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Electronic circuit Quiz 2

physics

 Finiya was observing an intrinsic semiconductor which has the electron current in it as $5.66\text{ }mA$ and the whole current as $3.23\text{ }mA$. What will be drift velocity in the semiconductor if the number of free electrons in it be $5\times {{10}^{5}}$ and the area of cross section be $1\text{ }m{{m}^{2}}$? 

 The current can be found by taking the product of the number of free electrons, electronic charge, area of cross section and the drift velocity of electrons. 

  The equation of total current in an intrinsic semiconductor, $I={{I}_{e}}+{{I}_{h}}$ $ \\ $ Substituting the values in it, $ \\ I=5.66+3.23 \\ \Rightarrow I=8.89\text{ }mA \\ $ The equation of current, $I=neA{{V}_{d}}$ $ \\ $ Rearranging the equation, ${{V}_{d}}=\dfrac{I}{neA}$ $ \\ $ Substituting the values in it, $ \\ {{V}_{d}}=\dfrac{8.89\times {{10}^{-3}}}{5\times {{10}^{5}}\times 1.6\times {{10}^{-19}}\times {{10}^{-6}}} \\ \therefore {{V}_{d}}=1.11\times {{10}^{17}}\text{ }V$

 Tintu was working with a silicon transistor where the current entering the transistor has been found to be $7.61\text{ }mA$ and the current leaving the transistor be $4.21\text{ }mA$. What will be the value of resistance of material if the base current from the semiconductor can be connected to this material and the potential difference occurring there be $2.34\text{ }V$? 

 The emitter current will be equivalent to the sum of the collector current and the base current.  

  In a transistor, The emitter current, ${{I}_{E}}={{I}_{B}}+{{I}_{C}}$ $ \\ $ Substituting the values in it, $ \\ 7.61={{I}_{B}}+4.21 \\ \Rightarrow {{I}_{B}}=3.4\text{ }mA \\ $ According to Ohm’s law, $V=IR$ $ \\ $ Rearranging, $R=\dfrac{V}{I}$ $ \\ $ Substituting the values in it, $ \\ R=\dfrac{V}{{{I}_{B}}} \\ \Rightarrow R=\dfrac{2.34}{3.4} \\ \therefore R=0.688\text{ }\Omega$

 The current amplification factor in a common base configuration has been found by different experiments as $3$. The emitter current in this transistor has been mentioned as $4.55\text{ }mA$. If the same circuit has been connected as a common collector after this then what will be the current amplification factor now?

 The ratio of the emitter current to the base current gives the current amplification factor in common collector configuration. The ratio of the collector current to the emitter current will give the current amplification factor in common base configuration. 

 In the common base configuration, $ \\ \alpha =\dfrac{{{I}_{C}}}{{{I}_{E}}} \\ $ Rearrange the equation to obtain the collector current, ${{I}_{C}}=\alpha {{I}_{E}}$ $ \\ $ Substituting the values, $ \\ {{I}_{C}}=0.3\times 4.55 \\ \Rightarrow {{I}_{C}}=1.365\text{ }mA \\ $ According to the definitions, ${{I}_{E}}={{I}_{B}}+{{I}_{C}}$ $ \\ $ Substituting and rearranging will give, $ \\ 4.55={{I}_{B}}+1.365 \\ \therefore {{I}_{B}}=3.185\text{ }mA \\ $ The current amplification factor in common collector configuration, $ \\ \gamma =\dfrac{{{I}_{E}}}{{{I}_{B}}} \\ $ Substituting the values, $ \\ \gamma =\dfrac{4.55}{3.185} \\ \therefore \gamma =1.43$

 The change in base emitter voltage in a common emitter configuration has been mentioned as $7\text{ V}$and the change in base current in the same configuration has been mentioned as $3.12\text{ }mA$.  What will be the percentage error in the value of input resistance of the common emitter configuration if the value Varun calculated is $4.32\text{ }\Omega $? 

 The input resistance in common emitter configuration will be the ratio of the change in emitter bias voltage to the change in base current. 

 The equation of input resistance in common emitter configuration, $\\ r=\dfrac{\Delta {{V}_{BE}}}{\Delta {{I}_{B}}} \\ $ Substituting the values in it, $ \\ r=\dfrac{7}{3.12} \\ \Rightarrow r=2.24\text{ }\Omega \\ $ The percentage error will be, $ \\ \text{percentage error}=\left| \dfrac{\text{actual value-observed value}}{\text{actual value}}\times 100 \right|\\ $ Substituting the values in it, $ \\ \text{percentage error}=\left| \dfrac{\text{2}\text{.24-4}\text{.32}}{2.24}\times 100 \right| \\ \Rightarrow \text{percentage error}=0.928\times 100 \\ \therefore \text{percentage error}=92.8\%$

 In a common emitter configuration, the collector current has been given as $2.1\text{ }mA$ and the emitter current has been mentioned as $7.7\text{ }mA$.  What will be the power gain of the transistor if it is connected to a load of resistance $4\text{ }\Omega $? $($Let us assume that the effective input resistance be $0.1\text{ }\Omega )$ 

 The current amplification factor in common emitter configuration is the ratio of the collector current to the base current. 

 The emitter current can be calculated by the equation, $ \\ {{I}_{E}}={{I}_{B}}+{{I}_{C}} \\ $ Substituting the values in it and rearranging, $ \\ 7.7={{I}_{B}}+2.1 \\ \Rightarrow {{I}_{B}}=5.6\text{ }mA \\ $ The current amplification factor, $ \\ \beta =\dfrac{{{I}_{C}}}{{{I}_{B}}} \\ $ Substituting the values in it, $ \\ \beta =\dfrac{2.1}{5.6} \\ \Rightarrow \beta =0.375 \\ $ The equation of current gain, $ \\ {{A}_{V}}=\dfrac{\beta {{R}_{L}}}{r} \\ $ Substituting the values in it, $ \\ {{A}_{V}}=\dfrac{0.375\times 4}{0.1} \\ \Rightarrow {{A}_{V}}=15 \\ $ The equation of power gain, $ \\ {{A}_{P}}=\beta \times {{A}_{V}} \\ $ Substituting the values, $ \\ {{A}_{P}}=0.375\times 15 \\ \therefore {{A}_{P}}=5.625$