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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]]>The post One’s Complement: What is it? and Its Examples | 1’s Complement: Convert Binary to 1’s Complement appeared first on Know Electronics.
]]>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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]]>The post Gray Code: What is it? And How to Convert Binary to Gray Code appeared first on Know Electronics.
]]>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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]]>The post Hartley Oscillator: What is it? Working and Its Applications appeared first on Know Electronics.
]]>The Hartley oscillator is a type of oscillator. Those are two inductive coils in series with a parallel capacitor to form its resonance tank circuit producing sinusoidal wave. A Hartley oscillator is harmonic type oscillators. The oscillation frequency is determined by an LC oscillators. It is typically tuned to produce waves in the radiofrequency band, which is why they are also known as RF oscillators.
The frequency is determined by LC tank circuit which is shown in above figure. In LC tank circuit the inductor is in series and the capacitor is connected across them in parallel. It is mostly use in radio frequency (RF) oscillator applications and the recommended frequency range is from 20 KHz to 30 MHz it can be also use in below 20 KHz frequency but at lower frequency the inductors value should be high and it has a practical limit.
The circuit diagram is shown below. In this circuit contain common emitter configuration resistor and tank circuit. Here the R_{C} is a collector resistor while R_{E} is emitter resistor to form stabilization. The resistor R_{1} and R_{2} are use for voltage divider biasing network for transistor in common emitter configuration.
The capacitor C_{i} and C_{o} are bock the DC voltage at the input and output side while C_{E} are emitter capacitor, that are use for bypass amplifier AC signal all the element are shown in figure.
Now let us discuss working of Hartley oscillators, first switch ON the power supply, the transistor starts to conduct and the collector current I_{C} increases. The collector current charge the capacitor C. when the capacitor are fully charge it starts discharging via inductors L_{1} and L_{2}. This process will continue. This charging and discharging result produce the damped oscillations in the tank circuit.
The oscillating current produces across inductors inductors L_{1} and L_{2} which are out of phase by 180.
From the figure, the output of amplifier is applied to L_{1} while the feedback voltage drown L_{2} is applied to the input of base. The amplifier produce 180 degree of phase shift and tank circuit also produce 180 degree phase shift, thus the total phase shift of amplifier is 360 or 0 degree.
The frequency of oscillations of Hartley oscillators is circuit is given as
But in Hartley oscillator, we consider two inductors in the tank circuit thus equivalent inductance will be given as
L_{eq} = L_{1} + L_{2}
The mutual inductance between the coils must be consider for calculation of equivalent inductance,
L_{eq} = L_{1} + L_{2} + 2M
Thus the oscillating frequency is given as
The advantages are listed below,
The disadvantages of Hartley oscillator are given below,
The applications of Hartley oscillator is discussed below,
What is Hartley oscillator used for?
It is an electron circuit that are use to generate sinusoidal with constant amplitude. It is use in radio frequency as a local oscillator.
What is the principle of Hartley oscillator?
It is made of two inductors and one capacitor. The principle of Hartley oscillators is based on charging and discharging of LC circuit.
What is oscillator Hartley oscillator?
It is an electronics circuit that generate sinusoidal wave. Frequency of this oscillator is determine by tune circuit consist of two inductors and one capacitor.
What is the frequency of Hartley oscillator?
20 kHz to 30 MHz
It is mostly use in radio frequency (RF). The frequencies lie in between 20 kHz to 30 MHz. it can be operate at low frequency but it required high value of inductor and it has a practical limit.
Where is Hartley oscillator commonly used?
Radio frequency (RF)
It is mostly use in radio frequency. The range of radio frequency is 20 kHz to 30 MHz.
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]]>The function of all different type of oscillator is same for generating continuous undamed output wave. But the main difference of oscillator lies in the method by the energy which is supplied to the tank circuit to meet up the losses. The common types of transistor oscillators mainly include tuned collector oscillator, Hartley oscillator, Colpitt’s oscillator, phase shift oscillator, Wein bridge oscillator and a crystal oscillator etc.
