Figure 5.2.LNA with metal resistance in series with its input. LNA 4kTRs 4kTRL Rs RL out Metal Line V2. n,in +4kTRL NFtot 1+ AKTRs R=(200/0.5)×40m2/▣=162 =1+ 哈m RL 4kTRs Rs NFtot =2.79 dB. NFLNA+ R Rs
Example 5.1: A student lays out an LNA and connects its input to a pad through a metal line 200 µm long. In order to minimize the input capacitance, the student chooses a width of 0.5 µm for the line. Assuming a noise figure of 2 dB for the LNA and a sheet resistance of 40 mΩ/□ for the metal line, determine the overall noise figure. Neglect the inputreferred noise current of the LNA. RL =(200/0.5) × 40 mΩ/□ = 16Ω
Gain Gain个NF Nonlinearity个 [View full size image] 1x+02x283x2 4kTRs Rs Rs out LO (a) b 1 NFtot A品(V匠M+4kTRs)+V .mix A 4kTRs 1 1 2 IP.tot P mixer 1 = NFLNA+ nmix A品 4kTRs )))
Gain Gain↑ NF ↓ Nonlinearity ↑
Input Return Loss Figure 5.4.Constant-r contours in the input impedance plane. Im(Zin)4 1.22+j0.703 502 Zin-Rs 12 -10dB r= Zin+Rs -20dB -15 dB 2.0 Re(Zin} 1.02 1.22 502 1.065
Input Return Loss
Stability the LNA must remain stable for all source impedances at all frequencies. One may think that the LNA must operate properly only in the frequency band of interest and not necessarily at other frequencies,but if the LNA begins to oscillate at any frequency,it becomes highly nonlinear and its gain is very heavily compressed. 1+1△2-1S112-1S222 2S21S12
Stability the LNA must remain stable for all source impedances at all frequencies. One may think that the LNA must operate properly only in the frequency band of interest and not necessarily at other frequencies, but if the LNA begins to oscillate at any frequency, it becomes highly nonlinear and its gain is very heavily compressed
Linearity In most applications,the LNA does not limit the linearity of the receiver. The linearity of the LNA also becomes critical in wideband receivers that may sense a large number of strong interferers. Figure 5.5.TX leakage to RX in a full-duplex system. Received -20 dBm Signal Transmit Receive Band Band RX LNA 3 Duplexer +30 dBm Transmitted Signal TX Transmit Receive Band Band )))
Linearity In most applications, the LNA does not limit the linearity of the receiver. The linearity of the LNA also becomes critical in wideband receivers that may sense a large number of strong interferers
Bandwidth Figure 5.6.Relationship between bandwidth and Q of a tank. z(o川 3 dB △0 7.(a)Band switching,(b)resulting frequency response. 00 VoD 1Z(o)川 L1意卡CR1 2 01 (a) (b) K
Bandwidth
Power Dissipation The LNA typically exhibits a direct trade-off among noise,linearity,and power dissipation. Nonetheless,in most receiver designs,the LNA consumes only a small fraction of the overall power. In other words,the circuit's noise figure generally proves much more critical than its power dissipation
Power Dissipation The LNA typically exhibits a direct trade-off among noise, linearity, and power dissipation. Nonetheless, in most receiver designs, the LNA consumes only a small fraction of the overall power. In other words, the circuit’s noise figure generally proves much more critical than its power dissipation
Problem of Input Matching Figure 5.8.Input admittance of a CS stage. Rp Is it possible to select the circuit o Vout parameters so as to obtain Re{yin} =1/(502)? CL For example, GS if CF 10 fF,CL 30 fF,gmRD in 4,and RD=1002, Equation 5.15 then Re{Yin}=(7.8 k)-1 at 5 Re(Ym)=RpCro2Cr+gmRp(CL+Cp) GHz,far from (50)-1. This is because CF introduces little R(CL CF)2@2+1 feedback at this frequency. Equation 5.16 RpCL(CL CF)@2+1+8mRp Im(Yin}CFo- RB(CL+CF)2@2+1
Problem of Input Matching Is it possible to select the circuit parameters so as to obtain Re{Yin} = 1/(50Ω)? For example, if CF = 10 fF, CL = 30 fF, gmRD = 4, and RD = 100Ω, then Re{Yin} = (7.8 kΩ)−1 at 5 GHz, far from (50Ω)−1. This is because CF introduces little feedback at this frequency