Question 1 |
For an ideal MOSFET biased in saturation, the magnitude of the small signal
current gain for a common drain amplifier is
0 | |
1 | |
100 | |
infinite |
Question 1 Explanation:
For ideal MOSFET, i_G=0
Therefore, Current gain, A_I=\frac{i_s}{i_G}=\infty
Therefore, Current gain, A_I=\frac{i_s}{i_G}=\infty
Question 2 |
The magnitude of the mid-band voltage gain of the circuit shown in figure is
(assuming h_{fe} of the transistor to be 100)
1 | |
10 | |
20 | |
100 |
Question 2 Explanation:
AC model,
\; \; \; \; Z_{i}=10\, k\Omega
Mid band voltage gain \; \; \; \; A_{v}=\frac{A_{I}R_{L}}{Z_{i}}
\; \; \; \; A_{v}=\frac{-h_{fe}R_{L}}{Z_{i}} =\frac{-100\times 10\, k\Omega }{10\, k\Omega }=-100
\; \; \; \; \left | A_{v} \right |=100
\; \; \; \; Z_{i}=10\, k\Omega
Mid band voltage gain \; \; \; \; A_{v}=\frac{A_{I}R_{L}}{Z_{i}}
\; \; \; \; A_{v}=\frac{-h_{fe}R_{L}}{Z_{i}} =\frac{-100\times 10\, k\Omega }{10\, k\Omega }=-100
\; \; \; \; \left | A_{v} \right |=100
Question 3 |
In the single-stage transistor amplifier circuit shown in figure, the capacitor C_{E} is removed. Then, the ac small-signal midband voltage gain of the amplifier
increases | |
decreases | |
is unaffected | |
drops to zero |
Question 3 Explanation:
\frac{A_{V_1}}{A_{V_1}}=1+\left ( \frac{1+h_{fe}}{h_{ie}} \right )R_{e}
\frac{A_{V_{1}}}{A_{V_{2}}}> 1
\Rightarrow \; A_{V_{2}} \lt A_{V_{1}}
\frac{A_{V_{1}}}{A_{V_{2}}}> 1
\Rightarrow \; A_{V_{2}} \lt A_{V_{1}}
There are 3 questions to complete.