## Chapter 7

## Q. 7.11

It is required to design the circuit of Fig. 7.48(c) to establish a dc drain current I_{D} = 0.5 mA. The MOSFET is specified to have V_{t} = 1 V and k_{n}^{′} W/L = 1 mA/V². For simplicity, neglect the channel-length modulation effect (i.e., assume λ = 0). Use a power-supply V_{D}_{D} = 15 V. Calculate the percentage change in the value of I_{D} obtained when the MOSFET is replaced with another unit having the same k_{n}^{′}W/L but V_{t} = 1.5 V.

## Step-by-Step

## Verified Solution

As a rule of thumb for designing this classical biasing circuit, we choose R_{D} and R_{S} to provide one-third of the power-supply voltage V_{D}_{D} as a drop across each of R_{D}, the transistor (i.e., V_{D}_{S}), and R_{S} . For V_{D}_{D} = 15 V, this choice makes V_{D} = +10 V and V_{S} = +5 V. Now, since I_{D} is required to be 0.5 mA, we can find the values of R_{D} and R_{S} as follows:

R_{D} = \frac{V_{DD} – V_{D}}{I_{D}} = \frac{15 – 10}{0.5} = 10 kΩ

R_{S} = \frac{V_{S}}{R_{S}} = \frac{5}{0.5 } = 10 kΩ

The required value of V_{GS} can be determined by first calculating the overdrive voltage V_{OV} from

I_{D} = \frac{1}{2}k_{n}^{′}(W/L)V_{OV}^{2}

0.5 = \frac{1}{2} × 1 × V_{OV}^{2}

which yields V_{OV} = 1 V, and thus,

V_{GS} = V_{t} + V_{OV} = 1 + 1 = 2 V

Now, since V_{S} = +5 V, V_{G} must be

V_{G} = V_{S} + V_{G}_{S} = 5 + 2 = 7 V

To establish this voltage at the gate we may select R_{G1} = 8 MΩ and R_{G2} = 7 MΩ. The final circuit is shown in Fig. 7.49. Observe that the dc voltage at the drain (+10 V) allows for a positive signal swing of +5 V (i.e., up to V_{D}_{D}) and a negative signal swing of 4 V [i.e., down to (V_{G} – V_{t} )].

If the NMOS transistor is replaced with another having V_{t} = 1.5 V, the new value of I_{D} can be found as follows:

I_{D} = \frac{1}{2} × 1 × (V_{GS} – 1.5)^{2} (7.138)

V_{G} = V_{GS} + I_{D}R_{S}

7 = V_{GS} + 10 I_{D} (7.139)

Solving Eqs. (7.138) and (7.139) together yields

I_{D} = 0.455 mA

Thus the change in I_{D} is

ΔI_{D} = 0.455 − 0.5 = −0.045 mA

which is \frac{-0.045}{0.5} × 100 = −9 % change.