Semiconductor Electronics Material Devices and Simple Circuits
150873
In the circuit shown for the given input voltage \(V_t\), the maximum output voltage \(V_0\) is:
1 \(0 \mathrm{~V}\)
2 \(5 \mathrm{~V}\)
3 \(10 \mathrm{~V}\)
4 \(\frac{5}{\sqrt{2}} \mathrm{~V}\)
Explanation:
B Given, For the half cycle \(D_1\) is reverse bias \& \(D_2\) is forward bias. \(\because\) Maximum voltage drop across \(\mathrm{V}_0\) will be \(\because \quad \mathrm{V}_{\mathrm{i}}=10 \mathrm{~V}\) \(\therefore \quad \mathrm{V}_{\mathrm{o}}=\frac{1}{2} \times 10=5 \mathrm{~V}\) \(\mathrm{V}_0=5 \mathrm{~V}\)
AP EAMCET-23.08.2021
Semiconductor Electronics Material Devices and Simple Circuits
150915
Current flowing in each of the following circuits \(A\) and \(B\) respectively are (neglecting the forward resistance of diodes).
1 \(1 \mathrm{~A}\) and \(2 \mathrm{~A}\)
2 \(2 \mathrm{~A}\) and \(1 \mathrm{~A}\)
3 \(4 \mathrm{~A}\) and \(2 \mathrm{~A}\)
4 \(2 \mathrm{~A}\) and \(4 \mathrm{~A}\)
Explanation:
C In figure A both diode is forward bias so \(4 \Omega\) resistor is in parallel. Then \(\mathrm{R}_{\mathrm{eq}}\) is \(2 \Omega\) Hence, \(\quad \mathrm{I}=\frac{8}{2}\) \(\mathrm{I}=4 \mathrm{~A}\) Similarly, in figure B upper diode is forward bias while lower diode is reverse bias therefore only upper diode is considered- \(\mathrm{I}=\frac{8}{4}=2 \mathrm{~A}\)
AP EAMCET(Medical)-2015]
Semiconductor Electronics Material Devices and Simple Circuits
150917
A full-wave \(p\)-n diode rectifier uses a load resistor of \(1500 \Omega\).No filter is used. The forward bias resistance of the diode is \(10 \Omega\). The efficiency of the rectifier is
1 \(81.2 \%\)
2 \(40.6 \%\)
3 \(80.4 \%\)
4 \(40.2 \%\)
Explanation:
C Given, Load resistance \(\left(\mathrm{R}_{\mathrm{L}}\right)=1500 \Omega\) forward bias resistance \(\left(\mathrm{R}_{\mathrm{f}}\right)=10 \Omega\) \(\eta_{\mathrm{o}}=0.812\) The efficiency of Rectifier \(\eta=\frac{\eta_o}{\left(1+\frac{R_f}{R_L}\right)} \times 100\) \(\eta =\frac{0.812}{\left(1+\frac{10}{1500}\right)} \times 100\) \(\eta =80.4 \%\)
AP EAMCET(Medical)-2009
Semiconductor Electronics Material Devices and Simple Circuits
150921
The two diodes \(A\) and \(B\) are biased as shown, then | $-5 V$ | $A$ | $-9 V$ | | :--- | :--- | :--- | | $-3 V$ | $B$ | $-6 V$ |
1 the diodes A and B are reverse biased
2 the diodes \(\mathrm{A}\) is forward biased and \(\mathrm{B}\) is reverse biased
3 the diodes \(\mathrm{B}\) is forward biased and diode \(\mathrm{A}\) is reverse biased
4 the diodes A and B are forward biased
Explanation:
D The diode A and B are forward biased because both are connected in positive terminal with pside and negative terminal with \(n\)-side.
Semiconductor Electronics Material Devices and Simple Circuits
150873
In the circuit shown for the given input voltage \(V_t\), the maximum output voltage \(V_0\) is:
1 \(0 \mathrm{~V}\)
2 \(5 \mathrm{~V}\)
3 \(10 \mathrm{~V}\)
4 \(\frac{5}{\sqrt{2}} \mathrm{~V}\)
Explanation:
B Given, For the half cycle \(D_1\) is reverse bias \& \(D_2\) is forward bias. \(\because\) Maximum voltage drop across \(\mathrm{V}_0\) will be \(\because \quad \mathrm{V}_{\mathrm{i}}=10 \mathrm{~V}\) \(\therefore \quad \mathrm{V}_{\mathrm{o}}=\frac{1}{2} \times 10=5 \mathrm{~V}\) \(\mathrm{V}_0=5 \mathrm{~V}\)
AP EAMCET-23.08.2021
Semiconductor Electronics Material Devices and Simple Circuits
150915
Current flowing in each of the following circuits \(A\) and \(B\) respectively are (neglecting the forward resistance of diodes).
