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:-
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:
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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:-
“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} ) and the load current I_{L} flowing through and load voltage, Using thevenin’s theorem. The circuit diagram are shown below.
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
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} +R_{L}
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.
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.
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;-
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:
There is slightly difference between optocoupler and solid state relays which is described below-
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:
The four types are shown below.
Optocouplers or of optocoupler are used in,
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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,
In diode current equation the two parameters require to be discussed in quite detail. These parameters are given below:
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-
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 × 10^{-19} C. Thus
q/KT = 1.6 × 10^{-19} / 1.38 × 10^{-23} × 300
= 0.003865 × 10^{4}
= 38.65 C J^{-1} or 38.65 V^{-1}
At room temperature the diode equation becomes as-
I = – I_{0} e^{-v/0.025}^{×}^{ η}
The diode has non liner device. The diode current is
I = – I_{0} e^{-v/0.025}^{×}^{ η}
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.
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-
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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]]>For example 1’s complement of 1011001 binary numbers is 0100110. We can also find out the 2’s complement of binary number by changing each 1 binary bits into 0 and 0 to 1 (0 to 1 and 1 to 0) than we added 1 to the least significant bit of 1’s complement. For example 2’s complement of 10010 binary number is (01101 + 1) = 01110.
There is simple method to convert a binary number into 1’s complements method. The conversions of 1’s complements we use NOT gate. The output of NOT gate is complements input. We use one NOT gate for one binary number. For example five binary bits we use five NOT gates. Implementation of logic circuit of 5-bit 1’s complements is given as following below.
Example 1: find out one’s complement of 11010.1101
We find out the 1’s complement of the given number, changing all 1’s to 0 and all 0’s into 1’s the 1’s complements 11010.1101 is 00101.0010.
Example 2: find out one’s complement of 100110.1001
We find out the 1’s complements of the given number, changing all 1’s to 0 and all 0’s into 1’s the 1’s complements 100110.1001 is 011001.0110.
1’s Complement Table
Binary Number | 1’s Complement |
0000 | 1111 |
0001 | 1110 |
0010 | 1101 |
0011 | 1100 |
0100 | 1011 |
0101 | 1010 |
0110 | 1001 |
0111 | 1000 |
1000 | 0111 |
1001 | 0110 |
1010 | 0101 |
1011 | 0100 |
1100 | 0011 |
1101 | 0010 |
1110 | 0001 |
1111 | 0000 |
The 1’s complements is mostly use to representation of binary signed number and it is also use in various arithmetic operations like addition and subtraction etc.
1’s complement is mostly use in binary sign bit representation. The positive number is simple represent as binary number. There is nothing for to do for representation of positive binary number. But in case of representation of negative binary number, we use 1’s complements. If we need to represent negative binary number we are using 1’s complement. For representation of negative number firstly we take positive binary number and then taking 1’s complement by changing 1 to 0 and 0 to 1.
Example: Let we are using 5 bits register. The representation of -5 and +5 will be as follows:
The representation of +5 is 0 0101. In this binary number the most significant bit 0 is use for + sign. Now we take the 1’s complement of 0 0101 is 1 1010. The 1 1010 represent -5. 1 1010 the most significant bit 1 shows the – sign.
Note the drawback of 1’s complements representation is that 0 has two different representation first is -0 (e.g., 1 1111) and second is -0 (e.g., 1 1111).
Now let’s discuss 1’s complement asthmatic operations like addition and substation.
There are some following steps for subtract to binary number using 1’s complements.
Eg – Evaluate 10101 – 00101
We take the 1’s complement of subtrahend 00101, which will be 11010, then add 10101 + 11010 =1 01111. Here the carry bit is 1 so this carry bit is added to the least significant bit of the given result i.e., 01111+1=10000 which is the answer.
Eg – Evaluate 11110 with 1110
Firstly we convert the subtrahend 11110, into 1’s complements, which will be 00011. Then add 11001 + 00011 =11100. In this addition result there is no carry bit so we take the 1’s complements of the given result which will be 00011, and this is negative number, i.e, 00011, which is the answer.
In this case we take the 1’s complements of negative number and the end around carry of the sum is added to the least significant bit.
Example: Add 1110 and -1101.
Firstly we take the 1’s complements of -1101, which will be 0010, and then we perform the addition operation. So, 1110 + 0010 = 1 0000. The carry bit 1 is again added to the least significant bit of the given result. The final result will be 0000 + 1 = 0001.
