(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.
Three photo diodes $D_1, D_2$ and $D_3$ are made of semiconductors having band gaps of $2.5 \mathrm{~eV}, 2 \mathrm{eV}$ and 3 eV , respectively. Which ones will be able to detect light of wavelength $6000 \mathop A\limits^o$ ?
Given, wavelength of light $\lambda=6000 \mathop A\limits^o=6000 \times 10^{-10} \mathrm{~m}$
Energy of the light photon
$$E=\frac{h c}{\lambda}=\frac{6.6 \times 10^{-34} \times 3 \times 10^8}{6000 \times 10^{-10} \times 1.6 \times 10^{-19}} \mathrm{eV}=2.06 \mathrm{eV}$$
The incident radiation which is detected by the photodiode having energy should be greater than the band-gap. So, it is only valid for diode $D_2$. Then, diode $D_2$ will detect this radiation.
If the resistance $R_1$ is increased (see figure), how will the readings of the ammeter and voltmeter change?
Consider the circuit in fig. (b) to find the change in reading As we know the formula for base current, $I_B=\frac{V_{\mathrm{BB}}-V_{\mathrm{BE}}}{R_i}$
As $R_i$ is increased, $I_B$ is decreased.
Now, the current in ammeter is collector current $I_C$. $I_C=\beta I_B$ as $I_B$ decreased $I_C$ also decreased and the reading of voltmeter and ammeter also decreased.
Two car garages have a common gate which needs to open automatically when a car enters either of the garages or cars enter both. Devise a circuit that resembles this situation using diodes for this situation.
As car enters in the gate, any one or both are opened.
The device is shown.
So, OR gate gives the desired output.
How would you set up a circuit to obtain NOT gate using a transistor?
The NOT gate is a device which has only one input and one output $i . e ., \bar{A}=Y$ means $Y$ equals NOT A.
This gate cannot be realised by using diodes. However it can be realised by making use of a transistor. This can be seen in the figure given below
Here, the base $B$ of the transistor is connected to the input $A$ through a resistance $R_b$ and the emitter $E$ is earthed. The collector is connected to 5 V battery. The output $Y$ is the voltage at $C$ w.r.t. earth.
The resistor $R_b$ and $R_c$ are so chosen that if emitter-base junction is unbiased, the transistor is in cut off mode and if emitter-base junction is forward biased by 5 V , the transistor is in saturation state.