Vehicle Number Plate Detection and Character Recognition using MATLAB

Before understanding the concept behind image reading and recognition, you should understand “Fourier Transform” and basic matrix operations in Matlab.

1) Let’s consider we have captured below the picture of the vehicle, in which number plate and its font is clear with the surrounding contrast.

Vehicle number plate detection using MATLAB
Vehicle number plate detection using MATLAB

2) To locate the car license plate and identify the characters, we use the Matlab platform to provide some image processing functions to Fourier transform the character template and the image to be processed as the core idea. The basic method is as follows:

  1. Read the image to be processed and convert it to a binary image. After converting, using the threshold of 0.2 near the license plate characters most clearly, the least points (as shown below)
    I = imread('car.jpg');
    I2 = rgb2gray(I);
    I4 = im2bw(I2, 0.2);
  2. Remove the area where the image is too small to be sure of the license plate.
    bw = bwareaopen(I4, 500);
  3. To locate the license plate, inflate the white area, corrode to unrelated small objects, including the license plate characters.
    se = strel('disk',15);
    bw = imclose(bw,se);

  4. At this point the number plate where the white connectivity domain is clearly visible, but outside the black area, there is a larger white connectivity domain and the license plate where the domain surrounded by. It is necessary to fill it.
    bw = imfill(bw,[1 1]);
  5. Now, let’s find the connectivity domain boundary. While retaining this graphic for later to mark it on it.
    [B,L] = bwboundaries(bw,4);
    imshow(label2rgb(L, @jet, [.5 .5 .5]))
    hold on
  6. Finding the one that is most likely to be the number plate in all connected domains. The standard of judgment is that the aspect ratio of the license plate is about 4.5: 1, and its area is related to the circumference:(4.5 × L × L) / (2 × (4.5 + 1) × L) 2 ≈ 1/27 ,Which is characterized by,metric = 27 * area / perimeter ^ 2as the matching degree of the connected domain, the closer it is to 1, the more likely that the corresponding connected domain is 4.5: 1.

    MATLAB Program

    stats = regionprops(L,'Area','Centroid');
    for k = 1:length(B)
      boundary = B{k};
      delta_sq = diff(boundary).^2;   
      perimeter = sum(sqrt(sum(delta_sq,2)));
      area = stats(k).Area;
      metric = 27*area/perimeter^2;
      metric_string = sprintf('%2.2f',metric);
    if metric >= 0.9 && metric <= 1.1
        centroid = stats(k).Centroid;
        plot(centroid(1),centroid(2),'ko');
        goalboundary = boundary;
        s = min(goalboundary, [], 1);
        e = max(goalboundary, [], 1);
       goal = imcrop(I4,[s(2) s(1) e(2)-s(2) e(1)-s(1)]);
    end
      text(boundary(1,2)-35,boundary(1,1)+13,...
      metric_string,'Color','g',...
    'FontSize',14,'FontWeight','bold');
    end
    for k = 1:length(B)
     boundary = B{k};
     plot(boundary(:,2),boundary(:,1),'w','LineWidth',2)
    end

    The rectangular matching degree is 0.99, which is the most probable area. Below is the license plate area in the binary image determined by it:

  7. The license plate image is highlighted and expanded to a square of 256 × 256 (left as shown below) so that the matrix rotation operation is performed in the following “Fourier transform”.
    goal = ~goal;
    goal(256,256) = 0;
    figure;
    imshow(goal);
  8. Read a character template from the file (with “P” as an example, the template image is taken directly from the above binary image). Calculate the Fourier descriptor for the image and calculate the descriptor with a predefined decision function. In the transformed image, the level of brightness indicates the degree to which the corresponding region matches the template (as shown in the figure below).
    w = imread('P.bmp');
    w = ~w;
    C= real(ifft2(fft2(goal).*fft2(rot90(w,2),256,256)));
  9. By checking the maximum value of C, the test determines a suitable threshold (where 240 is appropriate), showing the point where the brightness is greater than the threshold, that is, the highest degree of matching to the template (see below).
    thresh = 240;
    figure;
    imshow(C > thresh);

    In contrast to the left and right graphs, it can be shown that the character “P” is recognized and positioned. The same way that you can identify and locate other characters.

