The depletion layer in silicon diode is $1 \mu m$ wide and the knee potential is $0.6 V$, then the electric field in the depletion layer will be
A$0$
B$0.6 Vm$
C$6 \times 10^9 \mathrm{~V} / \mathrm{m}$
D$6 \times 10^5 \mathrm{~V} / \mathrm{m}$
Download our app for free and get started
D$6 \times 10^5 \mathrm{~V} / \mathrm{m}$
By using $E=\frac{V}{d}=\frac{0.6}{10^{-6}}=6 \times 10^5 \mathrm{~V} / \mathrm{m}$
Download our app
and get started for free
Experience the future of education. Simply download our apps or reach out to us for more information. Let's shape the future of learning together!No signup needed.*
In a triode amplifier, $\mu=25, r_p=40$ kilo ohm and load resistance $R_L=10$ kilo ohm. If the input signal voltage is $0.5$ volt, then output signal voltage will be