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## Q. 3.5

Objective: Design a MOSFET circuit biased with both positive and negative voltages to meet a set of specifications.

Specifications: The circuit configuration to be designed is shown in Figure 3.27. Design the circuit such that $I_{DQ} = 0.5 mA$ and $V_{DSQ} = 4 V$.

Choices: Standard resistors are to be used in the final design. A transistor with nominal parameters of $k´_{n} = 80 µA/V^{2} , (W/L) = 6.25,$ and $V_{T N} = 1.2 V$ is available.

## Verified Solution

Assuming the transistor is biased in the saturation region, we have $I_{DQ} = K_{n}(V_{GS} − V_{T N})^{2}$ . The conduction parameter is
$K_{n} = \frac{k´_{n}}{2} \cdot \frac{W}{L} = \left(\frac{0.080}{2} \right) (6.25) = 0.25 mA/V^{2}$

Solving for the gate-to-source voltage, we find the required gate-to-source voltage to induce the specified drain current.
$V_{GS} = \sqrt{\frac{I_{DQ}}{K_{n}}} + V_{T N} = \sqrt{\frac{0.5}{0.25}} + 1.2$
or
$V_{GS} = 2.614 V$
Since the gate current is zero, the gate is at ground potential. The voltage at the source terminal is then $V_{S} = − V_{GS} = −2.614 V$. The value of the source resistor is found from
$R_{S} = \frac{V_{S} − V^{−}}{I_{DQ}} = \frac{−2.614 − (−5)}{0.5}$
or
$R_{S} = 4.77 k \Omega$
The voltage at the drain terminal is determined to be
$V_{D} = V_{S} + V_{DS} = −2.614 + 4 = 1.386 V$
The value of the drain resistor is
$R_{D} = \frac{V^{+} − V_{D}}{I_{DQ}} = \frac{5 − 1.386}{0.5}$
or
$R_{D} = 7.23 k \Omega$
We may note that
$V_{DS} = 4 V \gt V_{DS}(sat) = V_{GS} − V_{T N} = 2.61 − 1.2 = 1.41 V$
which means that the transistor is indeed biased in the saturation region.
Trade-offs: The closest standard resistor values are $R_{S} = 4.7 k \Omega$ and $R_{D} = 7.5 k \Omega$.
We may find the gate-to-source voltage from
$V_{GS} + I_{D} R_{S} − 5 = 0$
where
$I_{D} = K_{n}(V_{GS} − V_{T N})^{2}$
Using the standard resistor values, we find $V_{GS} = 2.622 V, I_{DQ} = 0.506 mA$, and $V_{DSQ} = 3.83 V$. In this case, the drain current is within 1.2 percent of the design specification and the drain-to-source voltage is within 4.25 percent of the design specification.
Comment: It is important to keep in mind that the current into the gate terminal is zero. In this case, then, there is zero voltage drop across the $R_{G}$ resistor.
Design Pointer: In an actual circuit design using discrete elements, we need to choose standard resistor values that are closest to the design values. In addition, the discrete resistors have tolerances that need to be taken into account. In the final design, then, the actual drain current and drain-to-source voltage are somewhat different from the specified values. In many applications, this slight deviation from the specified values will not cause a problem.