Semiconductor Electronics Material Devices and Simple Circuits
151059
The output resistance of a common emitter transistor amplifier, if the input resistance is \(\mathbf{2 0 0} \Omega\left(\alpha=0.98\right.\) and power gain is \(\left.5 \times 10^6\right)\) is.
1 \(516 \mathrm{k} \Omega\)
2 \(216 \mathrm{k} \Omega\)
3 \(300 \mathrm{k} \Omega\)
4 \(416 \mathrm{k} \Omega\)
Explanation:
D Given, \(\mathrm{R}_{\text {in }}=200 \Omega\) \(\alpha=0.98\) Power gain \(=5 \times 10^6\) \(\because\) Current gain, \(\beta=\frac{\alpha}{1-\alpha}=\frac{0.98}{1-0.98}=49\) \(\because\) Voltage gain, \(A_v=\beta \times \frac{R_{\text {out }}}{R_{\text {in }}}\) \(A_v=49 \times \frac{R_{\text {out }}}{R_{\text {in }}}\) \(\therefore\) Power gain \(=\) Current gain \(\times\) Voltage gain \(5 \times 10^6=49 \times\left(49 \times \frac{\mathrm{R}_{\text {out }}}{200}\right)\) \(\mathrm{R}_{\text {out }}=\frac{5 \times 10^6 \times 200}{49 \times 49}=416.49 \times 10^3 \Omega\) \(\mathrm{R}_{\text {out }}=416 \mathrm{k} \Omega\)
JCECE-2015
Semiconductor Electronics Material Devices and Simple Circuits
151060
Two triodes having amplification factors 30 and 21 and plate resistance \(5 \mathrm{k} \Omega\) and 4 \(\mathrm{k} \Omega\) respectively are connected in parallel. The composite amplification factor of the system is
Semiconductor Electronics Material Devices and Simple Circuits
151061
For a transistor in common base, the current gain is 0.95 . If the load resistance is \(400 \mathrm{k} \Omega\) and input resistance is \(200 \Omega\), then the voltage gain and power gain will be
1 1900 and 1800
2 1900 and 1805
3 5525 and 3591
4 1805 and 1900
Explanation:
B Given, \(\boldsymbol{\alpha}=0.95\) \(\mathrm{R}_{\mathrm{L}}=400 \mathrm{k} \Omega, \mathrm{R}_{\mathrm{in}}=200 \Omega\) \(\because\) Voltage gain \(=\alpha \times \frac{R_L}{R_{\text {in }}}\) \(=0.95 \times \frac{400 \times 10^3}{200}\) \(=1900\) And, Power gain \(=\) voltage gain \(\times\) current gain \(=1900 \times 0.95\) \(=1805\)
JCECE-2013
Semiconductor Electronics Material Devices and Simple Circuits
151063
An amplifier has a voltage gain \(A_v=1000\). The voltage gain in \(\mathrm{dB}\) is
1 \(30 \mathrm{~dB}\)
2 \(60 \mathrm{~dB}\)
3 \(3 \mathrm{~dB}\)
4 \(20 \mathrm{~dB}\)
Explanation:
B Given, \(A_V=1000\) Voltage gain in \(\mathrm{dB}=20 \log _{10} \mathrm{~A}_{\mathrm{v}}\) \(=20 \log _{10} 1000\) \(=20 \log _{10}(10)^3\) \(=60 \log _{10} 10\) \(=60 \mathrm{~dB}\)
JCECE-2007
Semiconductor Electronics Material Devices and Simple Circuits
151064
For a transistor \(I_E=25 \mathrm{~mA}\) and \(I_B=1 \mathrm{~mA}\), the value of current gain \(\alpha\) will be :
Semiconductor Electronics Material Devices and Simple Circuits
151059
The output resistance of a common emitter transistor amplifier, if the input resistance is \(\mathbf{2 0 0} \Omega\left(\alpha=0.98\right.\) and power gain is \(\left.5 \times 10^6\right)\) is.
