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Q. 3.22

Objective: Design a circuit with an enhancement-mode MESFET.

Consider the circuit shown in Figure 3.65(a). The transistor parameters are: $V_{T N} = 0.24 V, K_{n} = 1.1 mA/V²$ , and λ = 0. Let $R_{1} + R_{2} = 50 kΩ$. Design the circuit such that $V_{GS} = 0.50 V$ and $V_{DS} = 2.5 V$.

Verified Solution

From Equation (3.38(a)) the drain current is
$I_{D} = K_{n}(V_{GS} − V_{T N})^{2} = (1.1)(0.5 − 0.24)^{2} = 74.4 µA$
From Figure 3.65(b), the voltage at the drain is
$V_{D} = V_{DD} − I_{D} R_{D} = 4 − (0.0744)(6.7) = 3.5 V$
Therefore, the voltage at the source is
$V_{S} = V_{D} − V_{DS} = 3.5 − 2.5 = 1 V$
The source resistance is then
$R_{S} = \frac{V_{S}}{I_{D}} = \frac{1}{0.0744} = 13.4 k \Omega$
The voltage at the gate is
$V_{G} = V_{GS} + V_{S} = 0.5 + 1 = 1.5 V$

Since the gate current is zero, the gate voltage is also given by a voltage divider equation, as follows:
$V_{G} = \left( \frac{R_{2}}{R_{1} + R_{2}} \right) (V_{DD})$
or

$1.5 = \left( \frac{R_{2}}{50} \right) (4)$
which yields
$R_{2} = 18.75 k\Omega$
and
$R_{1} = 31.25 k\Omega$
Again, we see that
$V_{DS} = 2.5 V \gt V_{GS} − V_{T N} = 0.5 − 0.24 = 0.26 V$
which confirms that the transistor is biased in the saturation region, as initially assumed.
Comment: The dc analysis and design of an enhancement-mode MESFET circuit is similar to that of MOSFET circuits, except that the gate-to-source voltage of the MESFET must be held to no more than a few tenths of a volt.