Semiconductor Electronics Material Devices and Simple Circuits
150740
Identify the mismatched pair from the following
1 Zener diode : voltage regulator
2 germanium doped : n-type semiconductor with phosphorous
3 semiconductor : band gap \(>3 \mathrm{eV}\)
4 p-n junction diode : rectifier
5 silicon doped with : p-type semiconductor aluminium
Explanation:
C Zener diodes are operated in the breakdown region, voltage across this zener diode always remain constant. We could say zener diode can be used as voltage regulator. Hence, option (a) is correct. Phosphorus being a pentavalent element, bring in free valence electron into germanium on being in it. Hence forming n-type semiconductor. Hence, option (b) is correct. For semiconductor, the band gap is less than three electron volt. Hence, option (c) is incorrect. In p-n junction diode can be used as half-wave and fullwave rectifier. Hence, option (d) is correct. Aluminium being the trivalent impurity forms holes hence forming p-type semiconductor. Hence, option (e) is correct. Hence, option (c) is mismatched pair.
Kerala CEE- 2013
Semiconductor Electronics Material Devices and Simple Circuits
150768
In a full wave rectifier, input voltage is represented by \(V=V_M \sin \omega t\), then peak inversion voltage of non conducting diode will be:
1 \(-\mathrm{V}_{\mathrm{M}}\)
2 \(\mathrm{V}_{\mathrm{M}} / 2\)
3 \(2 \mathrm{~V}_{\mathrm{M}}\)
4 0
Explanation:
C Given, \(\mathrm{V}=\mathrm{V}_{\mathrm{m}} \sin \omega \mathrm{t}\) So, \(\mathrm{V}_{\mathrm{m}}=\) maximum voltage We know that, in case of full wave rectifier the peak inversion voltage (PIV) \(=2 \times\) maximum voltage \(\mathrm{PIV}=2 \mathrm{~V}_{\mathrm{m}}\)
AIIMS-26.05.2018(M)
Semiconductor Electronics Material Devices and Simple Circuits
150772
In the half wave rectifier circuit operating from \(50 \mathrm{~Hz}\) mains frequency, the fundamental frequency in the ripple would be
1 \(25 \mathrm{~Hz}\)
2 \(50 \mathrm{~Hz}\)
3 \(70.7 \mathrm{~Hz}\)
4 \(100 \mathrm{~Hz}\)
Explanation:
B As output voltage obtained in half wave rectifier circuit has variation in 1 cycle of ac voltage. \(\therefore\) Fundamental frequency in ripple of output voltage \(=\) \(50 \mathrm{~Hz}\)
AIIMS-2007
Semiconductor Electronics Material Devices and Simple Circuits
150774
In a full wave rectifier circuit operating from \(50 \mathrm{~Hz}\) mains frequency, what is the fundamental frequency in the ripple?
1 \(50 \mathrm{~Hz}\)
2 \(100 \mathrm{~Hz}\)
3 \(70 \mathrm{~Hz}\)
4 \(25 \mathrm{~Hz}\)
Explanation:
B In full wave rectifier the ripple frequency \(=2\) \(\times\) input frequency \(=2 \times 50\) \(=100 \mathrm{~Hz}\)
NEET Test Series from KOTA - 10 Papers In MS WORD
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Semiconductor Electronics Material Devices and Simple Circuits
150740
Identify the mismatched pair from the following
1 Zener diode : voltage regulator
2 germanium doped : n-type semiconductor with phosphorous
3 semiconductor : band gap \(>3 \mathrm{eV}\)
4 p-n junction diode : rectifier
5 silicon doped with : p-type semiconductor aluminium
Explanation:
C Zener diodes are operated in the breakdown region, voltage across this zener diode always remain constant. We could say zener diode can be used as voltage regulator. Hence, option (a) is correct. Phosphorus being a pentavalent element, bring in free valence electron into germanium on being in it. Hence forming n-type semiconductor. Hence, option (b) is correct. For semiconductor, the band gap is less than three electron volt. Hence, option (c) is incorrect. In p-n junction diode can be used as half-wave and fullwave rectifier. Hence, option (d) is correct. Aluminium being the trivalent impurity forms holes hence forming p-type semiconductor. Hence, option (e) is correct. Hence, option (c) is mismatched pair.
