PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
365478
The current gain of transistor in common emitter mode is 49 . The change in collector current and emitter currect corresponding to change in base current by \(5.0\,\mu A\), will be
PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
365479
If \(\beta ,{R_L}\) and \(r\) are the \(AC\) current gain, load resistance and the input resistance of a transistor, respectively in \(CE\) configuration, the voltage and the power gains respectively are
1 \(\beta \frac{{{R_L}}}{r}\) and \({\beta ^2}\frac{{{R_L}}}{r}\)
2 \(\beta \frac{r}{{{R_L}}}\) and \({\beta ^2}\frac{r}{{{R_L}}}\)
3 \(\beta \frac{{{R_L}}}{r}\) and \(\beta {\left( {\frac{{{R_L}}}{r}} \right)^2}\)
4 \(\beta \frac{r}{{{R_L}}}\) and \(\beta {\left( {\frac{r}{{{R_L}}}} \right)^2}\)
Explanation:
Transistor as a common emitter amplifier, Voltage gain, \({A_V} = \frac{{\Delta {V_o}}}{{\Delta {V_i}}}\) \(\,\,\,\,\,\,\, = {\beta _{{\rm{AC}}}} \times \) Resistance gain \( = \beta \times \frac{{{R_L}}}{r}\) Power gain \( = \frac{{\Delta {P_0}}}{{\Delta {P_i}}}\) \( = \beta _{{\text{AC}}}^2 \times \) Resistance gain \( = {\beta ^2} \times \frac{{{R_L}}}{r}\)
PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
365480
For CE configuration \(n-p-n\) transistor. Which of the following statement is correct?
1 \(I_{C}=I_{E}+I_{B}\)
2 \(I_{B}=I_{E}+I_{C}\)
3 \(I_{E}=I_{C}+I_{B}\)
4 All of these
Explanation:
In \(CE\) configuration \(n-p-n\) transistor, emitter current \(\left(I_{E}\right)\) is equal to sum of collector current \(\left(I_{C}\right)\) and base current \(\left(I_{B}\right)\), i.e \(I_{E}=I_{C}+I_{B}\)
PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
365481
A \(p-n-p\) transistor having \(AC\) current gain of 50 is used to make an amplifier of a voltage gain of 5 . What will be the power gain of the amplifier?
1 125
2 178
3 250
4 354
Explanation:
For the amplifier, Power gain \(=\) Current gain \(\times\) Voltage gain \(=50 \times 5=250\)
PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
365478
The current gain of transistor in common emitter mode is 49 . The change in collector current and emitter currect corresponding to change in base current by \(5.0\,\mu A\), will be
PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
365479
If \(\beta ,{R_L}\) and \(r\) are the \(AC\) current gain, load resistance and the input resistance of a transistor, respectively in \(CE\) configuration, the voltage and the power gains respectively are
1 \(\beta \frac{{{R_L}}}{r}\) and \({\beta ^2}\frac{{{R_L}}}{r}\)
2 \(\beta \frac{r}{{{R_L}}}\) and \({\beta ^2}\frac{r}{{{R_L}}}\)
3 \(\beta \frac{{{R_L}}}{r}\) and \(\beta {\left( {\frac{{{R_L}}}{r}} \right)^2}\)
4 \(\beta \frac{r}{{{R_L}}}\) and \(\beta {\left( {\frac{r}{{{R_L}}}} \right)^2}\)
Explanation:
Transistor as a common emitter amplifier, Voltage gain, \({A_V} = \frac{{\Delta {V_o}}}{{\Delta {V_i}}}\) \(\,\,\,\,\,\,\, = {\beta _{{\rm{AC}}}} \times \) Resistance gain \( = \beta \times \frac{{{R_L}}}{r}\) Power gain \( = \frac{{\Delta {P_0}}}{{\Delta {P_i}}}\) \( = \beta _{{\text{AC}}}^2 \times \) Resistance gain \( = {\beta ^2} \times \frac{{{R_L}}}{r}\)
PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
365480
For CE configuration \(n-p-n\) transistor. Which of the following statement is correct?
