LED Driver Circuit Working & Applications

A Light Emitting Diode (LED) is a special type of diode that is used as an Optoelectronic device. It conducts when forward biased, just like a p-n junction diode. However, there is a special feature of this device which is its ability to emit energy in the visible range (visible light) of the electromagnetic spectrum.

Now, a major concern is that an LED needs a constant supply, but the supply that we receive is an alternating one. Thus to convert the AC supply to the required input for an LED (DC), we need a driver circuit. Many a time, an LED is driven using batteries or some controlled devices like microcontrollers. But these have their own disadvantages like low battery life, etc.

As told earlier, we need to convert the AC supply to DC. The most convenient and commonly used way to do this is by using the transformer. But for driving loads like LED would be costly and also producing low current is not possible.

LED Driver Circuit Principle

Keeping all the above factors in mind, let us design a simple and cost-effective circuit driving an LED from a household supply (230 volts). The basic principle behind the LED driver is a transformerless power supply. The main component is the rated AC capacitor, that reduces the supply current to the required amount. These capacitors are connected to high voltage AC circuits and thus line to line.

The capacitor used here reduces the current only and then rectification and regulation of voltage are done at later part of the circuit. This high voltage AC will be rectified using Full Wave Bridge Rectifier. The obtained high voltage DC is now rectified through Zener diode at a lower voltage value.

This rectified and regulated voltage-current combination is given at the input of an LED. Let us design a driver circuit for 230 volts supply which comes to our home.

Components Required for LED Driver Circuit

  • 390 kΩ resistor
  • 10 Ω resistor
  • 2.2 µF Polyester Film Capacitor
  • Bridge Rectifier
  • 22 KΩ
  • 4.7 µF / 400V Polarized Capacitor
  • 10 kΩ resistor
  • 4.7 V Zener diode
  • 47 µF / 25V Polarized Capacitor
  • 5mm LED

Design of 230V LED Driver Circuit

At first, a 2.2µF rated capacitor is connected in line with the mains. One thing to be noted here is that the voltage rating of the chosen capacitor must be greater than the supply voltage. Since the main supply gives 230V therefore, we have taken the 400V capacitor.

LED Driver Circuit for 230V mains supply

For the discharge of the capacitor charge while supply is off a 390kΩ resistor is connected in parallel. Also, a 10 Ω resistor acts as a fuse that is connected between the supply and the bridge rectifier. After this resistor, we have connected the full-wave bridge rectifier which is capable of handling 1.5 Amperes of current. 4.7 µF Capacitor is then used as a filter after rectification.

To regulate the DC output of the bridge rectifier, we use a Zener diode. A 4.7 volts Zener diode (IN4732A) is taken in use. Also, to limit the current value at Zener diode, we have used a 22 KΩ resistor in series. This controlled DC is given at LED after filtering it through 47 µF capacitors.

Working of 230V LED Driver Circuit

A transformerless, simple LED Driver circuit is made here. The key components of the circuit is the rated capacitor, the Zener diode and the resistor that lessens the current in the Zener diode.

Firstly, the 2.2µF  rated capacitor will limit the AC current from the mains supply. To calculate this current we need to use the capacitive reactance of the capacitor.

The formula for capacitive reactance is,

Capacitive Reactance X= (1/2πFC)

Now for, C = 2.2µF ; F = 50 Hertz

substituting values of F & C, we get

X= 1447.59

So from Ohm’s law, the current that capacitor allows is given by

I = V/R

Hence, the current through the capacitor is

Ic = 230/1447.59 = 158mA

This is the current which enters the bridge rectifier. Now a capacitor is used to filter the output of the rectifier. The selection of the appropriate voltage rating of this capacitor is important here.

The input at the bridge rectifier is 230V RMS. Therefore, the maximum voltage can be calculated as,

Vmax = VRMS x √2

= 230 x 1.414 = 325.26 volts

Thus, we use 400 volts rated filter capacitor. The output after the rectification of DC voltage is around 305 volts. To make it usable for lighting up the LED, the Zener diode is used.

Here we are using a 4.7 volts Zener diode. Also, three main things need to be kept in mind for the Zener diode which is used as a regulator here. A resistor in series, the power rating of the resistor and the power rating of the Zener diode.

First, the resistor will limit the amount of current going in the Zener diode. The formula to be used is,

Rs = VIN – VZ/(IL + IZ)

Here, VIN is the input voltage at Zener diode = 305 volts

VZ is the Zener voltage = 4.7 volts

IL is the load current, i.e., the current passing through LED = 5 mili Amperes

IZ is the Zener diode current = 10 mili Amperes

therefore, putting all the above values in RS, we get

RS = 20020Ω

Let us now calculate the power rating of RS. It is important to calculate because this tells about the amount of power that resistance can dissipate.

The power rating of RS = (VIN – VZ)2/RS

= (305-4.7)2/20020 = 4.5 W

Similarly, we have to calculate the power rating of Zener diode

The power rating of Zener Diode = ((VIN – VZ)*VZ)/RS

= ((305-4.7)*4.7)/20020 = 0.07 W

Based on the above calculations, one can choose the resistors, Zener diode, etc to make an efficient LED driver circuit.

Now, let us discuss on few advantages, disadvantages, and limitations of the circuit discussed so far.

Advantages of LED Driver Circuit

  • By using the above circuitry, we can drive LEDs directly from the mains supply.
  • The circuit is simple and cost-effective as it is a transformerless driver.

