Semiconductors and Electronic Devices MCQ Questions & Answers in Modern Physics | Physics

Figure shows the basic triode amplifier circuit. The weak signal $${e_g}$$ is applied in the grid circuit and useful output is obtained across the load $${R_L}$$ connected in the plate circuit. The bias battery $${E_C}$$ ensures the grid to be always negative w.r.t. cathode.
The weak signal voltage produces a large change in plate current. As the value of $${R_L}$$ is quite high, therefore a large voltage drop occurs across it. Thus, a weak signal applied in the grid circuit appears in the amplified form in the plate circuit.

The cohesive energy of solids is defined as the energy required for separating the condensed material into isolated free atoms.
The bond between two carbon atoms in diamond has a cohesive energy of $$7.3\,eV$$  with respect to separated neutral atoms. Due to such high cohesive energy diamond is very hard.

As one of the diodes (the lower one) will be in reverse bias, no current will pass through it, so the effective resistance of the circuit $$ = 10\,\Omega .$$

The $$p-n$$  junction diode can be shown as

If $$p$$-side of $$p-n$$  junction diode is given more positive potential or positive terminal of the battery and $$n$$-side is connected to less potential or one terminal of battery, then it is forward biased.
In option (D), $$p$$ -side is at $$0\,V$$  and $$n$$-side at $$-2\,V,$$  so $$p$$ is at higher potential. Hence, it is forward biased.
NOTE
$$p-n$$  junction diode is mainly used as a rectifier.

$$\beta = \frac{\alpha }{{1 - \alpha }} = \frac{{0.95}}{{1 - 0.95}} = \frac{{0.95}}{{0.05}} = 19$$

The Boolean expression expressed by the truth table may be written as
$$Y = \overline {A + B} $$
This is the expression of NOR gate (OR gate + NOT gate)
NOTE
The NOR gate may be shown as below

A transistor amplifier with proper positive feedback can act as an oscillator, i.e. it can generate oscillations without any external signal source. A positive feedback amplifier is one that produces a feedback voltage $$\left( {{V_f}} \right)$$  that is in phase with the original input signal. A phase shift of $${180^ \circ }$$ is produced by the amplifier and a further phase shift of $${180^ \circ }$$ is introduced by feedback network. Consequently, the signal is shifted by $${360^ \circ }$$ and fed to the input, i.e. feedback voltage is in phase with the input signal.

$$\eqalign{ & {I_C} = 5.488\,mA,{I_e} = 5.6\,mA,{I_B} = {I_E} - {I_C} \cr & \beta = \frac{{{I_c}}}{{{I_B}}} = \frac{{5.488}}{{5.6 - 5.485}} = 49 \cr} $$

$${\text{Here,}}\,\,E = 9\,V;{V_Z} = 6;{R_L} = 1000\,\Omega \,{\text{and}}\,{R_s} = 100\,\Omega $$
Potential drop across series resistor
$$V = E - {V_Z} = 9 - 6 = 3\,V$$
Current through series resistance $${R_S}$$ is
$$I = \frac{V}{R} = \frac{3}{{300}} = 0.03\,A$$
Current through load resistance $${R_L}$$ is
$${I_L} = \frac{{{V_Z}}}{{{R_L}}} = \frac{6}{{1000}} = 0.006\,A$$
Current through Zener diode is
$${I_Z} = I - {I_L} = 0.03 - 0.006 = 0.024\,amp.$$
Power dissipated in Zener diode is
$${P_Z} = {V_Z}{I_Z} = 6 \times 0.024 = 0.144\,Watt$$

$$Si$$ and $$Ge$$ are semiconductors but $$C$$ is an insulator. Also, the conductivity of $$Si$$ and $$Ge$$ is more than $$C$$ because the valence electrons of $$Si,\,Ge$$ and $$C$$ lie in third, fourth and second orbit respectively.