Objective: Determine the small-signal voltage gain of a JFET amplifier. Consider the circuit shown in Figure 4.55 with transistor parameters [latex]I_{DSS} = 12 mA, V_{P} = −4 V[/latex], and [latex]λ = 0.008 V^{−1}[/latex] . Determine the small-signal voltage gain [latex]A_{v} = v_{o}/v_{i}[/latex]

The dc quiescent gate-to-source voltage is determined from [latex]V_{GSQ} = \left( \frac{R_{2}}{R_{1} + R_{2}} \right) V_{DD} - I_{DQ} R_{S}[/latex] where [latex]I_{DQ} = I_{DSS} \left( 1 - \frac{V_{GSQ}}{V_{P}} \right)^{2}[/latex] Combining these two equations produces [latex]V_{GSQ} = \left( \frac{180}{180 + 420} \right) (20) - (12) (2.7) \left( 1 - \frac{V_{GSQ}}{(-4)} \right)^{2}[/latex] which reduces to [latex]0.025 V^{2}_{GSQ} + 17.25 V_{GSQ} + 26.4 = 0[/latex] The appropriate solution is [latex]V_{GSQ} = - 2.0 V[/latex] The quiescent drain current is [latex]I_{DQ} = I_{DSS} \left( 1 - \frac{V_{GSQ}}{V_{P}} \right)^{2} = (12) \left( 1 - \frac{(- 2.0)}{(-4)} \right)^{2} = 3.00 mA[/latex] The small-signal parameters are then [latex]g_{m} = \frac{2 I_{DSS}}{(- V_{P})} \left( 1 - \frac{V_{GS}}{V_{P}} \right) = \frac{2 (12)}{(4)} \left( 1 - \frac{(- 2.0)}{(-4)} \right) = 3.00 mA/V[/latex] and [latex]r_{o} = \frac{1}{\lambda I_{DQ}} = \frac{1}{(0.008)(3.00)} = 41.7 k \Omega[/latex] The small-signal equivalent circuit is shown in Figure 4.56. Since [latex]V_{gs} = V_{i} [/latex] the small-signal voltage gain is [latex]A_{v} = \frac{V_{o}}{V_{i}} = - g_{m}(r_{o} || R_{D} || R_{L})[/latex] or [latex]A_{v} = - (3.0)(41.7 || 2. 7|| 4) = - 4.66 [/latex] Comment: The voltage gain of JFET amplifiers is the same order of magnitude as that of MOSFET amplifiers

Question 4.18: Objective: Determine the small-signal voltage gain of a JFET...

Microelectronics Circuit Analysis and Design

Objective: Determine the small-signal voltage gain of a JFET amplifier.

Consider the circuit shown in Figure 4.55 with transistor parameters $I_{DSS} = 12 mA, V_{P} = −4 V$ , and $λ = 0.008 V^{−1}$ . Determine the small-signal voltage gain $A_{v} = v_{o}/v_{i}$

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The dc quiescent gate-to-source voltage is determined from

V_{GSQ} = \left( \frac{R_{2}}{R_{1}  +  R_{2}} \right) V_{DD}  –  I_{DQ} R_{S}

where

I_{DQ} = I_{DSS} \left( 1  –  \frac{V_{GSQ}}{V_{P}} \right)^{2}

Combining these two equations produces

V_{GSQ} = \left( \frac{180}{180  +  420} \right) (20)  –  (12) (2.7) \left( 1  –  \frac{V_{GSQ}}{(-4)} \right)^{2}

which reduces to

0.025 V^{2}_{GSQ} + 17.25 V_{GSQ} + 26.4 = 0

The appropriate solution is

V_{GSQ} = – 2.0  V

The quiescent drain current is

I_{DQ} = I_{DSS} \left( 1  –  \frac{V_{GSQ}}{V_{P}} \right)^{2} = (12) \left( 1  –  \frac{(- 2.0)}{(-4)} \right)^{2} = 3.00  mA

The small-signal parameters are then

g_{m} = \frac{2 I_{DSS}}{(- V_{P})} \left( 1  –  \frac{V_{GS}}{V_{P}} \right) = \frac{2 (12)}{(4)} \left( 1  –  \frac{(- 2.0)}{(-4)} \right) = 3.00  mA/V

and

r_{o} = \frac{1}{\lambda I_{DQ}} = \frac{1}{(0.008)(3.00)} = 41.7  k \Omega

The small-signal equivalent circuit is shown in Figure 4.56.
Since $V_{gs} = V_{i}$ the small-signal voltage gain is

A_{v} = \frac{V_{o}}{V_{i}} = – g_{m}(r_{o} || R_{D} || R_{L})

A_{v} =  – (3.0)(41.7 || 2. 7|| 4) = – 4.66

Comment: The voltage gain of JFET amplifiers is the same order of magnitude as that of MOSFET amplifiers

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Question 4.18: Objective: Determine the small-signal voltage gain of a JFET...

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