Limitations of 230 Volts LED Driver Circuit

  • This circuit can be dangerous as 230 volts AC mains supply is directly used here.
  • For single-phase supply generally in domestic applications, this circuit suites best. For a three-phase supply, if by chance any phase comes in contact with the input terminal then it can prove to be dangerous.
  • In the case of mains fluctuations, the capacitor can produce spikes.

Applications of 230 Volts LED Driver Circuit

  • It can be used for home lighting systems.
  • If needed, it can be used as an indicator circuit as well.

Many other applications can be done using this circuit, for instance, connecting it with the doorbell so as to make it comfortable for people having hearing issues.

Ajay Dheeraj

Technical Content Developer

Inductors in Series Explained in Detail

Inductors are one of the passive elements ( others being resistors & capacitors) used in an electric circuit. It stores energy in the form of magnetic field.

Now inductors can broadly be connected in two ways

  • Inductors in Series 
  • Inductors in Parallel

In this article, particularly we will be studying about the series connection of Inductors. We know that an inductor has a positive and a negative termial. So if the negative terminal of one is connected with the positive terminal of other then the connection is said to be made in series. Clearly, the current flowing through all inductors in series will be same.

Inductors in Series Circuit

If the current flowing in all the connected inductors are same then the connection is series connection. The equivalent inductance of the inductors in series is equivalent to the sum of all the individual inductors in series. 

As shown above, the current is same in all three inductors, 

i.e., I1 = I2 = I3

Also, we know that the voltage through an inductor is given by 

V= L (di/dt)

Equation of Equivalent Inductance

Applying Kirchoff’s law in the above circuit, we have

V = VL1 + VL2 + VL3

As, V = L (di/dt)

Therefore, V1 = L1 (dI1/dt)

V2 = L2 (dI2/dt)

V3 = L3 (dI3/dt)

putting the values of V1, V2 & V3 in the first equation, we get

V = L1 (dI1/dt) + L2 (dI2/dt) + L3 (dI3/dt)

Since current is same in all three, thus let I1 = I2 = I3 = I 

V = L1 (dI/dt) + L2 (dI/dt) + L3 (dI/dt)

   = (dI/dt) (L1 + L2 + L3)

   = Lequi (dI/dt)

therefore, Lequi = (L1 + L2 + L3)

In general, Lequi = (L1 + L2 + L3 + ……..Ln)   [for ‘n’ inductors in series]

‘n’ Inductors in Series

One point to remember here is that the equivalent inductance of two or more inductors connected in series is more than the largest inductance value among them.

Example for Inductors in Series 

Let four inductors are connected in series having values 10mH , 20mH, 30mH & 40mH, with no mutual inductance between them. Now to calculate the equivalent inductance of the combination is calculated by adding up all the individual inductances.

Therefore, Lequi = L1 + L2 + L3 + L4

Lequi = 10 + 20 + 30 + 40 = 100mH

Here also we can see that the largest value of individual inductance is 40mH and the equivalent is more that that (100mH). 

Mutually Connected Inductors in Series

When inductors are connected in series so that the magnetic field of one links with another, then the effect of mutual inductance either decreases or increases the total inductance depending on the magnetic coupling. The distance between the coils and their orientation to each other varies the effect of this mutual inductance.

Now, these mutually connected inductors can be broadly of two types:

  • Adding total inductance
  • Opposing total inductance

If the current flowing through the coil and the magnetic flux produced are in the same direction then the coils are said to be cumulatively coupled. If the current direction is opposite to that of the magnetic flux produced then the coils are said to be connected in Differentially coupled.

Cumulatively Coupled Series Inductors

As shown in fig below, when the current flowing between the two points through the cumulatively coupled coils has the same direction, therefore the voltage drops across each coil will be modified taking into consideration about the effect of mutual inducatance. The self inductances (L) will be the same here as well but there will be an addition of mutual inductance (M).

 
Cumulatively Coupled Coils

 

Therefore, the total emf induced in these coils is given by

V = L1 (dI/dt) + L2 (dI/dt) 2M (dI/dt) 

here, 2M represents the influence of L1 and L2 on each other.

As V = Lequi (di/dt)

putting it in above equation,

Lequi (di/dt) = L1 (dI/dt) + L2 (dI/dt) 2M (dI/dt) 

dividing the above equation by dI/dt we get,

Lequi = L1 + L2 + 2M

If one of the coil is reversed, so as the same current flows through each coil but oin the opposite direction, the mutual inductance then the existing mutual inductance will have the cancelling effect on each coil as explained below.

Differentially Coupled Series Inductors

 
Differentially Coupled Coils

The emf that is induced in coil one due to the effect of mutual inductance of coil two is in opposite direction when compared with the direction of self induced emf. This is because the same current now flows through each coil in opposite direction. Considering this cancelling effect, a minus sign is introduced with M when the magnetic field of two coils are differentially connected.

This gives us the final equation for calculating the total inductance of differentially connected coils as:

  Lequi = L1 + L2 – 2M

Therefore, the final equation for inductively coupled inductors in series is given as:

Lequi = L1 + L2 ± 2M

Example for Mutual Inductance Concept

Let two coils are connected in series having self inductances of 40mH and 50mH respectively. The total inductance of the combination was found to be 80mH. Find the value of mutual inductance that exists between the two coils between the two coils assuming that they are opposing each other.

we know that,

Lequi = L1 + L2 – 2M

substituting values for each variable, we have

80 = 40 + 50 -2(M)

2M = 90 – 80

therefore, M = 5mH

Let us take one more example to get more clear into the concept.

Suppose, two inductors of 30mH respectively are connected together in a series combination so that their magnetic fields aid each other giving cumulative coupling. Their mutual inductance is 5mH. Determine the total inductance of the series combination.