The tune collector oscillator is one of the simplest types of LC oscillator. It has a tune circuit consisting of inductor and capacitor and transistor to amplify the signal. The inductor and capacitor are store the energy in form of magnetic flux and electrostatic respectively. The tank circuit is connected to collector behaves like a load at resonance and decides the oscillator frequency.
One of the simplest and basic types of oscillator is tune collector oscillator. It has LC tune, this tune circuit is connected to collector of transistor. it behaves like a simple resistive load at resonance and inductor and capacitor decides oscillator frequency. This oscillator is use in to generate sinusoidal wave, TV, counter, frequency demodulators, mixers, etc.
The figure show circuit diagram of tune collector oscillator. That consist of transformer and the capacitor is connected to the collector side of the transistor. The oscillator here produces a sine wave. The transistor is bias by the voltage divider resistance R_{1} and R_{2}. _{2} is the bypass capacitor for resistor R_{2}. The R_{e} is an emitter resistor which provides thermal stability. C_{e} is used to bypass the amplified ac oscillations and is the emitter bypass capacitor. The primary of the transformer, L_{1} along with capacitor C_{1} forms the tank circuit.
Let us understand the working of tune collector oscillator, in this oscillator the transistor alter the 180 degree phase shift of input and tune circuit will also provide the 180 degree of phase shift then total phase shift will be 360 degree of 0 degree. The transformer provide positive feedback circuit and transistor amplify output signal.
When the power supply is ON the capacitor start charging. When capacitor fully charged, it starts discharge through inductor L_{1}. The energy in capacitor is electrostatic energy form. After fully charge capacitor it starts discharging through inductor L_{1} in from of electromagnetic energy. Once the capacitor fully discharges, the inductor starts charging the capacitor again. This is because inductors do not the current through them change quickly and hence it will change the polarity across itself and keep the current flowing in the same direction. The capacitor starts again charging in same manner. The charge polarities of capacitor are change periodically and hence we can get an output.
The coil L_{2} gets charged through electromagnetic induction and feeds this to the transistor. The transistor do own work, amplify the input signal, which is taken as the output. Part of the output is fed back to the system. The positive feedback is essential for oscillator because it sustained the oscillator output.
The output frequency is dependent on inductor and capacitor value and the frequency is:-
Where,
F = Frequency of the oscillation.
L_{1} = value of the inductance of primary of the transformer L_{1}.
C_{1} = value of capacitance of capacitor C_{1}.
The oscillator is electronics circuit that produces continuously sinusoidal output with positive feedback. In tune type of oscillator the tune circuit like inductor and capacitor.
Types of Tuned Circuit Oscillators
The oscillator are widely use in radio transmission and receiver. The LC oscillators are divided into following categories.
Tuned base Oscillator – Tune base oscillator use inductive feedback.
Hartley Oscillator – It have inductive feedback.
Colpitts Oscillator − It have capacitive feedback.
Clapp Oscillator − It have capacitive feedback.
The oscillator are use to generate sinusoidal wave with positive feedback circuit the example of tune frequency oscillator is Colpitts oscillator.
The oscillator can be made by using various type of component when we use inductor and capacitor is called tune base oscillator. The tune oscillator is also called Armstrong oscillator
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There are various application of oscillator like producing sine wave, square wave, saw tooth wave, triangular wave and pulse. There are various type of oscillator like LC oscillator, Hartley oscillator, Colpitts oscillator etc. the LC oscillator are mostly use in radio frequency because it is easy to implement and good phase noise characteristics.
An oscillator is an electronics circuit which produces a continuous wave without any input. It basically converts the unidirectional current flow of DC source into an AC waveform which is of the desired frequency, as decided by its circuit components. When an amplifier with positive feedback is called oscillator.
An oscillator is a self sustaining circuit, it produce a continuous wave at a single sinusoidal frequency. Thus for an electronics circuit to operate an oscillator it must have follow some characteristics.
