一个电容退化结构发大器的传递函数

wsbdave

电路结构如图所示 ，看论文给出的传递函数为 。小弟很不解，这个电路该如何分析？希望哪路大神给点提示。

semico_ljj

带通？

aa511950071

haonannana

iamshuang2013

Peaking?

laizhiguo

Since this is a perfectly symmetrical circuit, you can cut it to half and apply the small signal model resulting in a circuit like this:
- Vin at the gate, Vs at the source, Vout at the drain
- Rs/2 in parallel with 2Cs from Vs to ground
- gm(Vin-Vs) current source in parallel with RD from Vout to Vs
- RL in parallel with CL from Vout to ground

You can then write the two nodes equations at Vs and Vout. Cancelling out Vs, you should be able to get the transfer function Vout/Vin. It seems the transistor inherent capacitances (Cgs etc) are all ignored otherwise the equation would be even more complicated.

This is a low-pass filter from the equation. Without going through any equation: the transistors convert input voltage into output current, which is then injected into RL//CL to create output voltage -- at low frequency, RL dominates so you get some voltage; at high frequency, CL dominated so the voltage is small (the higher frequency, the smaller voltage) ==> low pass.

wsbdave

回复 2# semico_ljj

一个补偿高频衰减的peak,应该可以看作为一个带通滤波器。

wsbdave

回复 4# iamshuang2013
ｅｎ

wsbdave

回复 5# laizhiguo

Ｙｏｕｒ　ｒｅｐｌｙ　ｉｓ　ｕｓｅｆｕｌ　ｔｏ　ｍｅ．Ｔｈｘ！

semico_ljj

xuexi
xuexi

laizhiguo

CMOS VGA.pdf (284.42 KB, 下载次数: 60 )

2016-3-2 12:41 上传

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本来想好好把推导过程写一下，发现了这篇文章，完全一样的电路，感兴趣的同学自己照着这篇文章推导一下就好了。
对了，楼主原来的公式里面的omega_p2应该是1/(RL*CL)，不是1/（RD*CL）。