1 \(516 \mathrm{k} \Omega\)
2 \(216 \mathrm{k} \Omega\)
3 \(300 \mathrm{k} \Omega\)
4 \(416 \mathrm{k} \Omega\)
Explanation:
D Given, \(\mathrm{R}_{\text {in }}=200 \Omega\) \(\alpha=0.98\) Power gain \(=5 \times 10^6\) \(\because\) Current gain, \(\beta=\frac{\alpha}{1-\alpha}=\frac{0.98}{1-0.98}=49\) \(\because\) Voltage gain, \(A_v=\beta \times \frac{R_{\text {out }}}{R_{\text {in }}}\) \(A_v=49 \times \frac{R_{\text {out }}}{R_{\text {in }}}\) \(\therefore\) Power gain \(=\) Current gain \(\times\) Voltage gain \(5 \times 10^6=49 \times\left(49 \times \frac{\mathrm{R}_{\text {out }}}{200}\right)\) \(\mathrm{R}_{\text {out }}=\frac{5 \times 10^6 \times 200}{49 \times 49}=416.49 \times 10^3 \Omega\) \(\mathrm{R}_{\text {out }}=416 \mathrm{k} \Omega\)
JCECE-2015
Semiconductor Electronics Material Devices and Simple Circuits
151060
Two triodes having amplification factors 30 and 21 and plate resistance \(5 \mathrm{k} \Omega\) and 4 \(\mathrm{k} \Omega\) respectively are connected in parallel. The composite amplification factor of the system is
Semiconductor Electronics Material Devices and Simple Circuits
151061
For a transistor in common base, the current gain is 0.95 . If the load resistance is \(400 \mathrm{k} \Omega\) and input resistance is \(200 \Omega\), then the voltage gain and power gain will be
1 1900 and 1800
2 1900 and 1805
3 5525 and 3591
4 1805 and 1900
Explanation:
B Given, \(\boldsymbol{\alpha}=0.95\) \(\mathrm{R}_{\mathrm{L}}=400 \mathrm{k} \Omega, \mathrm{R}_{\mathrm{in}}=200 \Omega\) \(\because\) Voltage gain \(=\alpha \times \frac{R_L}{R_{\text {in }}}\) \(=0.95 \times \frac{400 \times 10^3}{200}\) \(=1900\) And, Power gain \(=\) voltage gain \(\times\) current gain \(=1900 \times 0.95\) \(=1805\)
JCECE-2013
Semiconductor Electronics Material Devices and Simple Circuits
151063
An amplifier has a voltage gain \(A_v=1000\). The voltage gain in \(\mathrm{dB}\) is
1 \(30 \mathrm{~dB}\)
2 \(60 \mathrm{~dB}\)
3 \(3 \mathrm{~dB}\)
4 \(20 \mathrm{~dB}\)
Explanation:
B Given, \(A_V=1000\) Voltage gain in \(\mathrm{dB}=20 \log _{10} \mathrm{~A}_{\mathrm{v}}\) \(=20 \log _{10} 1000\) \(=20 \log _{10}(10)^3\) \(=60 \log _{10} 10\) \(=60 \mathrm{~dB}\)
JCECE-2007
Semiconductor Electronics Material Devices and Simple Circuits
151064
For a transistor \(I_E=25 \mathrm{~mA}\) and \(I_B=1 \mathrm{~mA}\), the value of current gain \(\alpha\) will be :
Semiconductor Electronics Material Devices and Simple Circuits
151059
The output resistance of a common emitter transistor amplifier, if the input resistance is \(\mathbf{2 0 0} \Omega\left(\alpha=0.98\right.\) and power gain is \(\left.5 \times 10^6\right)\) is.
1 \(516 \mathrm{k} \Omega\)
2 \(216 \mathrm{k} \Omega\)
3 \(300 \mathrm{k} \Omega\)
4 \(416 \mathrm{k} \Omega\)
Explanation:
D Given, \(\mathrm{R}_{\text {in }}=200 \Omega\) \(\alpha=0.98\) Power gain \(=5 \times 10^6\) \(\because\) Current gain, \(\beta=\frac{\alpha}{1-\alpha}=\frac{0.98}{1-0.98}=49\) \(\because\) Voltage gain, \(A_v=\beta \times \frac{R_{\text {out }}}{R_{\text {in }}}\) \(A_v=49 \times \frac{R_{\text {out }}}{R_{\text {in }}}\) \(\therefore\) Power gain \(=\) Current gain \(\times\) Voltage gain \(5 \times 10^6=49 \times\left(49 \times \frac{\mathrm{R}_{\text {out }}}{200}\right)\) \(\mathrm{R}_{\text {out }}=\frac{5 \times 10^6 \times 200}{49 \times 49}=416.49 \times 10^3 \Omega\) \(\mathrm{R}_{\text {out }}=416 \mathrm{k} \Omega\)
JCECE-2015
Semiconductor Electronics Material Devices and Simple Circuits
151060
Two triodes having amplification factors 30 and 21 and plate resistance \(5 \mathrm{k} \Omega\) and 4 \(\mathrm{k} \Omega\) respectively are connected in parallel. The composite amplification factor of the system is
Semiconductor Electronics Material Devices and Simple Circuits
151061
For a transistor in common base, the current gain is 0.95 . If the load resistance is \(400 \mathrm{k} \Omega\) and input resistance is \(200 \Omega\), then the voltage gain and power gain will be
1 1900 and 1800
2 1900 and 1805
3 5525 and 3591
