The post Resistor Color Codes | Color Codes | Electronics Textbook appeared first on Know Electronics.

]]>The Resistor Color Codes consists of several colored bands, typically four or five, that are printed on the body of the resistor. The first two or three bands represent the resistance value, while the fourth band represents the tolerance. If there is a fifth band, it represents the temperature coefficient.

Color |
1st Band |
2nd Band |
3rd Band |
4th Band |
5th Band |

Black | 0 | 0 | 0 | – | – |

Brown | 1 | 1 | 1 | 1% | 100 ppm/K |

Red | 2 | 2 | 2 | 2% | 50 ppm/K |

Orange | 3 | 3 | 3 | – | 15 ppm/K |

Yellow | 4 | 4 | 4 | – | 25 ppm/K |

Green | 5 | 5 | 5 | 0.5% | – |

Blue | 6 | 6 | 6 | 0.25% | 10 ppm/K |

Violet | 7 | 7 | 7 | 0.1% | 5 ppm/K |

Gray | 8 | 8 | 8 | 0.05% | – |

White | 9 | 9 | 9 | – | – |

Gold | – | – | – | 5% | – |

Silver | – | – | – | 10% | – |

No color | – | – | – | 20% | – |

The resistor colour code is read from left to right as illustrated below:

**The Resistor Colour Code Table**

Colour | Digit | Multiplier | Tolerance |

Black | 0 | 1 | |

Brown | 1 | 10 | ± 1% |

Red | 2 | 100 | ± 2% |

Orange | 3 | 1,000 | |

Yellow | 4 | 10,000 | |

Green | 5 | 100,000 | ± 0.5% |

Blue | 6 | 1,000,000 | ± 0.25% |

Violet | 7 | 10,000,000 | ± 0.1% |

Grey | 8 | ± 0.05% | |

White | 9 | ||

Gold | 0.1 | ± 5% | |

Silver | 0.01 | ± 10% | |

None | ± 20% |

Then we can summerise the different weighted positions of each coloured band which makes up the resistors colour code above in the following table:

Number of Coloured Bands |
3 Coloured Bands (E6 Series) |
4 Coloured Bands (E12 Series) |
5 Coloured Bands (E48 Series) |
6 Coloured Bands (E96 Series) |

1^{st} Band |
1^{st} Digit |
1^{st} Digit |
1^{st} Digit |
1^{st} Digit |

2^{nd} Band |
2^{nd} Digit |
2^{nd} Digit |
2^{nd} Digit |
2^{nd} Digit |

3^{rd} Band |
Multiplier | Multiplier | 3^{rd} Digit |
3^{rd} Digit |

4^{th} Band |
– | Tolerance | Multiplier | Multiplier |

5^{th} Band |
– | – | Tolerance | Tolerance |

6^{th} Band |
– | – | – | Temperature Coefficient |

To calculate the value of a resistor using the color code, you need to identify the color of each band and their corresponding values.

The color code for a standard four-band resistor is as follows:

- The first band represents the first digit of the resistance value.
- The second band represents the second digit of the resistance value.
- The third band represents the multiplier for the resistance value.
- The fourth band represents the tolerance value.

Here are the steps to follow to calculate the value of a four-band resistor:

- Identify the color of the first band and match it to the corresponding number from the above table.
- Identify the color of the second band and match it to the corresponding number from the above table.
- Identify the color of the third band and match it to the corresponding multiplier from the above table. For example, if the third band is red, the multiplier is 100 ohms (10^2).
- Multiply the two-digit number formed by the first and second bands by the multiplier value from the third band. This gives you the resistance value in ohms.
- Identify the color of the fourth band and match it to the corresponding tolerance from the above table. For example, if the fourth band is gold, the tolerance is 5%.
- Calculate the tolerance range by multiplying the resistance value by the tolerance percentage. For example, if the resistance value is 1,000 ohms (1 kΩ) and the tolerance is 5%, the tolerance range is 50 ohms.
- Express the resistance value with its unit (ohms, kilohms, megohms) and tolerance range (if required).

Here’s an example:

Suppose you have a resistor with the color bands brown, black, red, and gold.

- Brown corresponds to 1.
- Black corresponds to 0.
- Red corresponds to 100 (10^2).
- Gold corresponds to ±5%.

Therefore, the resistor’s resistance value is (10 × 1) × 100 = 1,000 ohms (1 kΩ), and the tolerance range is ±5% of the resistance value, which is ±50 ohms. So the final value of the resistor would be expressed as 1 kΩ ± 50 Ω.

In addition to the color code, some resistors are also marked with a letter code that indicates the tolerance of the resistor. This code is used in place of the fourth band of the color code on some resistors.

Here is the letter code for resistor tolerances:

Letter code |
Tolerance |

F | ±1% |

G | ±2% |

J | ±5% |

K | ±10% |

M | ±20% |

Z | Special |

For example, if you see a resistor with the code “4.7KJ”, it indicates that the resistance value is 4.7 kilohms with a tolerance of ±5%. The “K” indicates that the third band is orange (multiplier of 1,000), and the “J” indicates a tolerance of ±5%.

Note that the letter code is not used on all resistors, and it is more commonly used on smaller, surface-mount resistors rather than larger through-hole resistors. If a resistor has both a letter code and a fourth band, the letter code takes precedence over the fourth band for indicating the tolerance.

Surface mount resistors (SMD resistors) are a type of electronic component that are designed to be mounted directly on the surface of a printed circuit board (PCB) without the need for leads or wires. They are typically smaller in size and have a lower profile than through-hole resistors, making them well-suited for high-density electronic circuits.

SMD resistors come in a variety of package sizes, each with its own code to indicate its dimensions. Some of the most common SMD resistor packages include 0402, 0603, 0805, and 1206, with the numbers indicating the length and width of the package in thousandths of an inch. For example, an 0402 package is 0.04 inches long and 0.02 inches wide.

SMD resistors also use a different coding system than through-hole resistors to indicate their resistance value and tolerance. The coding system typically consists of three or four digits, with the first two digits indicating the significant figures of the resistance value and the third digit (or fourth digit, if present) indicating the number of zeros to add to the end of the value to determine the multiplier. For example, a resistor marked “103” has a value of 10 x 10^3 ohms, or 10 kilohms.

SMD resistors may also include a letter code to indicate their tolerance, similar to through-hole resistors. Common letter codes for SMD resistors include F (±1%), G (±2%), J (±5%), and K (±10%).

In addition to their small size and low profile, SMD resistors offer other advantages over through-hole resistors, such as lower inductance and capacitance, and better high-frequency performance. However, they can be more difficult to work with due to their small size and lack of leads, and may require specialized equipment for placement and soldering.

Also read:- Summing amplifier, Norton theorem, High pass filter.

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]]>The post Capacitor and Capacitance – Formula, Uses, Factors Affecting & Application appeared first on Know Electronics.

