Semiconductor Electronics Material Devices and Simple Circuits
150771
If in a p-n junction diode, a square input signal of \(10 \mathrm{~V}\) is applied as shown Then the output signal across \(R_L\) will be
1
2
3
4
Explanation:
B Case \(\mathrm{I}-(-5 \mathrm{~V}\) to \(0 \mathrm{~V})\) \(\mathrm{V}_{\mathrm{P}} \lt \mathrm{V}_{\mathrm{N}}\) (Diode is reverse bias) Hence, it a open circuit. So, output voltage \(V_0=0\) Case II- \((0 \mathrm{~V}\) to \(5 \mathrm{~V})\) \(V_P>V_N\) (Diode is forward bias) \(\mathrm{V}_0=+5 \mathrm{~V}\) So, graph will be -
AIIMS-2008
Semiconductor Electronics Material Devices and Simple Circuits
150776
In a diode, when there is a saturation current, the plate resistance will be.
1 data insufficient
2 zero
3 some finite quantity
4 infinite quantity
Explanation:
D Plate resistance \(\left(\mathrm{r}_{\mathrm{p}}\right)=\frac{\mathrm{dV}}{\mathrm{dI}}\) Where \(\mathrm{dV}=\) change in voltage \(\mathrm{dI}=\) change in current At saturation, the current is to be constant i. e \(\mathrm{dI}=0\) \(\therefore r_p=\frac{d V}{0}\) \(r_p=\infty\)Hence, plate resistance will be infinitive quantity.
AIIMS-1997
Semiconductor Electronics Material Devices and Simple Circuits
150783
The depletion layer in a \(\mathbf{p}-\mathbf{n}\) junction diode is \(10^{-6} \mathrm{~m}\) wide and its knee potential is \(0.5 \mathrm{~V}\), then the inner electric field in the depletion region is
Semiconductor Electronics Material Devices and Simple Circuits
150785
In the given circuit, the current through the battery is-
1 \(1.5 \mathrm{~A}\)
2 \(2 \mathrm{~A}\)
3 \(3 \mathrm{~A}\)
4 \(5.33 \mathrm{~A}\)
Explanation:
C From the figure we can see that Diode \(\mathrm{D}_2\) is reverse biased, so it will not conduct. Diode \(D_1\) and \(D_3\) are forward biased, so they will conduct. So, \(R_{\text {net }} =\frac{(4+4) \times 16}{(4+4)+16}=\frac{8 \times 16}{8+16}\) \(=\frac{16}{3} \Omega\) Now, current through the battery is - \(I=\frac{V}{R_{\text {net }}}=\frac{16}{16 / 3}=3 \mathrm{~A}\)
Semiconductor Electronics Material Devices and Simple Circuits
150771
If in a p-n junction diode, a square input signal of \(10 \mathrm{~V}\) is applied as shown Then the output signal across \(R_L\) will be
1
2
3
4
Explanation:
B Case \(\mathrm{I}-(-5 \mathrm{~V}\) to \(0 \mathrm{~V})\) \(\mathrm{V}_{\mathrm{P}} \lt \mathrm{V}_{\mathrm{N}}\) (Diode is reverse bias) Hence, it a open circuit. So, output voltage \(V_0=0\) Case II- \((0 \mathrm{~V}\) to \(5 \mathrm{~V})\) \(V_P>V_N\) (Diode is forward bias) \(\mathrm{V}_0=+5 \mathrm{~V}\) So, graph will be -
AIIMS-2008
Semiconductor Electronics Material Devices and Simple Circuits
150776
In a diode, when there is a saturation current, the plate resistance will be.
1 data insufficient
2 zero
3 some finite quantity
4 infinite quantity
Explanation:
D Plate resistance \(\left(\mathrm{r}_{\mathrm{p}}\right)=\frac{\mathrm{dV}}{\mathrm{dI}}\) Where \(\mathrm{dV}=\) change in voltage \(\mathrm{dI}=\) change in current At saturation, the current is to be constant i. e \(\mathrm{dI}=0\) \(\therefore r_p=\frac{d V}{0}\) \(r_p=\infty\)Hence, plate resistance will be infinitive quantity.
AIIMS-1997
Semiconductor Electronics Material Devices and Simple Circuits
150783
The depletion layer in a \(\mathbf{p}-\mathbf{n}\) junction diode is \(10^{-6} \mathrm{~m}\) wide and its knee potential is \(0.5 \mathrm{~V}\), then the inner electric field in the depletion region is
Semiconductor Electronics Material Devices and Simple Circuits
150785
In the given circuit, the current through the battery is-
1 \(1.5 \mathrm{~A}\)
2 \(2 \mathrm{~A}\)
3 \(3 \mathrm{~A}\)
4 \(5.33 \mathrm{~A}\)
Explanation:
C From the figure we can see that Diode \(\mathrm{D}_2\) is reverse biased, so it will not conduct. Diode \(D_1\) and \(D_3\) are forward biased, so they will conduct. So, \(R_{\text {net }} =\frac{(4+4) \times 16}{(4+4)+16}=\frac{8 \times 16}{8+16}\) \(=\frac{16}{3} \Omega\) Now, current through the battery is - \(I=\frac{V}{R_{\text {net }}}=\frac{16}{16 / 3}=3 \mathrm{~A}\)
Semiconductor Electronics Material Devices and Simple Circuits
150771
If in a p-n junction diode, a square input signal of \(10 \mathrm{~V}\) is applied as shown Then the output signal across \(R_L\) will be
1
2
3
4
Explanation:
B Case \(\mathrm{I}-(-5 \mathrm{~V}\) to \(0 \mathrm{~V})\) \(\mathrm{V}_{\mathrm{P}} \lt \mathrm{V}_{\mathrm{N}}\) (Diode is reverse bias) Hence, it a open circuit. So, output voltage \(V_0=0\) Case II- \((0 \mathrm{~V}\) to \(5 \mathrm{~V})\) \(V_P>V_N\) (Diode is forward bias) \(\mathrm{V}_0=+5 \mathrm{~V}\) So, graph will be -
AIIMS-2008
Semiconductor Electronics Material Devices and Simple Circuits
150776
In a diode, when there is a saturation current, the plate resistance will be.
