The logic symbol is the same as the AND gate with an inversion bubble placed at the output side.
Learning Objectives:
Explain the working of nand gate
Introducing NAND gate implementation using
Switches
BJT
Working:
Case 1:
Input A = 0
Input B = 0
Output = 1
Case 2:
Input A = 0
Input B = 1
Output = 1
Case 3:
Input A = 1
Input B = 0
Output = q
Case 4:
Input A = 1
Input B = 1
Output = 0
In figure 1 the waveforms or timing diagram of the NAND gate are also given. The output pulse falls to 0V when both inputs are high.
 |
| Figure 1: The NAND Logic Expression, Truth Table and Timing Diagram |
 |
| Figure 2: The NAND gate simulation results |
Y = (A.B)C
Logical NAND Gate (Explain with the help of switches)
Figure 3: Implementation Of NAND Logic Using switches
This is the switch model of the NAND gate. Two switches are connected in series and an LED is connected in parallel.
When switch A is closed, and switch B is opened, the LED turns ON. Because there is no path for the current to flow.
When switch B is closed, the LED turns ON. There is no way for the current to flow.
When both switches are closed the current flows from input to output through closed switches.
Implementation Using Diode Logic
 |
| Figure 3: Implementation Of NAND Logic Using Diodes |
Case 1:
Look at switches S1 and S2
Both are connected to the ground. Both the
are forward biased. Current flows from higher potential level to ground level. Current through 5 V source divides into R2 and R2. And hence LED turns ON.
Case 2:
In this case the position of S1 changes from ground to a voltage source. While S2 remains in the same position. Current flowing from the voltage source to S2 and then grounded.
Case 3:
Same as case 2.
Case 4:
In this case, both switches are connected to a voltage source. Diodes turned off. No path for the current to flow. LED is not going to be turned on.
Implementation Using Transistor Logic
case 1:
Look at switches S1 and S2. Both are open. There is no voltage at the base. VBE = 0. The transistors are off. All the current will flow through the LED. Output is high.
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 approximately at the same potential level of VE. So, Q1 is in saturation.
Case 3:
It is the same as case 2.
Case 4:
Both switches are closed. Both transistors are in saturation mode.
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,
VCE = VCE(sat) = 0.2V
In saturation mode collector voltage VC and emitter voltage VE are approximately at the same potential. Current flows from both the transistors and then grounded. No current flows through the LED.
.jpg)
.jpg)