]]>A capacitors is an electrical component that stores electrical energy in an electric field between two conductive plates. Capacitors are used in a variety of electrical circuits to store and release electrical energy. They play an important role in filtering, buffering, and smoothing electrical signals, as well as in coupling and decoupling circuit stages. The amount of electrical energy that a capacitor can store is determined by its capacitance, which is measured in Farads. Capacitors are used in many electrical and electronic applications, including power supplies, radios, televisions, computers, and automotive systems.

A capacitor works by storing electrical energy in an electric field that is created between two conductive plates. The conductive plates are separated by an insulating material known as a dielectric. When a voltage is applied across the plates, an electric field is established and charges of opposite polarity are stored on the plates. The magnitude of the electric field, and thus the amount of energy stored in the capacitor, is proportional to the voltage applied and the capacitance of the capacitor. The capacitance of a capacitor is determined by the size and separation of the plates and the type of dielectric material used.

When a voltage is applied across the capacitor, a current flow into the positive plate and out of the negative plate until the voltage across the capacitor reaches the applied voltage. At this point, the current stops flowing and the energy stored in the electric field is at its maximum. If the voltage across the capacitor is now reduced to zero, the energy stored in the electric field will be slowly released back into the circuit. This property allows capacitors to be used for a variety of applications, such as filtering, smoothing, and buffering electrical signals, as well as for storing and releasing energy in pulse and power circuits.

The value of capacitance, which is measured in Farads (F), can be determined from the following formula:

C = ε * A / d

Where: C is the capacitance in Farads (F)

Ε is the permittivity of the dielectric material, which is a constant that depends on the material used and its properties

A Is the area of the conductive plates in square meters (m^2)

D is the separation between the conductive plates in meters (m)

So, in essence, the capacitance of a capacitor is proportional to the area of its conductive plates and inversely proportional to the distance between them. Increasing the plate area or reducing the plate separation will result in a higher capacitance. The type of dielectric material used will also affect the capacitance, with different materials having different permittivity values.

In practice, the value of capacitance is often indicated on the physical component itself or in the manufacturer’s datasheet. Capacitors are available in a range of values, from a fraction of a picofarad (pF) to several microfarads (µF) or even farads (F). The value of capacitance required for a particular application depends on the specific requirements of the circuit and the desired capacitance range.

The energy stored in a capacitor is given by the following equation:

U = ½ * C * V^{2}

Where: U is the energy stored in the capacitor in joules (J)

C is the capacitance of the capacitor in Farads (F)

V is the voltage across the capacitor in volts (V)

So, in essence, the energy stored in a capacitor is proportional to the square of the voltage across it and the capacitance of the capacitor. This means that if the voltage across the capacitor is increased, the energy stored in the capacitor will also increase. The capacitance of the capacitor also affects the amount of energy stored, with larger capacitances storing more energy than smaller capacitances for a given voltage.

It’s Important to note that the energy stored in a capacitor can be released rapidly, which makes it useful for applications that require a quick release of energy, such as in pulsed power supplies or voltage regulators. However, the energy stored in a capacitor can also be dangerous, as it can discharge quickly and cause harm to people or damage to equipment. Therefore, it’s important to handle capacitors safely and to follow proper safety precautions when working with them.

- Microfarad (μF)1μF = 1/1,000,000 = 0.000001 = 10
^{-6}F - Nanofarad (nF)1nF = 1/1,000,000,000 = 0.000000001 = 10
^{-9}F - Picofarad (pF)1pF = 1/1,000,000,000,000 = 0.000000000001 = 10
^{-12}F

Then using the information above we can construct a simple table to help us convert between pico-Farad (pF), to nano-Farad (nF), to micro-Farad (μF) and to Farads (F) as shown.

Pico-Farad (pF) | Nano-Farad (nF) | Micro-Farad (μF) | Farads (F) |

1,000 | 1.0 | 0.001 | |

10,000 | 10.0 | 0.01 | |

1,000,000 | 1,000 | 1.0 | |

10,000 | 10.0 | ||

100,000 | 100 | ||

1,000,000 | 1,000 | 0.001 | |

10,000 | 0.01 | ||

100,000 | 0.1 | ||

1,000,000 | 1.0 |

A parallel plate capacitor Is a basic type of capacitor that consists of two parallel conductive plates separated by a dielectric material. The conductive plates are typically made of metal and can be in the form of flat sheets, cylinders, or spheres. The dielectric material is an insulator that is placed between the plates to increase the capacitance of the capacitor.

When a voltage is applied across the parallel plates, an electric field is established between them, and charges of opposite polarity are stored on the plates. The magnitude of the electric field and the amount of energy stored in the capacitor are proportional to the voltage applied and the capacitance of the capacitor. The capacitance of a parallel plate capacitor is given by the following equation:

C = ε * A / d

Where: C is the capacitance in Farads (F)

Ε is the permittivity of the dielectric material, which is a constant that depends on the material used and its properties

A is the area of the conductive plates in square meters (m^{2})

D is the separation between the conductive plates in meters (m)

There are several factors that can affect the capacitance of a capacitor:

**Dielectric Material: **

The type and properties of the dielectric material used between the conductive plates can have a significant effect on the capacitance of a capacitor. Materials with high permittivity values increase the capacitance of the capacitor, while materials with low permittivity values decrease it.

**Plate Area: **

The surface area of the conductive plates also affects the capacitance of a capacitor. Larger plate areas result in higher capacitances, while smaller plate areas result in lower capacitances.

**Plate Separation:**

The distance between the conductive plates, also known as the plate separation or plate spacing, also affects the capacitance. Smaller plate separations result in higher capacitances, while larger plate separations result in lower capacitances.

Shape of the Plates: The shape of the conductive plates also affects the capacitance of a capacitor. For example, the capacitance of a cylindrical capacitor is different from that of a parallel plate capacitor, and the capacitance of a spherical capacitor is different from that of a cylindrical capacitor.

**Temperature: **

The temperature of the dielectric material can also affect the capacitance of a capacitor. In general, the capacitance of a capacitor decreases as the temperature increases, although the exact effect of temperature on capacitance depends on the specific dielectric material used.

It’s Important to consider these factors when selecting a capacitor for a specific application or when designing a circuit that uses capacitors. Understanding how these factors affect the capacitance of a capacitor can help ensure that the capacitor meets the requirements of the application and operates correctly within the circuit

Capacitors have a wide range of applications in various fields, including electronics, electrical engineering, and power systems. Some of the most common applications of capacitors are:

**Power supplies:**Capacitors are commonly used in power supplies to store energy and smooth out voltage fluctuations. Filtering: Capacitors are used in electronic circuits to filter signals and remove unwanted noise or interference.**Coupling:**Capacitors are used in coupling circuits to pass only the AC component of a signal from one circuit to another while blocking the DC component.**Energy storage:**Capacitors can store large amounts of electrical energy, making them useful in applications such as backup power supplies and high voltage energy storage systems.**Decoupling:**Capacitors are used in decoupling circuits to provide a stable supply voltage for electronic components, preventing fluctuations in the power supply from affecting the performance of the components. Timing: Capacitors are used in timing circuits to determine the timing of events in electronic systems.**Signal processing:**Capacitors are used in signal processing applications to store and release electrical charges, enabling the processing of signals in a specific manner.**Lighting:**Capacitors are used in lighting applications to store energy and discharge it in short bursts, providing the high voltage needed to ignite gas discharge lamps.**Motor starting:**Capacitors are used in motor starting circuits to provide the high current needed to start an electric motor.