When the negative number has greater value, we take the 1’s complements of negative number and perform the addition operation with the given number. Since, it has no end around carry bit, so take 1’s complements of the result and this result will be negative.
Example: Add 1010 and -1100.
We take the 1’s complements of the 01100 which will be 10011 and perform addition operation 01010 + 10011 = 11101. Now we take the 1’s complements of the given result which will be 00010. This is final and negative result.
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]]>Gray code is the arrangement of binary number system such that each incremental value can only differ by one bit. This code is also known as Reflected Binary Code (RBC), Cyclic Code and Reflected Binary (RB). In gray code when transverse from one step to another step the only one bit will be change of the group. This means that the two adjacent code numbers differ from each other by only one bit.
It is popular for unit distance code but it is not use from arithmetic operations. This code has some application like convert analog to digital, error correction in digital communication.
The conversion in between decimal to gray and binary to gray code is given below:
Decimal Number | Binary Number | Gray Code |
0 | 0000 | 0000 |
1 | 0001 | 0001 |
2 | 0010 | 0011 |
3 | 0011 | 0010 |
4 | 0100 | 0110 |
5 | 0101 | 0111 |
6 | 0110 | 0101 |
7 | 0111 | 0100 |
8 | 1000 | 1100 |
9 | 1001 | 1101 |
10 | 1010 | 1111 |
11 | 1011 | 1110 |
12 | 1100 | 1010 |
13 | 1101 | 1011 |
14 | 1110 | 1001 |
15 | 1111 | 1000 |
That logic circuit that converts binary to gray code is called binary to gray code converter. The binary to gray code conversion is given below;
Binary Number | Gray Code |
0000 | 0000 |
0001 | 0001 |
0010 | 0011 |
0011 | 0010 |
0100 | 0110 |
0101 | 0111 |
0110 | 0101 |
0111 | 0100 |
1000 | 1100 |
1001 | 1101 |
1010 | 1111 |
We have some binary number 010.01 which we wish to convert to gray code. Let us see the step how to convert binary to gray code?
You can convert n bit (b_{n}b_{(n-1)}…b_{2}b_{1}b_{0}) binary number to gray code (g_{n}g_{(n-1)}…g_{2}g_{1}g_{0}). for least significant bit b_{n}=g_{n}, and rest of the bit by XORing b_{(n-1)}=g_{(n-1)}⊕g_{n}, …. b_{1}=g_{1}⊕g_{2}⊕g_{3}…⊕g_{n} and b_{0}=g_{0}⊕g_{1}⊕g_{2}⊕g_{3}…⊕g_{n}.
Let b_{0} b_{1} b_{2} b_{3} are the binary bits representation. Where binary b_{0} is least significant bit (LSB) and binary b_{3} is most significant bit (MSB). And g_{0} g_{1} g_{2} g_{3} be the bits representation of gray codes. Where g_{0} is the least significant bit (LSB) and g_{3} is the most significant bit (MSB).
The truth table for binary to gray conversion is given below:
Binary bits | Gray bits | ||||||
b_{3} | b_{2} | b_{1} | b_{0} | g_{3} | g_{2} | g_{1} | g_{0} |
0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
0 | 0 | 0 | 1 | 0 | 0 | 0 | 1 |
0 | 0 | 1 | 0 | 0 | 0 | 1 | 1 |
0 | 0 | 1 | 1 | 0 | 0 | 1 | 0 |
0 | 1 | 0 | 0 | 0 | 1 | 1 | 0 |
0 | 1 | 0 | 1 | 0 | 1 | 1 | 1 |
0 | 1 | 1 | 0 | 0 | 1 | 0 | 1 |
0 | 1 | 1 | 1 | 0 | 1 | 0 | 0 |
1 | 0 | 0 | 0 | 1 | 1 | 0 | 0 |
1 | 0 | 0 | 1 | 1 | 1 | 0 | 1 |
1 | 0 | 1 | 0 | 1 | 1 | 1 | 1 |
1 | 0 | 1 | 1 | 1 | 1 | 1 | 0 |
1 | 1 | 0 | 0 | 1 | 0 | 1 | 0 |
1 | 1 | 0 | 1 | 1 | 0 | 1 | 1 |
1 | 1 | 1 | 0 | 1 | 0 | 0 | 1 |
1 | 1 | 1 | 1 | 1 | 0 | 0 | 0 |
To design the conversion of binary to gray digital logic circuit we will use the K-Map for each of the gray codes bits as output with all of the binary bits as input.
