The AND Gate:
The AND gate performs logical AND function. The output is HIGH (true) only when all the inputs are HIGH (true). Or in other words, if all inputs are non-zero, the output of this gate goes HIGH (non-zero). Or if any of the input is low the output remains low.
Learning Objectives:
Introduction to AND logic, and digital building block
Introducing AND gate implementation using
Working:
 |
| Figure 1 |
Case 1:
Input A = 0
Input B = 0
Output = 0
Case 2:
Input A = 0
Input B = 1
Output = 0
Case 3:
Input A = 1
Input B = 0
Output = 0
Case 4:
Input A = 1
Input B = 1
Output = 1
In figure 1 the waveforms or timing diagram of AND gate are also given. The output pulse rises from 0 to 1 when both inputs are high.
 |
| Figure 2 |
Logical Expression:
Y = A.B
Implementation Of AND Gate:
Logical AND Gate (Explain with the help of switches)
 |
| Figure 3 |
This is the switch model of AND logic. It is usually easy to understand the concept with the help of switches. In digital electronics diodes and transistors work as switches. So it is easy to grasp the concepts with the help of the switch model. Here you can see two switches connected in series. The input goes to the output only if both switches (switch A and switch B) are closed.
Implementation Using Diode Logic:
 |
| Figure 4 |
In modern electronics, it is ridiculous to implement this logic. But for beginners, it is necessary to understand the working of AND logic with the help of diodes.
Case 1:
It is seen in the schematic diagram, both inputs are tied to the ground. The positive side of the diode is connected to the positive side of the voltage supply. Both diodes are turned on and the current flows from the higher potential level to the lower potential level. The current will flow from the least resistive path. No current will flow from the LED and hence it remains off.
Case 2:
In this case diode, D2 is connected to the ground and D1 is connected to the 5V source. D2 is turned on. D1 is also tuned on. This is because the positive side of D1 is connected to the 7V source and the negative side is connected to the 5V source. Most of the current flow from D2 to ground and hence LED remains off.
Case 3:
The circuit working will be the same as in case 2.
Case 4:
In this case diode, D1 and D2 both are connected to a 5V source. Both the diodes are turned on. Only a little current flows from D1 and D2 because the potential difference is very low. The current flows from LED to ground. And hence LED turns on. The current and voltage values will be determined with the help of KVL.
Implementation Using Transistor Logic
In practical terms, BJTs are not suitable for switching. But we are interested to understand the logic or working of AND gate. It is easy to gain knowledge by practising different types of circuits on breadboards.
 |
| Figure 5 |
Case1:
Look at switches S1 and S2. Both are open. There is no voltage at the base. VBE = 0. The transistors are off. Output is low.
Case 2:
Look at switch S1, it is closed. Switch S2 is opened. Transistor Q2 is turned off. Now check the mode of each transistor. In digital switching devices BJT either works in saturation or cut off.
VBE > 0
The above condition is true. Transistor Q1 is ON Now check for the second condition.
VCE = VCE(sat) = 0.2V
VC is at a much higher potential than VE. So, Q1 is not in saturation.
Case 3:
It is the same as case 2.
Case 4:
Both switches are closed. Let's look at Q2, base is at 5V, the emitter is grounded.
VBE > 0
Q1's collector is connected to Q2's emitter.
Q1's Collector is connected to a 12 V source. Checking for the second condition,
VBC > 0
It is valid for both transistors.
.jpg)
.jpg)
