B Given that, \(\mathrm{V}_{\mathrm{s}}=\mathrm{V}_{\mathrm{m}} \sin \omega \mathrm{t}\) Which is \(A C\) voltage and \(V_m\) is the maximum value of voltage in the given circuit diode will only conduct current in forward biased so output across \(R_L\) will be the average voltage, \(\mathrm{V}_{\mathrm{av}}=\frac{\mathrm{V}_{\mathrm{m}}}{\pi}\) And, Average current, \(I_{\mathrm{av}}=\frac{\mathrm{V}_{\mathrm{av}}}{\mathrm{R}}\) Here, Total Resistance, \(R=R_S+R_L\) \(I_{\mathrm{av}}=\frac{\mathrm{V}_{\mathrm{m}}}{\left(\mathrm{R}_{\mathrm{S}}+\mathrm{R}_{\mathrm{L}}\right) \pi}\) Now, voltage across \(R_L\) will be - \(V_L=I_{a v} R_L\) \(V_L=\frac{V_m}{\left(R_S+R_L\right) \pi} \times R_L\) \(V_L=\left(\frac{R_L}{\left(R_L+R_S\right)}\right) \frac{V_m}{\pi}\)
TS- EAMCET-07.05.2018
Semiconductor Electronics Material Devices and Simple Circuits
150713
What is the current through the circuit and the potential difference across the diode shown in figure. The drift current for the diode is \(30 \mu \mathrm{A}\).
1 \(30 \mu \mathrm{A}, 5.99 \mathrm{~V}\)
2 \(30 \mu \mathrm{A}, 5 \mathrm{~V}\)
3 \(20 \mu \mathrm{A}, 6 \mathrm{~V}\)
4 \(20 \mu \mathrm{A}, 5.99 \mathrm{~V}\)
Explanation:
B Given that, \(\mathrm{I}_{\mathrm{D}}=30 \mu \mathrm{A}=30 \times 10^{-6} \mathrm{~A}\) Since both the resistor and diode are connected in series so the amount of current in both of them will be remains the same which is \(30 \mu \mathrm{A}\). Then potential drop across the resistor will be equal to, \(V=I R\) \(V=30 \times 10^{-6} \times 20=600 \times 10^{-6} \mathrm{~V}\) \(V=0.6 \mathrm{mV}\) Then potential drop across the diode is, \(V_D=\left[6-0.6 \times 10^{-3}\right]\) \(V_D=[6-0.0006]\) \(V_D=5.99 \mathrm{~V}\)Hence, \(\mathrm{I}_{\mathrm{D}}=30 \mu \mathrm{A}\) and \(\mathrm{V}_{\mathrm{D}}=5.99 \mathrm{~V}\)
Manipal UGET -2020
Semiconductor Electronics Material Devices and Simple Circuits
150716
Assuming that the junction diode is ideal, the current in the arrangement shown in
1 \(20 \mathrm{~mA}\)
2 \(10 \mathrm{~mA}\)
3 \(40 \mathrm{~mA}\)
4 \(30 \mathrm{~mA}\)
Explanation:
B Given, \(\mathrm{R}=100 \Omega\) We know that, \(V=I R\) \(I=\frac{V}{R}=\frac{3-1}{100}\) \(I=\frac{2}{100}=\frac{1}{50}\) \(I=20 \mathrm{~mA}\)
B Given that, \(\mathrm{V}_{\mathrm{s}}=\mathrm{V}_{\mathrm{m}} \sin \omega \mathrm{t}\) Which is \(A C\) voltage and \(V_m\) is the maximum value of voltage in the given circuit diode will only conduct current in forward biased so output across \(R_L\) will be the average voltage, \(\mathrm{V}_{\mathrm{av}}=\frac{\mathrm{V}_{\mathrm{m}}}{\pi}\) And, Average current, \(I_{\mathrm{av}}=\frac{\mathrm{V}_{\mathrm{av}}}{\mathrm{R}}\) Here, Total Resistance, \(R=R_S+R_L\) \(I_{\mathrm{av}}=\frac{\mathrm{V}_{\mathrm{m}}}{\left(\mathrm{R}_{\mathrm{S}}+\mathrm{R}_{\mathrm{L}}\right) \pi}\) Now, voltage across \(R_L\) will be - \(V_L=I_{a v} R_L\) \(V_L=\frac{V_m}{\left(R_S+R_L\right) \pi} \times R_L\) \(V_L=\left(\frac{R_L}{\left(R_L+R_S\right)}\right) \frac{V_m}{\pi}\)
TS- EAMCET-07.05.2018
Semiconductor Electronics Material Devices and Simple Circuits
150713
What is the current through the circuit and the potential difference across the diode shown in figure. The drift current for the diode is \(30 \mu \mathrm{A}\).