K-Map for g_{0} –
K-map for g_{1 }–
K-map for g_{2 }–
K-map for g_{3 }–
The corresponding Boolean expression for gray codes is-
The corresponding binary to gray codes converter digital circuit is shown below –
In gray to binary converter the input is gray codes and output is binary number. Let us we the four bits gray to binary converter. To design the four bits binary to gray to binary converter we first have to draw a gray codes conversion table, as shown below:
Gray Code | Binary Number |
0000 | 0000 |
0001 | 0001 |
0011 | 0010 |
0010 | 0011 |
0110 | 0100 |
0111 | 0101 |
0101 | 0110 |
0100 | 0111 |
1100 | 1000 |
1101 | 1001 |
1111 | 1010 |
In gray to binary conversion it is simple and easy process, only we have follow some following steps;
Below example of gray to binary conversions give will make your idea clear.
You can convert n-bit gray to (g_{n}g_{(n-1)}…g_{2}g_{1}g_{0}) to n-bit binary number (b_{n}b_{(n-1)}…b_{2}b_{1}b_{0}) for least significant bit g_{n} = b_{n} and rest of the bits XORing by g_{(n-1)}⊕ b_{1} b_{2}=b_{0}+g_{1(n-1)} and so on.. which is shown in above figure.
Let g_{0} g_{1} g_{2} g_{3} are the gray bits representation. Where g_{0} is the least significant bit (LSB) and g_{3} is the most significant bit. And the binary bits are b_{0} b_{1} b_{2} b_{3}. Where least significant bit is b_{0} and most significant bit is b_{3}.
The truth table for gray codes to binary bits conversion is given below:
Gray code | Binary bits | ||||||
g_{3} | g_{2} | g_{1} | g_{0} |
b_{3} |
b_{2} |
b_{1} |
b_{0} |
0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
0 | 0 | 0 | 1 | 0 | 0 | 0 | 1 |
0 | 0 | 1 | 1 | 0 | 0 | 1 | 0 |
0 | 0 | 1 | 0 | 0 | 0 | 1 | 1 |
0 | 1 | 1 | 0 | 0 | 1 | 0 | 0 |
0 | 1 | 1 | 1 | 0 | 1 | 0 | 1 |
0 | 1 | 0 | 1 | 0 | 1 | 1 | 0 |
0 | 1 | 0 | 0 | 0 | 1 | 1 | 1 |
1 | 1 | 0 | 0 | 1 | 0 | 0 | 0 |
1 | 1 | 0 | 1 | 1 | 0 | 0 | 1 |
1 | 1 | 1 | 1 | 1 | 0 | 1 | 0 |
1 | 1 | 1 | 0 | 1 | 0 | 1 | 1 |
1 | 0 | 1 | 0 | 1 | 1 | 0 | 0 |
1 | 0 | 1 | 1 | 1 | 1 | 0 | 1 |
1 | 0 | 0 | 1 | 1 | 1 | 1 | 0 |
1 | 0 | 0 | 0 | 1 | 1 | 1 | 1 |
For design digital logic circuit of the conversion of gray code to binary bits, we will use the K-Map for each of the binary bits as output with all of the gray codes as input.
K-map for b_{0} –
K-map for b_{1 }–
K-map for b_{2 }–
K-map for b_{3 }–
Corresponding Boolean expressions –
Corresponding digital circuit –
The gray codes has some specific applications, which is given below:-
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]]>The post Colpitts Oscillator: Circuit Diagram & How To Calculate Frequency of Colpitts Oscillator appeared first on Know Electronics.
]]>A colpitts oscillator is a one of the type of LC oscillator. The colpitts oscillator is a combination of inductor and capacitors to produce an oscillating frequency. In colpitts oscillator the feedback are taken from a voltage divider made of two capacitors in series across the inductor but in Hartley oscillator the feedback is taken out from voltage divider made of two inductors in series across the capacitor.
A colpitts oscillator is invented by American engineer Edwin H. Colpitts, in 1918. It is use to generate radio frequency. The colpotts oscillator is exact opposite of the Hartley oscillator as we look in previous tutorial.
The basic configuration of collpitts is same as Hartley oscillator but the difference this time is that the tank circuit is made of junction of two “capacitive voltage divider” network instead of two inductive voltages (tapped autotransformer) in the Hartley oscillator.