These are just a few examples of the many applications of capacitors. The versatility and usefulness of capacitors make them an essential component in a wide range of electronic and electrical systems.

Also read:- Binary number, Binary multiplications,

Here are answers to some frequently asked questions about capacitors and capacitance:

A capacitor is an electrical component that stores electrical energy in an electric field. It works by accumulating electrical charge on its conductive plates and using a dielectric material to separate the plates and maintain a separation between the charges. Capacitors are used in a wide range of electronic and electrical applications for tasks such as energy storage, filtering, coupling, decoupling, and timing.

Capacitance is measured in Farads (F), which is the unit of capacitance in the International System of Units (SI). The Farad is a large unit of capacitance, so smaller units such as the microfarad (µF) or picofarad (pF) are more commonly used in electronic applications.

The capacitance of a capacitor can be affected by several factors, including the type and properties of the dielectric material used between the conductive plates, the surface area of the plates, the distance between the plates, the shape of the plates, and the temperature of the dielectric material.

The capacitance of a capacitor can be calculated using a formula that depends on the type of capacitor and its specific geometry. For example, the formula for a parallel plate capacitor is given by C = ε * A / d,

where C is the capacitance, ε is the permittivity of the dielectric material,

A is the surface area of the conductive plates, and

d is the distance between the plates.

A capacitors can be charged by connecting it to a voltage source, such as a battery or a power supply. When the voltage is applied to the capacitors, electrical charge accumulates on the conductive plates, creating an electric field within the capacitors. The amount of charge stored on the plates and the electric field within the capacitors depend on the voltage applied and the capacitance of the capacitors.

A capacitors can be discharged by connecting it to a load, such as a resistor. When the load is connected to the capacitors, the electrical charge stored on the conductive plates will flow through the load and discharge the capacitors. The rate at which the capacitors discharges depends on the resistance of the load and the capacitance of the capacitor

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]]>The post What is Binary Coded Decimal or BCD and it’s Example? appeared first on Know Electronics.

]]>We have already seen the binary number that has n binary code represent in the 0 and 1 form. The advantage of binary decimal code is that the each decimal digit can be represented by a group of 4 binary digits. Or bit is much same as hexadecimal. So for the decimal digits (0-9) need four binary bits.

But don’t confuse in binary decimal code to hexadecimal both are not same, Whereas a 4-bit hexadecimal number is valid up to F_{16} representing binary 1111_{2}, (decimal 15), the binary code decimal stop at 9 binary 1001_{2}. This means that although 16 numbers can be represent using four bits for one and another four bits for six. The 16 representation in binary code decimal is 001 0110.

The main advantage of BCD is that it allow to easy covert to decimal, however the disadvantage is that BCD is wasteful as the states between 1010 (decimal 10), and 1111 (decimal 15) are not used. Nevertheless, binary coded decimal has many important applications especially using digital displays.

In BCD number the decimal number are separated in individual from and then each decimal number assign four bits of binary bits. So a 4 bits of binary digits display the individual of one decimal digit from 0000 for a zero to 1001 for a nine.

So for example, 357_{10} (Three Hundred and Fifty Seven) in decimal would be presented in Binary Coded Decimal as:

357_{10} = 0011 0101 0111 (BCD)

BCD is a weighted code and it is commonly called the **8421 code** as it forms the 4-bit binary representation of the relevant decimal digit.

Binary Power | 2^{3} |
2^{2} |
2^{1} |
2^{0} |

Binary Weight: | 8 | 4 | 2 | 1 |

The decimal weight of each decimal digit increase by factor of 10. In BCD code the binary weight of each digits increase by factor of 2 as shown in figure. Then the first digit has a weight of 1 ( 2^{0} ), the second digit has a weight of 2 ( 2^{1} ), the third a weight of 4 ( 2^{2} ), the fourth a weight of 8 ( 2^{3} ).

Then the relationship between decimal (denary) numbers and weighted binary coded decimal digits is given below.

Decimal Number |
BCD 8421 Code |

0 | 0000 0000 |

1 | 0000 0001 |

2 | 0000 0010 |

3 | 0000 0011 |

4 | 0000 0100 |

5 | 0000 0101 |

6 | 0000 0110 |

7 | 0000 0111 |

8 | 0000 1000 |

9 | 0000 1001 |

10 (1+0) | 0001 0000 |

11 (1+1) | 0001 0001 |

12 (1+2) | 0001 0010 |

… | … |

20 (2+0) | 0010 0000 |

21 (2+1) | 0010 0001 |

22 (2+2) | 0010 0010 |

etc, continuing upwards in groups of four |

Then we can see that **8421 BCD** code is nothing more than the weights of each binary digit, with each decimal (denary) number expressed as its four-bit pure binary equivalent.

As we already seen above, the conversions of decimal to binary code decimal is almost similar to the conversion of hexadecimal to binary from 0 to 9. The conversion of decimal to BCD, separate the decimal number into its weighted digits and write down the equivalent 4-bit 8421 BCD code representing each decimal digit as shown.

Using the above table, convert the following decimal (denary) numbers: 85_{10}, 572_{10} and 8579_{10} into their 8421 BCD equivalents.

85_{10} = 1000 0101 (BCD)

572_{10} = 0101 0111 0010 (BCD)

8579_{10} = 1000 0101 0111 1001 (BCD)

Note that the resulting binary number after the conversion will be a true binary translation of decimal digits. This is because the binary code translates as a true binary count.

The conversion of BCD to decimal is exactly opposite of the conversion of decimal to BCD. We can simply convert the BCD to decimal; firstly we create the group of four binary bits, starting with the least significant digit and then write the decimal digit represented by each 4-bit group. We add extra zero at the end if we need to make four bits of group. 110101_{2} would become: 0011 0101_{2} or 35_{10} in decimal.

Convert the following binary numbers: 1001_{2}, 1010_{2}, 1000111_{2} and 10100111000.101_{2} into their decimal equivalents.

1001_{2} = 1001_{BCD} = 9_{10}

1010_{2} = this will produce an error as it is decimal 10_{10} and not a valid BCD number

1000111_{2} = 0100 0111_{BCD} = 47_{10}

10100111000.101_{2} = 0101 0011 0001.1010_{BCD} = 538.625_{10}

The conversion of decimal to BCD or BCD to decimal is a relatively straight forward task but we need to remember the BCD and decimal is not a binary number. Even through them are representing in bits. The use of BCD is in microcomputer.

