1 1 vote The channel resistance of an $\text{N}$-channel $\text{JFET}$ shown in the figure below is $600 \; \Omega$ when the full channel thickness $\left(\mathrm{t}_{\mathrm{ch}}\right)$ of $10 \; \mu \mathrm{m}$ is available for conduction. The built-in voltage of the gate $\mathrm{P}^{+} \; \mathrm{N}$ junction $\left(\mathrm{V}_{\mathrm{b} i}\right)$ is $-1 \mathrm{~V}$. When the gate to source voltage $\left(\mathrm{V}_{\mathrm{GS}}\right)$ is $0 \mathrm{~V}$, the channel is depleted by $1 \; \mu \mathrm{m}$ on each side due to the built-in voltage and hence the thickness available for conduction is only $8 \; \mu \mathrm{m}$. The channel resistance when $\mathrm{V}_{\mathrm{GS}}=0 \mathrm{~V}$ is $480 \; \Omega$ $600 \; \Omega$ $750 \; \Omega$ $1000 \; \Omega$ BJT and MOSFET Amplifiers gate2011-ec analog-circuits semiconductor electronic-devices numerical-answers + – admin 238 views answer comment Share Follow Add Sync Questions Print 0 reply Please log in or register to add a comment.
