Semiconductor Electronics Material Devices and Simple Circuits
150924
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 In the given circuit, the diode is reverse biased. Only drift current flows through the diode. Consider the circuit, in which by applying KVL, we can write, \(\mathrm{V}_{\mathrm{A}}-6+\mathrm{I} \times 20=\mathrm{V}_{\mathrm{B}} \) \(\mathrm{V}_{\mathrm{A}}-\mathrm{V}_{\mathrm{B}}=6-\mathrm{I} \times 20 =6-30 \times 10^{-6} \times 20\) \(=6-0.0006\) \(=5.9994 \mathrm{~V}\)
CG PET- 2015
Semiconductor Electronics Material Devices and Simple Circuits
150928
In the figures shown below Fig.(a)
1 In both Fig. (a) and Fig. (b) the diodes are forward biased
2 In both Fig. (a) and Fig. (b) the diodes are reversed biased
3 In Fig. (a) the diode is forward biased and in Fig. (b) the diode is reverse biased
4 In Fig. (a) the diode is reversed biased and in Fig. (b) it is forward biased
Explanation:
C for forward bias \(V_p>V_n\) and for reverse bias \(\mathrm{V}_{\mathrm{n}}>\mathrm{V}_{\mathrm{p}}\) Forward Biasing: The forward bias means the positive region is connected to the p-terminal of the supply and the negative region is connected to the n-terminal of the supply in forward biasing the external voltage is applied across the PNjunction diode. Reverse Biasing: In reversed bias, the negative region is connected to the positive terminal of the battery and the positive region is connected to the negative terminal. It creates a high resistive path in which no current flows through the circuit.
AP EMCET(Medical)-2010
Semiconductor Electronics Material Devices and Simple Circuits
150930
In space charge limited region, the plate current in a diode is \(10 \mathrm{~mA}\) for plate voltage \(150 \mathrm{~V}\). If the plate voltage is increased to \(600 \mathrm{~V}\), then the plate current will be
1 \(10 \mathrm{~mA}\)
2 \(40 \mathrm{~mA}\)
3 \(80 \mathrm{~mA}\)
4 \(160 \mathrm{~mA}\)
Explanation:
C Given, \(\mathrm{V}_1=150\) volt, \(\mathrm{V}_2=600\) volt And, \(\quad i_{p_1}=10 \mathrm{~mA}\) We know that, in space charge limited region. The plate current is given by Child's law- \(\mathrm{i}_{\mathrm{p}}=\mathrm{kV}_{\mathrm{p}}^{3 / 2}\) Then, \(\frac{i_{p_2}}{i_{p_1}}=\left(\frac{V_{p_2}}{V_{p_1}}\right)^{3 / 2}\) \(\left(\frac{600}{150}\right)^{3 / 2}=\left(4^{3 / 2}\right)=8\) \(i_{p_2}=i_{p_1} \times 8=10 \times 8=80 \mathrm{~mA}\) \(i_{p_2}=80 \mathrm{~mA}\)
VITEEE-2010
Semiconductor Electronics Material Devices and Simple Circuits
150875
If the diodes are ideal potential drop across the \(10 \Omega\) resistor in the circuit shown is
1 \(22.5 \mathrm{~V}\)
2 \(7.5 \mathrm{~V}\)
3 \(5 \mathrm{~V}\)
4 \(25 \mathrm{~V}\)
Explanation:
B Given, If diodes are ideal the all diodes are forward bias condition so they will conduct. Current in circuit \((\mathrm{I})=\frac{30}{\mathrm{R}_{\mathrm{cq}}}\) \(R_{\text {eq }} =2.5+\frac{30 \times 10}{30+10}\) \(=2.5+\frac{300}{40}\) \(=2.5+7.5\) \(R_{\text {eq }} =10 \Omega\) \(\therefore \quad I=\frac{30}{10}=3 \mathrm{~A}\) \(\because\) Current through \(10 \Omega\) resistance apply current division rule. \(\mathrm{I}_{10 \Omega}=3 \times \frac{30}{10+30}=\frac{3 \times 30}{40}=\frac{9}{4} \mathrm{~A}\) \(\therefore\) Voltage drop across \(10 \Omega\) \(\mathrm{V}=\frac{9}{4} \times 10=\frac{90}{4}=22.5 \mathrm{~V}\) \(\mathrm{~V}=22.5 \mathrm{~V}\)
AP EAMCET-28.04.2017
Semiconductor Electronics Material Devices and Simple Circuits
150877
The circuit shown below contains two diodes \(D_1\) and \(D_2\) each with a forward resistance of 50 ohms and with infinite backward resistance. The current through the \(100 \mathrm{ohm}\) resistance (in amp) is
1 0
2 0.02
3 0.03
4 0.04
Explanation:
B Given, \(D_1\) is in forward bias and \(D_2\) is in reverse bias. So, new circuit is- Apply KVL in loop, \(6-50 \mathrm{I}-150 \mathrm{I}-100 \mathrm{I}=0\) \(300 \mathrm{I}=6\) \(\mathrm{I}=\frac{6}{300}=\frac{2}{100}=0.02 \mathrm{~A}\) \(\mathrm{I}=0.02 \mathrm{~A}\)This current will flow through \(100 \Omega\) Resistance is 0.02 A.