An oscillator input has summing of feedback amplifier. The gain products of inverting amplifier and feedback amplifier must be unity or garter than unity. It also have phase shift of 360 degree or zero phases shift. The amplifying device must be operational amplifier or Bipolar Transistor is required.
The block diagram of basic oscillator is shown below. In this block diagram contains one operational amplifier with positive feedback. The input of operational amplifier is the summation of feedback input and Vin input applied into operational amplifier.
Where: β is a feedback fraction.
Oscillator Gain without Feedback
Oscillator Gain with Feedback
Than oscillator is electronics circuit that produce continuous waveform with the help of feedback network. The feedback network is taken some output fraction of amplifier to feed into the input of amplifier in order to keep the circuit oscillating.
The feedback network is basically an attenuation circuit that has a voltage gain of less than one ( β <1 ). Oscillations start when Aβ >1 and then returns to unity ( Aβ =1 ) once oscillations are sustained.
The basic type of oscillator is:-
The oscillator circuit must have to follow the two criteria for continuous produce wave from, called as Barkhausen’s criteria. These two conditions are
In a LC oscillator, a tuned inductor-capacitor circuit alters the phase shift of 180 degree of input signal. Let us see the Barkhausen’s criteria with full explanation;
The input of signal is applied to inverting amplifier, it produce inverting output Vo which is 180 degree phase shift of input signal. The some fraction of output Vo is taken out by feedback as a input signal. The output of feedback is in out of phase 180 degree of Vo, this feedback output is summing with original input signal. This process will continue.
Let us consider the input voltage V_{i} is applied at the input of an inverting amplifier, then inverted output voltage will be:-
V_{0} = – A V_{i }
The output voltage of feedback circuit with feedback gain β is
V_{f} = – β V_{0}
The negative sign indicted phase shift of 180 degree of V_{0}.
Put V_{0 }in V_{f} = – β V_{0} , we get,
V_{f} = A β V_{i}
In oscillator the feedback output must drive the amplifier; hence the V_{f} must act as V_{i}. For achieving V_{f} = V_{i }, the A β in the above expression should be 1, i.e.,V_{f} = V_{i} when A β = 1.
That condition the product of amplifier gain and feedback gain is unity is called as Barkhausen criterion for oscillation.
The working principle of oscillator can be understood by analyzing the faction of LC tune circuit which is shown below. In LC circuit contains inductor and per charge capacitor. In figure the capacitor discharge via inductor, which result in the conversion of electrical charge to electromagnetic field. This electromagnetic field is store in inductor. Once the capacitor is fully discharge there will be no current flow from capacitor to inductor.
However by them, the stored electromagnetic field in inductor generate back electro motive force (EMF) which result the current start flow form inductor to capacitor. This flow of current will continue until the electromagnetic field collapses which result in the back conversion of electromagnetic to electrical charge, causing the cycle to repeat. However the capacitor is charge with opposite polarity, due to which get an oscillating wave as the input.
The electrical energy is lose in real resistance of the inductor, in the dielectric of the capacitor due to this loses the oscillation steadily decreases until they die away completely and the process stops.
Once the power supply is turn “ON” oscillation start. The gain of oscillator is
Where the A is voltage gain of amplifier and gain of feedback network is β
Condition for Oscillation: Aβ < 1: the output signal will die out (Damped oscillations)
Aβ > 1: the output signal will build up.
Aβ = 1: the output signal will be steady, undamped oscillations (stable oscillations)
In other word the gain product of amplifier and feedback is greater than unit the output will be distorted; if gain product is less than unity the output will be Damped oscillations. If gain product is equal to unity the output will be constant amplitude are called undamped oscillations or stable oscillations.
There is much type of oscillation, but it broadly classified into two main categories
The main types of Oscillators include:
The oscillation can be also classified into various types depending upon the parameter like shape of wave, feedback etc. These types of classification are given below:-
The design of oscillators is simple and cheap. It is the easiest way to specific Frequency of a signal. For example, LC oscillators is used to generate a High Frequency signal, RC oscillators is used to generate a Low Frequency signal and Op-Amp based oscillators is used to generate a stable frequency.