Lequi = L1 + L2 + 2M

          = 30 + 30 + 2(5)

          = 70mH  

Inductors in Series Summary

We know that the inductors can be connected in series to produce a total inductance value, Lequi  which is equal to the sum of the individuals inductors connected (same as in the case of resistors in series). However, when these are connected together then the value gets affected by the mutual inductance as well.

Mutual inductance case can be of two types, series aiding and series opposing. It depends on whether the coils are coupled cumulatively (in the same direction) or differentially (in the opposite direction). Accordingly the formula for the total inductance will change in both the cases.

AJAY DHEERAJ

Technical Content Developer

Half Wave Rectifier & Applications

A rectifier can be a simple diode or a group of diodes that converts the AC (Alternating Current) to DC (Direct Current). As the diode allows electric current only in one direction and blocks in another direction, therefore, this principle is used to construct the various types of rectifiers. Broadly, rectifiers are classified as Half Wave Rectifiers & Full Wave Rectifiers.

Half Wave Rectifier

An HWR (Half Wave Rectifier) circuit is the one that allows only one cycle of input of the AC signal and blocks the other. In general, we can say that it converts the positive half cycle of the sinusoidal wave of the input to pulsating DC output signal, though the conversion of the positive or negative cycle depends on the way the diode is connected.

Construction of Half Wave Rectifier

In HWR, we use only one diode which is more than sufficient to do the desired work. As we want the DC at the output for the AC sinusoidal signal given at the input, so the single diode placed in series does the work for us.

This is not all but when we talk about the whole construction of  Half Wave Rectifier circuit, it consists of  mainly three components (without filter):

  1. A Transformer (step down)
  2. A Resistive Load
  3. A Diode

Theory of Half Wave Rectifier

Now, we’ll see how the HWR circuit converts the AC voltage to DC voltage. First, high voltage AC voltage is applied at the primary side of the step-down transformer and correspondingly a low voltage is obtained at the secondary which is given to the diode.
The diode will be in Forward Biased mode during the positive half cycle of the AC voltage, thus the current flows through it. During the next half-cycle, i,e., negative cycle, the diode becomes reverse biased and blocks the current through it. Thus, when seen at the final output we can see that input has been traced as output for the positive half cycle only as shown in the figure below.


Let us try to understand this concept in a more convenient way by taking a sinusoidal voltage instead of the step down transformer.


For the positive half cycle, the circuit looks like:



This is because during the positive half cycle, the diode is in forward bias and allows the current to pass through it (diode acts as a short circuit) and we get the same voltage as at input.

For the negative half cycle, the circuit becomes open circuit as the diode becomes reverse biased and bocks the current, thus the output voltage is zero as shown below:


The input-output waveform of the above-mentioned situation is shown by the waveform below. This happens very fast depending on the frequency of the incoming voltage (for 50 hertz, 20ms of time).

The above graph shows the positive half-wave rectifier which allows only the positive cycle and blocks the negative one.
Similarly, if the polarity of the diode is reversed then the same rectifier becomes the negative half-wave rectifier which allows only the negative cycle and blocks the positive one.

Half Wave Rectifier with Capacitor Filter 

The output waveform obtained from the half-wave rectifier circuit without filter explained above is a pulsating DC waveform.

Now, as we know that all the circuits that we use practically need a constant DC and not the pulsating one thus we use filters to get the desired form of DC signal. Filters do so by suppressing the DC ripples in the waveform.

Hence to get smoother DC output waveform we can either use a capacitor or an inductor, but HWR (Half Wave Rectifier) along with capacitive filter is most commonly used. The below diagram shows how  the capacitor filter smoothens the waveform. Capacitor is connected in parallel with the resistive load.

                                                                             HWR with capacitor filter

Let us now see a few formulas of Half Wave Rectifier based on the above explanations and waveforms.

Ripple Factor of Half Wave Rectifier

When converting the AC voltage waveform to DC, the remaining unwanted AC component is called ripple. Even after all the filtration, we are still left out with some AC component which pulsates the DC waveform. This unwanted AC component is called ripple.

Ripple Factor (represented by ‘ɣ’) is used to quantify the quality of the conversion from AC voltage to DC voltage. The Ripple Factor is given by the ratio of the RMS value of the AC voltage (at input) to the DC voltage at output of the rectifier.

The formula for the ripple factor goes like:

 ɣ = √[(Vrms/VDC)2 – 1]

Alternatively,  ɣ = (I2RMS – I2DC )/IDC    = 1.21(for sinusoidal waveform)

Actually, for a good rectifier the ripple factor should be as less as possible that is why capacitor or inductor filter(s) are used to die out the ripples in the circuit.

Efficiency of Half Wave Rectifier

Rectifier efficiency (ɳ) is the ratio of output DC power to the input AC power, the formula goes like:

 ɳ = (Pdc/Pac)

The efficiency of HWR is 40.6% (ɳmax = 40.6%)

RMS value of Half Wave Rectifier

To find the rms value of the half-wave rectifier, we need to calculate the current across the load. If the instantaneous load current, iL = Imsinwt, then the average of load current (Idc) equals to:

  Idc = (1/2 π) ∫0π  Imsinwt = (Im / π)

Here Im represents the peak instantaneous current across the load (Imax). This the DC current obtained across the load (output) is

 IDC = Imax/ π        ; where Imax =maximum amplitude of dc current

For a half-wave rectifier, the RMS load current, Irms is equal to the average current, Idc multiplied with π/2. Thus,  Irms = Im/4

Where Imax = Im which is equal to the peak instantaneous current across the load.