4 1805 and 1900
Explanation:
B Given, \(\boldsymbol{\alpha}=0.95\) \(\mathrm{R}_{\mathrm{L}}=400 \mathrm{k} \Omega, \mathrm{R}_{\mathrm{in}}=200 \Omega\) \(\because\) Voltage gain \(=\alpha \times \frac{R_L}{R_{\text {in }}}\) \(=0.95 \times \frac{400 \times 10^3}{200}\) \(=1900\) And, Power gain \(=\) voltage gain \(\times\) current gain \(=1900 \times 0.95\) \(=1805\)
JCECE-2013
Semiconductor Electronics Material Devices and Simple Circuits
151063
An amplifier has a voltage gain \(A_v=1000\). The voltage gain in \(\mathrm{dB}\) is
1 \(30 \mathrm{~dB}\)
2 \(60 \mathrm{~dB}\)
3 \(3 \mathrm{~dB}\)
4 \(20 \mathrm{~dB}\)
Explanation:
B Given, \(A_V=1000\) Voltage gain in \(\mathrm{dB}=20 \log _{10} \mathrm{~A}_{\mathrm{v}}\) \(=20 \log _{10} 1000\) \(=20 \log _{10}(10)^3\) \(=60 \log _{10} 10\) \(=60 \mathrm{~dB}\)
JCECE-2007
Semiconductor Electronics Material Devices and Simple Circuits
151064
For a transistor \(I_E=25 \mathrm{~mA}\) and \(I_B=1 \mathrm{~mA}\), the value of current gain \(\alpha\) will be :
NEET Test Series from KOTA - 10 Papers In MS WORD
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Semiconductor Electronics Material Devices and Simple Circuits
151059
The output resistance of a common emitter transistor amplifier, if the input resistance is \(\mathbf{2 0 0} \Omega\left(\alpha=0.98\right.\) and power gain is \(\left.5 \times 10^6\right)\) is.
1 \(516 \mathrm{k} \Omega\)
2 \(216 \mathrm{k} \Omega\)
3 \(300 \mathrm{k} \Omega\)
4 \(416 \mathrm{k} \Omega\)
Explanation:
D Given, \(\mathrm{R}_{\text {in }}=200 \Omega\) \(\alpha=0.98\) Power gain \(=5 \times 10^6\) \(\because\) Current gain, \(\beta=\frac{\alpha}{1-\alpha}=\frac{0.98}{1-0.98}=49\) \(\because\) Voltage gain, \(A_v=\beta \times \frac{R_{\text {out }}}{R_{\text {in }}}\) \(A_v=49 \times \frac{R_{\text {out }}}{R_{\text {in }}}\) \(\therefore\) Power gain \(=\) Current gain \(\times\) Voltage gain \(5 \times 10^6=49 \times\left(49 \times \frac{\mathrm{R}_{\text {out }}}{200}\right)\) \(\mathrm{R}_{\text {out }}=\frac{5 \times 10^6 \times 200}{49 \times 49}=416.49 \times 10^3 \Omega\) \(\mathrm{R}_{\text {out }}=416 \mathrm{k} \Omega\)
JCECE-2015
Semiconductor Electronics Material Devices and Simple Circuits
151060
Two triodes having amplification factors 30 and 21 and plate resistance \(5 \mathrm{k} \Omega\) and 4 \(\mathrm{k} \Omega\) respectively are connected in parallel. The composite amplification factor of the system is
Semiconductor Electronics Material Devices and Simple Circuits
151061
For a transistor in common base, the current gain is 0.95 . If the load resistance is \(400 \mathrm{k} \Omega\) and input resistance is \(200 \Omega\), then the voltage gain and power gain will be
1 1900 and 1800
2 1900 and 1805
3 5525 and 3591
4 1805 and 1900
Explanation:
B Given, \(\boldsymbol{\alpha}=0.95\) \(\mathrm{R}_{\mathrm{L}}=400 \mathrm{k} \Omega, \mathrm{R}_{\mathrm{in}}=200 \Omega\) \(\because\) Voltage gain \(=\alpha \times \frac{R_L}{R_{\text {in }}}\) \(=0.95 \times \frac{400 \times 10^3}{200}\) \(=1900\) And, Power gain \(=\) voltage gain \(\times\) current gain \(=1900 \times 0.95\) \(=1805\)
JCECE-2013
Semiconductor Electronics Material Devices and Simple Circuits
151063
An amplifier has a voltage gain \(A_v=1000\). The voltage gain in \(\mathrm{dB}\) is
1 \(30 \mathrm{~dB}\)
2 \(60 \mathrm{~dB}\)
3 \(3 \mathrm{~dB}\)
4 \(20 \mathrm{~dB}\)
Explanation:
B Given, \(A_V=1000\) Voltage gain in \(\mathrm{dB}=20 \log _{10} \mathrm{~A}_{\mathrm{v}}\) \(=20 \log _{10} 1000\) \(=20 \log _{10}(10)^3\) \(=60 \log _{10} 10\) \(=60 \mathrm{~dB}\)
JCECE-2007
Semiconductor Electronics Material Devices and Simple Circuits
151064
For a transistor \(I_E=25 \mathrm{~mA}\) and \(I_B=1 \mathrm{~mA}\), the value of current gain \(\alpha\) will be :
Semiconductor Electronics Material Devices and Simple Circuits
151059
The output resistance of a common emitter transistor amplifier, if the input resistance is \(\mathbf{2 0 0} \Omega\left(\alpha=0.98\right.\) and power gain is \(\left.5 \times 10^6\right)\) is.