Semiconductor Electronics Material Devices and Simple Circuits
150924
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 In the given circuit, the diode is reverse biased. Only drift current flows through the diode. Consider the circuit, in which by applying KVL, we can write, \(\mathrm{V}_{\mathrm{A}}-6+\mathrm{I} \times 20=\mathrm{V}_{\mathrm{B}} \) \(\mathrm{V}_{\mathrm{A}}-\mathrm{V}_{\mathrm{B}}=6-\mathrm{I} \times 20 =6-30 \times 10^{-6} \times 20\) \(=6-0.0006\) \(=5.9994 \mathrm{~V}\)
CG PET- 2015
Semiconductor Electronics Material Devices and Simple Circuits
150928
In the figures shown below Fig.(a)
1 In both Fig. (a) and Fig. (b) the diodes are forward biased
2 In both Fig. (a) and Fig. (b) the diodes are reversed biased
3 In Fig. (a) the diode is forward biased and in Fig. (b) the diode is reverse biased
4 In Fig. (a) the diode is reversed biased and in Fig. (b) it is forward biased
Explanation:
C for forward bias \(V_p>V_n\) and for reverse bias \(\mathrm{V}_{\mathrm{n}}>\mathrm{V}_{\mathrm{p}}\) Forward Biasing: The forward bias means the positive region is connected to the p-terminal of the supply and the negative region is connected to the n-terminal of the supply in forward biasing the external voltage is applied across the PNjunction diode. Reverse Biasing: In reversed bias, the negative region is connected to the positive terminal of the battery and the positive region is connected to the negative terminal. It creates a high resistive path in which no current flows through the circuit.
AP EMCET(Medical)-2010
Semiconductor Electronics Material Devices and Simple Circuits
150930
In space charge limited region, the plate current in a diode is \(10 \mathrm{~mA}\) for plate voltage \(150 \mathrm{~V}\). If the plate voltage is increased to \(600 \mathrm{~V}\), then the plate current will be
1 \(10 \mathrm{~mA}\)
2 \(40 \mathrm{~mA}\)
3 \(80 \mathrm{~mA}\)
4 \(160 \mathrm{~mA}\)
Explanation:
C Given, \(\mathrm{V}_1=150\) volt, \(\mathrm{V}_2=600\) volt And, \(\quad i_{p_1}=10 \mathrm{~mA}\) We know that, in space charge limited region. The plate current is given by Child's law- \(\mathrm{i}_{\mathrm{p}}=\mathrm{kV}_{\mathrm{p}}^{3 / 2}\) Then, \(\frac{i_{p_2}}{i_{p_1}}=\left(\frac{V_{p_2}}{V_{p_1}}\right)^{3 / 2}\) \(\left(\frac{600}{150}\right)^{3 / 2}=\left(4^{3 / 2}\right)=8\) \(i_{p_2}=i_{p_1} \times 8=10 \times 8=80 \mathrm{~mA}\) \(i_{p_2}=80 \mathrm{~mA}\)
VITEEE-2010
Semiconductor Electronics Material Devices and Simple Circuits
150875
If the diodes are ideal potential drop across the \(10 \Omega\) resistor in the circuit shown is
1 \(22.5 \mathrm{~V}\)
2 \(7.5 \mathrm{~V}\)
3 \(5 \mathrm{~V}\)
4 \(25 \mathrm{~V}\)
Explanation:
