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.
Explain why elemental semiconductor cannot be used to make visible LEDs.
In elemental semiconductor, the band gap is such that the emission are in infrared region and not in visible region.
Write the truth table for the circuit shown in figure given below. Name the gate that the circuit resembles.
The circuit resemble AND gate. The boolean expression of this circuit is, $V_0=A . B$ i.e., $V_0$ equals A AND B. The truth table of this gate is as given below
A Zener of power rating 1 W is to be used as a voltage regulator. If Zener has a breakdown of 5 V and it has to regulate voltage which fluctuated between 3 V and 7 V , what should be the value of $R_s$ for safe operation (see figure)?
Given,
$$\begin{aligned} \text { power } & =1 \mathrm{~W} \\ \text { Zener breakdown } V_z & =5 \mathrm{~V} \\ \text { Minimum voltage } V_{\min } & =3 \mathrm{~V} \\ \text { Maximum voltage } V_{\max } & =7 \mathrm{~V} \\ \text { Current } I_{Z_{\max }} & =\frac{P}{V_Z}=\frac{1}{5}=0.2 \mathrm{~A} \end{aligned}$$
The value of $R_s$ for safe operation $R_s=\frac{V_{\max }-V_z}{I_{Z_{\max }}}=\frac{7-5}{0.2}=\frac{2}{0.2}=10 \Omega$