PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
365517
In the following circuit, a voltmeter \(V\) is connected across a lamp \(L\). What change would occur in voltmeter reading if the resistance \({R_1}\) is reduced in value?
1 Decreases
2 Increases
3 Remains same
4 None of these
Explanation:
Here the emitter base junction of \(N{\rm{ - }}P{\rm{ - }}N\) transistor is forward biased with battery \({V_{BB}}\) through resistance \(R.\) When the value of \(R\) is reduced, then the emitter current \({i_e}\) will increase. As a result the collector current will also increase \(({i_c} = {i_e} - {i_b}).\) Due to increase in \({i_c},\) the potential difference across \(L\) increases and hence the reading of voltmeter will increases.
PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
365518
In the given circuit the value of \(\beta \) is 100. When \({I_C} = 1.5mA\) then the transistor is operating in
1 Normal Active mode
2 Saturation mode
3 Inverse Active mode
4 Cut off mode
Explanation:
\({I_b} = \frac{{{I_C}}}{\beta } = \frac{{1.5 \times {{10}^{ - 3}}}}{{100}} = 1.5 \times {10^{ - 5}}A\) Applying \(KVL\) to the closed part of the circuit containing and emitter base junction, we have \({V_{cc}} = {V_{be}} + {I_b}{R_b}\) \({V_{be}} = {V_{cc}} - {I_b}{R_b}\) \( = 24 - \left( {1.5 \times {{10}^{ - 5}}} \right)\left( {220 \times {{10}^3}} \right) = 20.7\,V\) Applying \(KVL\) to the closed part of the circuit \({R_b},{R_c}\) containing \({R_b}\) and base collector junction, we have \({V_{BC}} - {I_b}{R_b} + {I_c}{R_c} = 0\) \({V_{BC}} = {I_b}{R_b} - {I_c}{R_c} = 3.75V\) As both \({V_{BE}}\& {V_{BC}}\) are in forward biasing condition, transistor is in saturation state.
PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
365519
In a \(n\)-\(p\)-\(n\) transistor \({10^{10}}\) electron enter the emitter in \({10^{ - 6}}s\). If 2% of the electrons are lost in the base, the current transfer ratio and the current amplification factor are
PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
365517
In the following circuit, a voltmeter \(V\) is connected across a lamp \(L\). What change would occur in voltmeter reading if the resistance \({R_1}\) is reduced in value?
1 Decreases
2 Increases
3 Remains same
4 None of these
Explanation:
Here the emitter base junction of \(N{\rm{ - }}P{\rm{ - }}N\) transistor is forward biased with battery \({V_{BB}}\) through resistance \(R.\) When the value of \(R\) is reduced, then the emitter current \({i_e}\) will increase. As a result the collector current will also increase \(({i_c} = {i_e} - {i_b}).\) Due to increase in \({i_c},\) the potential difference across \(L\) increases and hence the reading of voltmeter will increases.
PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
365518
In the given circuit the value of \(\beta \) is 100. When \({I_C} = 1.5mA\) then the transistor is operating in
1 Normal Active mode
2 Saturation mode
3 Inverse Active mode
4 Cut off mode
Explanation:
\({I_b} = \frac{{{I_C}}}{\beta } = \frac{{1.5 \times {{10}^{ - 3}}}}{{100}} = 1.5 \times {10^{ - 5}}A\) Applying \(KVL\) to the closed part of the circuit containing and emitter base junction, we have \({V_{cc}} = {V_{be}} + {I_b}{R_b}\) \({V_{be}} = {V_{cc}} - {I_b}{R_b}\) \( = 24 - \left( {1.5 \times {{10}^{ - 5}}} \right)\left( {220 \times {{10}^3}} \right) = 20.7\,V\) Applying \(KVL\) to the closed part of the circuit \({R_b},{R_c}\) containing \({R_b}\) and base collector junction, we have \({V_{BC}} - {I_b}{R_b} + {I_c}{R_c} = 0\) \({V_{BC}} = {I_b}{R_b} - {I_c}{R_c} = 3.75V\) As both \({V_{BE}}\& {V_{BC}}\) are in forward biasing condition, transistor is in saturation state.
PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
365519
In a \(n\)-\(p\)-\(n\) transistor \({10^{10}}\) electron enter the emitter in \({10^{ - 6}}s\). If 2% of the electrons are lost in the base, the current transfer ratio and the current amplification factor are
PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
365517
In the following circuit, a voltmeter \(V\) is connected across a lamp \(L\). What change would occur in voltmeter reading if the resistance \({R_1}\) is reduced in value?