B Given, If diodes are ideal the all diodes are forward bias condition so they will conduct. Current in circuit \((\mathrm{I})=\frac{30}{\mathrm{R}_{\mathrm{cq}}}\) \(R_{\text {eq }} =2.5+\frac{30 \times 10}{30+10}\) \(=2.5+\frac{300}{40}\) \(=2.5+7.5\) \(R_{\text {eq }} =10 \Omega\) \(\therefore \quad I=\frac{30}{10}=3 \mathrm{~A}\) \(\because\) Current through \(10 \Omega\) resistance apply current division rule. \(\mathrm{I}_{10 \Omega}=3 \times \frac{30}{10+30}=\frac{3 \times 30}{40}=\frac{9}{4} \mathrm{~A}\) \(\therefore\) Voltage drop across \(10 \Omega\) \(\mathrm{V}=\frac{9}{4} \times 10=\frac{90}{4}=22.5 \mathrm{~V}\) \(\mathrm{~V}=22.5 \mathrm{~V}\)
AP EAMCET-28.04.2017
Semiconductor Electronics Material Devices and Simple Circuits
150877
The circuit shown below contains two diodes \(D_1\) and \(D_2\) each with a forward resistance of 50 ohms and with infinite backward resistance. The current through the \(100 \mathrm{ohm}\) resistance (in amp) is
1 0
2 0.02
3 0.03
4 0.04
Explanation:
B Given, \(D_1\) is in forward bias and \(D_2\) is in reverse bias. So, new circuit is- Apply KVL in loop, \(6-50 \mathrm{I}-150 \mathrm{I}-100 \mathrm{I}=0\) \(300 \mathrm{I}=6\) \(\mathrm{I}=\frac{6}{300}=\frac{2}{100}=0.02 \mathrm{~A}\) \(\mathrm{I}=0.02 \mathrm{~A}\)This current will flow through \(100 \Omega\) Resistance is 0.02 A.
Semiconductor Electronics Material Devices and Simple Circuits
150924
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 In the given circuit, the diode is reverse biased. Only drift current flows through the diode. Consider the circuit, in which by applying KVL, we can write, \(\mathrm{V}_{\mathrm{A}}-6+\mathrm{I} \times 20=\mathrm{V}_{\mathrm{B}} \) \(\mathrm{V}_{\mathrm{A}}-\mathrm{V}_{\mathrm{B}}=6-\mathrm{I} \times 20 =6-30 \times 10^{-6} \times 20\) \(=6-0.0006\) \(=5.9994 \mathrm{~V}\)
CG PET- 2015
Semiconductor Electronics Material Devices and Simple Circuits
150928
In the figures shown below Fig.(a)
1 In both Fig. (a) and Fig. (b) the diodes are forward biased
2 In both Fig. (a) and Fig. (b) the diodes are reversed biased
3 In Fig. (a) the diode is forward biased and in Fig. (b) the diode is reverse biased
4 In Fig. (a) the diode is reversed biased and in Fig. (b) it is forward biased
Explanation:
C for forward bias \(V_p>V_n\) and for reverse bias \(\mathrm{V}_{\mathrm{n}}>\mathrm{V}_{\mathrm{p}}\) Forward Biasing: The forward bias means the positive region is connected to the p-terminal of the supply and the negative region is connected to the n-terminal of the supply in forward biasing the external voltage is applied across the PNjunction diode. Reverse Biasing: In reversed bias, the negative region is connected to the positive terminal of the battery and the positive region is connected to the negative terminal. It creates a high resistive path in which no current flows through the circuit.
