陕西理工学院毕业论文
致谢
在本次毕业论文即将完成之时,我首先要向我的指导老师贾建科老师致以最崇高的敬意和最真挚的感谢!老师在百忙之中抽出时间为我们解答疑惑,正是老师拼搏的敬业精神,严谨的治学态度,丰富的知识经验,帮助我解决了一个又一个的难题。本文的工作从方案论证,到电路仿真调试都是在老师的指导下完成的,老师深厚的理论功底,丰富的实际经验,孜孜不倦的教诲都使我受益非浅,也促成了本文的顺利完成。在此,请允许我再次向老师致以崇高的敬意和最真挚的感谢!
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陕西理工学院毕业论文
附录A 外文文献
David M.Pozar. Microwave Engineering Page 628 Low-noise amplifier design
Beside stability and gain,another important design consideration for a microwave amplifier is its noise figure.In receiver applications especially,it is often required to have a preamplifier with as low a noise figure as possible since,as we saw in section 10.1,the first stage of a receiver front end has the dominant effect on the noise performance of the overall system.Generally it is not possible to obtain both minimum noise figure and maximum gain for an amplifier,so some sort of compromise must be made.This can be done by using constant gain circles and circles of constant noise figure to select a usable trade-off between noise figure and gain.Here we will drive the equations for constant noise figure circles,and show how they are used in transistor amplifier design.
As derived in references [4] and [5],the noise figure of a two-port amplifier can be expressed as
F?Fmin?Where the following definition apply:
RN|YS?Yopt|2, 11.54 GSYs?GS?jBS=source admittance presented to transistor.
Yopt=optimum source admittance that results in minimum noise figure. Fmin=minimum noise figure of transistor,attained when YS?Yopt.
RN=equivalent noise resistance of transistor. GS=real part of source admittance.
Instead of the admittance YSand Yopt,we can use the reflection coefficients ?S and ?opt, Where
YS?11??S 11.55a
Z01??S Yopt?11??opt 11.55b
Z01??opt?S is the source reflection coefficient defined in Figure 11.8. The quantities Fmin,?opt and RN are
characteristics of the particular transistor being uesd,and are called the noise parameters of the device:they may be given by the manufacturer,or measured.
2 Using(11.55),the quantity |YS?Yopt|can be expressed in terms of ?S and ?opt: 2|???|4Sopt |YS?Yopt|2?2 11.56 2Z0|1??S||1??opt|2Also.
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陕西理工学院毕业论文
*11??S1??S11?|?S|2 GS?Re{YS}? 11.57 (?)?*22Z01??S1??SZ0|1??S|Using these result in(11.54)gives the noise figure as
|?S??opt|24RN F?Fmin? 11.58 22Z0(1?|?S|)|1??opt|For a fixed noise figure. F, we can show that this result defines a circle in the ?S plane.First define the noise figure parameter, N,as
N?|?S??opt|21?|?S|2*?F?Fmin|1??opt|2 11.59
4RN/Z0*2Which is a constant, for a given noise figure and set of noise parameters. Then rewrite(11.59)as (?S??opt)(?S??opt)?N(1?|?S|) ?S?S?(?S?opt??S?opt)??opt?opt?N?N|?S| ?S?S?22***(?S?opt??S?opt)****2N?1?N?|?opt|2N?1
Now add |?opt|/(N?1) to both sides to complete the square to obtain
|?S??optN?1|?N(N?1?|?opt|2)(N?1) 11.60
This result defines circles of constant noise figure with centers at CF?And radii of
RF??optN?1 11.61a
N(N?1?|?opt|2)N?1 11.61b
EXAMPLE 11.5 Low-Noise Amplifier Design
A GaAs FET is biased for minimum noise figure,and has the following S parameters and noise parameters at 4 GHz(Z0?50Ω):
S11?0.6??60?,S21?1.9?81?,S12?0.05?26?,S22?0.5??60?;Fmin=1.6dB,?opt?0.62?100?,
RN=20Ω.For design puposes,assume the device is unilateral,and calculate the maximum error in GT
resulting from this assumotion. Then design an amplifier having a 2.0 dB noise figure with the maximum gain that is compatiable with this noise figure.
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陕西理工学院毕业论文
Solution
We first compute the unilateral figure of merit from (11.47): U?|S12S21S11S22|?0.059 22(1?|S11|)(1?|S22|)Then from (11.46) the ratio GT/GTU is bounded as
1GT1 ??22(1?U)GTU(1?U)GT?1.130 GTUor 0.891?In dB,
?0.50?GT?GTU?0.53dB
where GT and GTU are now in dB. Thus,we should expect less than about ?0.5dB error in gain.
Next,we use (11.59) and (11.61) to compute the center and radius of the 2dB noise figure circle: N?F?Fmin1.58?1.445|1??opt|2?|1?0.62?10?0|2?0.098 64RN/Z04(20/50)?optN?1?0.56?10?0
CF? RF?N(N?1?|?opt|2)N?1?0.24
This noise figure circle is plotted in Figure 11.15a. Minimum noise figure (Fmin?1.6dB) occurs for
?S??opt?0.62?100?.
Next we calculate data for several input section constant gaincircles. From(11.52),
Gs(dB) 1.0 1.5 1.7
These circles are also plotted in Figure 11.15a. We see that the Gs=1.7dB gain circle just intersects the
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gs 0.805 0.904 0.946 Cs Rs 0.300 0.205 0.150 0.52?60? 0.56?60? 0.58?60?
陕西理工学院毕业论文
F=2dB noise figure circle, and that any higher gain will result in a worse noise figure. From the Smith chart the optimum solution is then ?S?0.53?75?,yielding Gs=1.7dB and F=2.0dB.
*For the output section we choose ?L?S22?0.5?60?for a maximum GLof
GL?The transistor gain is
1?1.33?1.25dB 21?|S22| G0?|S21|2?3.61?5.58dB
Circuit design for the transistor amplifier of Example 11.5.(a) Constant gain and noise figure circles.(b) RF circuit.
So the overall transducer gain will be
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