1 \(30 \mu \mathrm{A}, 5.99 \mathrm{~V}\)
2 \(30 \mu \mathrm{A}, 5 \mathrm{~V}\)
3 \(20 \mu \mathrm{A}, 6 \mathrm{~V}\)
4 \(20 \mu \mathrm{A}, 5.99 \mathrm{~V}\)
Explanation:
B Given that, \(\mathrm{I}_{\mathrm{D}}=30 \mu \mathrm{A}=30 \times 10^{-6} \mathrm{~A}\) Since both the resistor and diode are connected in series so the amount of current in both of them will be remains the same which is \(30 \mu \mathrm{A}\). Then potential drop across the resistor will be equal to, \(V=I R\) \(V=30 \times 10^{-6} \times 20=600 \times 10^{-6} \mathrm{~V}\) \(V=0.6 \mathrm{mV}\) Then potential drop across the diode is, \(V_D=\left[6-0.6 \times 10^{-3}\right]\) \(V_D=[6-0.0006]\) \(V_D=5.99 \mathrm{~V}\)Hence, \(\mathrm{I}_{\mathrm{D}}=30 \mu \mathrm{A}\) and \(\mathrm{V}_{\mathrm{D}}=5.99 \mathrm{~V}\)
Manipal UGET -2020
Semiconductor Electronics Material Devices and Simple Circuits
150716
Assuming that the junction diode is ideal, the current in the arrangement shown in
1 \(20 \mathrm{~mA}\)
2 \(10 \mathrm{~mA}\)
3 \(40 \mathrm{~mA}\)
4 \(30 \mathrm{~mA}\)
Explanation:
B Given, \(\mathrm{R}=100 \Omega\) We know that, \(V=I R\) \(I=\frac{V}{R}=\frac{3-1}{100}\) \(I=\frac{2}{100}=\frac{1}{50}\) \(I=20 \mathrm{~mA}\)
B Given that, \(\mathrm{V}_{\mathrm{s}}=\mathrm{V}_{\mathrm{m}} \sin \omega \mathrm{t}\) Which is \(A C\) voltage and \(V_m\) is the maximum value of voltage in the given circuit diode will only conduct current in forward biased so output across \(R_L\) will be the average voltage, \(\mathrm{V}_{\mathrm{av}}=\frac{\mathrm{V}_{\mathrm{m}}}{\pi}\) And, Average current, \(I_{\mathrm{av}}=\frac{\mathrm{V}_{\mathrm{av}}}{\mathrm{R}}\) Here, Total Resistance, \(R=R_S+R_L\) \(I_{\mathrm{av}}=\frac{\mathrm{V}_{\mathrm{m}}}{\left(\mathrm{R}_{\mathrm{S}}+\mathrm{R}_{\mathrm{L}}\right) \pi}\) Now, voltage across \(R_L\) will be - \(V_L=I_{a v} R_L\) \(V_L=\frac{V_m}{\left(R_S+R_L\right) \pi} \times R_L\) \(V_L=\left(\frac{R_L}{\left(R_L+R_S\right)}\right) \frac{V_m}{\pi}\)
TS- EAMCET-07.05.2018
Semiconductor Electronics Material Devices and Simple Circuits
150713
What is the current through the circuit and the potential difference across the diode shown in figure. The drift current for the diode is \(30 \mu \mathrm{A}\).
