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 $V_0=5 \mathrm{~V}$.
In a CE transistor amplifier, there is a current and voltage gain associated with the circuit. In other words there is a power gain. Considering power a measure of energy, does the circuit violate conservation of energy?
In CE transistor amplifier, the power gain is very high.
In this circuit, the extra power required for amplified output is obtained from DC source. Thus, the circuit used does not violet the law of conservation.
(i) Name the type of a diode whose characteristics are shown in figure. (a) and (b).
(ii) What does the point $P$ in fig. (a) represent?
(iii) What does the points $P$ and $Q$ in fig. (b) represent?
(i) The characteristic curve (a) is of Zener diode and curve (b) is of solar cell.
(ii) The point $P$ in fig. (a) represents Zener break down voltage.
(iii) In fig. (b), the point $Q$ represents zero voltage and negative current. It means light falling on solar cell with atleast minimum threshold frequency gives the current in opposite direction to that due to a battery connected to solar cell. But for the point $Q$, the battery is short circuited. Hence represents the short circuit current.
In fig. (b), the point $P$ represents some positive voltage on solar cell with zero current through solar cell.
It means, there is a battery connected to a solar cell which gives rise to the equal and opposite current to that in solar cell by virtue of light falling on it.
As current is zero for point $P$, hence we say $P$ represents open circuit voltage.