The Transistor — Question 9

Question 9

An insulated gate controls a field-effect transistor through overdrive

An ideal long-channel nMOS transistor has threshold $V_{th}=1.00\,\mathrm V$ and parameter $k=2.00\,\mathrm{mA/V^2}$. In saturation, $I_D=\tfrac12k(V_{GS}-V_{th})^2$ when $V_{DS}\ge V_{GS}-V_{th}>0$. Ignore channel-length modulation.

1. At $V_{GS}=2.50\,\mathrm V$ and $V_{DS}=3.00\,\mathrm V$, verify the region and find current and transconductance.

2. At the same gate voltage but $V_{DS}=.500\,\mathrm V$, use $I_D=k[(V_{GS}-V_{th})V_{DS}-V_{DS}^2/2]$ to find the current and identify the operating region.

Question 9 - Solution

An insulated gate controls a field-effect transistor through overdrive

(a) Method and calculation

$$V_{ov}=V_{GS}-V_{th}=1.50\,\mathrm V<3.00\,\mathrm V\Rightarrow\boxed{\text{saturation}}.$$$$\boxed{I_D=\tfrac12(2.00)(1.50)^2=2.25\,\mathrm{mA},\quad g_m=kV_{ov}=3.00\,\mathrm{mS}}.$$

(b) Interpretation and further calculation

$$V_{DS}=.500\,\mathrm V<V_{ov}\Rightarrow\boxed{\text{triode region}}.$$$$\boxed{I_D=2.00[1.50(.500)-(.500)^2/2]=1.25\,\mathrm{mA}}.$$

Physical check

MOS saturation means channel pinch-off and approximately constant current with drain voltage. It is a different operating condition from BJT saturation. The ideal insulated gate draws no dc current.