Peak Inverse Voltage of Half Wave Rectifier

It is the maximum voltage that the diode can withstand during the reverse bias condition. If a voltage is applied more than the PIV then the diode will be destroyed.

Form Factor of Half Wave Rectifier

Form factor is the ratio of rms value to the average value.

F.F = RMS value/ Average value

The form factor of HWR is 1.57, i.e., FF = 1.57

Output DC Voltage

The output voltage (VDC) across the load resistor is denoted by

 VDC = Vsmax/ π  , where Vsmax is the maximum amplitude of secondary voltage

Advantages of Half Wave Rectifier

  • Simple circuit with less number of components
  • Economical at initial state. Although there is a higher cost over time due to more power losses

Disadvantages of Half Wave Rectifier

  • Converts only one cycle of the sinusoidal input given to it and the other cycle gets wasted. Thus, giving more power loss.
  • HWR produces lower output voltage.
  • The output current thus obtained is not purely DC and it still contains a lot of ripple (i.e. it has a high ripple factor)

 

Applications of Half Wave Rectifier

In day-to-day life, the half-wave rectifier is mostly used in low power applications because of its major disadvantage being the output amplitude which is less than the input amplitude. Thus power is wasted and output is pulsated DC resulting in excessive ripple.

Some of the uses and applications of rectifiers are in :

  • Appliances
  • Used with transformers
  • Soldering
  • AM radio
  • Pulse generated circuits
  • Single demodulation
  • Voltage multiplier

Use of Rectifier for powering appliances

As we know that all electrical appliances use DC power supply to function thus using a rectifier in the power supply helps in converting AC to DC power supply. Bridge rectifiers are used widely for large appliances, where these are capable of converting high AC voltage to lower DC voltage.

Used with Transformer

With the help of half wave rectifier we can achieve the desired DC voltage with the use of step up or step down transformers. Even Full Wave Rectifiers are used for powering up motors an LEDs that works on DC voltage.

Uses of Rectifier While Soldering

Half Wave Rectifiers are used in soldering iron types of circuit and is also used in mosquito repellent to drive the lead for fumes. In electric welding, rectifiers with bridge configurations are used to supply steady and polarized DC voltage.

Used in AM Radio

Half wave diode rectifiers are used in AM radio as a detector because the output contains an audio signal. Due to the less intensity of the current, it is of very little use to the more complex rectifier.

Uses of Rectifier in Circuits

Pulse generating circuits and firing circuits use half-wave rectifiers.

Used for Modulation

In a modulating signal, for demodulating the amplitude, a half-wave rectifier is used. To detect the amplitude of modulating the signal, in a radio signal, a full-wave bridge rectifier is used.

Used in Voltage Multiplier

For the purpose of voltage multiplier circuit, a half-wave rectifier circuit is used.

3 Phase Half Wave Rectifier

Although, the principle and theory of 3 phase HWR are the same as single-phase HWR but the characteristics are different. The waveform, ripple factor, efficiency, and output RMS values are not the same.

The three-phase half-wave (diode) rectifier is used for the conversion of 3 phase AC power to the DC power. Since the diodes are used as switches here hence are uncontrolled switches, implying that there is no way controlling the ON OFF times of these switches.

Generally, the 3-phase half-wave diode rectifier is constructed with a three-phase supply connected to a three-phase transformer where the secondary winding of the transformer is always connected as a star connection. This is done because of the reason that the neutral point is required to connect the load back to the transformer secondary windings, providing a return path for the flow of power.

A typical 3 phase transformer supplying a purely resistive load is shown below. Here, each phase of the transformer is used as an individual alternating source. The measurement and simulation of voltages are as shown in the fig below. Also, we have connected individual voltmeters across each source as well as across the load.

So from the above waveform that the diode D1 conducts when the R phase has the voltage value that is higher than the value of the voltage of the other two phases and the stated condition starts when the R phase is at 300 repeats itself after every complete cycle. So, D1 conducts next at 3900. Similarly, diode D2 starts conducting at 1500 as voltage in B phase becomes maximum (when compared with other two phases) at that instant. Therefore, each diode conducts for 1500 – 300 = 1200.

The average of the output voltage across the resistive load is given by

Vo = (3/2π) Vm line

Where Vm line = √6 Vphase

The RMS value of the output voltage can be given by

Vo rms = 0.84068 Vm phase

And the voltage ripple factor equals

Vr/Vo = 0.151/0.827 = 0.186 = 18.26%

Thus voltage ripple is significant and thus undesirable as it leads to power loss.

Efficiency, ɳ = (Po/Pi) = 0.968 = 96.8%

Even after better efficiency also, a three-phase half-wave diode rectifier is not commonly used as the power loss here is more significant.

Ajay Dheeraj

Technical Content Developer

Full Wave Bridge Rectifier Working and Application

The full-wave bridge rectifier is a circuit consisting of four diodes arranged in a bridge-type structured figure as shown. This circuit gives full-wave rectification and is cost-effective as well, thus used in many applications.

 

Construction Of Full Wave Rectifier

Full Wave Bridge Rectifier

Four diodes are used in the bridge rectifier. All the four diodes are connected in the form of a diamond-shape to the transformer and the load as in the shown circuit diagram.

Working of Full Wave Rectifier

The working of the full-wave rectifier is simple. The primary winding of the transformer is supplied with a sinusoidal supply. Whereas, the secondary of the transformer is diametrically connected to the two diametrically opposite points. The load Rload is connected to the remaining two points of the bridge.