3) This method is generally easier to understand, Matlab function hides the Fourier transform and other complex calculations. Disadvantages: In the positioning of the license plate, the program specifically according to the characteristics of the given image design, there is no universality. In character recognition, only characters that are substantially consistent with a given template are recognized. License plate size, angle, light, integrity, clarity changes, it can not be identified. At the same time for “8” and “B” similar characters, recognition is often confused.

Download

Click here to download MATLAB program and related files.

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Voltage Controlled Oscillator – Working Principle

What is an oscillator circuit?

Assume an electrical circuit produces the following waveform output (voltage or current output).

Voltage Controlled Oscillator Waveform
Voltage Controlled Oscillator Waveform

This output is a square wave. It can be considered to be a sequence of repeating the following wave at an interval of time period 4.

Voltage Controlled Oscillator
Voltage Controlled Oscillator

This circuit which is producing a waveform by repeating a wave after a specific time interval is an oscillator circuit. Another example can be of a circuit producing continuous sine wave by repeating one cycle of a sine wave.

What is voltage controlled oscillator (VCO)?

The produced continuous waveform produced by the oscillator circuit has a frequency. A circuit in which the frequency of the produced output can be varied by the magnitude of a separately applied external voltage (other than the main supply voltage VCC) is known as voltage controlled oscillator.

Types of VCO:

  1. Linear or harmonic oscillator: This type of oscillator produces a sine wave. It consists of an LC tank circuit or crystal oscillator. The frequency of a tank circuit can be varied by changing the value of the capacitor. Now, a varactor diode’s capacitance can be varied by varying the applied voltage across it. So a varactor diode if used in an LC circuit converts it to a VCO.
  2. Relaxation oscillator: The output signal is a saw tooth or triangularwaveform. This circuit employs the charging and discharging of a capacitor through a resistance. The output frequency depends on the time of charging and discharging of the capacitor.If it is desired to produce a square wave, a triangular wave can be differentiated to produce so. Also a periodic waveform can be passed through a Schmitt trigger to produce a square wave.

IC 566

The IC 566 (or LM566) is an integrated circuit that produces a triangular wave and a square wave output from two different output pins. It is an 8 pin IC shown below:

Pin Configuration

IC 566
IC 566

frequency fo = (2/(R1C1))*((Vcc-Vc)/Vcc)

  1. Ground
  2. No connection
  3. Square wave output
  4. Triangular wave output
  5. Modulating/Control voltage VC
  6. Timing resistor R1 (connected between pin 6 to supply voltage VCC)
  7. Timing capacitor C1 (connected between pin 7 to ground)
  8. Supply voltage VCC

A rough internal circuit is shown below:

Voltage Controlled Oscillator
Voltage Controlled Oscillator

Basically, the principle of operation is as such:

The Schmitt trigger switches the current source from charging and discharging the capacitor.

The IC charges and discharges the external capacitor C1 through the resistor R1. A triangular waveform is obtained by passing the voltage waveform across the capacitor C1 through Buffer Amplifier 2 and obtained as output through pin 4.

The voltage waveform across the capacitor when passed through a Schmitt trigger, produces a square wave which is passed through the Buffer Amplifier 1 and obtained as output through pin 4.

Modulating voltage VC should be in the range of  (3/4)Vcc < Vc < Vcc where VCC is the supply voltage.

VCC should be within 10 to 24 Volts.

The frequency modulation (by applying a varying modulating voltage VC) can be done in 10:1 ratio.

The frequency of the output waveform is f0 = (2/R1C1)*((Vcc – Vc)/Vcc) .

An example circuit is shown below:

Voltage Controlled Oscillator
Voltage Controlled Oscillator

Applications of VCO

  1. Tone generators
  2. Frequency modulation
  3. Function generator
  4. Phase Locked Loop

This article is written by Sayantan Roychowdhury.

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Series Voltage Regulator – Working Principle

We assume that the voltage across a zener diode remains constant.

i.e. DVZ = 0.

