NOR Gate using Diodes and Transistor on Breadboard

I am sure you must be familiar with a NOR Gate, it’s Truth Table, Logic symbol, and its working. But do you know, you can build your own NOR Gate on Breadboard?  So in this post, you will learn how to make a NOR Gate using Diodes and Transistor on Breadboard.

 

**Read Similar Article: And Gate using Diodes

 

What is a NOR Gate?

 

NOR GATE Symbol
NOR GATE Symbol

 

NOR Gate is a universal Logic gate. It is simply a NOT Gate connected ahead of an OR Gate. That means we get an inverted output of OR gate out of a Nor gate. So the output is 1 only when both the inputs are zero i.e,  just opposite of what happens in case of an OR gate.

We are going to make this NOR gate by combining OR gate and NOT gate. Click to See How to build an OR gate and How to build a NOT gate.

 

 

You can also watch the video below for quick reference:

 

Truth Table:

 

Truth Table of NOR Gate
Truth Table of NOR Gate

 

 

Components required:

1. 1N4007 Diode X 2

2. BC547 Transistor

3. 1k Resistor

4. 100-ohm Resistor X 2

5. LED

6. 9 Volt Battery

7. Connecting Wires

 

SCHEMATIC:

 

Circuit of a NOR Gate
Circuit of a NOR Gate

 

This is the circuit we are going to build on Breadboard.

1. N terminal of diodes is connected together.

2. Logic inputs are given to P terminal of diodes.

3. End terminals of the Resistance are connected to N- terminals of the diode and base of the transistor respectively.

4. One 100 ohm resistor is connected from +VE terminal of the battery to the collector terminal of transistor.

5. Output indicator i.e LED is connected from collector terminal to 100-ohm Resistor to Ground.

6. Emitter terminal of the transistor is connected to Ground.

7. Ground- Negative terminal of the battery.

 

WORKING:

 

NOTE: When P-terminal of a Diode is at a higher potential (>.7v) then its n terminal, it acts like a short circuit. And when n terminal is at a higher potential, Diode act as an open circuit.

NOTE: Transistor acts as a switch in this circuit. Whenever there is some threshold voltage at the base, transistor acts like a closed switch from collector to emitter. And whenever there is no or less voltage then threshold voltage at the base, it acts like an open circuit from collector to emitter.

**from the circuit diagram

 

CASE 1: Input A=0, Input B= 0

In this case, p-terminal of both diodes is at 0 volts w.r.t n-terminal which is  <.7v and hence acts like an open circuit. Thus no current flow from 1k resistor to the base terminal of transistor. In this case, there is no voltage at the base of the transistor. So it acts like an open switch from collector to emitter i.e, infinite resistance between collector and emitter. And due to this no current flow from collector to Ground. All current flow from the 100-ohm resistors to LED and then to ground i.e, through the path of minimum resistance. Hence LED turns on i.e, 1 at the output. 

 

CASE 2: Input A=0, Input B= 1

In this case, p-terminal of diode A is at 0 volts w.r.t n-terminal which is  <.7v and hence acts like an open circuit. But p-terminal of diode B is at a higher potential than its n-terminal which is >.7v. Thus some current flow from 1k resistor to the base of transistor. In this case, there is some threshold voltage at the base of the transistor. So it acts like a closed switch from collector to emitter i.e, almost zero resistance between collector and emitter. And due to this no current flow from 100-ohm resistors to LED and then to ground. All current flow from collector to Ground i.e, through the path of minimum resistance. Hence LED turns off i.e, 0 at the output. 

 

CASE 3: Input A=1, Input B= 0

In this case, p-terminal of diode B is at 0 volts w.r.t n-terminal which is  <.7v and hence acts like an open circuit. But p-terminal of diode A is at a higher potential than its n-terminal which is >.7v. Thus some current flow from 1k resistor to the base of transistor. In this case, there is some threshold voltage at the base of the transistor. So it acts like a closed switch from collector to emitter i.e, almost zero resistance between collector and emitter. And due to this no current flow from 100-ohm resistors to LED and then to ground. All current flow from collector to Ground i.e, through the path of minimum resistance. Hence LED turns off i.e, 0 at the output. 

 

CASE 4: Input A=1, Input B= 1

In this case, p-terminal of both diodes is at higher potential w.r.t n-terminal which is  >.7v and hence acts like short circuits. Thus some current flow from 1k resistor to the base of transistor. In this case, there is some threshold voltage at the base of the transistor. So it acts like a closed switch from collector to emitter i.e, almost zero resistance between collector and emitter. And due to this no current flow from 100-ohm resistors to LED and then to ground. All current flow from collector to Ground i.e, through the path of minimum resistance. Hence LED turns off i.e, 0 at the output. 