Now when the AC input is given, during the first half cycle, the upper part of the transformer becomes positive when compared to the lower half of the same. Thus during the first half diodes, D4 and D1 are in forward biased. The currents follow the path 1-2 and enter the load. While returning, the path is 4-3. Similarly, diodes D2 and D3 are reverse biased. Thus, no current flows through the path 2-3 and 1-4.

During the negative half-cycle, i.e. in the next cycle, the upper portion of the transformer becomes negative when compared with the lower portion of the same. Thus diodes D2 & D3 are forward biased and the currents follow the path as 3-2-4-1. The other two, i.e., D1 & D4 are reversed biased for this cycle. So no current flows through the path 1-2 and 3-4. Hence negative cycle is rectified as it appears across the load.

The waveform of the rectified input will look like

 

 

Maximum of secondary Voltage = Peak Inverse Voltage (PIV) of the bridge rectifier

Thus in the bridge rectifier circuit, two diodes out of four conduct during one half-cycle and henceforward resistance becomes double that is 2RF.

Peak Current in Full Wave Rectifier

The value of the instantaneous applied voltage to the rectifier is given as

Vs = Vsmaxsin wt

Let us suppose that the diode has a forward resistance of Rf ohms and the reverse resistance equals to infinity, hence current through the RL load can be given as :

  • I1 = Imax Sin wt and I2 = 0 for first half-cycle and
  • I1 = 0 and I2 = Imax Sin wt for the second half-cycle.

The sum of currents I1 and I2 flows through the load resistance RL which is given as:

I = I1 + I2 = Imax Sin wt for complete cycle

The peak value of load current flowing through the load resistance RL is :

Imax = Vsmax/(2RF + RL)

Output Current of Full Wave Rectifier

The current through the load RL is the sum of the two halves of the ac cycles. The magnitude of the direct current Idc is equal to the average value of the ac current and is obtained by integrating the current I1 between 0 & Π or current I2 between Π & 2Π.

Idc = 1 / Π ∫ ∫0Π  I1 d(wt)

= 1/Π ∫ ∫0Π  Imax sin wt d(wt)

= 2Imax / Π

DC Output voltage

DC or average value of voltage across the load is given by

Vdc = Idc RL = 2/ (Π ImaxRL)

Root Mean Square (RMS) Value of Current

Effective or RMS value of the current flowing through the load resistance RL can be given as:

I2RMS = 1/ Π ∫ ∫0Π I1 d (wt) = I2max / 2

IRMS   = Imax / √2

Rectification Efficiency of Full Wave Rectifier

Now the amount of power delivered to the load

Pdc = I2dc RL = 2Im/Π =RLoad

= (4/Π2)I2RL

Alternating (AC) power input to the transformer = power dissipated at the diode + power dissipated from load resistance RL

I2RMSRf + I2RMSRLoad = (I2m/2) (Rf + RLoad)

Rectification efficiency,

n = Pdc / Pac = ((4/Π2)I2m RLoad) / ((I2m/2) (Rf + RLoad)

=0.812(1+Rf / RLoad)

Ripple Factor of Full Wave Rectifier

Rectified output voltage, form factor is given as:

Kf = IRMS/ Iavg = (Im/√2) / (2Im/ Π) = 1.11

Thus ripple factor, γ = √(1.112 – 1) = 0.482

Regulation of Full Wave Rectifier

The DC output voltage is given as

Vdc = Idc RL= 2/(ImRL)

= (2VsmRL / (Rf + RLoad))

= (2Vsm / Π) – (IdcRf)

Applications of Full Wave Rectifier

Conversion of AC Power to DC Power

In many of the electronics applications, regulated DC Power supply is used. One of the most convenient and reliable ways is to convert the AC supply to DC supply. The said conversion is done through a rectifier which uses a system of diodes arranged in a particular way.

Now if two the diodes are used then the arrangement made is a half-wave rectifier which rectifies only half of the AC signal. Whereas, when four diodes are arranged in a particular way then the circuit is the Full Wave Rectifier which rectifies both the cycles of the AC signal. Also with the help of a center-tapped transformer and with two diodes, we can have a full-wave rectifier.

Here the bridge rectifier is used for the application. The arrangement consists of 4 diodes arranged in a way that the anodes of the two adjacent diodes are connected together to give the positive supply to output and the cathodes of the other two are connected together to give the negative supply to the output. The cathode and anode of two adjacent diodes are connected together at the positive of AC supply whereas cathode and anode of another two diodes are connected at the negative of the AC supply. Hence, the four diodes are arranged in a bridge-type configuration such that in each half-cycle two alternate diodes conduct producing a DC voltage with some repels.

Uses of Rectifier while Soldering

Half-wave rectifiers are used in soldering iron type of circuit and are also used in mosquito repellent to drive the lead for the fumes. In electric welding, the bridge rectifier circuit is used to supply polarized and steady DC voltage.

In AM Radio

A half-wave rectifier is used in AM radio as a detector because the output consists of an audio signal. Due to the less intensity of the current, it is if very less use to more complex rectifiers.

For Modulation

For demodulating amplitude of a modulated signal, a half-wave rectifier is used. In a radio signal, for detecting the amplitude of a modulating signal, a full-wave bridge rectifier is used.

Also, a half-wave rectifier is used for the purpose of voltage multiplier.

