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.

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

]]>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}.

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

]]>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 post Superposition Theorem – Steps and Circuit Analysis with Solved Example appeared first on Know Electronics.

]]>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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]]>The post What is Diode Current Equation? appeared first on Know Electronics.

]]>The diode current equation shows the relationship in between the current flow through diode as a function of the voltage applied across it. The current through the diode is does not change linearly with respect to applied voltage. The relation in between voltage and current has the exponential. The region behind of non-linear current is that the resistance of diode is temperature dependent. When the temperature increase the diode current is increases. The mathematically the diode current equation can be expressed as:

…………………(1)

Where,

- I is the diode current
- I
_{0}is the dark saturation current, - q is the charge on the electron,
- V is the voltage applied across the diode,
- η is the (exponential) ideality factor.
- K is the Boltzmann constant which is equal to K = 1.38 x 10
^{-23 }J/K - T is absolute temperature in Kelvin.

In diode current equation the two parameters require to be discussed in quite detail. These parameters are given below:

- Dark Saturation Current, I
_{0}. - Exponential Ideality Factor, η.

When the diode is in reverse bias the current flow through the diode is called reverse saturation current. The region behind of flow of reverse current is due to minority carrier. The range of this current in μA to nA. It is an important parameter of a diode characteristic and indicates the amount of recombination which occurs within it.

That is the value of I_{0} will be high when recombination rate is high and vice versa. The dark saturation current is directly proportional to the absolute temperature and inversely proportional to the material quality.

The exponential identical factor is the nearness of ideal diode, how accurately the diode follows the ideal diode equation. If the identical factor is 1 the diode is almost same as ideal diode. The identical factor for germanium is 1 diode and 2 silicon diode. This factor are depends on the following factor which are mention below-

- Electron Drift
- Diffusion
- Carrier Combination in the depletion region
- Doping Level
- Manufacturing Process
- Purity of the material

The value of η is “1” for Silicon diode and “2” for Germanium diode

When the diode is forward bias, the current through diode will be high in the range of milli ampere and the diode current equation becomes

On the other hand when the diode is reverse bias, the exponential term of the diode current equation is neglected and the current becomes:

I = – I_{0}

Now let us understand the mode of **diode current equation**. When diode are operate at room temperature. In this case, T = 300 K, also, k= 1.38 × **10 ^{-23} Jk^{-1}** and q = 1.6 ×

q/KT = 1.6 ×

= 0.003865 ×

At room temperature the diode equation becomes as-

I = – I_{0} **e ^{-v/0.025}**

The diode has non liner device. The diode current is

I = – I_{0} **e ^{-v/0.025}**

The diode currant is show the relationship between the current flowing through the diode when we applied the voltage across the diode.

Initially when we applied the voltage through diode the current flow through it and the graph of current is slowly at first, then more quickly, and eventually very quickly. This occurs because **the relationship between a diode’s forward voltage and its forward current is exponential rather than linear**.

“The thermal voltage VT is approximately **25.8563 mV at 300 K (27 °C; 80 °F)**.

Fundamentally, a diode is a component that **permits current to flow in a single direction and blocks it in the other direction**. Diodes allow current to flow in one direction without the effects of any impedance, while entirely blocking all flow of current flow in the other.

What is a Current Limiting Diode? A Current Limiting Diode, also known as a “Current Regulating Diode” or a “Constant Current Diode”, per- forms quite a unique function. Similar to a Zener diode, which regulates voltage at a particular current, **the CLD limits or regulates current over a wide voltage range**.

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]]>The post What is Two’s Complement? – And its Examples | How To Convert Binary Number Into 2’s Complement appeared first on Know Electronics.

]]>The logic circuit of 2’s complement are designed by using AND, OR and NOT gates. The logic circuit of two’s complement of five bit binary number is as follows:

**Two’s Complement Table**

Binary Number |
1’s Complement |
2’s complement |

0000 | 1111 | 1111 +1 = 0000 |

0001 | 1110 | 1110 +1 = 1111 |

0010 | 1101 | 1101 +1 = 1110 |

0011 | 1100 | 1100 +1 = 1101 |

0100 | 1011 | 1011 +1 = 1100 |

0101 | 1010 | 1010+1 = 1011 |

0110 | 1001 | 1001+1 = 1010 |

0111 | 1000 | 1000+1 = 1001 |

1000 | 0111 | 0111+1 = 1000 |

1001 | 0110 | 0110+1 = 0111 |

1010 | 0101 | 0101+1 = 0110 |

1011 | 0100 | 0100+1 = 0101 |

1100 | 0011 | 0011+1 = 0100 |

1101 | 0010 | 0010+1 = 0011 |

1110 | 0001 | 0001+1 = 0010 |

1111 | 0000 | 0000+1 = 0001 |

There is simple method to convert a binary number to Two’s complement. For finding the Two’s complement and any binary number, simply we convert the binary number into 1’s complement and add 1 to the least significant bit (LSB) of the 1’s complement. For good understanding of two’s complemented we discuss the four bit two’s complement examples.

**Eg **– Obtained the 2’s complement of binary bits 10101110.

First we convert binary bit 10101110 into 1’s complement is 01010001. Then add 1 to the least significant bit of the 1’s complement 01010001+1= 01010010 (2’s complement)

**Eg **− Obtained the 2’s complement of point binary bits 10001.001.

First we convert binary bit 10001.001 into 1’s complement is 01110.110. Then add 1 to the least significant bit of the 1’s complement 01110.110 +1= 01110.111 (2’s complement)

** **The use of behind of 2’s complement is that to perform the subtraction operation of two binary bits in digital computer. The computer only understands the binary number, it doesn’t understand the negative number is binary number system, but it is absolutely necessary to represent a negative number using binary number. The representation of sign binary number is done by 2’s complement.