Kerala CEE- 2013
Semiconductor Electronics Material Devices and Simple Circuits
150768
In a full wave rectifier, input voltage is represented by \(V=V_M \sin \omega t\), then peak inversion voltage of non conducting diode will be:
1 \(-\mathrm{V}_{\mathrm{M}}\)
2 \(\mathrm{V}_{\mathrm{M}} / 2\)
3 \(2 \mathrm{~V}_{\mathrm{M}}\)
4 0
Explanation:
C Given, \(\mathrm{V}=\mathrm{V}_{\mathrm{m}} \sin \omega \mathrm{t}\) So, \(\mathrm{V}_{\mathrm{m}}=\) maximum voltage We know that, in case of full wave rectifier the peak inversion voltage (PIV) \(=2 \times\) maximum voltage \(\mathrm{PIV}=2 \mathrm{~V}_{\mathrm{m}}\)
AIIMS-26.05.2018(M)
Semiconductor Electronics Material Devices and Simple Circuits
150772
In the half wave rectifier circuit operating from \(50 \mathrm{~Hz}\) mains frequency, the fundamental frequency in the ripple would be
1 \(25 \mathrm{~Hz}\)
2 \(50 \mathrm{~Hz}\)
3 \(70.7 \mathrm{~Hz}\)
4 \(100 \mathrm{~Hz}\)
Explanation:
B As output voltage obtained in half wave rectifier circuit has variation in 1 cycle of ac voltage. \(\therefore\) Fundamental frequency in ripple of output voltage \(=\) \(50 \mathrm{~Hz}\)
AIIMS-2007
Semiconductor Electronics Material Devices and Simple Circuits
150774
In a full wave rectifier circuit operating from \(50 \mathrm{~Hz}\) mains frequency, what is the fundamental frequency in the ripple?
1 \(50 \mathrm{~Hz}\)
2 \(100 \mathrm{~Hz}\)
3 \(70 \mathrm{~Hz}\)
4 \(25 \mathrm{~Hz}\)
Explanation:
B In full wave rectifier the ripple frequency \(=2\) \(\times\) input frequency \(=2 \times 50\) \(=100 \mathrm{~Hz}\)
Semiconductor Electronics Material Devices and Simple Circuits
150740
Identify the mismatched pair from the following
1 Zener diode : voltage regulator
2 germanium doped : n-type semiconductor with phosphorous
3 semiconductor : band gap \(>3 \mathrm{eV}\)
4 p-n junction diode : rectifier
5 silicon doped with : p-type semiconductor aluminium
Explanation:
C Zener diodes are operated in the breakdown region, voltage across this zener diode always remain constant. We could say zener diode can be used as voltage regulator. Hence, option (a) is correct. Phosphorus being a pentavalent element, bring in free valence electron into germanium on being in it. Hence forming n-type semiconductor. Hence, option (b) is correct. For semiconductor, the band gap is less than three electron volt. Hence, option (c) is incorrect. In p-n junction diode can be used as half-wave and fullwave rectifier. Hence, option (d) is correct. Aluminium being the trivalent impurity forms holes hence forming p-type semiconductor. Hence, option (e) is correct. Hence, option (c) is mismatched pair.
Kerala CEE- 2013
Semiconductor Electronics Material Devices and Simple Circuits
150768
In a full wave rectifier, input voltage is represented by \(V=V_M \sin \omega t\), then peak inversion voltage of non conducting diode will be:
1 \(-\mathrm{V}_{\mathrm{M}}\)
2 \(\mathrm{V}_{\mathrm{M}} / 2\)
3 \(2 \mathrm{~V}_{\mathrm{M}}\)
4 0
Explanation:
C Given, \(\mathrm{V}=\mathrm{V}_{\mathrm{m}} \sin \omega \mathrm{t}\) So, \(\mathrm{V}_{\mathrm{m}}=\) maximum voltage We know that, in case of full wave rectifier the peak inversion voltage (PIV) \(=2 \times\) maximum voltage \(\mathrm{PIV}=2 \mathrm{~V}_{\mathrm{m}}\)
AIIMS-26.05.2018(M)
Semiconductor Electronics Material Devices and Simple Circuits
150772
In the half wave rectifier circuit operating from \(50 \mathrm{~Hz}\) mains frequency, the fundamental frequency in the ripple would be
1 \(25 \mathrm{~Hz}\)
2 \(50 \mathrm{~Hz}\)
3 \(70.7 \mathrm{~Hz}\)
4 \(100 \mathrm{~Hz}\)
Explanation:
B As output voltage obtained in half wave rectifier circuit has variation in 1 cycle of ac voltage. \(\therefore\) Fundamental frequency in ripple of output voltage \(=\) \(50 \mathrm{~Hz}\)
AIIMS-2007
Semiconductor Electronics Material Devices and Simple Circuits
150774
In a full wave rectifier circuit operating from \(50 \mathrm{~Hz}\) mains frequency, what is the fundamental frequency in the ripple?