1 \(1 \mathrm{~A}\) and \(2 \mathrm{~A}\)
2 \(2 \mathrm{~A}\) and \(1 \mathrm{~A}\)
3 \(4 \mathrm{~A}\) and \(2 \mathrm{~A}\)
4 \(2 \mathrm{~A}\) and \(4 \mathrm{~A}\)
Explanation:
C In figure A both diode is forward bias so \(4 \Omega\) resistor is in parallel. Then \(\mathrm{R}_{\mathrm{eq}}\) is \(2 \Omega\) Hence, \(\quad \mathrm{I}=\frac{8}{2}\) \(\mathrm{I}=4 \mathrm{~A}\) Similarly, in figure B upper diode is forward bias while lower diode is reverse bias therefore only upper diode is considered- \(\mathrm{I}=\frac{8}{4}=2 \mathrm{~A}\)
AP EAMCET(Medical)-2015]
Semiconductor Electronics Material Devices and Simple Circuits
150917
A full-wave \(p\)-n diode rectifier uses a load resistor of \(1500 \Omega\).No filter is used. The forward bias resistance of the diode is \(10 \Omega\). The efficiency of the rectifier is
1 \(81.2 \%\)
2 \(40.6 \%\)
3 \(80.4 \%\)
4 \(40.2 \%\)
Explanation:
C Given, Load resistance \(\left(\mathrm{R}_{\mathrm{L}}\right)=1500 \Omega\) forward bias resistance \(\left(\mathrm{R}_{\mathrm{f}}\right)=10 \Omega\) \(\eta_{\mathrm{o}}=0.812\) The efficiency of Rectifier \(\eta=\frac{\eta_o}{\left(1+\frac{R_f}{R_L}\right)} \times 100\) \(\eta =\frac{0.812}{\left(1+\frac{10}{1500}\right)} \times 100\) \(\eta =80.4 \%\)
AP EAMCET(Medical)-2009
Semiconductor Electronics Material Devices and Simple Circuits
150921
The two diodes \(A\) and \(B\) are biased as shown, then | $-5 V$ | $A$ | $-9 V$ | | :--- | :--- | :--- | | $-3 V$ | $B$ | $-6 V$ |
1 the diodes A and B are reverse biased
2 the diodes \(\mathrm{A}\) is forward biased and \(\mathrm{B}\) is reverse biased
3 the diodes \(\mathrm{B}\) is forward biased and diode \(\mathrm{A}\) is reverse biased
4 the diodes A and B are forward biased
Explanation:
D The diode A and B are forward biased because both are connected in positive terminal with pside and negative terminal with \(n\)-side.
Semiconductor Electronics Material Devices and Simple Circuits
150873
In the circuit shown for the given input voltage \(V_t\), the maximum output voltage \(V_0\) is:
1 \(0 \mathrm{~V}\)
2 \(5 \mathrm{~V}\)
3 \(10 \mathrm{~V}\)
4 \(\frac{5}{\sqrt{2}} \mathrm{~V}\)
Explanation:
B Given, For the half cycle \(D_1\) is reverse bias \& \(D_2\) is forward bias. \(\because\) Maximum voltage drop across \(\mathrm{V}_0\) will be \(\because \quad \mathrm{V}_{\mathrm{i}}=10 \mathrm{~V}\) \(\therefore \quad \mathrm{V}_{\mathrm{o}}=\frac{1}{2} \times 10=5 \mathrm{~V}\) \(\mathrm{V}_0=5 \mathrm{~V}\)
AP EAMCET-23.08.2021
Semiconductor Electronics Material Devices and Simple Circuits
150915
Current flowing in each of the following circuits \(A\) and \(B\) respectively are (neglecting the forward resistance of diodes).
1 \(1 \mathrm{~A}\) and \(2 \mathrm{~A}\)
2 \(2 \mathrm{~A}\) and \(1 \mathrm{~A}\)
3 \(4 \mathrm{~A}\) and \(2 \mathrm{~A}\)
4 \(2 \mathrm{~A}\) and \(4 \mathrm{~A}\)
Explanation:
C In figure A both diode is forward bias so \(4 \Omega\) resistor is in parallel. Then \(\mathrm{R}_{\mathrm{eq}}\) is \(2 \Omega\) Hence, \(\quad \mathrm{I}=\frac{8}{2}\) \(\mathrm{I}=4 \mathrm{~A}\) Similarly, in figure B upper diode is forward bias while lower diode is reverse bias therefore only upper diode is considered- \(\mathrm{I}=\frac{8}{4}=2 \mathrm{~A}\)
AP EAMCET(Medical)-2015]
Semiconductor Electronics Material Devices and Simple Circuits
150917
A full-wave \(p\)-n diode rectifier uses a load resistor of \(1500 \Omega\).No filter is used. The forward bias resistance of the diode is \(10 \Omega\). The efficiency of the rectifier is
1 \(81.2 \%\)
2 \(40.6 \%\)
3 \(80.4 \%\)
4 \(40.2 \%\)
Explanation:
C Given, Load resistance \(\left(\mathrm{R}_{\mathrm{L}}\right)=1500 \Omega\) forward bias resistance \(\left(\mathrm{R}_{\mathrm{f}}\right)=10 \Omega\) \(\eta_{\mathrm{o}}=0.812\) The efficiency of Rectifier \(\eta=\frac{\eta_o}{\left(1+\frac{R_f}{R_L}\right)} \times 100\) \(\eta =\frac{0.812}{\left(1+\frac{10}{1500}\right)} \times 100\) \(\eta =80.4 \%\)
AP EAMCET(Medical)-2009
Semiconductor Electronics Material Devices and Simple Circuits
150921
The two diodes \(A\) and \(B\) are biased as shown, then | $-5 V$ | $A$ | $-9 V$ | | :--- | :--- | :--- | | $-3 V$ | $B$ | $-6 V$ |
1 the diodes A and B are reverse biased
2 the diodes \(\mathrm{A}\) is forward biased and \(\mathrm{B}\) is reverse biased
3 the diodes \(\mathrm{B}\) is forward biased and diode \(\mathrm{A}\) is reverse biased
4 the diodes A and B are forward biased
Explanation:
D The diode A and B are forward biased because both are connected in positive terminal with pside and negative terminal with \(n\)-side.