1 \(516 \mathrm{k} \Omega\)
2 \(216 \mathrm{k} \Omega\)
3 \(300 \mathrm{k} \Omega\)
4 \(416 \mathrm{k} \Omega\)
Explanation:
D Given, \(\mathrm{R}_{\text {in }}=200 \Omega\) \(\alpha=0.98\) Power gain \(=5 \times 10^6\) \(\because\) Current gain, \(\beta=\frac{\alpha}{1-\alpha}=\frac{0.98}{1-0.98}=49\) \(\because\) Voltage gain, \(A_v=\beta \times \frac{R_{\text {out }}}{R_{\text {in }}}\) \(A_v=49 \times \frac{R_{\text {out }}}{R_{\text {in }}}\) \(\therefore\) Power gain \(=\) Current gain \(\times\) Voltage gain \(5 \times 10^6=49 \times\left(49 \times \frac{\mathrm{R}_{\text {out }}}{200}\right)\) \(\mathrm{R}_{\text {out }}=\frac{5 \times 10^6 \times 200}{49 \times 49}=416.49 \times 10^3 \Omega\) \(\mathrm{R}_{\text {out }}=416 \mathrm{k} \Omega\)
JCECE-2015
Semiconductor Electronics Material Devices and Simple Circuits
151060
Two triodes having amplification factors 30 and 21 and plate resistance \(5 \mathrm{k} \Omega\) and 4 \(\mathrm{k} \Omega\) respectively are connected in parallel. The composite amplification factor of the system is
Semiconductor Electronics Material Devices and Simple Circuits
151061
For a transistor in common base, the current gain is 0.95 . If the load resistance is \(400 \mathrm{k} \Omega\) and input resistance is \(200 \Omega\), then the voltage gain and power gain will be
1 1900 and 1800
2 1900 and 1805
3 5525 and 3591
4 1805 and 1900
Explanation:
B Given, \(\boldsymbol{\alpha}=0.95\) \(\mathrm{R}_{\mathrm{L}}=400 \mathrm{k} \Omega, \mathrm{R}_{\mathrm{in}}=200 \Omega\) \(\because\) Voltage gain \(=\alpha \times \frac{R_L}{R_{\text {in }}}\) \(=0.95 \times \frac{400 \times 10^3}{200}\) \(=1900\) And, Power gain \(=\) voltage gain \(\times\) current gain \(=1900 \times 0.95\) \(=1805\)
JCECE-2013
Semiconductor Electronics Material Devices and Simple Circuits
151063
An amplifier has a voltage gain \(A_v=1000\). The voltage gain in \(\mathrm{dB}\) is
1 \(30 \mathrm{~dB}\)
2 \(60 \mathrm{~dB}\)
3 \(3 \mathrm{~dB}\)
4 \(20 \mathrm{~dB}\)
Explanation:
B Given, \(A_V=1000\) Voltage gain in \(\mathrm{dB}=20 \log _{10} \mathrm{~A}_{\mathrm{v}}\) \(=20 \log _{10} 1000\) \(=20 \log _{10}(10)^3\) \(=60 \log _{10} 10\) \(=60 \mathrm{~dB}\)
JCECE-2007
Semiconductor Electronics Material Devices and Simple Circuits
151064
For a transistor \(I_E=25 \mathrm{~mA}\) and \(I_B=1 \mathrm{~mA}\), the value of current gain \(\alpha\) will be :