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Midterm Exam 3 -- E84, Fall 2008

  1. Problem 1. (33 points)

    An inductive load is connected to a AC voltage source of 220 Volts (RMS) and 50 Hz. It is known that the average power is 1.1 kW and the power factor is $\lambda=0.5$. If the power factor is to be improved to 0.866 by including a capacitor in paralle with the load, what is the its capacitance?

    final08fa.gif

    Solution: Given the power factor $\lambda=\cos\phi=0.5$, the phase angle of the inductive impedance is $\phi=\cos^{-1}0.5=60^\circ$. Also, given the average power

    \begin{displaymath}P_{av}=VI\cos\phi =VI \lambda \end{displaymath}

    we can get the current:

    \begin{displaymath}I=\frac{P}{V\lambda}=\frac{1100}{220\times 0.5}=10\;A \end{displaymath}

    The impedance is therefore

    \begin{displaymath}Z_{RL}=R+j\omega L=\vert Z\vert e^{j\angle Z}=\sqrt{R^2+(\omega L)^2}\angle 60^\circ \end{displaymath}

    As

    \begin{displaymath}\vert Z\vert=\frac{V}{I}=\frac{220}{10}=22 \end{displaymath}

    we have

    \begin{displaymath}\left\{ \begin{array}{l}
\vert Z\vert^2=R^2+(\omega L)^2=22^...
... \frac{\omega L}{R}=\tan 60^\circ=\sqrt{3} \end{array} \right. \end{displaymath}

    Solving these equations we get

    \begin{displaymath}R=11\Omega,\;\;\;\;\;\omega L=11\sqrt{3}\;\Omega \end{displaymath}

    With a capacitor in parallel with the inductive load $R+j\omega L$, the total impedance becomes:

    \begin{displaymath}Z_{RCL}=\frac{1}{j\omega C} \vert\vert (R+j\omega L)
=\frac{(...
...L)+1/j\omega C}
=\frac{R+j\omega L}{(1-\omega^2LC)+j\omega RC} \end{displaymath}

    In order to satisfy the required power factor $\lambda=\cos \phi=0.866$, the phase angle of the total impedance $Z_{RCL}$ has to be $\phi=\cos^{-1} 0.866=30^\circ$, i.e.,

    \begin{displaymath}\angle N-\angle D=30^\circ \end{displaymath}

    where $\angle N$ is the phase of the numerator $R+j\omega L$, known to be $60^\circ$, therefore the angle of the denominator has to be $30^\circ$ as well:

    \begin{displaymath}\angle [(1-\omega^2LC)+j\omega RC]
=\tan^{-1}\frac{\omega RC}{1-\omega^2 LC}=30^\circ \end{displaymath}

    i.e.,

    \begin{displaymath}
\frac{\omega RC}{1-\omega^2 LC}=\tan 30^\circ =\frac{1}{\sqr...
...;\;\;\;\;\mbox{i.e.}\;\;\;\;\;
\sqrt{3}\omega RC=1-\omega^2 LC
\end{displaymath}

    Substituting $R=11$ and $\omega L=11\sqrt{3} $ and solving for $C$ we get

    \begin{displaymath}C=\frac{1}{2\times314\times 11\sqrt{3}}=83.6\;\mu F \end{displaymath}

  2. Problem 2. (33 points)

    Find the DC operating point $(I_C, V_{CE})$ of the transistor circuit given below, where $R_1=100K\Omega$, $R_2=300K\Omega$, $R_C=R_E=2K\Omega$, $V_{CC}=12V$, and $\beta=100$. If you find the DC operating point is not in the middle of the linear region of the output characteristic plot, modify $R_2$ so that the DC operating point is in the middle of the linear region (to maximize the dynamic range of the AC output).

    transistorbiasingb.gif

    Solution:

    Find the load line: When $I_C=0$, $V_{CE}=V_{CC}=12V$, when $V_{CE}=0$, $I_C=V_{CC}/(R_C+R_E)=3mA$


    \begin{displaymath}R_{B}=\frac{R_1 R_2}{R_1+R_2}=75K,\;\;\;V_{BB}=12\frac{R_2}{R_1+R_2}=9V \end{displaymath}


    \begin{displaymath}I_B=\frac{V_{BB}-V_{BE}}{R_B+(\beta+1)R_E}=\frac{9-0.7}{75+202}=30\mu A \end{displaymath}


    \begin{displaymath}I_C=\beta I_B=3 mA, \;\;\;\;V_{CE}=V_{CC}-(R_C+R_E)I_C=0V \end{displaymath}

    The DC operating point is in saturation region.

    Now we modify $R_2$ to move Q-point to the middel point where $I_C=1.5mA$.

    \begin{displaymath}V_{BB}=12R_2/(100+R_2),\;\;\;\;R_B=100R_2/(100+R_2) \end{displaymath}

    We then plug these into

    \begin{displaymath}I_C=\beta \frac{V_{BB}-V_{BE}}{(\beta+1) R_E+R_B}=1.5 \end{displaymath}

    and solve it for $R_2$ to get $R_2=55K$.

  3. Problem 3. (34 points)

    The circuit shown below is called Darlington transistor amplifier which is composed of two transistors $T_1$ and $T_2$ with their collectors connected and the emitter of $T_1$ connected to the base of $T_2$. Assume $V_{CC}=20V$ and both transistors have $\beta=50$.

    1. Give the expressions of $I_{C1}$, $I_{E1}=I_{B2}$, $I_{C2}$, $I_{E2}$ all in terms of $\beta$ and $I_{B1}$.
    2. Given $R_C=1 k\Omega$, find $R_B$ so that the DC operating point $Q$ is in the middle of the linear region of the output V-I characteristic plot.
    3. Given $R_B= 4.6 M\Omega$ and the same $R_C$ as before, find the proper DC voltage $V_{cc}$ so that the Q point is in the middle of the linear region.
    For simplicity, assume for both transistors $v_{be}=0.7V$.

    midterm3g.gif

    Solution:

    1. $I_{C1}=\beta I_{B1}$, $I_{E1}=I_{B2}=(\beta+1) I_{B1}$, $I_{C2}=\beta I_{B2}=\beta (\beta+1) I_{B1}$, $I_{E2}=(\beta+1) I_{B2}=(\beta+1)^2 I_{B1}=(\beta^2+2\beta+1)I_{B1}$,

    2. In order for $V_C=V_{CC}-R_C (I_{C1}+I_{C2})=10V$, we need to have $I_{C1}+I_{C2}=\beta I_{B1}+\beta (\beta+1) I_{B1}=2500 I_{B1}=10 mA$, i.e., $I_{B1}=10 mA/2600=3.8 \mu A$.

      $R_B=(V_{CC}-2V_{be})/I_{B1}=18.6/3.8=4.89 M\Omega$


    3. \begin{displaymath}\beta I_{B1}+\beta (\beta+1) I_{B1}
=(2\beta+\beta^2)\;\frac...
...rac{V_{cc}-1.4}{4.6\times 10^6}=\frac{1}{2}\frac{V_{cc}}{10^3} \end{displaymath}

      Solving this we get

      \begin{displaymath}V_{cc}=12\;V \end{displaymath}




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Ruye Wang 2013-07-31