Other applications are as follows:

  • Because of the low cost (when compared with center-tapped), these are widely used in power supply circuits.
  • It can be used to detect the amplitude of the modulated radio signals.
  • To supply polarised voltage in welding, the bridge rectifiers can be used.
  • A bridge rectifier with a filter is ideal for many general power supply applications like charging a battery, powering a dc device, maybe LED, motor, etc

Full Wave Bridge Rectifier with Capacitor Filter

As we know, the output voltage of the full-wave rectifier is not constant, it is always pulsating and thus can’t be used in real-life applications. In other words, we require a DC supply with a constant output voltage. This need can be fulfilled by using an adequate filter with an inductor or a capacitor to make the output voltage smooth and constant.

Here capacitor is connected in parallel to the load resistance in a linear power supply. The capacitor is used to increase the DC voltage and to reduce the ripple voltage components of the output obtained. This capacitor is also called a reservoir or smoothing capacitor. Generally, this is followed by a voltage regulator which eliminates the remaining ripples so that the required output can be achieved.

Till the time, the rectifier conducts and the potential is higher than the charge across the capacitor, the capacitor would store the energy from the transformer. But when the output of the rectifier falls below the charge across the capacitor, the capacitor would discharge its energy into the circuit. As the rectifier conducts current only in the forward direction, all the energy discharged by the capacitor will flow into the load. This makes the output a sawtooth wave which is a convenient linear approximation to the actual waveform as shown below:

Advantages of Full Wave Bridge Rectifier

  • In terms of rectification, the efficiency of the full-wave rectifier is double that of a half-wave.
  • Higher transformer utilization factor, higher output power and higher voltage in case of a full-wave rectifier.
  • In the case of the bridge rectifier, the transformer used is simple when compared to the center-tapped one and even the transformer can be eliminated if voltage up and down is not required.
  • The ripple voltage is lower and of high frequency in case of the full-wave rectifier, so even simple filtering is enough.
  • For a certain power output, a smaller power transformer can be used in the case of a bridge rectifier because the current in both primary and secondary winding of the transformer flows for the entire AC cycle.

Disadvantages of Full Wave Bridge Rectifier

  • The requirement of four diodes (two in case of a center-tapped)
  • Additional voltage drops because of the two extra diodes thereby reducing the output voltage.

 

Ajay Dheeraj

(Technical Content Developer)

Wheatstone Bridge: WORKING PRINCIPLE AND APPLICATION

Wheatstone Bridge _1

The Wheatstone bridge was formerly invented by Samuel Hunter Christie in the year 1833 and later improvised by
Sir Charles Wheatstone in 1843 after which it became popular.

A Wheatstone bridge is an electrical circuit used to measure an unknown resistance value by maintaining a balance between two legs of a bridge circuit. The primary advantage of using the Wheatstone bridge is its accuracy in finding the unknown (electrical resistance) value when compared to instruments like a simple voltage divider.

Construction of Wheatstone Bridge

Wheatstone Bridge _1

Wheatstone bridge is designed in a bridge type structure having four resistances, three known and one unknown.

Here R1, R2 are known resistances, R3 is variable (adjustable) and R2 is the one which needs to be measured. Apart from the resistance, there is a voltmeter connected between the points C & B and the DC supply is connected between A & D.

Working Principle of Wheatstone Bridge

As explained above, in a Wheatstone Bridge R1, R2 and R3 are the resistances of known value and the resistance R2 is the one which has to be adjusted until no current flows through the galvanometer V.

This condition where the current through Galvanometer is zero and thus voltage between two midpoints B & C also comes to be zero is called balanced bride condition of Wheatstone bridge.

Therefore the ratio of two resistances in one leg (R1/ R2) is equal to the ratio of another two resistances in the other leg (Rx/ R3). If somehow bridge is unbalanced, the direction of the current indicates whether R2 is too high or too low.

At bridge balanced condition,

R 2 /R 1 =R x /R 3

and hence Rx= R 3 *(R 2 /R 1 ) could be measured easily.


This detection of zero current in galvanometer can be done with high precision, thus if R1, R2, and R3 are known with high precision then Rx measured will be of high precision as well. Also, even a small change in the value of Rx would disrupt the balance and could easily be detected. Alternatively, if R2 is not variable then voltage difference across or current flowing through the meter can be used to calculate Rx. This method is faster for measuring the unknown resistance of the Wheatstone bridge.

Example of Wheatstone Bridge

Wheatstone Bridge example _1

Consider the circuit is shown above, where the Wheatstone bridge is an unbalanced condition and output the voltage across C & D and the value of R 4 are to be measured for a balanced bridge condition.

Now as per the voltage division law,

Vc = (R2 /( R1 + R2 ))*Vs

Putting R2= 12 ohms, R2 = 8 ohms, Vs= 10 volts

Vc = (12/(8+12))*10
= 6 volts

In second arm, applying voltage division law,

Vd = (R4 /( R3 + R4 ))*Vs

V2 = (16/(48+16))*10
= 2.5 volts

The voltage between points C & D can be calculated as

Vcd = Vc – Vd

Vout = 6 – 2.5 = 3.5 volts.

Now to calculate R 4, for balanced bridge condition can be measured as R4 = (R3 /R1 )*R2 = 72 ohms.

So here we can conclude that the Wheatstone bridge acts like a 2 port network having 2 inputs (A & B) and 2 outputs (C & D). Also if the voltage across the output terminal is 0 volts then the bridge is called to be in a balanced condition, while in an unbalanced condition voltage may have any value (either positive, negative) depending on the circuit parameters.

Applications of Wheatstone Bridge

Strain Measurement 

Generally, for strain measurement, strain gauges are used whose resistance varies with proportionate to strain present in the device. Practically, strain gauge resistance ranges from 30 ohms to 300 ohms. Since the change in resistance may be only a fraction of full scale thus a highly precise and accurate measuring instrument is required
and Wheatstone bridge perfectly suites for it.