1 \(I_{C}=I_{E}+I_{B}\)
2 \(I_{B}=I_{E}+I_{C}\)
3 \(I_{E}=I_{C}+I_{B}\)
4 All of these
Explanation:
In \(CE\) configuration \(n-p-n\) transistor, emitter current \(\left(I_{E}\right)\) is equal to sum of collector current \(\left(I_{C}\right)\) and base current \(\left(I_{B}\right)\), i.e \(I_{E}=I_{C}+I_{B}\)
PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
365481
A \(p-n-p\) transistor having \(AC\) current gain of 50 is used to make an amplifier of a voltage gain of 5 . What will be the power gain of the amplifier?
1 125
2 178
3 250
4 354
Explanation:
For the amplifier, Power gain \(=\) Current gain \(\times\) Voltage gain \(=50 \times 5=250\)
PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
365478
The current gain of transistor in common emitter mode is 49 . The change in collector current and emitter currect corresponding to change in base current by \(5.0\,\mu A\), will be
PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
365479
If \(\beta ,{R_L}\) and \(r\) are the \(AC\) current gain, load resistance and the input resistance of a transistor, respectively in \(CE\) configuration, the voltage and the power gains respectively are
1 \(\beta \frac{{{R_L}}}{r}\) and \({\beta ^2}\frac{{{R_L}}}{r}\)
2 \(\beta \frac{r}{{{R_L}}}\) and \({\beta ^2}\frac{r}{{{R_L}}}\)
3 \(\beta \frac{{{R_L}}}{r}\) and \(\beta {\left( {\frac{{{R_L}}}{r}} \right)^2}\)
4 \(\beta \frac{r}{{{R_L}}}\) and \(\beta {\left( {\frac{r}{{{R_L}}}} \right)^2}\)
Explanation:
Transistor as a common emitter amplifier, Voltage gain, \({A_V} = \frac{{\Delta {V_o}}}{{\Delta {V_i}}}\) \(\,\,\,\,\,\,\, = {\beta _{{\rm{AC}}}} \times \) Resistance gain \( = \beta \times \frac{{{R_L}}}{r}\) Power gain \( = \frac{{\Delta {P_0}}}{{\Delta {P_i}}}\) \( = \beta _{{\text{AC}}}^2 \times \) Resistance gain \( = {\beta ^2} \times \frac{{{R_L}}}{r}\)
PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
365480
For CE configuration \(n-p-n\) transistor. Which of the following statement is correct?
1 \(I_{C}=I_{E}+I_{B}\)
2 \(I_{B}=I_{E}+I_{C}\)
3 \(I_{E}=I_{C}+I_{B}\)
4 All of these
Explanation:
In \(CE\) configuration \(n-p-n\) transistor, emitter current \(\left(I_{E}\right)\) is equal to sum of collector current \(\left(I_{C}\right)\) and base current \(\left(I_{B}\right)\), i.e \(I_{E}=I_{C}+I_{B}\)
PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
365481
A \(p-n-p\) transistor having \(AC\) current gain of 50 is used to make an amplifier of a voltage gain of 5 . What will be the power gain of the amplifier?
1 125
2 178
3 250
4 354
Explanation:
For the amplifier, Power gain \(=\) Current gain \(\times\) Voltage gain \(=50 \times 5=250\)
PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
365478
The current gain of transistor in common emitter mode is 49 . The change in collector current and emitter currect corresponding to change in base current by \(5.0\,\mu A\), will be
PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
365479
If \(\beta ,{R_L}\) and \(r\) are the \(AC\) current gain, load resistance and the input resistance of a transistor, respectively in \(CE\) configuration, the voltage and the power gains respectively are
1 \(\beta \frac{{{R_L}}}{r}\) and \({\beta ^2}\frac{{{R_L}}}{r}\)
2 \(\beta \frac{r}{{{R_L}}}\) and \({\beta ^2}\frac{r}{{{R_L}}}\)
3 \(\beta \frac{{{R_L}}}{r}\) and \(\beta {\left( {\frac{{{R_L}}}{r}} \right)^2}\)
4 \(\beta \frac{r}{{{R_L}}}\) and \(\beta {\left( {\frac{r}{{{R_L}}}} \right)^2}\)
Explanation:
Transistor as a common emitter amplifier, Voltage gain, \({A_V} = \frac{{\Delta {V_o}}}{{\Delta {V_i}}}\) \(\,\,\,\,\,\,\, = {\beta _{{\rm{AC}}}} \times \) Resistance gain \( = \beta \times \frac{{{R_L}}}{r}\) Power gain \( = \frac{{\Delta {P_0}}}{{\Delta {P_i}}}\) \( = \beta _{{\text{AC}}}^2 \times \) Resistance gain \( = {\beta ^2} \times \frac{{{R_L}}}{r}\)
PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
365480
For CE configuration \(n-p-n\) transistor. Which of the following statement is correct?
