Why are elemental dopants for Silicon or Germanium usually chosen from group XIII or group XV?
The size of the dopant atom should be such that their presence in the pure semiconductor does not distort the semiconductor but easily contribute the charge carriers on forming covalent bonds with Si or Ge atoms, which are provided by group XIII or group XV elements.
$\mathrm{Sn}, \mathrm{C}$ and $\mathrm{Si}, \mathrm{Ge}$ are all group XIV elements. Yet, Sn is a conductor, C is an insulator while Si and Ge are semiconductors. Why?
A material is a conductor if in its energy band diagram, there is no energy gap between conduction band and valence band. For insulator, the energy gap is large and for semiconductor the energy gap is moderate.
The energy gap for Sn is 0 eV, for C is 5.4 eV , for Si is 1.1 eV and for Ge is 0.7 eV , related to their atomic size. Therefore Sn is a conductor, C is an insulator and Ge and Si are semiconductors.
Can the potential barrier across a $p-n$ junction be measured by simply connecting a voltmeter across the junction?
We cannot measure the potential barrier across a $p-n$ junction by a voltmeter because the resistance of voltmeter is very high as compared to the junction resistance.
Draw the output waveform across the resistor in the given figure.
As we know that the diode only works in forward biased, so the output is obtained only when positive input is given, so the output waveform is
The amplifiers $X, Y$ and $Z$ are connected in series. If the voltage gains of $X, Y$ and $Z$ are 10, 20 and 30 , respectively and the input signal is 1 mV peak value, then what is the output signal voltage (peak value)
(i) if DC supply voltage is 10 V ?
(ii) if DC supply voltage is 5 V ?
Given,
$$\begin{aligned} A v_x & =10, A v_y=20, A v_z=30 ; \\ \Delta V_i & =1 \mathrm{mV}=10^{-3} \mathrm{~V} \end{aligned}$$
Now, $\quad \frac{\text { Output Signal Voltage }\left(\Delta V_0\right)}{\text { Input Singal Voltage }\left(\Delta V_i\right)}=$ Total voltage amplification
$$\begin{aligned} & =A v_x \times A v_y \times A v_z \\ \Rightarrow\quad\Delta V_0 & =A v_x \times A v_y \times A v_z \times \Delta V_i \\ & =10 \times 20 \times 30 \times 10^{-3}=6 \mathrm{~V} \end{aligned}$$
(i) If DC supply voltage is 10 V , then output is 6 V , since theoretical gain is equal to practical gain, i.e., output can never be greater than 6 V.
(ii) If DC supply voltage is 5 V , i.e., $V_{c c}=5 \mathrm{~V}$. Then, output peak will not exceed 5 V . Hence $V_0=5 \mathrm{~V}$.