AP EMCET(Medical)-2010
Semiconductor Electronics Material Devices and Simple Circuits
150930
In space charge limited region, the plate current in a diode is \(10 \mathrm{~mA}\) for plate voltage \(150 \mathrm{~V}\). If the plate voltage is increased to \(600 \mathrm{~V}\), then the plate current will be
1 \(10 \mathrm{~mA}\)
2 \(40 \mathrm{~mA}\)
3 \(80 \mathrm{~mA}\)
4 \(160 \mathrm{~mA}\)
Explanation:
C Given, \(\mathrm{V}_1=150\) volt, \(\mathrm{V}_2=600\) volt And, \(\quad i_{p_1}=10 \mathrm{~mA}\) We know that, in space charge limited region. The plate current is given by Child's law- \(\mathrm{i}_{\mathrm{p}}=\mathrm{kV}_{\mathrm{p}}^{3 / 2}\) Then, \(\frac{i_{p_2}}{i_{p_1}}=\left(\frac{V_{p_2}}{V_{p_1}}\right)^{3 / 2}\) \(\left(\frac{600}{150}\right)^{3 / 2}=\left(4^{3 / 2}\right)=8\) \(i_{p_2}=i_{p_1} \times 8=10 \times 8=80 \mathrm{~mA}\) \(i_{p_2}=80 \mathrm{~mA}\)
VITEEE-2010
Semiconductor Electronics Material Devices and Simple Circuits
150875
If the diodes are ideal potential drop across the \(10 \Omega\) resistor in the circuit shown is
1 \(22.5 \mathrm{~V}\)
2 \(7.5 \mathrm{~V}\)
3 \(5 \mathrm{~V}\)
4 \(25 \mathrm{~V}\)
Explanation:
B Given, If diodes are ideal the all diodes are forward bias condition so they will conduct. Current in circuit \((\mathrm{I})=\frac{30}{\mathrm{R}_{\mathrm{cq}}}\) \(R_{\text {eq }} =2.5+\frac{30 \times 10}{30+10}\) \(=2.5+\frac{300}{40}\) \(=2.5+7.5\) \(R_{\text {eq }} =10 \Omega\) \(\therefore \quad I=\frac{30}{10}=3 \mathrm{~A}\) \(\because\) Current through \(10 \Omega\) resistance apply current division rule. \(\mathrm{I}_{10 \Omega}=3 \times \frac{30}{10+30}=\frac{3 \times 30}{40}=\frac{9}{4} \mathrm{~A}\) \(\therefore\) Voltage drop across \(10 \Omega\) \(\mathrm{V}=\frac{9}{4} \times 10=\frac{90}{4}=22.5 \mathrm{~V}\) \(\mathrm{~V}=22.5 \mathrm{~V}\)
AP EAMCET-28.04.2017
Semiconductor Electronics Material Devices and Simple Circuits
150877
The circuit shown below contains two diodes \(D_1\) and \(D_2\) each with a forward resistance of 50 ohms and with infinite backward resistance. The current through the \(100 \mathrm{ohm}\) resistance (in amp) is
1 0
2 0.02
3 0.03
4 0.04
Explanation:
B Given, \(D_1\) is in forward bias and \(D_2\) is in reverse bias. So, new circuit is- Apply KVL in loop, \(6-50 \mathrm{I}-150 \mathrm{I}-100 \mathrm{I}=0\) \(300 \mathrm{I}=6\) \(\mathrm{I}=\frac{6}{300}=\frac{2}{100}=0.02 \mathrm{~A}\) \(\mathrm{I}=0.02 \mathrm{~A}\)This current will flow through \(100 \Omega\) Resistance is 0.02 A.
Semiconductor Electronics Material Devices and Simple Circuits
150924
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 In the given circuit, the diode is reverse biased. Only drift current flows through the diode. Consider the circuit, in which by applying KVL, we can write, \(\mathrm{V}_{\mathrm{A}}-6+\mathrm{I} \times 20=\mathrm{V}_{\mathrm{B}} \) \(\mathrm{V}_{\mathrm{A}}-\mathrm{V}_{\mathrm{B}}=6-\mathrm{I} \times 20 =6-30 \times 10^{-6} \times 20\) \(=6-0.0006\) \(=5.9994 \mathrm{~V}\)
CG PET- 2015
Semiconductor Electronics Material Devices and Simple Circuits
150928
In the figures shown below Fig.(a)
1 In both Fig. (a) and Fig. (b) the diodes are forward biased
2 In both Fig. (a) and Fig. (b) the diodes are reversed biased
3 In Fig. (a) the diode is forward biased and in Fig. (b) the diode is reverse biased
4 In Fig. (a) the diode is reversed biased and in Fig. (b) it is forward biased
Explanation:
C for forward bias \(V_p>V_n\) and for reverse bias \(\mathrm{V}_{\mathrm{n}}>\mathrm{V}_{\mathrm{p}}\) Forward Biasing: The forward bias means the positive region is connected to the p-terminal of the supply and the negative region is connected to the n-terminal of the supply in forward biasing the external voltage is applied across the PNjunction diode. Reverse Biasing: In reversed bias, the negative region is connected to the positive terminal of the battery and the positive region is connected to the negative terminal. It creates a high resistive path in which no current flows through the circuit.