The feedback circuit uses a capacitive voltage divider network. Two capacitors, C_{1} and C_{2} are in series and these series capacitors are parallel with inductor. The C_{1}, C_{2} and L create a tank circuit. The condition for oscillations being: X_{C1} + X_{C2} = X_{L}, is same as for the Hartley oscillator circuit.
The main advantage of this type of capacitive circuit is that with less mutual inductance and self inductance within the tank circuit, the frequency stability of this oscillator is improve with a more simple design.
The circuit diagram of colpitts oscillator are shown below, the circuit contains common emitter amplifier circuit and tank circuit. Resistors R_{1} and R_{2 }is a voltage divide to bias amplifier. R_{C} is a collector resistance and R_{E} in emitter resistance. The capacitors C_{i} and C_{o }are input and output capacitors for blocking the DC current and C_{E }is a bypass capacitor, it bypass the alternating current. The one end of capacitors is joining together and other end of C_{1} is connected to collector of transistor via C_{0 }and C_{2 }is connected to base of transistor. the feedback output are taken out through this path.
Now let us discuss the working of colpitts oscillator. Firstly, switch ON the power supply, transistor starts conduction. The collector current I_{c} increases due to which the capacitors C_{1} and C_{2} get charged. When acquiring the maximum charge feasible these capacitors are start discharge through inductor.
During this process electrostatics energy store in capacitor gets converts in to magnetic flux. This is store in inductor L. When inductor is fully charge it start discharging through capacitor. And again capacitor will start charging. Likewise, the charging and discharging of tank circuit element will continue the oscillations.
From the figure the output is present is across C_{1} and thus is in-phase with the tank circuit’s voltage and makes-up for the energy lost by re-supplying it.
On the other hand, the voltage feedback to the transistor is obtained across the capacitor C_{2}, which means the feedback signal is out-of-phase with the voltage at the transistor by 180^{o}.
The voltage across C_{1} and C_{2} are opposite in polarity as the point where they join is grounded.
Further the signal are provided 180 degree phase shift by transistor. The result of total phase shift is 0 degree or 360 degree. Which is satisfied the Barkhausen principle.
The frequency of colpitts oscillator is determined by the tank LC circuit. The formula of oscillating frequency is given as:
where C_{T} is the total capacitance of C_{1} and C_{2} connected in series and is given as:
The transistor use common emitter configuration with the output is 180 degree phase shift on input. Another 180 degree phase shift provide by feedback circuit which is two capacitors are in series with the inductor. The resultant phase shift is 0 degree of 360 degree.
The amount of feedback is depends on the value of C_{1} and C_{2}. The voltage across the C_{1} is same as the oscillator’s output voltage and voltage across C_{2 }is the oscillator feedback voltage. Then the voltage across C_{2} will be much lesser than that across C_{1}.
Therefore the changing the value of capacitors we can adjust the amount of feedback voltage returned to the tank circuit. However the large amount of feedback may cause of distorted sinusoidal wave, while small amount of feedback may not allow to oscillator circuit.
The amount of feedback of colpitts oscillator is depends on capacitors C_{1} and C_{2}. The ratio of C_{1} and C_{2} is called feedback fraction and it is given as:
In non inverting amplifier configuration the ratio of R_{2}/ R_{1} sets the amplifiers gain. The minimum gain required of oscillation is 2.9. In the figure the resistor R_{3} is provide the feedback path to the LC tank circuit.
The advantage of colpitts oscillator over the Hartley oscillator is that the colpitts oscillator produces a more pure sinusoidal waveform due to the low impedance paths of the capacitors at high frequencies. Also due these capacitive reactance properties the FET based Colpitts oscillator can operate at very high frequencies. Of course any operational and field effect transistor is use as a amplifying device must be able to operate at the required high frequencies.
The colpitts oscillator consist of two capacitors are connected in series and this series capacitors are in parallel with inductor. The midpoint of series is connected to ground and one end of capacitor C_{1 }is connected to output amplifier and C_{2 }end of capacitor provide feedback to input of amplifier.
The common emitter configuration provides 180 degree phase shift and another phase shift provides by feedback circuit. Hence totoal phase shift is 0 degree or 360 degree. The pure sine wave is determined by the LC tank circuit.
The advantages of Colpitts oscillator are given below −
The drawback of Hartley oscillator is removing by using colpitts oscillator. The colpitts oscillator provides constant amplitude over fixed frequency range.
The applications of colpitts oscillator are given as:
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