PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
365311
At \(0.3 {~V}\) and \(0.7 {~V}\), the diodes \({Ge}\) and \({Si}\) become conductor respectively. In given figure, if ends of diode Ge are reversed, the change in potential \({V}_{0}\) will be
1 \(0.4\,V\)
2 \(0.9\,V\)
3 \(0.1\,V\)
4 \(0.7\,V\)
Explanation:
Initially \(Ge\) and \({Si}\) are both forward biased. So, current will effectively pass through Ge diode with a voltage drop of \(0.3 {~V}\). \(\therefore\) Initial output voltage, \(V_{0}=10-0.3=9.7 {~V}\) If \({Ge}\) is reversed biased, then only Si diode will conduct. In this condition, output voltage \({V}_{0}=10-0.7=9.3 {~V}\)
\(\therefore\) Change in output voltage \(=9.7-9.3=0.4 {~V}\)
PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
365312
Assuming that the junction diode is ideal, the current in the arrangement shown in figure is
1 \(30\,mA\)
2 \(40\,mA\)
3 \(20\,mA\)
4 \(10\,mA\)
Explanation:
In the given circuit, the diode is forward biased, hence the current will flow through it Now, applying KVL in the above circuit \( + 3 - 100i - ( + 1) = 0\) \( \Rightarrow 3 - 1 = 100I\) \( \Rightarrow i = \frac{2}{{100}} = 0.02A\) \(i = 20\,mA\) Hence the current flow in the circuit is 20\(mA\).
PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
365313
The current in the circuit will be
1 \(\frac{5}{{40}}\;A\)
2 \(\frac{5}{{50}}\;A\)
3 \(\frac{5}{{10}}\;A\)
4 \(\frac{5}{{20}}\;A\)
Explanation:
In the given circuit, diode in the lower branch is forward biased so it conducts and diode in the upper branch is reverse biased so it does not conduct. \(\therefore\) The total resistance in the circuit \(=20\, \Omega+30\, \Omega=50\, \Omega\) The current in the circuit is \(I = \frac{{5\;V}}{{50\,\Omega }} = \frac{5}{{50}}\;A\)
PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
365314
A diode connected to an external resistance and an e.m.f. Assuming that the barrier potential developed in diode is \(0.5\;V\). What will be the value of current in the circuit in milliampere?
1 \(40\,mA\)
2 \(4\,mA\)
3 \(0.4\;mA\)
4 \(400\;mA\)
Explanation:
Effective voltage in the circuit \(V = 4.5 - 0.5 = 4\;V\) Current in the circuit \(I = \frac{V}{R} = \frac{4}{{100}} = 0.04\;A\) \(I = 0.04A = 40\;mA.\)
PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
365315
Find the current passing through the battery.
1 \(1\,A\)
2 \(2\,A\)
3 \(0.5\,A\)
4 \(4\,A\)
Explanation:
The lower diode is reverse biased and hence current does not pass through it. The equivalent resistance is \(15\Omega \). The net current flows through the battery is \(i = \frac{{15}}{{15}} = 1A\)
PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
365311
At \(0.3 {~V}\) and \(0.7 {~V}\), the diodes \({Ge}\) and \({Si}\) become conductor respectively. In given figure, if ends of diode Ge are reversed, the change in potential \({V}_{0}\) will be
1 \(0.4\,V\)
2 \(0.9\,V\)
3 \(0.1\,V\)
4 \(0.7\,V\)
Explanation:
Initially \(Ge\) and \({Si}\) are both forward biased. So, current will effectively pass through Ge diode with a voltage drop of \(0.3 {~V}\). \(\therefore\) Initial output voltage, \(V_{0}=10-0.3=9.7 {~V}\) If \({Ge}\) is reversed biased, then only Si diode will conduct. In this condition, output voltage \({V}_{0}=10-0.7=9.3 {~V}\)
\(\therefore\) Change in output voltage \(=9.7-9.3=0.4 {~V}\)
PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
365312
Assuming that the junction diode is ideal, the current in the arrangement shown in figure is
1 \(30\,mA\)
2 \(40\,mA\)
3 \(20\,mA\)
4 \(10\,mA\)
Explanation:
In the given circuit, the diode is forward biased, hence the current will flow through it Now, applying KVL in the above circuit \( + 3 - 100i - ( + 1) = 0\) \( \Rightarrow 3 - 1 = 100I\) \( \Rightarrow i = \frac{2}{{100}} = 0.02A\) \(i = 20\,mA\) Hence the current flow in the circuit is 20\(mA\).
PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
365313
The current in the circuit will be
1 \(\frac{5}{{40}}\;A\)
2 \(\frac{5}{{50}}\;A\)
3 \(\frac{5}{{10}}\;A\)
4 \(\frac{5}{{20}}\;A\)
Explanation:
In the given circuit, diode in the lower branch is forward biased so it conducts and diode in the upper branch is reverse biased so it does not conduct. \(\therefore\) The total resistance in the circuit \(=20\, \Omega+30\, \Omega=50\, \Omega\) The current in the circuit is \(I = \frac{{5\;V}}{{50\,\Omega }} = \frac{5}{{50}}\;A\)
PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
365314
A diode connected to an external resistance and an e.m.f. Assuming that the barrier potential developed in diode is \(0.5\;V\). What will be the value of current in the circuit in milliampere?