There are some common example of oscillators is listed below:-
The basic requirements are as follow:-
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]]>A monostable is an electronics circuit, as the name implies, has only one stable state. In this type of multivibratior, one transistor conducts and second transistor is non-conduct. A stable state is that state where the transistor remains without being altered, unless disturbed by external trigger pulse. As monostable work on the same principle so it is also called one-short multivibratior. It is use in oscillator, timer and flip flops.
A multivibrator produces a symmetrical and asymmetrical square wave and that’s why a multivibrator is mostly use in square wave generator. It belongs to an oscillator family so it is commonly called Relaxation Oscillators.
Generally, a multivibrator is the cross coupling of two transistor so that one or more of its output are fed back to input to the other transistor with a resistor and capacitor network connected across them to produce the feedback tank circuit.
A multivibrator is a two different state i.e. electrical “high” and electrical “low” state giving them either a stable or quasi-stable state depending upon the type of multivibrator.
The monostable multivibrator has only one stable state (their name “mono”), mans produce a single pulse in output when triggered is applied. It returns back to original and stable state after a time period. The time period is determine by the resistor and capacitor (RC) network.
The two transistors TR_{1} and TR_{2} are cross coupled with each other. The collector of TR_{1} is connected to base of TR_{2} with capacitor C_{T} and the collector of TR_{2} is connected to base of TR_{1} with resistor R_{3}. The trigger pulse is applied to the base of TR_{1 }through the capacitor C_{1 }to change its state. The load resistor of TR_{1} and TR_{2} is R_{1 }and R_{2} respectively.
The monostable multivibrator circuit diagram is given below:-
Let us suppose, initially the transistor Q_{1} will be in OFF state and Q_{2} will be in ON state and this is a stable state. As the transistor Q_{1} is off, the collector voltage of Q_{1} will be V_{CC} and the capacitor C_{1 }gets charge with supply voltage. This charge capacitor provides the transistor TR_{1} stay at ON position. When external trigger pulse are provided through the capacitor C_{1} and diode. The transistor TR_{1 }is become ON and starts conduction. Due to turn ON TR_{1 }will decreases the collector voltage which turn OFF the TR_{2} and the capacitor start discharging. The positive collector voltage is of TR_{1 }is applied to the base of TR_{1}, it remain turn ON TR_{1.} This is called Meta-stable state or quasi-stable state.
The monostable multivibrator can produce a short and longer rectangular pulse. The shape of rectangular pulse is depends upon the trigger pulse and RC networks. The leading edge and trailing edge of rectangular wave is depends on the trigger pulse and RC network time constant respectively. The time constant may be varied, which are shown below.
The time constant of monstable multivibrator may be change by varying the value of RC network. The pulse width output is dependent on resistor and capacitor network “RC” the pulse width duration formula is T=0.69 R_{1}C_{1}
This multivibratiorn is use where the wider pulse width is required or to produce a time delay within a circuit.
The advantages of Monostable Multivibrator are given below:
The major drawback of the monostable multivibraitor is that the timing of trigger pulse is greater than the RC time constant of the circuit.
It is use in television circuit and control system circuit. It is also use in oscillator, counter and flip flop.
Applications of monostable multivibrator.
It is use in delay circuit, temporary memory, trigger another pulse generator and regenerating old and worn out pulses.
This vibrator is a only one stable state and produced a constant output when it is not applied trigger pulse. If trigger pulse is applied it goes to unstable stage. After removing trigger pulse it comes on original stage after a time period is depend on RC networks.
An Astable multivibrator in which the circuit has no stable sate. It is continuously switches state from one to other. It functions as a relaxation oscillator. While the Monostable multivibrator, in which have only one stable state and other is unstable state.
Multivibertor is and electronics circuit that are use to generate non sinusoidal wave like square and rectangular wave and it is use in various application like oscillator, timer and flip flop. The multivibrator are three types: – astable multivibrator, bistable multivibrator and monostable multivibrator.
