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I have been trying to learn how JFETs work. The first amplifier that I read about was the CS amplifier:

enter image description here

I found that for a typical common source amplifier configuration at low frequency, the input impedance would just be RG since the input impedance of the JFET is very high.

What happens to the input impedance at high frequency?

There is a free book called Semiconductor Devices. The author there mentions on page 298 that the input impedance will be affected by the gate to source capacitance Ciss. He was mentioning that this capacitance will appear in parallel with RG and lowers the input impedance. Here is a link to the free book: https://www2.mvcc.edu//users/faculty/jfiore/books/SemiconductorDevices.pdf

What about Cgd? does the gate to drain capacitance play no role in changing this impedance? Does Cgd appear in parallel with Cgs and RG?

After that I started reading an article on JFETs. The link to the article is below:

https://audioxpress.com/article/JFETs-The-New-Frontier-Part-1

This article seems to explain that the input capacitance seen by the source will be: Cin = Ciss - AV Crss.

This article seems to take into account the drain to gate capacitance. Then the article author continues to talk about the Miller effect and how we can reduce that.

why did the book author only consider Ciss and not consider this miller effect in his calculation?

was he maybe just trying to over simplify the topic?

Also, what about the common drain amplifier, does the above equation for Cin still applies? if it does, then that would make Cin very small for the common drain amplifier since Cin would just be:

Cin = Ciss-Crss

I'm confused on this topic and hoping somebody here can help me understand this or give me some sources to read on the topic.

Thanks!

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    \$\begingroup\$ This was just an introduction chapter. The more details about JFET HF response you will find on page 456. \$\endgroup\$
    – G36
    Commented Mar 1 at 20:30
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    \$\begingroup\$ As for the common drain amplifier, Cin will be equal to Cin = Cgs*(1 - Av) + Cgd. Where Av is a low fequancy common drain stage voltage gain Av = (Rs||R_L)/(1/gm + Rs||R_L) \$\endgroup\$
    – G36
    Commented Mar 1 at 20:41
  • \$\begingroup\$ Just to make sure I'm understanding this correctly, are you saying that if I design my common drain amplifier or source follower to have a gain of 1, then Cin will just be equal to Cgd alone? \$\endgroup\$
    – Analog
    Commented Mar 1 at 21:37
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    \$\begingroup\$ Yes, this is the case due to Miller's effect, but this time the gain is positive and and less than one, thus we have a positive feedback, Cgs is bootstrapped in a source follower stage. electronics.stackexchange.com/questions/234349/… \$\endgroup\$
    – G36
    Commented Mar 1 at 21:58

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