It is easy to code and decode. It is not too efficient to store numbers. In BCD the number for representation of decimal no need more binary bits which creates long length due to this it is not efficient to store.

- It is using integer arithmetic.
- It is enables easy conversion between machine-readable and human-readable numerals.
- Comparing with binary number system it is easy to decode and encode.
- It offers a fast and efficient system to convert decimal numbers into binary numbers.
- It is use in seven segment display, where it can be difficult to manipulate or display large numbers.
- It is also used in some currency applications where floating point representations are not completely accurate.

It has certain limitation which are given below-

- It is not efficient to store in computer memory.
- BCD increasers circuits complexity.
- It is wasteful above 10 to 16.

There are two types of binary-coded decimal.

In unpacked binary-coded decimal numbers, each 4-bit binary-coded decimal group is stored in a separate register inside a computer. The drawback here is that, if the registers are 8 bits or wider, space is wasted storing the binary-coded decimal numbers.

In packed binary-coded decimal numbers, a single 8-bit register stores two binary-coded decimal digits. By shifting the number in the upper register to the left four times and then adding the numbers in the upper and lower registers, the process combines two binary-coded decimal digits, which enables storage in a single register.

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]]>The post Summing Amplifier: Circuit Diagram and Its Applications. appeared first on Know Electronics.

]]>An operational amplifier an amplifier which can perform various types of operations like invert input signal, summing, integrating and differentiating operations. In this tutorial we are going to study about summing amplifier. What are the circuit configurations and applications of summing amplifier?

In electronics circuit, most of the time required two or more than two voltage required to add these voltages into a single signal. One of the best examples to add two input voltages is summing amplifier. The best example of use of summing amplifier is the Music Recording and Broadcasting applications.

This is where the Summing Amplifier comes handy, as it combines several inputs into one common signal without noise or interference. For this reason, the Summing Amplifier is also called as Voltage Adder as its output is the addition of voltages present at its input terminal.

As we know that the inverting operational amplifier has single input voltage (Vin) at inverting input. If we add more input resistance, each resistance have same value to the original input resistance (Rin) we end up with another operational amplifier circuit called a **Summing Amplifier**, “*summing inverter*” or even a “*voltage adder*” circuit as shown above.

From the circuit diagram, the output voltage is proportional to the sum of inputs voltage V_{1}, V_{2}, V_{3}, etc. than we can modify the original equation of inverting amplifier to take account of these new inputs thus:

If all input resistance is equal we can simplify the output voltage of:

We have an operational amplifier circuit that will amplify each individual voltage and produce an output voltage which is proportional to algebraic sum of the individual voltage of the three individual input voltages V_{1}, V_{2} and V_{3}. We can also add more voltage as per required.

The input voltages are effectively isolated from each other by the “virtual earth” node at the inverting input of the op-amp. A direct voltage addition can also be obtained when all the resistances are of equal value and Rƒ is equal to Rin.

Note when the summing point is connected to the inverting input of operational amplifiers the circuit output will produce negative sum of any number of input voltage. Likewise, when the summing point is connected to the non-inverting input of operational amplifier the circuit output will be positive sum of input voltage.

The scaling summing amplifier can be made, if individual input voltage is not equal than the summing equation would have to be modified to:

We can rearrange the above formula, the output voltage is

The summing amplifiers is very useful of when two or more signal needs to added or combine like in audio mixing applications. The sounds have a different musical instrument can be converted in a specific voltage level using transducer, and connected as a input to a summing amplifiers.

These different signals are combining in summing amplifier and sent to audio amplifier. Below the circuit is shown the a summing amplifiers as audio mixer

Digital to analog is use to converts the binary data to its analog voltage output. Now a day, industry use for control application is microcomputer. These microcomputers are digital data needs to convert to an analog voltage to drive the motors, relays, actuators, etc.

The circuit diagram of digital to analog convert using summing amplifier and weighted resistor network are shown below.

The circuit has four inputs Q_{A}, Q_{B}, Q_{C} and Q_{D}, the 5v represents the logic 1 and 0v shows the logical 0.

The summing amplifiers is also use for level shifter. The two inputs summing amplifiers is act as a level shifter, which one is AC and another input is DC.

The AC Signal will be offset by the input DC Signal. One of the major applications of such level shifter is in Signal Generators for DC Offset Control.

The summing amplifiers is the type of operational amplifiers that are use to combine two or more than two signal into a single output signal.

The summing amplifiers is a type of operational amplifiers that combine the two or more the two inputs voltage over a single output voltage. If we provide the input at the inverting terminal of summing amplifiers the output will be phase shift of 180 degree of input signal.

The output of summing is the combinations of two or more than two outputs. The phase shift of output will be depends upon the which terminal we applied input i.e. Inverting or non inverting terminal.

The summing is use for various types of application like audio signal processing, digital to analog converter and level shifter.

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]]>The post What is Cutoff Frequency? – formula and How to Find it Formula appeared first on Know Electronics.

]]>In an electronics system the cut-off frequency either below or above which the power output of a circuit, such as a line, amplifier, or electronic filter (e.g. a high pass filter) has fallen to a given proportion of the power in the pass band.

Most frequently this proportion is one-half the pass band power, also referred to as the 3 dB point since a fall of 3 dB corresponds approximately to half power. As a voltage ratio, this is a fall to approximately 0.707.

The filter circuits are electronics circuit that allows certain frequency and block remain frequency. The filters are various types. Here three type of filter are given below

As the name suggests itself, a low pass filter pass the low frequency signal whose range from the 0 Hz to a cutoff frequency and attenuate high frequency signal.

A high pass filter pass the high frequency signal whose range from the above the cutoff frequency and attenuate the low frequency which is below cut-off frequency.

A band pass filter pass the certain band of frequency which lie in between of high pass and low pass filters.

In signal the bandwidth is defined as the range or difference between two frequencies i.e. upper cut-off frequency and low cut-off frequency. The frequency F_{u} is the upper frequency and F_{L} is a lower frequency. We can also name these two frequencies as half –Power frequencies because at this range of frequency the voltage gain drops to 70.7 % of the maximum value.

This represents the power level of one–half of the power at reference frequency in the mid-range frequency. Since the change is not noticeable, the audio amplifier has a flat response from f1 to f2.

The formula for cut-off frequency is given below:

**F _{C} = 1/ 2πRC**

Where,

R → Resistance.

C → Capacitance.

For example the cut-off (3dB point) frequency of RC low pass filter is defined as when the resistance is the same magnitude as the capacitive reactance.

The gain is usually express in decibel. The unit decibel comes from logarithmic response of the human ear to intensity of sound. The decibel is defined as logarithmic measurement the ratio of output power to input power. It can also express in term of voltage.