AP EMCET(Medical)-2010
Semiconductor Electronics Material Devices and Simple Circuits
150930
In space charge limited region, the plate current in a diode is \(10 \mathrm{~mA}\) for plate voltage \(150 \mathrm{~V}\). If the plate voltage is increased to \(600 \mathrm{~V}\), then the plate current will be
1 \(10 \mathrm{~mA}\)
2 \(40 \mathrm{~mA}\)
3 \(80 \mathrm{~mA}\)
4 \(160 \mathrm{~mA}\)
Explanation:
C Given, \(\mathrm{V}_1=150\) volt, \(\mathrm{V}_2=600\) volt And, \(\quad i_{p_1}=10 \mathrm{~mA}\) We know that, in space charge limited region. The plate current is given by Child's law- \(\mathrm{i}_{\mathrm{p}}=\mathrm{kV}_{\mathrm{p}}^{3 / 2}\) Then, \(\frac{i_{p_2}}{i_{p_1}}=\left(\frac{V_{p_2}}{V_{p_1}}\right)^{3 / 2}\) \(\left(\frac{600}{150}\right)^{3 / 2}=\left(4^{3 / 2}\right)=8\) \(i_{p_2}=i_{p_1} \times 8=10 \times 8=80 \mathrm{~mA}\) \(i_{p_2}=80 \mathrm{~mA}\)
VITEEE-2010
Semiconductor Electronics Material Devices and Simple Circuits
150875
If the diodes are ideal potential drop across the \(10 \Omega\) resistor in the circuit shown is
1 \(22.5 \mathrm{~V}\)
2 \(7.5 \mathrm{~V}\)
3 \(5 \mathrm{~V}\)
4 \(25 \mathrm{~V}\)
Explanation:
B Given, If diodes are ideal the all diodes are forward bias condition so they will conduct. Current in circuit \((\mathrm{I})=\frac{30}{\mathrm{R}_{\mathrm{cq}}}\) \(R_{\text {eq }} =2.5+\frac{30 \times 10}{30+10}\) \(=2.5+\frac{300}{40}\) \(=2.5+7.5\) \(R_{\text {eq }} =10 \Omega\) \(\therefore \quad I=\frac{30}{10}=3 \mathrm{~A}\) \(\because\) Current through \(10 \Omega\) resistance apply current division rule. \(\mathrm{I}_{10 \Omega}=3 \times \frac{30}{10+30}=\frac{3 \times 30}{40}=\frac{9}{4} \mathrm{~A}\) \(\therefore\) Voltage drop across \(10 \Omega\) \(\mathrm{V}=\frac{9}{4} \times 10=\frac{90}{4}=22.5 \mathrm{~V}\) \(\mathrm{~V}=22.5 \mathrm{~V}\)
AP EAMCET-28.04.2017
Semiconductor Electronics Material Devices and Simple Circuits
150877
The circuit shown below contains two diodes \(D_1\) and \(D_2\) each with a forward resistance of 50 ohms and with infinite backward resistance. The current through the \(100 \mathrm{ohm}\) resistance (in amp) is
1 0
2 0.02
3 0.03
4 0.04
Explanation:
B Given, \(D_1\) is in forward bias and \(D_2\) is in reverse bias. So, new circuit is- Apply KVL in loop, \(6-50 \mathrm{I}-150 \mathrm{I}-100 \mathrm{I}=0\) \(300 \mathrm{I}=6\) \(\mathrm{I}=\frac{6}{300}=\frac{2}{100}=0.02 \mathrm{~A}\) \(\mathrm{I}=0.02 \mathrm{~A}\)This current will flow through \(100 \Omega\) Resistance is 0.02 A.
Semiconductor Electronics Material Devices and Simple Circuits
150924
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 In the given circuit, the diode is reverse biased. Only drift current flows through the diode. Consider the circuit, in which by applying KVL, we can write, \(\mathrm{V}_{\mathrm{A}}-6+\mathrm{I} \times 20=\mathrm{V}_{\mathrm{B}} \) \(\mathrm{V}_{\mathrm{A}}-\mathrm{V}_{\mathrm{B}}=6-\mathrm{I} \times 20 =6-30 \times 10^{-6} \times 20\) \(=6-0.0006\) \(=5.9994 \mathrm{~V}\)
CG PET- 2015
Semiconductor Electronics Material Devices and Simple Circuits
150928
In the figures shown below Fig.(a)
1 In both Fig. (a) and Fig. (b) the diodes are forward biased
2 In both Fig. (a) and Fig. (b) the diodes are reversed biased
3 In Fig. (a) the diode is forward biased and in Fig. (b) the diode is reverse biased
4 In Fig. (a) the diode is reversed biased and in Fig. (b) it is forward biased
Explanation:
C for forward bias \(V_p>V_n\) and for reverse bias \(\mathrm{V}_{\mathrm{n}}>\mathrm{V}_{\mathrm{p}}\) Forward Biasing: The forward bias means the positive region is connected to the p-terminal of the supply and the negative region is connected to the n-terminal of the supply in forward biasing the external voltage is applied across the PNjunction diode. Reverse Biasing: In reversed bias, the negative region is connected to the positive terminal of the battery and the positive region is connected to the negative terminal. It creates a high resistive path in which no current flows through the circuit.
