A common-gate amplifier and a something else

a)

An answer:

pict

b)

An answer:

\begin{array}{rcl} a_{V} & = \frac{v_{out}}{v_{mic}} \\ v_{out} = - g_{m} \cdot v_{gs} \cdot (R_{D} ∕ ∕ r_{in, next}) \\ v_{gs} = - v_{mic} \\ v_{out} = + g_{m} \cdot v_{mic} \cdot (R_{D} ∕ ∕ r_{in, next}) \\ a_{V} & = \frac{v_{out}}{v_{mic}} = g_{m} \cdot (R_{D} ∕ ∕ r_{in, next}) \end{array}

c)

An answer:

r_{in} = R_{S} ∕ ∕ \frac{1}{g_{m}} \approx \frac{1}{g_{m}}

d)

An answer:

Recycling the results of the previous question and combining that with one of the equations for $g_{m}$ of an MOS transistor:

\begin{array}{rcl} g_{m} & = \sqrt{2 K I_{D}} = K (V_{GS} - V_{T}) = \frac{2 I_{D}}{V_{GS} - V_{T}} \to \\ I_{D} & = \frac{g_{m}^{2}}{2 K} \\ g_{m} = \frac{1}{250 Ω} \\ K = 40 mA ∕ V^{2} \to \\ I_{D} & = 0.2 mA \end{array}

e)

An answer:

pict

f)

An answer:

\begin{array}{rcl} a_{v} & = \frac{v_{out}}{v_{in}} \\ v_{out} = - g_{m} v_{gs} R_{D} \\ v_{gs} = v_{in} - V_{S} \\ v_{S} = g_{m} v_{gs} R_{S 1} \\ \dots \\ a_{v} & = - \frac{g_{m} R_{D}}{1 + g_{m} R_{S 1}} \end{array}

g)

An answer:
A derivation can be:

\begin{array}{rcl} R_{S 2} & = \frac{V_{S}}{I_{S}} - R_{S 1} \\ I_{S} = I_{D} \\ V_{S} = V_{G} - V_{GS} \\ V_{GS} = V_{T} + \sqrt{\frac{2 I_{D}}{K}} \\ V_{G} = \frac{R_{G 2}}{R_{G 1} + R_{G 2}} \cdot V_{DD} \end{array}

For which back substitution yields:

\begin{array}{rcl} V_{S} = \frac{R_{G 2}}{R_{G 1} + R_{G 2}} \cdot V_{DD} - V_{T} - \sqrt{\frac{2 I_{D}}{K}} \\ R_{S 2} & = \frac{\frac{R_{G 2}}{R_{G 1} + R_{G 2}} \cdot V_{DD} - V_{T} - \sqrt{\frac{2 I_{D}}{K}}}{I_{D}} - R_{S 1} \end{array}

h)

An answer:
Some numerical values are:

\begin{array}{rcl} g_{m} = \sqrt{2 k I_{D}} = 0.02 A ∕ V \\ V_{GS} = 1.5 V \\ V_{G} = 6 V \\ V_{S} = 4.5 V \\ R_{S 2} = 650 Ω \end{array}

i)

An answer:

a_{V} = - \frac{g_{m} R_{D}}{1 + g_{m} R_{S 1}} = - 4

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