In all cases, we indicate load resistance by RL

Series voltage regulator circuit diagram
Series voltage regulator circuit diagram
  • Assume supply voltage increases by DVS

Current through resistance R is IR = (VS-Vz )/R

or, DIR = DVS / R                                (Equation 1.1)

Also IR = IB + IZ ;  or, DIR = DIB + DIZ ;            (Equation 1.2)

From Equation 1.1 and Equation 1.2, an increase in VS increases base current IB and zener current IZ. Since collector current IC = ß*IB so IC also increases.

We know IE = IB + IC;

or,          (since IB is very small compared to IC)

DIE = DIC

As IC increases, IE will also increase through load RL, thus voltage output VO = IE*RL tends to increase.      (Effect 1: VO tends to increase)

Using KVL in output circuit, VO + VBE – VZ = 0; where VBE = base emitter voltage, VZ = zener voltage

DVO = -DVBE        (Equation 1.3)

Equation 1.3 suggests, an increase in output voltage VO decrease base emitter voltage VBE. As decrease in VBE causes decrease in IC , and conversely IEthrough RL , hence output voltage VO tends to decrease.             (Effect 2: VO tends to decrease)

Effect 1 and Effect 2 neutralise each other and VO is constant.

Opposite happens when VS decreases.

  • Assume supply current IS increases by DIS (keeping VS constant)

IS = IC + IR; or, DIS = DIC + DIR         (Equation 2.1)

IR = IB + IZ; or, DIR = DIB + DIZ         (Equation 2.2)

From Equation 2.1, IS increases IC (also IR). Also this is evident as from Equation 2.2, that increase in IR increases IB and hence increases IC.

As IE˜IC so IE increases through RL. Hence output voltage VO = IE*RL tends to increase.

(Effect 1: VO tends to increase)

Remaining analysis is similar to previous case. VO tends to increase decreasing VBE (like Equation 1.3). This decreases IC, consequently decreasing IE and this tends to decrease VO.

(Effect 2: VO tends to decrease)

Effect 1 and Effect 2 neutralise each other and VO is constant.

Short circuit protection

To prevent short circuit i.e. to prevent an excessive high flow of current, the following arrangement is made.

Short circuit protection
Short circuit protection

A very small resistance RSC is connected in series with the load. The base emitter terminals of BJT Q2 are connected across this RSC resistance. When a high current flows across the load, an appreciable amount of voltage is developed across RSC. Hence base emitter voltage of Q2 increases, collector current of Q2 increases, so IB is shunted away from Q1. As IE˜IC= ß*IB , hence IE decreases and a large current flow is prevented.

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Shunt Voltage Regulator – Working Principle

A zener diode forms an integral part of any voltage regulator. Before we go ahead, we know, that in all cases, the voltage across a zener diode will remain constant. i.e. ?VZ = 0. In all cases, we indicate load resistance by RL.

Regulator using zener diode only

Regulator using zener diode only
Regulator using zener diode only
  • Across RL we have: VO­ = VZ = ILRL                                            (Equation 1)
  • From current law: I = IZ + IL                                                          (Equation 2)
  • From KVL along indicated path: VS = I*R + VZ                             (Equation 3)

Equation 1 tells that output voltage VO will always be constant = VZ.

Assume two cases:

  • Assume supply current I change by dI
    From Equation 2: ?I = ?IZ + ?IL
    From Equation 1: ?VZ = ?ILRL ; or, ?IL = 0 (since ?VZ = 0)
    Thus ?I = ?IZ. This shows that excess current is bifurcated through the zener diode.
  • Assume load RL changes by ?RL (with VS constant)
    Output voltage VO will remain constant, but change in IL will be compensated by change in IZ
    From Equation 3: ?VS = ?I*R+ ?VZ ;      or, 0 = ?I*R + 0 ;               or, ?I = 0
    From Equation 2: ?I = ?IZ + ?IL ;            or, 0 = ?IZ + ?IL  ;             or, ?IL = – ?IZ

Thus if IL increases, IZ decreases and vice versa.

Regulator using transistor and zener diode

Regulator using transistor and zener diode
Regulator using transistor and zener diode

Few points:

Correlating VO and indicated path from point X to GND: VO = VX = VZ + VBE          (Equation 1)

I = IB + IC + IL ;   or, I = IC + IL(since IB is very small)                                                     (Equation 2)

The increase in VBE causes more collector current IC to flow.

