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## Q. 4.19

Objective: Design a JFET source-follower circuit with a specified small-signal voltage gain.

For the source-follower circuit shown in Figure 4.57, the transistor parameters are: $I_{DSS} = 12 mA, V_{P} = −4 V$, and $λ = 0.01 V^{−1}$ . Determine $R_{S}$ and $I_{DQ}$ such that the small-signal voltage gain is at least $A_{v} = v_{o}/v_{i} = 0.90$. ## Verified Solution

The small-signal equivalent circuit is shown in Figure 4.58. The output voltage is
$V_{o} = g_{m} V_{gs}(R_{S} || R_{L} || r_{o})$
Also
$V_{i} = V_{gs} + V_{o}$
or
$V_{gs} = V_{i} − V_{o}$
Therefore, the output voltage is
$V_{o} = g_{m}(V_{i} − V_{o})(R_{S} || R_{L} || r_{o})$
The small-signal voltage gain becomes
$A_{v} = \frac{V_{o}}{V_{i}} = \frac{g_{m} (R_{S} || R_{L} || r_{o})}{1 + g_{m}(R_{S} || R_{L} || r_{o})}$
As a first approximation, assume $r_{o}$ is sufficiently large for the effect of $r_{o}$ to be neglected.
The transconductance is
$g_{m} = \frac{2 I_{DSS}}{(−V_{P})} \left(1 − \frac{V_{GS}}{V_{P}} \right) = \frac{2(12)}{4} \left(1 − \frac{V_{GS}}{(−4)} \right)$
If we pick a nominal transconductance value of $g_{m} = 2 mA/V$, then $V_{GS} = −2.67 V$ and the quiescent drain current is
$I_{DQ} = I_{DSS} \left(1 − \frac{V_{GS}}{V_{P}} \right)^{2} = (12) \left(1 − \frac{(−2.67)}{(−4)} \right)^{2} = 1.335 mA$
The value of $R_{S}$ is then determined from
$R_{S} = \frac{−V_{GS} − (−10)}{I_{DQ}} = \frac{2.67 + 10}{1.335} = 9.49 k \Omega$
Also, the value of $r_{o}$ is
$r_{o} = \frac{1}{λ I_{DQ}} = \frac{1}{(0.01)(1.335)} = 74.9 k \Omega$

The small-signal voltage gain, including the effect of $r_{o}$, is
$A_{v} = \frac{g_{m} (R_{S} || R_{L} || r_{o})}{1 + g_{m}(R_{S} || R_{L} || r_{o})} = \frac{(2)(9.49||10||74.9)}{1 + (2)(9.49||10||74.9)} = 0.902$
Comment: This particular design meets the design criteria, but the solution is not unique. 