 

** Hence we get 1(LED is ON) at the output only when both the inputs are 0 as given in Truth Table.

 

Simulation:

 

Note: Before building the circuit on a breadboard, it is first simulated on “Every circuit” app. You can download this app from here.

** The circuit is exactly similar to what is given above.

 

CIRCUIT OF NOR GATE
CIRCUIT OF NOR GATE

 

 

 

** Slideshow

CASE 1: Input A=0, Input B= 0 , Output=0 (LED is OFF)

CASE 2: Input A=0, Input B= 1, Output=1 (LED is ON)

CASE 3: Input A=1, Input B= 0, Output=1 (LED is ON)

CASE 4: Input A=1, Input B= 1, Output=1 (LED is ON)

This slideshow requires JavaScript.

 

Let’s Make it!

 

1. Place the two diodes as shown below. N-terminal is the silver portion, P-terminal is the Black portion.

**Slideshow

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2. Connect two jumper wires, each at respective p-terminal of the diodes indicating input Logic:

 

 

 

3. Connect  N terminal of the Diodes together using wire as shown below:

 

4. Place Transistor on breadboard as show:

 

 

5. Connect 1k ohm resistor from n terminal of Diodes to the base terminal of transistor.

**Slideshow

This slideshow requires JavaScript.

 

6. Now connect Emitter terminal of the transistor to ground(battery’s -ve terminal).

**Slideshow

 

This slideshow requires JavaScript.

 

7. Now connect a 100-ohm resistor from collector terminal of transistor to the +ve terminal of battery.

**Slideshow

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8. Now connect a 100-ohm resistor in series with LED to Ground as shown below:
**Slideshow

This slideshow requires JavaScript.

 

CASE 1: Input A=0, Input B= 0 , Output=1 (LED is ON)

 

 

In this case, p-terminal of both diodes is at 0 volts w.r.t n-terminal which is  <.7v and hence acts like an open circuit. Thus no current flow from 1k resistor to the base terminal of transistor. In this case, there is no voltage at the base of the transistor. So it acts like an open switch from collector to emitter i.e, infinite resistance between collector and emitter. And due to this no current flow from collector to Ground. All current flow from the 100-ohm resistors to LED and then to ground i.e, through the path of minimum resistance. Hence LED turns on i.e, 1 at the output. 

 

 

CASE 2: Input A=0, Input B= 1, Output=0 (LED is OFF)

 

 

n this case, p-terminal of diode B is at 0 volts w.r.t n-terminal which is  <.7v and hence acts like an open circuit. But p-terminal of diode A is at a higher potential than its n-terminal which is >.7v. Thus some current flow from 1k resistor to the base of transistor. In this case, there is some threshold voltage at the base of the transistor. So it acts like a closed switch from collector to emitter i.e, almost zero resistance between collector and emitter. And due to this no current flow from 100-ohm resistors to LED and then to ground. All current flow from collector to Ground i.e, through the path of minimum resistance. Hence LED turns off i.e, 0 at the output. 

 

 

CASE 3: Input A=1, Input B= 0, Output=0 (LED is OFF)

 

n this case, p-terminal of diode B is at 0 volts w.r.t n-terminal which is  <.7v and hence acts like an open circuit. But p-terminal of diode A is at a higher potential than its n-terminal which is >.7v. Thus some current flow from 1k resistor to the base of transistor. In this case, there is some threshold voltage at the base of the transistor. So it acts like a closed switch from collector to emitter i.e, almost zero resistance between collector and emitter. And due to this no current flow from 100-ohm resistors to LED and then to ground. All current flow from collector to Ground i.e, through the path of minimum resistance. Hence LED turns off i.e, 0 at the output. 

 

 

CASE 4: Input A=1, Input B= 1, Output=0 (LED is OFF)

 

 

In this case, p-terminal of both diodes is at higher potential w.r.t n-terminal which is  >.7v and hence acts like short circuits. Thus some current flow from 1k resistor to the base of transistor. In this case, there is some threshold voltage at the base of the transistor. So it acts like a closed switch from collector to emitter i.e, almost zero resistance between collector and emitter. And due to this no current flow from 100-ohm resistors to LED and then to ground. All current flow from collector to Ground i.e, through the path of minimum resistance. Hence LED turns off i.e, 0 at the output. 

** Hence we get 1(LED is ON) at the output only when both the inputs are 0 as given in Truth Table.

 

You can also watch the video below for quick reference:

 

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