1 \(50 \mathrm{~Hz}\)
2 \(100 \mathrm{~Hz}\)
3 \(70 \mathrm{~Hz}\)
4 \(25 \mathrm{~Hz}\)
Explanation:
B In full wave rectifier the ripple frequency \(=2\) \(\times\) input frequency \(=2 \times 50\) \(=100 \mathrm{~Hz}\)
Semiconductor Electronics Material Devices and Simple Circuits
150740
Identify the mismatched pair from the following
1 Zener diode : voltage regulator
2 germanium doped : n-type semiconductor with phosphorous
3 semiconductor : band gap \(>3 \mathrm{eV}\)
4 p-n junction diode : rectifier
5 silicon doped with : p-type semiconductor aluminium
Explanation:
C Zener diodes are operated in the breakdown region, voltage across this zener diode always remain constant. We could say zener diode can be used as voltage regulator. Hence, option (a) is correct. Phosphorus being a pentavalent element, bring in free valence electron into germanium on being in it. Hence forming n-type semiconductor. Hence, option (b) is correct. For semiconductor, the band gap is less than three electron volt. Hence, option (c) is incorrect. In p-n junction diode can be used as half-wave and fullwave rectifier. Hence, option (d) is correct. Aluminium being the trivalent impurity forms holes hence forming p-type semiconductor. Hence, option (e) is correct. Hence, option (c) is mismatched pair.
Kerala CEE- 2013
Semiconductor Electronics Material Devices and Simple Circuits
150768
In a full wave rectifier, input voltage is represented by \(V=V_M \sin \omega t\), then peak inversion voltage of non conducting diode will be:
1 \(-\mathrm{V}_{\mathrm{M}}\)
2 \(\mathrm{V}_{\mathrm{M}} / 2\)
3 \(2 \mathrm{~V}_{\mathrm{M}}\)
4 0
Explanation:
C Given, \(\mathrm{V}=\mathrm{V}_{\mathrm{m}} \sin \omega \mathrm{t}\) So, \(\mathrm{V}_{\mathrm{m}}=\) maximum voltage We know that, in case of full wave rectifier the peak inversion voltage (PIV) \(=2 \times\) maximum voltage \(\mathrm{PIV}=2 \mathrm{~V}_{\mathrm{m}}\)
AIIMS-26.05.2018(M)
Semiconductor Electronics Material Devices and Simple Circuits
150772
In the half wave rectifier circuit operating from \(50 \mathrm{~Hz}\) mains frequency, the fundamental frequency in the ripple would be
1 \(25 \mathrm{~Hz}\)
2 \(50 \mathrm{~Hz}\)
3 \(70.7 \mathrm{~Hz}\)
4 \(100 \mathrm{~Hz}\)
Explanation:
B As output voltage obtained in half wave rectifier circuit has variation in 1 cycle of ac voltage. \(\therefore\) Fundamental frequency in ripple of output voltage \(=\) \(50 \mathrm{~Hz}\)
AIIMS-2007
Semiconductor Electronics Material Devices and Simple Circuits
150774
In a full wave rectifier circuit operating from \(50 \mathrm{~Hz}\) mains frequency, what is the fundamental frequency in the ripple?
1 \(50 \mathrm{~Hz}\)
2 \(100 \mathrm{~Hz}\)
3 \(70 \mathrm{~Hz}\)
4 \(25 \mathrm{~Hz}\)
Explanation:
B In full wave rectifier the ripple frequency \(=2\) \(\times\) input frequency \(=2 \times 50\) \(=100 \mathrm{~Hz}\)
