Deduce the expression for the electrostatic energy stored in a capacitor of capacitance‘C’ and having charge ‘Q’.How will the 

(i) Energy stored and

(ii) The electric field inside the capacitor be affected when it is completely filled with a dielectric material of dielectric constant ‘K’?

                                                           OR

a) Give a circuit diagram of a common emitter amplifier using an n-p-n transistor. Draw the input and output waveforms of the signal. Write the expression for its voltage gain. 

b) Explain with the help of a circuit diagram how a Zener diode works as a DC voltage regulator. Draw its I – V characteristics.

Potential of capacitor $=\frac{\mathrm{q}}{\mathrm{C}}$

Small amount of work done in giving an additional charge dq to the capacitor.

$\mathrm{dW}=\frac{\mathrm{q}}{\mathrm{C}}\times \mathrm{dq}$

Total work done in giving a charge Q to the capacitor

 $\mathrm{W}=\underset{\mathrm{q}=0}{\overset{\mathrm{q}=\mathrm{Q}}{\int }}\frac{\mathrm{q}}{\mathrm{C}}=\frac{1}{2}{\left[\frac{{\mathrm{q}}^2}{\mathrm{C}}\right]}_{\mathrm{q}=0}^{\mathrm{q}=\mathrm{Q}} \therefore \mathrm{W}=\frac{1 {\mathrm{Q}}^2}{\mathrm{C} 2}$

As electrostatic force is conservative, thus work is stored in the form of protential energy (U) of the capacitor.

$$\begin{gathered} \mathrm{U}=\mathrm{W}=\frac{1 {\mathrm{Q}}^2}{2 \mathrm{C} } \\ \mathrm{Put} \mathrm{Q}=\mathrm{CV} \\ \therefore \mathrm{U}=\frac{1}{2}{\mathrm{CV}}^2 \\ \left(\mathrm{i}\right) \mathrm{C}={\mathrm{C}}_0\mathrm{K} \\ \mathrm{where} \left[{\mathrm{E}}_0=\frac{1}{2}{\mathrm{C}}_0\mathrm{V}\right] \end{gathered}$$

  $$\begin{gathered} \mathrm{Energy} \mathrm{stored}=\frac{1}{2}{\mathrm{CV}}^2=\frac{1}{2}{\mathrm{KC}}_0{\mathrm{V}}^2 \\ \boxed{\underset{\downarrow }{\mathrm{E} = {\mathrm{KE}}_0}} \end{gathered}$$

Energy stored will become K times the initial energy

$$\begin{gathered} \left(\mathrm{ii}\right) \mathrm{q}=\mathrm{CV} \Rightarrow \mathrm{V}=\frac{\mathrm{q}}{\mathrm{C}} {\mathrm{\epsilon }}_0=\frac{\mathrm{q}}{{\mathrm{C}}_0\mathrm{d}} \\ \mathrm{V}=\frac{\mathrm{q}}{{\mathrm{KC}}_0} \mathrm{\epsilon }=\frac{\mathrm{V}}{\mathrm{d}}=\frac{\mathrm{q}}{{\mathrm{KC}}_0\mathrm{d}} \\ \boxed{\stackrel{\mathrm{\epsilon } = \frac{{\mathrm{\epsilon }}_0}{\mathrm{K}}}{\downarrow }} \\ \mathrm{electric} \mathrm{field} \\ \therefore \mathrm{Electric} \mathrm{field} \mathrm{will} \mathrm{become} \frac{1}{\mathrm{k}} \mathrm{times} \mathrm{its} \mathrm{initial} \mathrm{value}. \end{gathered}$$

     OR

a)   Common emitter configuration of n-p-n transistor

(ii)   Transistor as an amplifier (C.E. configuration): The circuit diagram of a common emitter amplifier using n-p-n transistor is given below:

The input (base-emitter) circuit is forward baised and the output circuit collector-emitter) is reverse baised.

the input (base-emitter) circuit is forward biased and the output circuit (collector-emitter) is reverse biased. When no a.c. signal is applied, the potential difference ${\mathrm{V}}_{\mathrm{CC}}$ between the collector and emitter is given by

${\mathrm{V}}_{\mathrm{CC}}={\mathrm{V}}_{\mathrm{CE}}+{\mathrm{I}}_{\mathrm{C}} {\mathrm{R}}_{\mathrm{C}}$ 

when an a a.c. signal is fed to the input circuit, the forward bias increases during the positive half cycle of the input. This results in increase in ${\mathrm{I}}_{\mathrm{C}}$ and ${\mathrm{V}}_{\mathrm{CC}}$. Thus during positive half cycle of the input, the collector become less positive.

During the negative half cycle of the input, the forward bias is decreased resulting in decrease in ${\mathrm{I}}_{\mathrm{E}}$ and hence ${\mathrm{I}}_{\mathrm{C}}$. Thus ${\mathrm{V}}_{\mathrm{CC}}$ would increase making the collector more positive. Hence in a common-emitter amplifier, the output voltage ${180}^{\circ }$ out of phase with the input voltage.

${\mathrm{A}}_{\mathrm{V}}=\frac{{\mathrm{V}}_0}{{\mathrm{V}}_{\mathrm{i}}}=\frac{{\mathrm{I}}_{\mathrm{C}}{\mathrm{R}}_{\mathrm{C}}}{{\mathrm{I}}_{\mathrm{B}}{\mathrm{R}}_{\mathrm{B}}}=\mathrm{\beta }\left(\frac{{\mathrm{R}}_{\mathrm{C}}}{{\mathrm{R}}_{\mathrm{B}}}\right) [\because \mathrm{\beta }=\frac{{\mathrm{I}}_{\mathrm{C}}}{{\mathrm{I}}_{\mathrm{B}}}$

  b)

    Zener diode is fabricated by heavily doping both p and n-side. Due to this, depletion region formed is very thin (<10^-6 n and the electric field of the junction is extremely high (~5x10⁶ V/m) even for a small reverse bias voltage of 5 volts. It is seen that when the applied reverse bias voltage (V) reaches the breakdown voltage (Vz) of the Zener diode, there is a large change in the current. After the breakdown voltage Vz, a large change in the current can be produced by almost insignificant change in the reverse bias voltage. In other words. Zener voltage remains constant even though current through the. Zener diode varies over a wide range. This property of the zener diode is used for regulating voltages so that they are constant. Semiconductor diode as a half wave Rectifier : The junction diode D, supplies rectified current to the band during one half of the alternating input voltage and is always in the same direction. During the first half cycles of the alternating input voltage and is always in the same direction. During the first half cycles of the alternating input voltage, junction diodes, D₁ will conduct each permitting current to flow during one half cycle whenever its p-terminal is positive with respect to the n-terminal.  

The resulting output current is a series of unidirectional pulses with alternate gaps.