  • Assume current I increase by dI keeping VS constant (opposite analysis will take place of I decreases)?I is positive. VS – I*R = VX ;         or, 0 – ?I*R = ?VX ; (since VS is constant)
    i.e. VX = VO  decreases on increase in I.                                                                  (Effect 1: VO tends to decrease)Next, from Equation 1: ?VO = ?VZ + ?VBE ;            or, ?VO = 0 + ?VBE ;
    i.e VBE also decreases on decrease in VO
    As VBE decreases, IC decreases.From Equation 2: ?I = ?IC + ?IL
    If ?I = positive (assumed);           ?IC = negative (as VBE decrease);           so IL must increase.
    The voltage across load VL = IL*RL increases.                                                       (Effect 2: VO tends to increase)The Effect 1 and Effect 2 neutralize and VO is constant.
  • Assume supply voltage VS is increased keeping current I constantThe analysis will take place just as done previously.
    VS – I*R = VX ;    or, ?VS – 0 = ?VX ; (since I is constant)
    i.e. VX = VO increases on increase in VS.                                                                 (Effect 1: VO tends to increase)Next, from Equation 1: ?VO = ?VZ + ?VBE ;            or, ?VO = 0 + ?VBE ;
    i.e VBE also increases on increase in VO As VBE decreases, IC increases.From Equation 2: ?I = ?IC + ?IL
    If ?I = 0 (assumed);         ?IC = positive (as VBE increase);  so IL must decrease.
    The voltage across load VL = IL*RL decreases.                                                     (Effect 2: VO tends to decrease)
    The Effect 1 and Effect 2 neutralize and VO is constant.

This post is written by Sayantan Roychowdhury. If you like this article, please share it with your friends and like or facebook page for future updates. Subscribe to our newsletter to get notifications about our updates via email. If you have any queries, feel free to ask in the comments section below. Have a nice day!

Flash ADC or Parallel ADC and its Working Principle

Another type of ADC is parallel ADC. Parallel ADC is called as Flash ADC. Its response is very fast. it converts analog signal into digital signal using parallel set of comparators. As its conversion time is very fast it is called as flash ADC.

Following figure shows circuit diagram of parallel ADC or flash ADC.

Flash ADC or Parallel ADC
Flash ADC or Parallel ADC

n-bit Flash ADC consist of parallel combination of 2^n-1 comparators. Outputs of all comparators are connected to an encoder.

Working Principle of flash ADC

Analog voltage is applied to non inverting terminals of all comparators using a single line. Reference voltage is applied to inverting terminals of comparators using divider circuit.

Each comparator produces digital output in the form of 1 or 0. If unknown analog voltage is greater than reference voltage comparator produces high logic. If analog voltage is less than reference voltage then comparator produces low logic i.e. 0.

Thus all parallel comparator produces digital representation of analog voltage in the form of zero and one. These outputs of comparator are then applied to the fast encoder. Encoder converts those zeros and once into binary number and produces digital binary output.

For example, see below table. When unknown voltage is 5 i.e. lies between 4.375 &5.625 is applied to the flash ADC, first four encoders produces output ‘1’ and last three encoders produces output ‘0’. Encoder converts this ‘1111000’ comparator output into ‘100’ binary number as digital output.

Table shows the outputs of comparators and encoder for a 3 bit flash ADC. The range of operation is given as 0-10V.

Analog input

(V)

Comparator Output Encoder Output
C1 C2 C3 C4 C5 C6 C7 D2 D1 D0
0.000-0.625 0 0 0 0 0 0 0 0 0 0
0.625-1.875 1 0 0 0 0 0 0 0 0 1
1.875-3.125 1 1 0 0 0 0 0 0 1 0
3.125-4.375 1 1 1 0 0 0 0 0 1 1
4.375-5.625 1 1 1 1 0 0 0 1 0 0
5.625-6.875 1 1 1 1 1 0 0 1 0 1
6.875-8.125 1 1 1 1 1 1 0 1 1 0
8.125-10.000 1 1 1 1 1 1 1 1 1 1

 

As the number of bits of ADC increases its resolution increases. But such high bit converter is bulky and expensive.