Semiconductor Electronics Material Devices and Simple Circuits
150873
In the circuit shown for the given input voltage \(V_t\), the maximum output voltage \(V_0\) is:
1 \(0 \mathrm{~V}\)
2 \(5 \mathrm{~V}\)
3 \(10 \mathrm{~V}\)
4 \(\frac{5}{\sqrt{2}} \mathrm{~V}\)
Explanation:
B Given, For the half cycle \(D_1\) is reverse bias \& \(D_2\) is forward bias. \(\because\) Maximum voltage drop across \(\mathrm{V}_0\) will be \(\because \quad \mathrm{V}_{\mathrm{i}}=10 \mathrm{~V}\) \(\therefore \quad \mathrm{V}_{\mathrm{o}}=\frac{1}{2} \times 10=5 \mathrm{~V}\) \(\mathrm{V}_0=5 \mathrm{~V}\)
AP EAMCET-23.08.2021
Semiconductor Electronics Material Devices and Simple Circuits
150915
Current flowing in each of the following circuits \(A\) and \(B\) respectively are (neglecting the forward resistance of diodes).
1 \(1 \mathrm{~A}\) and \(2 \mathrm{~A}\)
2 \(2 \mathrm{~A}\) and \(1 \mathrm{~A}\)
3 \(4 \mathrm{~A}\) and \(2 \mathrm{~A}\)
4 \(2 \mathrm{~A}\) and \(4 \mathrm{~A}\)
Explanation:
C In figure A both diode is forward bias so \(4 \Omega\) resistor is in parallel. Then \(\mathrm{R}_{\mathrm{eq}}\) is \(2 \Omega\) Hence, \(\quad \mathrm{I}=\frac{8}{2}\) \(\mathrm{I}=4 \mathrm{~A}\) Similarly, in figure B upper diode is forward bias while lower diode is reverse bias therefore only upper diode is considered- \(\mathrm{I}=\frac{8}{4}=2 \mathrm{~A}\)
AP EAMCET(Medical)-2015]
Semiconductor Electronics Material Devices and Simple Circuits
150917
A full-wave \(p\)-n diode rectifier uses a load resistor of \(1500 \Omega\).No filter is used. The forward bias resistance of the diode is \(10 \Omega\). The efficiency of the rectifier is
1 \(81.2 \%\)
2 \(40.6 \%\)
3 \(80.4 \%\)
4 \(40.2 \%\)
Explanation:
C Given, Load resistance \(\left(\mathrm{R}_{\mathrm{L}}\right)=1500 \Omega\) forward bias resistance \(\left(\mathrm{R}_{\mathrm{f}}\right)=10 \Omega\) \(\eta_{\mathrm{o}}=0.812\) The efficiency of Rectifier \(\eta=\frac{\eta_o}{\left(1+\frac{R_f}{R_L}\right)} \times 100\) \(\eta =\frac{0.812}{\left(1+\frac{10}{1500}\right)} \times 100\) \(\eta =80.4 \%\)
AP EAMCET(Medical)-2009
Semiconductor Electronics Material Devices and Simple Circuits
150921
The two diodes \(A\) and \(B\) are biased as shown, then | $-5 V$ | $A$ | $-9 V$ | | :--- | :--- | :--- | | $-3 V$ | $B$ | $-6 V$ |
1 the diodes A and B are reverse biased
2 the diodes \(\mathrm{A}\) is forward biased and \(\mathrm{B}\) is reverse biased
3 the diodes \(\mathrm{B}\) is forward biased and diode \(\mathrm{A}\) is reverse biased
4 the diodes A and B are forward biased
Explanation:
D The diode A and B are forward biased because both are connected in positive terminal with pside and negative terminal with \(n\)-side.
