Basic Electronics (3110016)

Question-5b

BE | Semester-1 Winter-2019 | 06-01-2020

Q5) (b)

This simple enhancement-mode common source MOSFET amplifier configuration uses a single supply at the drain and generates the required gate voltage, VG using a resistor divider.
As we know that for a MOSFET, no current flows into the gate terminal, we can make the following basic assumptions about the MOSFET amplifiers DC operating conditions

$V_{DD} = I_{D} R_{D} + V_{DS} + I_{D} R_{S}$

$\Rightarrow V_{DD} = I_{D} (R_{D} + R_{S}) + V_{DS}$

$\Rightarrow R_{D} + R_{S} = \frac{V_{DD} - V_{DS}}{I_{D}}$

So, $R_{D} = \frac{V_{DD} - V_{DS}}{I_{D}} & R_{S} = \frac{V_{s}}{I_{D}}$

and the MOSFET gate-to-source voltage, VGS is given as

$V_{GS} = V_{G} - I_{S} R_{S}$
For proper operation of the MOSFET, this gate-source voltage must be greater than the threshold voltage of the MOSFET, that is $V_{GS} > V_{TH}$ .
Since
$I_{S} = I_{D}$ , the gate voltage

$V_{GS} = V_{G} - I_{D} R_{S}$

$\Rightarrow V_{G} = V_{GS} + I_{D} R_{S}$

$\Rightarrow V_{G} = V_{GS} + V_{S}$
To set the MOSFET amplifier gate voltage to this value we select the values of the resistors, $R_{1}$ and $R_{2}$ within the voltage divider network to the required values.
Since no current flows into the gate terminal of a MOSFET device so the formula for voltage division is given as

$V_{G} = V_{DD} (\frac{R_{2}}{R_{1} + R_{2}})$