Generally the voltage gain of amplifier is express in term of decibel. The voltage gain is given by 20log** A _{v}**. The power of amplifier is also express in term of decibel (dB). The power gain is given by 10log

When voltage gain **A _{v }**is greater than one, we can say the gain is positive. It shows the amplification. When voltage gain

In amplification under few circumstances, the 0 dB gain use as reference. Which are use to compare with another value of gain.

The amplifier has maximum gain at mid frequency range and minimum at low frequency. The maximum gain is called mid frequency range gain.

There is various ways to find out cut-off frequency.

Analysis of a circuit with an altering frequency of sinusoidal sources is termed as the frequency response of a circuit. The ratio of transfer functions of output to input voltage in s domain.

**H(s) = Vo(s) / Vi(s)**

When using a sinusoidal source, the transfer function will be given as the magnitude and phase of the output voltage to the magnitude and phase of the input. In such case jω will be use in place of “s”.

**H(s) = Vo(**jω**) / Vi(**jω**)**

For example, consider the transfer function

**H(s) = 20(s+10) / (s+100)**

To obtain the corner frequency from the above equation, H(s) can be replaced as

So from this equitation, the cut-off frequency ω1 is 10 rad/s and ω2 is 100 rad/s

The above graph shows the bode plot. This graph is commonly use control system engineering for determine the stability of system. The graph plotted between phase (degree) and angular frequency (radian per second). In the bode plot, the corner frequency is the frequency at which the two asymptotes meet each other or cut each other.

The transfer function **H(s) = Vo(**s**) / Vi(**s) of a system carries extensive information of gain and stability. Bode plots give an estimated picture of a given **H(s) **from

The low pass filter can only allow low frequency, it cannot allow high frequency. The low pass filter have certain cut-off frequency, above the cutoff frequency the voltage drops below 70.7% of its input voltage. . The frequency, at which the magnitude response is 3 dB lower than the value at 0 Hz, is known as Cut-off Frequency of a low pass filter.

The high pass filter only passes the high frequency. It blocks the low frequency below the cut-off frequency. The graph of high pass filter is shown below.

The band pass filter consists of two cut-off frequencies. This filer made of low pass filter and high pass filter. The first cutoff frequency is from high pass filter is known as high cutoff frequency and second cutoff frequency is taken from low pass filter which is known as low cutoff frequency.

The cut-off frequency is a specific frequency at which the output voltage drops by a factor of 70%. The low pass filter allows frequency in between 0 Hz to cut-off frequency. Above the cut-off frequency low pass filter attenuate the output voltage.

The decibel is a logarithmic scale expressed as 20 log(Output power/Input power). A -3 dB gain corresponds to an (Output power/Input power) ratio of 0.5, i.e., **the output power of the circuit reduces by a factor of half**. This serves as a standard reference for defining the boundary in the frequency response of a system.

The cutoff frequency is a frequency either below and above the signal are pass.

The cutoff frequency is defined as the frequency where the amplitude of H(jω) is 1√2 times the DC amplitude (approximately -3dB, half power point). Solve it for ωc (cutoff angular frequency), you’ll get 1RC. Divide that by 2π and you get the cutoff frequency fc.

**The frequency at which the magnitude response is 3 dB lower than the value at 0 Hz**, is known as Cutoff Frequency of a low pass filter.

A cut-off frequency refers to **the frequency at which the impulse response of a filter starts to fall off**. A resonant frequency refers to a frequency at which there’s a strong peak in the impulse response so that tone stands out from its neighbors. For example, think of a tuning fork.

It’s **because decibels are logarithmic, and the log (base 10) of 3 is about 50% power**. So the 3 decibel cutoff is where power drops off by a half. 3 dB implies 1/2 the power and since the power is proportional to the square of voltage, the voltage will be 0,707 of the pass band voltage.

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]]>The post What is High Pass Filter? – Circuit Diagram, Characteristics & Applications appeared first on Know Electronics.

]]>The high pass filters is electronics circuits that can pass frequencies are higher than cutoff frequencies and below cutoff frequencies attenuated. The values of cutoff frequency are defined by circuit elements.

The high pass filters circuit has one capacitive element and resistive element. The circuit of high pass filters is shown below. In this arrangement the reactance of capacitor is high for low frequency so the capacitor act as a open circuit and block input signal but in high frequency the reactance of capacitor are low now the circuit act as a short circuit allowing all of the input signal to pass directly to the output as shown below in the filters response curve.

The frequency response or bode plot of high pass filter is exactly opposite to the low pass filter. From the graph which are shown above the low frequency signal are attenuated with the output increasing at +20dB/Decade until the frequency reaches the cut-off point ( ƒc ). At cutoff frequency the reactance of capacitor and resistor becomes same .i.e. R = Xc. The gain before the cutoff frequency is called stop band and above the cutoff frequency i.e. -3 dB points is known as the **passband**. At cutoff frequency, point output voltage amplitude will be 70.7% value or -3dB (20 log (Vout/Vin)) of the input value of the input voltage.

Gain is calculated as **Gain (dB) = 20 log (Vout/Vin).**

The phase angle (Φ) of high pass filter leads of input by **+45 ^{o}** at frequency ƒc. The phase angle of high pass filter is calculated by:-

**∅ ****= arctan (1/2πfRC)**

The formula for Cut-off Frequency of high pass filter is same equation as that of the low pass filter and the formula for phase shift high pass filter is given below:

Circuit gain Av, is Vout/Vin and is calculated as:

Find out cutoff frequency ( ƒc ) of high pass consisting of an 82pF capacitor connected in series with a 240kΩ resistor.

The second order high pass filter can be formed by cascading two first order filter the figure of second order filter are shown in figure.

The above figure show the second order of filter which is made by connecting two first order filters in series manner. Using this process the first order filter is converted into second order filter and It is also called two-pole high pass network. Due to this result a slope of 40dB/decade (12dB/octave).

The cutoff frequency of second order is calculated by as follows:-

In practical, the cascading of two filter produce large order filters. However to reduce the loading effect we make the impedance of each following stage 10x the previous stage, so R_{2} = 10*R_{1} and C_{2} = 1/10th of C_{1}.

A high pass filter that pass high frequency or above the cutoff frequency and attenuate the low frequency. The RC high pass is constructing with the help of resistor and capacitor. The input is applied to capacitor. The circuit diagram of high pass RC is given below.

**Low pass :** The low pass is a type of filters that can pass only low frequency or below cutoff frequency. And it attenuates high frequency. It is use for smoothing the image.

**High pass :** The high pass are one of the types of filters that are use to attenuate the low pass and allow to high frequency, or above the cutoff frequency. It is also use for sharpening the image.

