## Chapter 4

## 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.

## Step-by-Step

## 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.