1 \(30 \mu \mathrm{A}, 5.99 \mathrm{~V}\)
2 \(30 \mu \mathrm{A}, 5 \mathrm{~V}\)
3 \(20 \mu \mathrm{A}, 6 \mathrm{~V}\)
4 \(20 \mu \mathrm{A}, 5.99 \mathrm{~V}\)
Explanation:
B Given that, \(\mathrm{I}_{\mathrm{D}}=30 \mu \mathrm{A}=30 \times 10^{-6} \mathrm{~A}\) Since both the resistor and diode are connected in series so the amount of current in both of them will be remains the same which is \(30 \mu \mathrm{A}\). Then potential drop across the resistor will be equal to, \(V=I R\) \(V=30 \times 10^{-6} \times 20=600 \times 10^{-6} \mathrm{~V}\) \(V=0.6 \mathrm{mV}\) Then potential drop across the diode is, \(V_D=\left[6-0.6 \times 10^{-3}\right]\) \(V_D=[6-0.0006]\) \(V_D=5.99 \mathrm{~V}\)Hence, \(\mathrm{I}_{\mathrm{D}}=30 \mu \mathrm{A}\) and \(\mathrm{V}_{\mathrm{D}}=5.99 \mathrm{~V}\)
Manipal UGET -2020
Semiconductor Electronics Material Devices and Simple Circuits
150716
Assuming that the junction diode is ideal, the current in the arrangement shown in
1 \(20 \mathrm{~mA}\)
2 \(10 \mathrm{~mA}\)
3 \(40 \mathrm{~mA}\)
4 \(30 \mathrm{~mA}\)
Explanation:
B Given, \(\mathrm{R}=100 \Omega\) We know that, \(V=I R\) \(I=\frac{V}{R}=\frac{3-1}{100}\) \(I=\frac{2}{100}=\frac{1}{50}\) \(I=20 \mathrm{~mA}\)
B Given that, \(\mathrm{V}_{\mathrm{s}}=\mathrm{V}_{\mathrm{m}} \sin \omega \mathrm{t}\) Which is \(A C\) voltage and \(V_m\) is the maximum value of voltage in the given circuit diode will only conduct current in forward biased so output across \(R_L\) will be the average voltage, \(\mathrm{V}_{\mathrm{av}}=\frac{\mathrm{V}_{\mathrm{m}}}{\pi}\) And, Average current, \(I_{\mathrm{av}}=\frac{\mathrm{V}_{\mathrm{av}}}{\mathrm{R}}\) Here, Total Resistance, \(R=R_S+R_L\) \(I_{\mathrm{av}}=\frac{\mathrm{V}_{\mathrm{m}}}{\left(\mathrm{R}_{\mathrm{S}}+\mathrm{R}_{\mathrm{L}}\right) \pi}\) Now, voltage across \(R_L\) will be - \(V_L=I_{a v} R_L\) \(V_L=\frac{V_m}{\left(R_S+R_L\right) \pi} \times R_L\) \(V_L=\left(\frac{R_L}{\left(R_L+R_S\right)}\right) \frac{V_m}{\pi}\)
TS- EAMCET-07.05.2018
Semiconductor Electronics Material Devices and Simple Circuits
150713
What is the current through the circuit and the potential difference across the diode shown in figure. The drift current for the diode is \(30 \mu \mathrm{A}\).
1 \(30 \mu \mathrm{A}, 5.99 \mathrm{~V}\)
2 \(30 \mu \mathrm{A}, 5 \mathrm{~V}\)
3 \(20 \mu \mathrm{A}, 6 \mathrm{~V}\)
4 \(20 \mu \mathrm{A}, 5.99 \mathrm{~V}\)
Explanation:
B Given that, \(\mathrm{I}_{\mathrm{D}}=30 \mu \mathrm{A}=30 \times 10^{-6} \mathrm{~A}\) Since both the resistor and diode are connected in series so the amount of current in both of them will be remains the same which is \(30 \mu \mathrm{A}\). Then potential drop across the resistor will be equal to, \(V=I R\) \(V=30 \times 10^{-6} \times 20=600 \times 10^{-6} \mathrm{~V}\) \(V=0.6 \mathrm{mV}\) Then potential drop across the diode is, \(V_D=\left[6-0.6 \times 10^{-3}\right]\) \(V_D=[6-0.0006]\) \(V_D=5.99 \mathrm{~V}\)Hence, \(\mathrm{I}_{\mathrm{D}}=30 \mu \mathrm{A}\) and \(\mathrm{V}_{\mathrm{D}}=5.99 \mathrm{~V}\)
Manipal UGET -2020
Semiconductor Electronics Material Devices and Simple Circuits
150716
Assuming that the junction diode is ideal, the current in the arrangement shown in
1 \(20 \mathrm{~mA}\)
2 \(10 \mathrm{~mA}\)
3 \(40 \mathrm{~mA}\)
4 \(30 \mathrm{~mA}\)
Explanation:
B Given, \(\mathrm{R}=100 \Omega\) We know that, \(V=I R\) \(I=\frac{V}{R}=\frac{3-1}{100}\) \(I=\frac{2}{100}=\frac{1}{50}\) \(I=20 \mathrm{~mA}\)