Low pass |
High pass |

Smoothing image. | Sharpening image. |

It attenuates high frequency or above cutoff frequency. | It attenuates the low frequency or below cutoff frequency. |

Low frequency is preserved in it. | High frequency is preserved in it. |

It allows the frequencies below cut off frequency | It allows the frequencies above cut off frequency. |

The input applied to resistor that is followed by capacitor. | The input applied to capacitor that is followed by a resistor. |

It is use for removal of aliasing effect. | It is use for removal of noise. |

G(u, v) = H(u, v) . F(u, v) | H(u, v) = 1 – H'(u, v) |

The high pass filters are use in many applications:-

- It is used in speakers for amplification.
- It is used to remove unwanted sounds near to the lower end of the audible range.
- It is used for AC-coupling.
**For Image Processing**: High pass filters are used in image processing for sharpening the details.

The function of high pass filter is reverse of low pass filter. This filters attenuate signal below cutoff frequency and pass above cutoff frequency. The voltage gain at cut-off frequency is 70.7% or **-3dB** that allow to passes.

The frequency below the cutoff is known as stop band while above the cutoff frequency is known as pass band. The cutoff frequency are calculated by using ƒc = 1/(2πRC). The phase angle of output signal**+45 ^{o}** at cutoff frequency (ƒc). Generally the high pass has low distortion then low pass due to the higher operating frequencies. It is commonly use in audio amplifier as a coupling between two stages of amplifiers, image processing etc.

The output voltage are depends upon the time constant and frequency of input signal. When we applied an alternating current to input side of circuit, it behaves like a simple first order high pass. But when we applied square wave to the input side of circuit, the shape of output is like a vertical step input.

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]]>The post What is Nortons Theorem? – How to Find a Norton Equivalent Circuit and Solved Examples appeared first on Know Electronics.

]]>The Nortons theorem on the other hand the circuit reduces a single resistance in parallel with single current source. In this tutorial we are going to discuss the concept of Norton theorem with example.

It is state that “ Any linear network and complex that contain many voltage and resistive element can be replace with a simple circuit that have a constant current source in parallel with a Single Resistor“.

The Norton’s theorem is just a current source transformation of Thevenin’s theorem and dual with Thevenin’s theorem.

The Norton circuit contain the load resistance R_{L} is concerned this single resistance R_{S} the value of R_{S} measure from the terminal A and B by short circuit the voltage source and open for current source, (the same as Thevenin).

Let us understand the theorem, in below circuit consist of the voltage source and more than two resistances.

After conversion into Norton

In above circuit that contain current source whose provide a constant amount of current, outputting as much or as little voltage necessary to maintain that constant current.

In Thevenin’s theorem “Any linear and complex circuit that contain many voltage sources or resistances are replace with simple circuit. And that simple circuit have single voltage source in series with single resistance” and in Norton theorem’s we find out the equivalent current source (I_{Norton}) and equivalent resistance (R_{Norton}).

In Norton theorem firstly we identify the load and remove the load from the circuit diagram

After that we find the Norton current (for the current source in the Norton equivalent circuit) by short circuit connection to load and find out the current. Note that this staple is exactly opposite to the Thevenin’s Theorem, where we replaced the load resistor with a break (open circuit):

There is no voltage drop across the load, the current through R_{1} is strictly a function of B_{1}‘s voltage and R_{1}‘s resistance: 7 amps (I=E/R) and the current through R_{3 }is now strictly a function of B_{2}‘s voltage and R_{3}‘s resistance: 7 amps (I=E/R). The total current through short circuit branch through short circuit branch:

I_{R1} + I_{R2 }= 7 Amp +7 Amp

= 14 Amp

The 14 Amp becomes the Norton current and shown in above figure

For finding the Norton resistance we do exactly same thing as we did for calculating Thevenin resistance R_{th}. Take the original circuit, short circuit voltage source and open circuit the current source. Calculate the equivalent resistance across the load. And figure total resistance from one load connection point to the other:

The final Norton equivalent circuit looks like this:

For finding the voltage across the load resistance we reconnect our original resistance 2 Ω, and analysis the circuit as a simple circuit.

Find out the Norton equivalent circuit.

For Find out the Norton equivalent of the above circuit, firstly we remove the load resistor 40Ω and terminals A and B to give us the following circuit.

When the terminal A and B is short circuit the two resistances is 10 Ω and 20 Ω are parallel across their two respective voltage sources. And the current through each resistor as well as the total short circuit current can now be calculated as:

**with A-B Shorted Out**

The two voltage source are short circuited and open A and B terminal. The two resistances are now effetely connected in parallel. The total resistance calculated across the A and B terminal.

**Find the Equivalent Resistance (Rs)**

Now draw the Norton from the Norton equivalent circuit, connecting the current source in parallel with equivalent resistance.

**Nortons equivalent circuit**

Now we connect the load 40Ω across the A and B terminal.

Again, the two resistors are connected in parallel and calculate the total resistance.

Calculate the voltage across the terminal A and B with load resistor connected is given as:

Then the current flowing in the 40Ω load resistor can be found as:

The basic procedure for solving a circuit using **Norton’s Theorem** is as follows:

- Remove the load resistor R
_{L}. - Find R
_{n }by shorting all voltage sources or by open circuiting all the current sources. - Find I
_{n }by placing a shorting link on the output terminals A and B. - Find the current flowing through the load resistor R
_{L}.

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]]>The post Thevenin Theorem: What is it? Solved Example with Step by Step Procedure appeared first on Know Electronics.

]]>The thevenin’s theorem states that “Any liner network consisting several voltages and resistances can be replace by a single equivalent voltage in series with a single resistance connected to load in series combination”. In other word the thevenin theorem are simplify any complex network, to replace by simple equivalent two terminal voltage just by single voltage and equivalent single resistor in series with load. The thervenin’s complex and equivalent circuit bock diagram are shown below:-

This theorem is also applicable for analysis of power, but in superposition theorem power cant analysis through it except, three exceptions that exception are given below:-

- Sinusoidal source have different frequency on the network.
- Two sinusoidal sources have same frequency but phase difference is 90 degree operating on network.
- In DC network, power is individual sum.

“In any liner complex network is equivalent to one voltage in series with a single resistance connected to load in series combination”.

In thevenin’s circuit the load resistance R_{L} is concerned, in any complex “one-port” network consisting a multiple voltage and resistance element can be replace by single equivalent R_{s} and single equivalent voltage source V_{s}. R_{s} is the source resistance value looking back into the circuit and V_{s} is the open circuit voltage at the terminals.

Let us understand the thevenin’s theorem with an example

**Example: **find out the thevenin voltage and resistance. the circuit diagram are shown below.

**Step 1: **The analysis of the above circuit diagram using Thevenin’s theorem, firstly we remove the load resistance, in this case, 40 Ω.

**Step 2: **Remove all voltage sources internal resistance by shorting all the voltage sources connected to the circuit, i.e. v = 0. If any current sources are present in circuit, then remove the internal resistance by open circuiting the sources.

**Step 3: **Find the value of equivalent resistance, Rs is found by calculating the total resistance looking back from the terminals A and B with all the voltage sources shorted and the load resistance removed. We then get the following circuit.