Also see: Sigma Delta ADC or noise eliminating ADC.

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Sigma Delta ADC & its Working Principle

Hi friends, today we will see Sigma Delta ADC and its working. Sigma Delta ADC is widely used in communication system, professional audio system and high precision measurement system. Sigma Delta ADC has characteristics like, high resolution, low cost and low conversion speed.

Following figure shows block diagram of Sigma Delta ADC.

Sigma Delta ADC
Sigma Delta ADC

Sigma Delta ADC consists of two main blocks, Sigma Delta modulator and digital decimeter. Sigma Delta modulator consist of difference amplifier, integrator, comparator and 1-bit DAC. Digital decimeter is used for digital filtering and down sampling.

Working Principle

Analog signal which is to be digitized is applied to the non inverting terminal of difference amplifier. Inverting terminal of difference amplifier is connected with either +Vref or –Vref depends on output of 1-bit DAC. Output of difference amplifier is integrated using integrator as shown in diagram.

Output of integrator is applied to the non inverting terminal of comparator. In this case comparator works as 1-bit ADC and produces output as 1 or 0.

Output of comparator is connected to the 1-bit DAC. DAC is used to connect either +Vref or –Vref to the inverting terminal of difference amplifier.

DACs output is them again subtracted from analog input. This process is continuous in closed loop. After each loop 1-bit ADC produces 1 or 0. Density of ‘1’ depends on analog voltage supplied. If analog voltage is high then density will be high and if analog signal is low density of ‘1’ will be low.

Output of 1-bit ADC is also connected to the digital decimeter. It is used for digital filtering and down sampling. It produces n-bit digital output in binary format.

Advantages of Sigma Delta ADC

  • Sigma Delta ADC is inexpensive since all circuitry within the converter is digital.
  • The output of sigma delta ADC is inherently linear but it has little differential non linearity.
  • It do not require sample and hold circuit. It is because due to high sampling rate and low precision.

Disadvantages

  • It is limited to high resolution and very low frequency applications.
  • It takes quite long time for producing first digital output because of digital filtering and down sampling.
  • It is not possible to use Sigma Delta ADC for multiplexed ac input signals.

Note: Sigma Delta ADC is also known as Delta Sigma ADC, oversampling ADC or noise shaping ADC.

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Weighted Resistor DAC and its Operation

HI friends, today we are going to learn weighted resistor DAC. Many times we require to convert a digital output into its equivalent analog voltage. To perform this operation we are going to use weighted resistor digital to analog converter.

The following figure shows the circuit diagram of weighted resistor DAC. This DAC circuit uses weighted values of resistor like 2R, 4R, 6R, 8R and so on depending on the digital inputs available therefore such type of network is known as weighted resistor DAC.

Weighted Resistor DAC Circuit
Weighted Resistor DAC Circuit

This circuit consists of a transistor switch (shown by the upward arrow) which turns on the switch when the digital input is ‘1’ and if digital input becomes ‘0’ it will open the switch. When transistor switch gets closed, current flows through the weighted resistor due to the reference voltage as shown in circuit diagram.

When all such currents from different weighted resistors get added at summing point (which is also known as a virtual ground) of the operational amplifier it will produce a proportional voltage as its output.

For a 4 bit DAC, the output V0 is given as follows

Where S3, S2, S1 and S0 represents the status of the switches i.e. on or off (1 or 0).

If resistors are in binary weights i.e. R3=2Rf, R2=4Rf, R1=8Rf and R0=16Rf, the above equation can be written as,

From the above discussion, we can say that for a 4 bit DAC 4 switches produces 16 different combinations of output and hence produces 16 different output voltage.in general n-bit DAC produces 2^n different discrete analog voltages.

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Superposition Theorem with Solved Examples

Hello friends, in this article, we are going to learn a superposition theorem. We will also solve some simple examples using superposition theorem.

Statement of Superposition Theorem

Superposition theorem states that the response in any element of LTI linear bilateral network containing more than one sources is the sum of the responses produced by the sources each acting independently.

The superposition theorem is not applicable for the power, as power is directly proportional to the square of the current which is not a linear function.

