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Explain high frequency response of JFET amplifer. OR Explain high frequency analysis of CS amplifier.
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Figure 1 : CS JFET Amplifier

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Figure 2 : High frequency equivalent circuit

  • Figure 1 shows the implementation of CS amplifier and Figure 2 shows the high frequency equivalent circuit. At high frequency gain decreases due to stray and wiring capacitance. The coupling capacitors and bypass capacitor are act as a short circuit for high frequencies.
  • Since rds is very large than Rd||RL, hence neglecting rds.
  • Miller theorem may be used to replace the series capacitance Cgd with the capacitance seen in the input and output side as follow :

$C_M1$ = Cgd ( 1 - $A_V$ ) ….(i/p side)

$C_M2$ = Cgd ( 1 - $\frac{1}{A_V}$ ) …(o/p side)

  • The capacitance at the input side are parallel i.e. Cgs, Cwi and $C_M1$ which are combined to form single equivalent capacitance Cin,

Cin = Cgs + $C_M1$ + Cwi

  • Similarly, capacitance at the output side,

Cout = Cds + $C_M2$ + Cwo

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Figure 3 : Simplified high frequency equivalent circuit of CS amplifier

  • As XCout = $\frac{1}{2*\pi*fCout}$, at high frequency XCout is comparable with Rd||RL. In such case drain current will pass through XCout and Rd||RL .

  • As input frequency increases XCout becomes lesser than Rd||RL. Hence more current passes through XCout which decreases output voltage and voltage gain.

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