1 \(I_{C}=I_{E}+I_{B}\)
2 \(I_{B}=I_{E}+I_{C}\)
3 \(I_{E}=I_{C}+I_{B}\)
4 All of these
Explanation:
In \(CE\) configuration \(n-p-n\) transistor, emitter current \(\left(I_{E}\right)\) is equal to sum of collector current \(\left(I_{C}\right)\) and base current \(\left(I_{B}\right)\), i.e \(I_{E}=I_{C}+I_{B}\)
PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
365481
A \(p-n-p\) transistor having \(AC\) current gain of 50 is used to make an amplifier of a voltage gain of 5 . What will be the power gain of the amplifier?
1 125
2 178
3 250
4 354
Explanation:
For the amplifier, Power gain \(=\) Current gain \(\times\) Voltage gain \(=50 \times 5=250\)
PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
365478
The current gain of transistor in common emitter mode is 49 . The change in collector current and emitter currect corresponding to change in base current by \(5.0\,\mu A\), will be
PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
365479
If \(\beta ,{R_L}\) and \(r\) are the \(AC\) current gain, load resistance and the input resistance of a transistor, respectively in \(CE\) configuration, the voltage and the power gains respectively are
1 \(\beta \frac{{{R_L}}}{r}\) and \({\beta ^2}\frac{{{R_L}}}{r}\)
2 \(\beta \frac{r}{{{R_L}}}\) and \({\beta ^2}\frac{r}{{{R_L}}}\)
3 \(\beta \frac{{{R_L}}}{r}\) and \(\beta {\left( {\frac{{{R_L}}}{r}} \right)^2}\)
4 \(\beta \frac{r}{{{R_L}}}\) and \(\beta {\left( {\frac{r}{{{R_L}}}} \right)^2}\)
Explanation:
Transistor as a common emitter amplifier, Voltage gain, \({A_V} = \frac{{\Delta {V_o}}}{{\Delta {V_i}}}\) \(\,\,\,\,\,\,\, = {\beta _{{\rm{AC}}}} \times \) Resistance gain \( = \beta \times \frac{{{R_L}}}{r}\) Power gain \( = \frac{{\Delta {P_0}}}{{\Delta {P_i}}}\) \( = \beta _{{\text{AC}}}^2 \times \) Resistance gain \( = {\beta ^2} \times \frac{{{R_L}}}{r}\)
PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
365480
For CE configuration \(n-p-n\) transistor. Which of the following statement is correct?
1 \(I_{C}=I_{E}+I_{B}\)
2 \(I_{B}=I_{E}+I_{C}\)
3 \(I_{E}=I_{C}+I_{B}\)
4 All of these
Explanation:
In \(CE\) configuration \(n-p-n\) transistor, emitter current \(\left(I_{E}\right)\) is equal to sum of collector current \(\left(I_{C}\right)\) and base current \(\left(I_{B}\right)\), i.e \(I_{E}=I_{C}+I_{B}\)
PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
365481
A \(p-n-p\) transistor having \(AC\) current gain of 50 is used to make an amplifier of a voltage gain of 5 . What will be the power gain of the amplifier?
1 125
2 178
3 250
4 354
Explanation:
For the amplifier, Power gain \(=\) Current gain \(\times\) Voltage gain \(=50 \times 5=250\)