For example represent the -5 and +5.

+5 are represented as using sign magnitude method but the representation of -5 using the following steps.

**Step 1** – Firstly +5 converted in binary using sign magnitude method. The +5 is 0 0101.

**Step 1** – Take 2’s complement of 0 0101 and the result of 2’s complement is 1 1011. The most significant bit of 2’s complement is 1 which indicates that number is negative. In the negative number, the MSB is always 1.

The advantage of 2’s complement is that the 0 has only one representation for -0 and +0 is always consider +0 in 2’s complement representation. The 2’s complement is most popular as 1’s complement because it has unique or unambiguous representation.

Now let’s see the some arithmetic operation of 2’s complement.

There are some following steps for the subtraction of two binary number using 2’s complement.

- Step 1 – Take 2’s complement of the subtrahend
- Step 2 – Add with minuend
- If the result not produces any carry then we take the 2’s complement of the result and it will be negative.
- If the result produces a carry bit 1, then we discard the carry bit and after this the result will be positive number.

Note that subtrahend is numbers that are use to subtracted from another number, i.e. minuend. And also note the adding end-around-carry-bit is only occurs in 1’s complement arithmetic operations but not 2’s complement arithmetic operations.

**Example** − Evaluate 10101 – 00101

Firstly we take the 2’s complement of subtrahend 00101, which will be 11011, and then we perform the addition operation. So, 10101 + 11011 = 1 10000. Since this result has carry bit so we discard the carry bit and the final result will be 10000 will be positive number.

**Example** − Evaluate 11001 – 11100

Firstly we take the 2’s complement of subtrahend 11110, which will be 00100, and then we perform the addition operation. So, 11001 + 00100 = 11101. Since in this operation there is no generation of any carry bit, so we take the 2’s complement of 11101 is 00011. 00011 is a negative number, which is the answer. Similarly, we can subtract two mixed (with fractional part) binary numbers.

There are three different case of addition of 2’s complement which is explained below:

When the positive number is greater than negative number, then we take the 2’s complement of negative number, and perform the addition operation. The carry bit is discarded and the** **result will be positive number, i.e., +0001.

**Example** −Add 1110 and -1101.

Take the 2’s complement of 1101, which is 0010 and adding 1110 + 0010 = 1 0001. The carry bit 1 discard and this result will be positive number, i.e., +0001.

When the negative number has grater then positive number, we firstly convert the negative number into 2’s complement and we perform the addition number with positive number. So there will be no any end around carry bit, so we convert the result into 2’s complement and this is the final result which is negative.

**Example** −Add 01010 and -01100

In this example there are negative number is -12 which is greater than the positive number. So, we convert the negative number 01100 into 2’s complement, which is 10100 and add with positive number i.e. 01010+10100=11110. After this addition operation we will convert again this result into 2’s complement, which will be 00010 and this will be negative number, i.e., -00010, which is the answer.

In the addition of two negative numbers, we convert both numbers into 2’s complement. Since there will be always end around carry, so we dropped the carry bit and again take the 2’s complement of previous result, and it will be result of addition of two negative numbers.

**Example** − add -01010 and -00101

We take the 2’s complement of 01010 and 00101 and this will be 10110 & 11011 respectively. After this, we add the10110+11011 =1 10001. In this result there is a carry bit so we dropped it. And again we convert 2’s complement of pervious result. And it is a final result which is negative number.

Note- The difficulty level of 2’s complement operation is low than 1’s complement, because there is no extra addition operation of *end-around-carry-bit.*

Here, the two’s complement are mostly use in representation of negative number and subtraction operation. There are some advantages of two’s complement which are given below-

- It is use for representation of negative numbers
- It is use for subtraction for two binary numbers
- It is easy to accomplish with larger circuit.
- There is no need operation on end around carry bit as like as in one’s complement.

To get the Two’s complement of any number, firstly we convent in One’s complement of given binary number after this we will add 1 to lest significant bit. For example Two’s complement of binary number 10010 is convert into One’s which is 01101 and then add 1 in LSB 01101 + 1 = 01110.

The example of **2’s complement of “01000” is “11000”**. To find out of 2’s complement we convent the given binary number into 1’s complement and then we add 1 in least significant bit.

The 2’s complement of the number – 33 is **(1101 1111) _{2}**.

Number of Bits: Enter decimal value: Enter a decimal number between -128 to 127.

**Two’s complement Table.**

2’s complement of -15 |
00001111 |

2’s complement of -45 | 00101101 |

2’s complement of -19 | 00010011 |

2’s complement of – 50 i | 00110010 |

**-8**

Interestingly, if you take the Two’s complement of 1000, you get 1000 Remember 1000 is **-8**, and not +8, since the MSB is 1 . in 2’s complement the -0 and +0 has unique representation which is 0000 but in 1’s complement “0” has two representation.

The 2’s complement of -17 is (**1110 1111)**.

** **In Two’s complement notation of positive number is same as ordinary binary representation. In Two’s complement 8-bit number can only represent **positive integers that starts from 0 to 127 (01111111)**, the region behind is that the negative number representation use binary “1” in most significant bit which is not possible because the total bit becomes “9”.

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