Find the Equivalent Resistance (Rs)

** **

**Step 4: **Find the equivalent voltage.

To calculate the equivalent voltage source, We now need to reconnect the two voltages back into the circuit, and as V_{S} = V_{AB} the current flowing around the loop is calculated as:

This current of 0.33 amperes (330mA) is same for 20Ω resistor or the 10Ω resistor and it is determine by as follow:

V_{AB} = 20 – (20Ω x 0.33amps) = 13.33 volts.

or

V_{AB} = 10 + (10Ω x 0.33amps) = 13.33 volts, the same.

**Step 5: **Draw the Thevenin’s equivalent circuit. The Thevenin’s equivalent circuit element will be 6.67 Ω series resistance and a 6.67 Ω voltage source. In below circuit diagram show it.

The current flowing in the above circuit will be calculated as:

In an AC and DC circuit the thevenin’s theorem can be apply. But it should be a liner circuit.

Find out the* thevenin voltage (V _{TH}), thevenin resistance (R_{TH}*

**Solution:**

**Step 1: **firstly we remove the load resistance 5 kΩ from the circuit.

**Step 2: **calculate the open-circuit voltage. This open circuit voltage wills you the Thevenin’s voltage (V_{TH}).

**Step 3: **find out the current that follow thought resistor 12 kΩ and 4 kΩ resistors by applying current division rule.

**I = 48 V /( 12 kΩ + 4 kΩ) = 3 mA**

The voltage across the 4 kΩ resistors will be:

= 3 mA x 4 kΩ = 12 V

There is no current follow through 8 kΩ resistor, so there is no voltage drop across it and hence the voltage across the terminals AB is same as the voltage across the 4 kΩ resistor. Therefore, AB terminal voltage is12 V. Hence, the Thevenin’s voltage, V_{TH} = 12 V.

**Step 4: **now we calculate the thevenin resistor across the AB terminal.

**From figure, **the 8 kΩ resistor is in series with the parallel connection of 12 kΩ and 4 kΩ resistors. So thevenin resistor will be calculated as follow:

8kΩ + (4k Ω || 12kΩ)

R_{TH} = 8 kΩ + [(4 kΩ x 12 kΩ) / (4 kΩ + 12 kΩ)]

R_{TH }= 8 kΩ + 3 kΩ

R_{TH} = 11 kΩ

**Step 6: **Now, from a new circuit diagram by connect the** **R_{TH} in series with Voltage Source V_{TH} and the load resistor.

**Step 7: the final stage is to calculate the voltage and current across the load by using Ohm’s law as follow:** I_{L} = V_{TH} / (R_{TH} + R_{L})

I_{L} = 12 V / (11 kΩ + 5 kΩ) = 12 V/16 kΩ = 0.75 mA

The load voltage is determined as follows:

V_{L} = 0.75 mA x 5 kΩ = 3.75 V

- It is use for analysis of power systems.
- It is used to measurement resistance using the Wheatstone bridge and source modeling.

- It is only applicable for liner circuit.
- The power dissipation applicable for calculating power dissipation.

**What is Thevenin’s theorem formula?**

The thenenin voltage or open circuit voltage is **V _{Th}=V_{oc} **across the load terminal. The thenenin voltage formula is:

**I=V _{th}/R_{th}**

**What is basic difference between Norton’s theorem and Thevenin’s theorem?**

The basic difference between thenenin and Norton is – the Norton’s theorem is using for a current source, whereas Thevenin’s theorem uses a voltage source. In thevenin’s theorem the thevenin resistor in series with load but in Norton’s theorem is the resistor is use in parallel with load.

**What is Thevenin’s theorem used for?**

By using thevenin’s theorem any liner complex circuit is transfer into a simple circuit, which typically has a load that changes value during the analysis process.

**What are the limitations of Thevenin theorem?**

Limitations of Thevinen’s Theorem

**it is not applicable for non linear elements, **also to the unilateral networks it is not applicable. There should not be magnetic coupling between the load and circuit to be replaced with the thevinen’s equivalent.

**What is Z in Thevenin’s theorem?**

The Thevenin impedance is the impedance that is to be calculated by short circuit voltage source and open circuit for current source. Z_{Th} = **ohms at degrees**.

**What is relationship between Thevenin and Norton Theorem?**

Thevenin and Norton’s are dual. The Norton’s theorem is the current source transformation of thevenin’s volage. The resistor of both theorems is same. The thevenin voltage is equal to the multiplication of Norton current and Norton resistance. And Norton current is equal to Thevenin.

**What is Thevenin’s resistance?**

The thenenin resistor is the part of thevenin’s theorem. The thevenin resistor is calculated by short circuit the voltage and open circuit for currant. Look the all series and parallel resistor in front from load.

**What is Thevenin’s voltage Vth?**

The thevenin voltage is to be calculated across the load using Kirchhoff’s law. And that voltage will be Thevenin’s voltage Vth.

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]]>The superposition theorem is use for solve the network where more than one source are present.

Superposition theorem states the following:

“If more than one independent source is present in an electrical circuit, then the current through any one branch of the circuit is the algebraic sum of the individual effect of source at one time.”

In this method, we will consider only independent source at a time. So we have to element the other source from the circuit. We can element the voltage source by short circuiting of two terminals and the current source are eliminated by opening their two terminals.

- When you sum the individual contribution of each source, you should be careful while assigning signs to the quantities. The reference direction is use for assign positive and negative sign to each unknown quantity. If contribution source is in same direction as the reference direction, take positive sign. And if it is opposite direction, than take negative sign.
- It is only applicable for liner network.
- The superposition theorem is not applied for power because the power is not a liner quantity.

- Select the one source from the multiple sources present in the bilateral network. And eliminate the remaining independent source present in network.
- Except the selected source, all source must be replace by their internal impedance.
- Evaluate the current or voltage drop across a particular element in a network.
- Upon obtaining the respective response for individual source, and perform the algebraic sum of all individual drop or current in particular branch.

Let us understand the superposition theorem with solve example.

** Example: **Using the superposition theorem find the current flowing through 20 Ω.

**Solution****:**

**Step 1: **firstly we consider the 20V and find out the current in circuit. The current source can be open-circuited. The modified circuit is given below.

**Step 2:** Using node analysis at V_{1}.

The nodal equation is:

The current in 20 Ω resistor is:

I_{1} = V_{1} /10+20

Put V_{1} = 12 v in the above equation, we get

*I*_{1} = 0.4 A

Therefore, the currant in 20 ohm resistor is 0.4A.

**Step 3: **now considering the 4 ampere current in a circuit and eliminate the 20 volt by short circuiting of two terminals. The new circuit is shown below.

The above circuit diagram, the resistors 5 Ω and 10 Ω are parallel to each other, and this parallel combination are series with 10 Ω resistances. Then the equivalent resistance will be R_{AB} =40/3

Now, the new circuit is shown as follows:

The current in 20 Ω resistor can be determined by current division rule.