Steps

  1. Select any one source and short all other voltage sources and open all current sources if the internal impedance is not known. If known replace them by their impedance.
  2. Find out the current or voltage across the required element, due to the source under consideration.
  3. Repeat the above steps for all other sources.
  4. Add all the individual effects produced by individual sources to obtain the total current in or across the voltage element.

Example

Using the superposition theorem, determine the voltage drop and current across the resistor 3.3K as shown in the figure below.

superposition theorem 1

Solution

Step 1: Remove the 8V power supply from the original circuit, such that the new circuit becomes as the following and then measure the voltage across a resistor.

superposition theorem 2

Here 3.3K and 2K are in parallel, therefore resultant resistance will be 1.245K.

Using voltage divider rule voltage across 1.245K will be

V1= [1.245/(1.245+4.7)]*5 = 1.047V

Step 2: Remove the 5V power supply from the original circuit such that the new circuit becomes as the following and then measure the voltage across a resistor.

superposition theorem 3

Here 3.3K and 4.7K are in parallel, therefore resultant resistance will be 1.938K.

Using voltage divider rule voltage across 1.938K will be

V2= [1.938/(1.938+2)]*8 = 3.9377V

Therefore voltage drop across a 3.3K resistor is V1+V2 = 1.047+3.9377=4.9847

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Tags: superposition theorem solved problems, superposition theorem examples, superposition principle.

RC and LC Filters – Circuit Diagram, Waveforms and Working Principle

Hi friends, today we are going to learn some basic filter circuits like RC filter and LC filter.

RC Filter

RC filter
RC filter

In the above figure two sections of RC filter are shown. These are connected between the input capacitor and the load resistor. The value of R should be at least 10 times greater than the capacitive reactance Xc. Therefore the ripple is dropped across the series resistors instead of across the load resistor. Each section reduces the ripple by a factor of 10. Therefore ac components are removed and at the output we get a steady dc voltage.

The main disadvantage of RC filter is the loss of dc voltage across each R. Therefore RC filter is suitable only for light loads. i.e. small load current.

LC Filters

LC filter
LC filter

In this type inductor L is in series and capacitor C is in shunt with load. The choke (L) allows the dc component to pass through easily because its dc resistance R is very small. The capacitive reactance Xc is very high for dc and it acts as open circuit. All dc current passes through across which dc output voltage is obtained.

The inductive reactance XL = 2pfL is high for ac components. Therefore the ripples are reduced. Even if any ac current passes through L, it flows through the capacitor because of its low capacitive reactance.

Advantages of LC Filter

  1. In choke input filter, current flows continuously. Therefore the transformer is used more efficiently.
  2. Ripple content at the output is low.
  3. It is less dependent on the load current.
  4. DC voltage drop across L is much smaller because its de resistance R is very small.

Disadvantages of LC Filter

  1. Large size and weight of inductors,
  2. More cost,
  3. External hold is produced by inductor.

For providing smoothest output voltage p type (capacitor input) filter can be used.

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Series Inductor Filter – Circuit diagram, Waveforms and Working Principle

This post provides an information about series inductor filter. Before going to study series inductor filter it is necessary to understand what is meant by filter circuit?

Filter Circuit – Most of the electronic circuits require a dc voltage that is constant, similar to the voltage from a battery but the rectifiers cannot provide ripple free dc voltage. They provide a pulsating dc. The circuit used for filtering or smoothing out the AC variations from the rectified voltage is called as `Filter circuit.

Series Inductor Filter

The circuit diagram and waveforms of series inductor filter are shown below:

Series Inductor Filter
Series Inductor Filter

Series Inductor Filter – An inductor opposes any change in the current flowing through it. Whenever the current through an inductor tends to change, a back emf is induced in it. This prevents the change in current.

Inductive reactance XL = 2p* f*L. For dc, f = 0, therefore direct current easily passes through the inductor to the load. The only opposition to dc is due to internal resistance of choke. The reactance increases with frequency. Therefore AC component is opposed. The output waveform shows a large dc component and a small AC component.

The operation of a series inductor filter depends upon the current through it. The higher the current flowing through it, the better is its filtering action. An increase in load current reduces the ripples.

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