1 \(40\,mA\)
2 \(4\,mA\)
3 \(0.4\;mA\)
4 \(400\;mA\)
Explanation:
Effective voltage in the circuit \(V = 4.5 - 0.5 = 4\;V\) Current in the circuit \(I = \frac{V}{R} = \frac{4}{{100}} = 0.04\;A\) \(I = 0.04A = 40\;mA.\)
PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
365315
Find the current passing through the battery.
1 \(1\,A\)
2 \(2\,A\)
3 \(0.5\,A\)
4 \(4\,A\)
Explanation:
The lower diode is reverse biased and hence current does not pass through it. The equivalent resistance is \(15\Omega \). The net current flows through the battery is \(i = \frac{{15}}{{15}} = 1A\)
PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
365311
At \(0.3 {~V}\) and \(0.7 {~V}\), the diodes \({Ge}\) and \({Si}\) become conductor respectively. In given figure, if ends of diode Ge are reversed, the change in potential \({V}_{0}\) will be
1 \(0.4\,V\)
2 \(0.9\,V\)
3 \(0.1\,V\)
4 \(0.7\,V\)
Explanation:
Initially \(Ge\) and \({Si}\) are both forward biased. So, current will effectively pass through Ge diode with a voltage drop of \(0.3 {~V}\). \(\therefore\) Initial output voltage, \(V_{0}=10-0.3=9.7 {~V}\) If \({Ge}\) is reversed biased, then only Si diode will conduct. In this condition, output voltage \({V}_{0}=10-0.7=9.3 {~V}\)
\(\therefore\) Change in output voltage \(=9.7-9.3=0.4 {~V}\)
PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
365312
Assuming that the junction diode is ideal, the current in the arrangement shown in figure is
1 \(30\,mA\)
2 \(40\,mA\)
3 \(20\,mA\)
4 \(10\,mA\)
Explanation:
In the given circuit, the diode is forward biased, hence the current will flow through it Now, applying KVL in the above circuit \( + 3 - 100i - ( + 1) = 0\) \( \Rightarrow 3 - 1 = 100I\) \( \Rightarrow i = \frac{2}{{100}} = 0.02A\) \(i = 20\,mA\) Hence the current flow in the circuit is 20\(mA\).
PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
365313
The current in the circuit will be
1 \(\frac{5}{{40}}\;A\)
2 \(\frac{5}{{50}}\;A\)
3 \(\frac{5}{{10}}\;A\)
4 \(\frac{5}{{20}}\;A\)
Explanation:
In the given circuit, diode in the lower branch is forward biased so it conducts and diode in the upper branch is reverse biased so it does not conduct. \(\therefore\) The total resistance in the circuit \(=20\, \Omega+30\, \Omega=50\, \Omega\) The current in the circuit is \(I = \frac{{5\;V}}{{50\,\Omega }} = \frac{5}{{50}}\;A\)
PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
365314
A diode connected to an external resistance and an e.m.f. Assuming that the barrier potential developed in diode is \(0.5\;V\). What will be the value of current in the circuit in milliampere?
1 \(40\,mA\)
2 \(4\,mA\)
3 \(0.4\;mA\)
4 \(400\;mA\)
Explanation:
Effective voltage in the circuit \(V = 4.5 - 0.5 = 4\;V\) Current in the circuit \(I = \frac{V}{R} = \frac{4}{{100}} = 0.04\;A\) \(I = 0.04A = 40\;mA.\)
PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
365315
Find the current passing through the battery.
1 \(1\,A\)
2 \(2\,A\)
3 \(0.5\,A\)
4 \(4\,A\)
Explanation:
The lower diode is reverse biased and hence current does not pass through it. The equivalent resistance is \(15\Omega \). The net current flows through the battery is \(i = \frac{{15}}{{15}} = 1A\)
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PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
365311
At \(0.3 {~V}\) and \(0.7 {~V}\), the diodes \({Ge}\) and \({Si}\) become conductor respectively. In given figure, if ends of diode Ge are reversed, the change in potential \({V}_{0}\) will be
1 \(0.4\,V\)
2 \(0.9\,V\)
3 \(0.1\,V\)
4 \(0.7\,V\)
Explanation:
Initially \(Ge\) and \({Si}\) are both forward biased. So, current will effectively pass through Ge diode with a voltage drop of \(0.3 {~V}\). \(\therefore\) Initial output voltage, \(V_{0}=10-0.3=9.7 {~V}\) If \({Ge}\) is reversed biased, then only Si diode will conduct. In this condition, output voltage \({V}_{0}=10-0.7=9.3 {~V}\)
\(\therefore\) Change in output voltage \(=9.7-9.3=0.4 {~V}\)
PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
365312
Assuming that the junction diode is ideal, the current in the arrangement shown in figure is
1 \(30\,mA\)
2 \(40\,mA\)
3 \(20\,mA\)
4 \(10\,mA\)
Explanation:
In the given circuit, the diode is forward biased, hence the current will flow through it Now, applying KVL in the above circuit \( + 3 - 100i - ( + 1) = 0\) \( \Rightarrow 3 - 1 = 100I\) \( \Rightarrow i = \frac{2}{{100}} = 0.02A\) \(i = 20\,mA\) Hence the current flow in the circuit is 20\(mA\).
PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
365313
The current in the circuit will be
1 \(\frac{5}{{40}}\;A\)
2 \(\frac{5}{{50}}\;A\)
3 \(\frac{5}{{10}}\;A\)
4 \(\frac{5}{{20}}\;A\)
Explanation:
In the given circuit, diode in the lower branch is forward biased so it conducts and diode in the upper branch is reverse biased so it does not conduct. \(\therefore\) The total resistance in the circuit \(=20\, \Omega+30\, \Omega=50\, \Omega\) The current in the circuit is \(I = \frac{{5\;V}}{{50\,\Omega }} = \frac{5}{{50}}\;A\)
PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
365314
A diode connected to an external resistance and an e.m.f. Assuming that the barrier potential developed in diode is \(0.5\;V\). What will be the value of current in the circuit in milliampere?
1 \(40\,mA\)
2 \(4\,mA\)
3 \(0.4\;mA\)
4 \(400\;mA\)
Explanation:
Effective voltage in the circuit \(V = 4.5 - 0.5 = 4\;V\) Current in the circuit \(I = \frac{V}{R} = \frac{4}{{100}} = 0.04\;A\) \(I = 0.04A = 40\;mA.\)
PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
365315
Find the current passing through the battery.
1 \(1\,A\)
2 \(2\,A\)
3 \(0.5\,A\)
4 \(4\,A\)
Explanation:
The lower diode is reverse biased and hence current does not pass through it. The equivalent resistance is \(15\Omega \). The net current flows through the battery is \(i = \frac{{15}}{{15}} = 1A\)
PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
365311
At \(0.3 {~V}\) and \(0.7 {~V}\), the diodes \({Ge}\) and \({Si}\) become conductor respectively. In given figure, if ends of diode Ge are reversed, the change in potential \({V}_{0}\) will be
1 \(0.4\,V\)
2 \(0.9\,V\)
3 \(0.1\,V\)
4 \(0.7\,V\)
Explanation:
Initially \(Ge\) and \({Si}\) are both forward biased. So, current will effectively pass through Ge diode with a voltage drop of \(0.3 {~V}\). \(\therefore\) Initial output voltage, \(V_{0}=10-0.3=9.7 {~V}\) If \({Ge}\) is reversed biased, then only Si diode will conduct. In this condition, output voltage \({V}_{0}=10-0.7=9.3 {~V}\)
\(\therefore\) Change in output voltage \(=9.7-9.3=0.4 {~V}\)
PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
365312
Assuming that the junction diode is ideal, the current in the arrangement shown in figure is
1 \(30\,mA\)
2 \(40\,mA\)
3 \(20\,mA\)
4 \(10\,mA\)
Explanation:
In the given circuit, the diode is forward biased, hence the current will flow through it Now, applying KVL in the above circuit \( + 3 - 100i - ( + 1) = 0\) \( \Rightarrow 3 - 1 = 100I\) \( \Rightarrow i = \frac{2}{{100}} = 0.02A\) \(i = 20\,mA\) Hence the current flow in the circuit is 20\(mA\).
PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
365313
The current in the circuit will be
1 \(\frac{5}{{40}}\;A\)
2 \(\frac{5}{{50}}\;A\)
3 \(\frac{5}{{10}}\;A\)
4 \(\frac{5}{{20}}\;A\)
Explanation:
In the given circuit, diode in the lower branch is forward biased so it conducts and diode in the upper branch is reverse biased so it does not conduct. \(\therefore\) The total resistance in the circuit \(=20\, \Omega+30\, \Omega=50\, \Omega\) The current in the circuit is \(I = \frac{{5\;V}}{{50\,\Omega }} = \frac{5}{{50}}\;A\)
PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
365314
A diode connected to an external resistance and an e.m.f. Assuming that the barrier potential developed in diode is \(0.5\;V\). What will be the value of current in the circuit in milliampere?
1 \(40\,mA\)
2 \(4\,mA\)
3 \(0.4\;mA\)
4 \(400\;mA\)
Explanation:
Effective voltage in the circuit \(V = 4.5 - 0.5 = 4\;V\) Current in the circuit \(I = \frac{V}{R} = \frac{4}{{100}} = 0.04\;A\) \(I = 0.04A = 40\;mA.\)
PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
365315
Find the current passing through the battery.
1 \(1\,A\)
2 \(2\,A\)
3 \(0.5\,A\)
4 \(4\,A\)
Explanation:
The lower diode is reverse biased and hence current does not pass through it. The equivalent resistance is \(15\Omega \). The net current flows through the battery is \(i = \frac{{15}}{{15}} = 1A\)
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