In this application, the unknown resistance is replaced with a strain gauge. Here R1 and R3 have the same value and R2 is a variable one. Now without applying force to the strain gauge, a rheostat is varied until voltmeter indicates null deflection. This indicated the bridge is balanced and thus no strain on the gauge.

Light Detector Circuit

Light Detector Using Wheatstone Bridge Circuit

It is one of the simplest applications of the Wheatstone bridge using the photoresistive device. A light dependent resistor is placed in the place of the unknown resistor in the Wheatstone bridge. An LDR which is a passive resistive sensor is used for converting visible light levels into a change in resistance and afterward a voltage. LDR has around 900 ohms resistance in dark light (at a light intensity of 100 lux) and as less as 30 ohms in bright light.

Therefore, the connection of light dependent resistor in Wheatstone bridge would help in measuring and monitoring the changes in the light levels.

The Wheatstone bridge goes with the concept of a difference measurement, which could be highly accurate.

Some variations on the Wheatstone bridge could be used to measure capacitance, inductance, impedance, etc. Some other applications can be as follows:

  • Light detector using Wheatstone Bridge
  • Light measurement using a photoresistive device (a light dependent resistor is placed as a substitute for one of the resistors)
  • Measuring strain with the help of Bridge (strain gauge is used in place of the variable resistor) 
  • It is also used for sensing mechanical and electrical quantities.
  • The circuit is used to measure the changes in pressure.
  • It is used in thermometers for temperature measurements with high accuracy.

Limitations of Wheatstone bridge

  • Susceptibility to high DC current is not there.
  • Under the unbalanced condition, readings might be inaccurate.
  • Resistance measured by Wheatstone bridge ranges from “few” ohms to “mega” ohms
  • With the help of applied voltage (emf), the upper range of the bridge can be increased while the lower range can be limited by connecting a lead at the binding post.

Now, if we summarize whatever we have studied till now then we can say that the Wheatstone bridge is the most common and simplest bridge network for finding out the resistance. With its ability to measure precise changes, these are mostly used for sensor applications, where a resistance change is converted to voltage change (for a transducer).

The combination of operational amplifier along with the Wheatstone bridge is used extensively in industries for various sensors and transducers. Also as said earlier, any changes in quantities like temperature, light intensity, strain, electrical and mechanical sensing, pressure, etc. could be measured in a most accurate and precise manner.

The only change required to calculate these values is to replace the unknown resistance in the bridge circuit with the required quantity (mentioned above). This is the reason why being the simplest circuit also, Wheatstone bridge is one of the most used bridge circuits.


Ajay Dheeraj
Technical Content Developer

How to use Transistors in DCACLAB (both NPN and PNP)

using pnp transistor in DCACLab

In this Post you will know that how to use Transistor in DCACLab (both NPN and PNP).

BJT (Bipolar Junction Transistor) also Generally known as a transformer is an active electronic component used in either signal amplification or logic switches in signal processing.

The transistor is a semiconductor device just like a PN Junction diode but unlike to that it has two PN junctions with different width of the depletion layers.

How to use Transistors in DCACLAB and Customize their Properties

Steps to use PNP Transistor in Lab

Step 1 : Navigate through the list of components and click on the PNP Transistor as shown in the below screen shot.

Step 2 : A PNP Transistor will appear in the board as shown in the screenshot below. You can easily change the position of  Emitter, Base and Collector from the property box of the PNP transistor.

pnp transistor

Step 3 : Now use the transistor in your desired circuit. 🙂

How to use Transistors in dcaclab

Steps to use NPN Transistor in Lab

Step 1 : Navigate through the list of components and click on the NPN Transistor as shown in the below screen shot.

Step 2 : A NPN Transistor will appear in the board as shown in the screenshot below. You can easily change the position of  Emitter, Base and Collector from the property box of the NPN transistor.

npn transistor

Step 3 : Connect the Emitter, Base and Collector as your need in your circuit.

How to use Transistors in dcaclab

Controlling Properties of PNP and NPN Transistor

While using the transistor in a circuit whether its PNP Transistor or NPN Transistor. Click on it and a gear icon will appear near to the transistor.

gear sign will appear on npn transistor
Gear sign on NPN transistor

gear sign will appear on pnp transistor
Gear sign PNP transistor

Now, click on the gear sign shown in the above screenshots. The property box will appear near to the respective transistor as shown in the below screenshots.

how to use the properties of pnp transistor in dcaclab
Property box of pnp transistor

how to use the properties of npn transistor in dcaclab
Property box of npn transistor

In the property box of the transistors you can Swap the position of Emitter and Collector. You can also rotate the transistor.

Components in Properties of PNP and NPN Transistor

properties of pnp transistor in dcaclab 1
Properties of pnp transistor

properties of npn transistor in dcaclab 2
Properties of npn transistor

Both the property box for PNP transistor as well as NPN Transistors work the same way.

  1. Current Gain or the Amplification factor of the transistor can be controlled from this section.
  2. Select or deselect this checkbox in order to swap the position of emitter and collector. It Basically flips the transistor.
  3. In this you can rotate the battery, you can rotate it from 0° to 90°, 180° and 270°.

Also see the logical switch made by the transistor.

How to use logic gates

how to use logic gate without wires in dcaclab

Logic gates are the Digital Electronic circuit which are used to perform logical operations and tasks.

The Logic Gates operate on Boolean functions to give the logical output for one or more than one logical inputs. The Boolean function uses the one or more than one binary input and produces a new binary output according to the function applied.

