## Chapter 3

## Q. 3.6

Objective: Design a circuit with a p-channel MOSFET that is biased with both negative and positive supply voltages and that contains a source resistor R_{S} to meet a set of specifications.

Specifications: The circuit to be designed is shown in Figure 3.29. Design the circuit such that I_{DQ} = 100 µA , V_{SDQ} = 3 V, and V_{RS} = 0.8 V. Note that V_{RS} is the voltage across the source resistor R_{S} . The value of the larger bias resistor, either R_{1} or R_{2}, is to be 200 kΩ.

Choices: A transistor with nominal parameter values of K_{p} = 100 µA/V^{2} and V_{T P} = −0.4 V is available. The conduction parameter may vary by ±5 percent.

## Step-by-Step

## Verified Solution

Assuming that the transistor is biased in the saturation region, we have I_{DQ} = K_{p}(V_{SG} + V_{T P})^{2} . Solving for the source-to-gate voltage, we find the required value of source-to-gate voltage to be

V_{SG} = \sqrt{\frac{I_{DQ}}{K_{p}}} – V_{T P} = \sqrt{\frac{100}{100}} − (−0.4)

or

V_{SG} = 1.4 V

We may note that the design value of

V_{SDQ} = 3 V \gt V_{SDQ} (sat) = V_{SGQ} + V_{T P} = 1.4 − 0.4 = 1 V

so that the transistor will be biased in the saturation region.

The voltage at the gate with respect to ground potential is found to be

V_{G} = V^{+} − V_{RS} − V_{SG} = 2.5 − 0.8 − 1.4 = 0.3 V

With V_{G} \gt 0, the resistor R_{2} will be the larger of the two bias resistors, so set R_{2} = 200 k\Omega. The current through R_{2} is then

I_{Bias} = \frac{V_{G} − V^{−}}{R_{2}} = \frac{0.3 − (−2.5)}{200} = 0.014 mA

Since the current through R_{1} is the same, we can find the value of R_{1} to be

R_{1} = \frac{V^{+} − V_{G}}{I_{Bias}} = \frac{2.5 − 0.3}{0.014}

which yields

R_{1} = 157 k \Omega

The source resistor value is found from

R_{S} = \frac{V_{RS}}{I_{DQ}} = \frac{0.8}{0.1}

or

R_{S} = 8 k\Omega

The voltage at the drain terminal is

V_{D} = V^{+} − V_{RS} − V_{SD} = 2.5 − 0.8 − 3 = −1.3 V

Then the drain resistor value is found as

R_{D} = \frac{V_{D} − V^{−}}{I_{DQ}} = \frac{−1.3 − (−2.5)}{0.1}

or

R_{D} = 12 k \Omega

Trade-offs: If the conduction parameter K_{p} varies by ±5%, the quiescent drain current I_{DQ} and the source-to-drain voltage V_{SDQ} will change. Using the resistor values found in the previous design, we find the following:

K_{p} | V_{SGQ} | I_{DQ} | V_{SDQ} |

95 µA/V² | 1.416 V | 98.0 µA | 3.04 V |

105µA/V² | 1.385 V | 101.9 µA | 2.962 V |

±5% | ±1.14% | ±2% | ±1.33% |

Comment: We may note that the variation in the Q-point values is smaller that the variation in K_{p}. Including the source resistor R_{S} tends to stabilize the Q-point.