1 data insufficient
2 zero
3 some finite quantity
4 infinite quantity
Explanation:
D Plate resistance \(\left(\mathrm{r}_{\mathrm{p}}\right)=\frac{\mathrm{dV}}{\mathrm{dI}}\) Where \(\mathrm{dV}=\) change in voltage \(\mathrm{dI}=\) change in current At saturation, the current is to be constant i. e \(\mathrm{dI}=0\) \(\therefore r_p=\frac{d V}{0}\) \(r_p=\infty\)Hence, plate resistance will be infinitive quantity.
AIIMS-1997
Semiconductor Electronics Material Devices and Simple Circuits
150783
The depletion layer in a \(\mathbf{p}-\mathbf{n}\) junction diode is \(10^{-6} \mathrm{~m}\) wide and its knee potential is \(0.5 \mathrm{~V}\), then the inner electric field in the depletion region is
Semiconductor Electronics Material Devices and Simple Circuits
150785
In the given circuit, the current through the battery is-
1 \(1.5 \mathrm{~A}\)
2 \(2 \mathrm{~A}\)
3 \(3 \mathrm{~A}\)
4 \(5.33 \mathrm{~A}\)
Explanation:
C From the figure we can see that Diode \(\mathrm{D}_2\) is reverse biased, so it will not conduct. Diode \(D_1\) and \(D_3\) are forward biased, so they will conduct. So, \(R_{\text {net }} =\frac{(4+4) \times 16}{(4+4)+16}=\frac{8 \times 16}{8+16}\) \(=\frac{16}{3} \Omega\) Now, current through the battery is - \(I=\frac{V}{R_{\text {net }}}=\frac{16}{16 / 3}=3 \mathrm{~A}\)
Semiconductor Electronics Material Devices and Simple Circuits
150771
If in a p-n junction diode, a square input signal of \(10 \mathrm{~V}\) is applied as shown Then the output signal across \(R_L\) will be
1
2
3
4
Explanation:
B Case \(\mathrm{I}-(-5 \mathrm{~V}\) to \(0 \mathrm{~V})\) \(\mathrm{V}_{\mathrm{P}} \lt \mathrm{V}_{\mathrm{N}}\) (Diode is reverse bias) Hence, it a open circuit. So, output voltage \(V_0=0\) Case II- \((0 \mathrm{~V}\) to \(5 \mathrm{~V})\) \(V_P>V_N\) (Diode is forward bias) \(\mathrm{V}_0=+5 \mathrm{~V}\) So, graph will be -
AIIMS-2008
Semiconductor Electronics Material Devices and Simple Circuits
150776
In a diode, when there is a saturation current, the plate resistance will be.
1 data insufficient
2 zero
3 some finite quantity
4 infinite quantity
Explanation:
D Plate resistance \(\left(\mathrm{r}_{\mathrm{p}}\right)=\frac{\mathrm{dV}}{\mathrm{dI}}\) Where \(\mathrm{dV}=\) change in voltage \(\mathrm{dI}=\) change in current At saturation, the current is to be constant i. e \(\mathrm{dI}=0\) \(\therefore r_p=\frac{d V}{0}\) \(r_p=\infty\)Hence, plate resistance will be infinitive quantity.
AIIMS-1997
Semiconductor Electronics Material Devices and Simple Circuits
150783
The depletion layer in a \(\mathbf{p}-\mathbf{n}\) junction diode is \(10^{-6} \mathrm{~m}\) wide and its knee potential is \(0.5 \mathrm{~V}\), then the inner electric field in the depletion region is
Semiconductor Electronics Material Devices and Simple Circuits
150785
In the given circuit, the current through the battery is-
1 \(1.5 \mathrm{~A}\)
2 \(2 \mathrm{~A}\)
3 \(3 \mathrm{~A}\)
4 \(5.33 \mathrm{~A}\)
Explanation:
C From the figure we can see that Diode \(\mathrm{D}_2\) is reverse biased, so it will not conduct. Diode \(D_1\) and \(D_3\) are forward biased, so they will conduct. So, \(R_{\text {net }} =\frac{(4+4) \times 16}{(4+4)+16}=\frac{8 \times 16}{8+16}\) \(=\frac{16}{3} \Omega\) Now, current through the battery is - \(I=\frac{V}{R_{\text {net }}}=\frac{16}{16 / 3}=3 \mathrm{~A}\)