I_{2} = I_{s} (R_{1} /R_{1}+R_{2})

I_{2} = 1.6 A

Therefore the current following through 4A current source is 1.6 A.

**Step 4: **The algebraic sum of currents I_{1} and I_{2} will give us the current flowing through the 20 Ω resistor.

Mathematically, this is represented as follows:

I = I_{1} + I_{2}

Put I_{1} and I_{2} value in the above equation, we get

I = 0.4+1.6 = 2 A

The current flowing through the resistor 20 Ω is 2 A.

- It is not applicable for non-liner network circuit.
- It is not suitable for find out power.
- It is only applicable of more than one source.

**The current in a liner network is equal to the algebraic sum of the current produce by individual source.** To evaluate the superposition theorem consider only one source and other source replace all other voltage sources by short circuits and all other current sources by open circuits.

**The step for solve circuit using superposition theorem;-**

- Only consider one EMF or current source at a time.
- The voltage source, replace it with a short circuit.
- The current source, replace it with an open circuit.

The superposition theorem is simple and it is use when two or more the two voltage or current source. It is mainly used **to shorten the calculations of the circuit**.

The superposition theorem, is state that one than one independent source in any liner network branch is algebraic sum of individual source at a time.

**Also read**:- Optocoupler/optoisolator, Diode current equation.

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]]>The post Optocoupler or Optoisolator – Construction, Working Principle and How it works? appeared first on Know Electronics.

]]>The optocoupler consist a light emitting diode (LED) and phototransistor in the same opaque package. The other types of optocouple are the combination of LED-photodiode, LED-LASCR, and lamp-photoresistor pairs. The optoisolator are use for transfer digital and analog signals.

As we know transformer has no physical connection between primary and secondary side. In step down Transformer they provide electrical isolation between the higher voltage on the primer side and low voltage on secondary side.

The function of transform is based on mutual induction between two cores. This means transformer isolate the primer input voltage from secondary output voltage using electromagnetic coupling and this is achieved using the magnetic flux circulating within their laminated iron core.

Beside the transform we can also provide the electrical isolation between input and output side by using light and phototransistor and it is called optocoupler.

Isolation circuits are that circuit which does not have any physical common conductor in between one side to another side and proper isolation is maintained.

We know the massage signal highly contain distortion and noise in it which can be beyond the tolerance limit of the logic circuit at the output end during transmission. In this case the optocoupler is use to prevent the noise and distortion. The optocoupler can work on DC and AC high current.

The photocoupler or optocoupler contain two basic thing which is described below-

Light emitter: the light emitter is on the input side which takes electrical signal and this electrical convert into light signal. Typically the light emitter is a light emitting diode.

Light detector: the function of light detector within an optocoupler or photocoupler is converting the incoming light signal form light emitting diode into original electrical signal. The light detector may be photo diode, photo transistor or photodarlington etc.

The light emitter and light detector are tailored to match one another, having matching wavelengths so that the maximum coupling is achieved.

The symbol of optocoupler is use for indication of structure and function. The symbol contains the light emitter element LED and light detector such as phototransistor, photodiode and light sensitive device etc. the symbol are shown below.

The optocoupler use in AC power application is based around a diac.

The above figure shows the internal structure of optocoupler. The input pins are 1 and 2. It is a light emitting diode terminal. This led emit **infrared light** to **photosensitive transistor** on the right side. The output is taken form pin 3 and 4 which is the collector and emitter terminal of phototransistor. The output is depends on the light emitting diode. The LED is control by an external circuit. There is no any conducting material present in between photodiode and phototransistor. It is electrically isolated. The hole structure is packed in glass or transparent plastic, the electrical isolation is much higher, typically **10 kV** or higher.

The photosensor is the output circuit that can detect the light, the output will be DC and AC. Firstly the current is applied to the LED that emit the infrared light proportional to the current going through the device. When light fall on photosenstor a current flow, and switched ON. When input is interrupted, the IR beam is cut-off, causing the photosensor to stop conducting.

There are several specifications that need to be taken into account when using opto-couplers and opto-isolators:

The current transformer ratio is a ratio of output current in photosenstor to input current in LED. The current transfer ratio (CTR) are vary according to the type of optocoupler is use in output side. The photdarlington type has much higher than ordinary phototransistor.*Current transfer ratio, CTR:*In order to understand the maximum data rate in it. It is necessary to know the bandwidth. The phototransistor type bandwidth is 250 kHz and photodarlingtons bandwidth is around tenth digits.*Bandwidth:*It is current required for LED. This current is control by using resistor.*Input current:*-
The optocoupler use photosensor using transistor. The maximum output is equal to V*Output device maximum voltage:*_{CE(max)}for the transistor.

There is slightly difference between optocoupler and solid state relays which is described below-

- The solid state relay are use as a electronics switch to control the AC and DC power.
- The solid sate relay provides the high resistance or isolation between input and output. It is also based on opto-coupling technology
- The main difference between optocoupler and solid state switch is that optocoupler is use for low power application and solid state relay are use for high power application up to 100 of volts
- The optocoupler are pack in thin IC. However the solid state relay is high package and it use heat sink.

Device type |
Source of light |
Sensor type |
Speed |
Current transfer ratio |

Resistive opto-isolator (Vactrol) |
Incandescent light bulb | CdS or CdSe photoresistor (LDR) | Very low | <100% |

Neon lamp | Low | |||

GaAs infrared LED | Low | |||

Diode opto-isolator | GaAs infrared LED | Silicon photodiode | Highest | 0.1–0.2% |

Transistor opto-isolator | GaAs infrared LED | Bipolar silicon phototransistor | Medium | 2–120% |

Darlington phototransistor | Medium | 100–600% | ||

Opto-isolated SCR | GaAs infrared LED | Silicon-controlled rectifier | Low to medium | >100% |

Opto-isolated triac | GaAs infrared LED | TRIAC | Low to medium | Very high |

Solid-state relay | Stack of GaAs infrared LEDs | Stack of photodiodes driving a pair of MOSFETs or an IGBT |
Low to high | Practically unlimited |

The above figure shows the phototransistor. This contains the LED and phototransistor. At input terminal we provide voltage the LED produce infra ray. This infra ray active the phototransistor and gave some output. If the input current interrupts, there is no output current.

**The optocoupler is available in four type,** each one having an infra-red LED source but with different photo-sensitive devices. The four optocouplers are the:

*Photo-transistor**Photo-darlington**Photo-SCR**Photo-triac*

The four types are shown below.

- It allow easy interfacing with logic circuits.
- It provides Electrical isolation to for circuit protection.
- It is suitable for wideband signal transmission.
- It is small in size and lightweight device.

- Its operation is slow.
- It is not suitable for high power applications.

Optocouplers or of optocoupler are used in,

- It is use for Lamp Ballasts
- In Light Dimmers
- It is use in Valve or Motor Controllers
- Microcontrollers for interfacing with High Voltage Circuits.

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