Monostable – A monostable multivibrator has only single stable state. It use two cross coupled transistor to generate output. It changes the state when trigger is applied and after removing of trigger pulse it return back to original state. Bistable – In bistable multivibrator has two stable states. And its state is depending of the trigger pulses.
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]]>Generally this circuit stays high until a trigger pulse is not applied, if when the trigger pulse is applied, it changes the output state. Bistable multivibratior is also known as flip flop or latches because the flip flop behave same operation like bistable multivibratior. This is also called binary or flip flop circuit.
There are few types in Bistable Multivibrators, which are shown below in figure from.
The collector coupled further divide in two types
A bistable is the one of the type of multivibratior device similar to the monostable, we discus about monostable in previous article but the difference this time is that BOTH states are stable.
It has two stable states and maintains the state until the external trigger is not applied. This means the output will be shifted from one stage to another stage by applying trigger pulse. It required two external pulses to return original state. As it has two stable stages they are known as latches and flip flop.
The bistable multivibrator has two state non-regenerative devices. The circuit configuration is a cross coupling of two transistors one is in ON and OFF switching. That means one transistor is in cut-off region and other transistor is in saturation region. The bistable circuit is capable in either stable state without trigger pulse.
To change the stage from one stable state to another state it required external trigger pulse; when we applied two trigger pulses to bistable circuit it return to original position. It is also known as flip flop or latch circuit. The circuit diagram is shown below.
The circuit diagram of bistable is shown above is stable in both state. In this circuit shows two transistors, one transistor is in cut off region and other in saturation region. Let’s suppose the base of transistor TR_{1} is connected to the ground which is shown in figure, and it is cut off region producing output at Q. That would mean that transistor TR_{2} is saturation region. The base of transistor TR_{2 }is connected to Vcc with the series combination of R_{1 }& R_{2}. As transistor TR_{2} is “ON” there will be zero output at Q, the opposite or inverse of Q.
Now if we applied a trigger pulse at point “B”. the transistor TR_{2 }will switch “OFF” and transistor TR_{1} will switch “ON” through the combination of resistors R3 and R4 resulting in an output at Q and zero output at Q the reverse of above. Then we can say that the stable state exist when TR_{1} is ON and TR_{2} is OFF, switching position A. and other stable state is exist, TR_{2} is ON and TR_{1} is OFF, switching position B.
In monostable multivibrator whose output is dependent on the RC time constant with feedback but in bistable the output is depend on the applications of two individual trigger with changing switching position from A to B.
The bistable multivibrator waveform depends upon the output of transistors. The leading edge of waveform depends on applied trigger pulse and trailing edge is on second trigger this is shown in figure.
Manually switching pulse between state produce bistable multivibrator circuit, and it is not practically possible. So we use trigger circuit for trigging.
The advantages of bistable Multivibrator are given below
The disadvantage are given below
Bistable Multivibrators are used in applications such as pulse generation and digital operations like counting, flip flop and storing of binary information.
The bistable multivibrator is an electronics circuit, it has two stable states. It can change state from one state to another state by applying external trigger pulse. It is also known as flip flop and latch. It is use in digital computer for store binary bits.
The bistable is motley use in digital communication, counter and reversing to the supply to a given circuit at regular intervals.
There are three types of multivibrator monostable, astable, and bistable. The bistable multivibrator is a pulse detector, when it detects pulse it changes the states from high to low or low to high. There are few types in Bistable Multivibrators, which are shown below.
The collector coupled further divide in two types
The bistable multivibratior has two states and both states are stable without trigger pulse. The word “BI” means two.
A bistable multivibrator is two stable states. This circuit stays in any one stable state when no external trigger is applied it. The trigger pulse changes its state. Bistable multivibrator is also called flip flop.
The advantage is Stores the previous output unless disturbed and circuit design is easy and simple. Disadvantage is required two pulses for change stage from high to low or low to high.
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