AP EMCET(Medical)-2010
Semiconductor Electronics Material Devices and Simple Circuits
150930
In space charge limited region, the plate current in a diode is \(10 \mathrm{~mA}\) for plate voltage \(150 \mathrm{~V}\). If the plate voltage is increased to \(600 \mathrm{~V}\), then the plate current will be
1 \(10 \mathrm{~mA}\)
2 \(40 \mathrm{~mA}\)
3 \(80 \mathrm{~mA}\)
4 \(160 \mathrm{~mA}\)
Explanation:
C Given, \(\mathrm{V}_1=150\) volt, \(\mathrm{V}_2=600\) volt And, \(\quad i_{p_1}=10 \mathrm{~mA}\) We know that, in space charge limited region. The plate current is given by Child's law- \(\mathrm{i}_{\mathrm{p}}=\mathrm{kV}_{\mathrm{p}}^{3 / 2}\) Then, \(\frac{i_{p_2}}{i_{p_1}}=\left(\frac{V_{p_2}}{V_{p_1}}\right)^{3 / 2}\) \(\left(\frac{600}{150}\right)^{3 / 2}=\left(4^{3 / 2}\right)=8\) \(i_{p_2}=i_{p_1} \times 8=10 \times 8=80 \mathrm{~mA}\) \(i_{p_2}=80 \mathrm{~mA}\)
VITEEE-2010
Semiconductor Electronics Material Devices and Simple Circuits
150875
If the diodes are ideal potential drop across the \(10 \Omega\) resistor in the circuit shown is
1 \(22.5 \mathrm{~V}\)
2 \(7.5 \mathrm{~V}\)
3 \(5 \mathrm{~V}\)
4 \(25 \mathrm{~V}\)
Explanation:
B Given, If diodes are ideal the all diodes are forward bias condition so they will conduct. Current in circuit \((\mathrm{I})=\frac{30}{\mathrm{R}_{\mathrm{cq}}}\) \(R_{\text {eq }} =2.5+\frac{30 \times 10}{30+10}\) \(=2.5+\frac{300}{40}\) \(=2.5+7.5\) \(R_{\text {eq }} =10 \Omega\) \(\therefore \quad I=\frac{30}{10}=3 \mathrm{~A}\) \(\because\) Current through \(10 \Omega\) resistance apply current division rule. \(\mathrm{I}_{10 \Omega}=3 \times \frac{30}{10+30}=\frac{3 \times 30}{40}=\frac{9}{4} \mathrm{~A}\) \(\therefore\) Voltage drop across \(10 \Omega\) \(\mathrm{V}=\frac{9}{4} \times 10=\frac{90}{4}=22.5 \mathrm{~V}\) \(\mathrm{~V}=22.5 \mathrm{~V}\)
AP EAMCET-28.04.2017
Semiconductor Electronics Material Devices and Simple Circuits
150877
The circuit shown below contains two diodes \(D_1\) and \(D_2\) each with a forward resistance of 50 ohms and with infinite backward resistance. The current through the \(100 \mathrm{ohm}\) resistance (in amp) is
1 0
2 0.02
3 0.03
4 0.04
Explanation:
B Given, \(D_1\) is in forward bias and \(D_2\) is in reverse bias. So, new circuit is- Apply KVL in loop, \(6-50 \mathrm{I}-150 \mathrm{I}-100 \mathrm{I}=0\) \(300 \mathrm{I}=6\) \(\mathrm{I}=\frac{6}{300}=\frac{2}{100}=0.02 \mathrm{~A}\) \(\mathrm{I}=0.02 \mathrm{~A}\)This current will flow through \(100 \Omega\) Resistance is 0.02 A.