1 Decreases
2 Increases
3 Remains same
4 None of these
Explanation:
Here the emitter base junction of \(N{\rm{ - }}P{\rm{ - }}N\) transistor is forward biased with battery \({V_{BB}}\) through resistance \(R.\) When the value of \(R\) is reduced, then the emitter current \({i_e}\) will increase. As a result the collector current will also increase \(({i_c} = {i_e} - {i_b}).\) Due to increase in \({i_c},\) the potential difference across \(L\) increases and hence the reading of voltmeter will increases.
PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
365518
In the given circuit the value of \(\beta \) is 100. When \({I_C} = 1.5mA\) then the transistor is operating in
1 Normal Active mode
2 Saturation mode
3 Inverse Active mode
4 Cut off mode
Explanation:
\({I_b} = \frac{{{I_C}}}{\beta } = \frac{{1.5 \times {{10}^{ - 3}}}}{{100}} = 1.5 \times {10^{ - 5}}A\) Applying \(KVL\) to the closed part of the circuit containing and emitter base junction, we have \({V_{cc}} = {V_{be}} + {I_b}{R_b}\) \({V_{be}} = {V_{cc}} - {I_b}{R_b}\) \( = 24 - \left( {1.5 \times {{10}^{ - 5}}} \right)\left( {220 \times {{10}^3}} \right) = 20.7\,V\) Applying \(KVL\) to the closed part of the circuit \({R_b},{R_c}\) containing \({R_b}\) and base collector junction, we have \({V_{BC}} - {I_b}{R_b} + {I_c}{R_c} = 0\) \({V_{BC}} = {I_b}{R_b} - {I_c}{R_c} = 3.75V\) As both \({V_{BE}}\& {V_{BC}}\) are in forward biasing condition, transistor is in saturation state.
PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
365519
In a \(n\)-\(p\)-\(n\) transistor \({10^{10}}\) electron enter the emitter in \({10^{ - 6}}s\). If 2% of the electrons are lost in the base, the current transfer ratio and the current amplification factor are
PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
365517
In the following circuit, a voltmeter \(V\) is connected across a lamp \(L\). What change would occur in voltmeter reading if the resistance \({R_1}\) is reduced in value?
1 Decreases
2 Increases
3 Remains same
4 None of these
Explanation:
Here the emitter base junction of \(N{\rm{ - }}P{\rm{ - }}N\) transistor is forward biased with battery \({V_{BB}}\) through resistance \(R.\) When the value of \(R\) is reduced, then the emitter current \({i_e}\) will increase. As a result the collector current will also increase \(({i_c} = {i_e} - {i_b}).\) Due to increase in \({i_c},\) the potential difference across \(L\) increases and hence the reading of voltmeter will increases.
PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
365518
In the given circuit the value of \(\beta \) is 100. When \({I_C} = 1.5mA\) then the transistor is operating in
1 Normal Active mode
2 Saturation mode
3 Inverse Active mode
4 Cut off mode
Explanation:
\({I_b} = \frac{{{I_C}}}{\beta } = \frac{{1.5 \times {{10}^{ - 3}}}}{{100}} = 1.5 \times {10^{ - 5}}A\) Applying \(KVL\) to the closed part of the circuit containing and emitter base junction, we have \({V_{cc}} = {V_{be}} + {I_b}{R_b}\) \({V_{be}} = {V_{cc}} - {I_b}{R_b}\) \( = 24 - \left( {1.5 \times {{10}^{ - 5}}} \right)\left( {220 \times {{10}^3}} \right) = 20.7\,V\) Applying \(KVL\) to the closed part of the circuit \({R_b},{R_c}\) containing \({R_b}\) and base collector junction, we have \({V_{BC}} - {I_b}{R_b} + {I_c}{R_c} = 0\) \({V_{BC}} = {I_b}{R_b} - {I_c}{R_c} = 3.75V\) As both \({V_{BE}}\& {V_{BC}}\) are in forward biasing condition, transistor is in saturation state.
PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
365519
In a \(n\)-\(p\)-\(n\) transistor \({10^{10}}\) electron enter the emitter in \({10^{ - 6}}s\). If 2% of the electrons are lost in the base, the current transfer ratio and the current amplification factor are