Below is the method which shows that how to use logic gates to build your logic circuit and also how to test them with their corresponding truth table.

Steps to use a logic gate in Lab

In the below step we’re taking an or gate in order to demonstrate that how to use a logic gate in our virtual lab.

Step 1 : Scroll through the list of the components on the above and navigate to the logic gates as shown in the screen shots below.

where to find logic gates in dcaclab

Step 2 : Click on the logic gate you want to use from the list. The logic gate will appear on the board as shown in the screenshot below.

or gate

Step 3 : Now you will need to grab the input as well as the output for the logic gate

how to give input to logic gate in dcaclab
logical input for logic gate

how to give output to logic gate in dcaclab
Logic output for the logic gate

Just similar to the step 2 if you click on the input and the out icon, they will appear on the board.

Step 4 : Connect the input as well as the output to your logic gates. There are the two ways you can give them a connection.

  1. By directly connecting the input as well as the output to the logic gate. how to use logic gate without wires in dcaclab
  2. By using the wires to connect the input as well as the output to the logic gate.how to use logic gate with wires in dcaclab

Truth tables for the different logic gates

Below are the list of logic gate truth tables which might be helpful for you to keep it up while working with the logic gates.

Truth Table for AND Gate
A (Input 1) B (Input 2) A.B = Y (Output)
0 0 0
0 1 0
1 0 0
1 1 1

 

Truth Table for OR Gate
A (Input 1) B (Input 2) A+B = Y (Output)
0 0 0
0 1 1
1 0 1
1 1 1

 

Truth Table for NOT Gate
A (Input 1) A (Output)
0 1
1 0

 

Truth Table for NAND Gate
A (Input 1) B (Input 2) A.B = Y (Output)
0 0 0
0 1 1
1 0 1
1 1 1

 

Truth Table for NOR Gate
A (Input 1) B (Input 2) A+B = Y (Output)
0 0 1
0 1 0
1 0 0
1 1 0

 

Truth Table for EXOR Gate
A (Input 1) B (Input 2) A⊕B = Y (Output)
0 0 0
0 1 1
1 0 1
1 1 0

 

Truth Table for EXNOR Gate
A (Input 1) B (Input 2) A⊕B = Y (Output)
0 0 1
0 1 0
1 0 0
1 1 1

 

How to use Transformers

gear sign will appear on the transformer

The Transformer is an static electrical energy which transfers the electrical energy from one circuit to another with the help of electromagnetic induction.

The most general use of transformer is to either increase or decrease the AC voltage. However, they too are used to isolate a part of circuit or other components.

In this post you are going to learn that how to use Transformers as per your need. Also how you can increase or decrease the voltage using the transformer in DCAClab.

Steps to use the transformer in Lab

Step 1 : Navigate through the list of devices and components and locate the transformer as shown in the screenshot below.

where to find transformer in dcaclab

Step 2 : Click on the icon of the transformers as shown in the above. The transformer will appear on the board as shown in the screenshot below.

transformer in dcaclab

Step 3 : Define the Primary as well as secondary winding and connect them as per your need.

using transformer in dcaclab

Controlling the properties of Transformer

While working with the circuit that has a transformer, click on the Core part of the transformer. You will see a gear sign as shown in the image below.

gear sign will appear on the transformer

Click on the Gear sign shown as the above screenshot and a box with a label “Transformer properties” will appear as shown below.

how to use the peoperties of transformer in dcaclab

From the Transformer properties box you can control the various factors such as ratio between the two winding, their inductance and the coupling coefficient.

Components in Transformer properties

properties of transformer in dcaclab

  1. In this section you can set the value of inductance of the coil used in in order to make the primary as well as secondary winding. You can either use the scroll bar or the input field in order to define the value of the inductance.
  2. You can also control the inductance value from the range of µ as well as mili.
  3. In this you can set the ratio of number of wire turns between the two winding. For your ease of use in the DCACLAB you can follow the rule(ø) at the end of this post.
  4. The Coupling coefficient (i.e. K) has an ideal condition value of “1”, however with this option you can set the coupling coefficient of the two winding to the more practical value.

Rule(ø) : Assume the Left winding as A and the right winding as B.

  • The value of the ratio must be a natural number in order to make the number of winding in B bigger then A.
  • The value of the ratio must be a fractional number in between 0 & 1 in order to make the number of winding in A more then B.

Saving circuit when not logged in

NEVER LOOSE YOUR CIRCUIT AGAIN

We have a update which is going to enhance your virtual lab experience !

As we always try to provide you the best of DCACLab experience to you.

It is very common and you certainly might have an experienced it earlier.

You’ve worked through an experiment or made the amazing circuit.

But, You remember that either you are not logged in or you do not have a DCACLab account.

It is very painful, even for us.

We have to login, make the circuit and calibrate those components again.

Not after this update !

Now you can save your work whether if you’re not logged in.

All you have to do is to follow the simple steps below.

Steps to save the circuit in DCAClab when the user is not logged in

After the new update, you will be able to save the experiments temporarily if you’re not logged in to your account.

Step 1 : Head over to the top left of the page.

Optional : Check on the Public box if you want your experiment to be shown as a public or leave it as it if you want to keep it private (you can change it later).

Step 2 : Click on the Save button as shown in the screenshot below.

click on the save button
Step 3 :
Enter the name of your experiment/circuit and hit the OK button as shown in the screenshot below.

enter the name of your circuit or experiment

Step 4 : You’ll see a message asking for weather you want to login/signup to save your work permanently.  Click on yes and it will redirect you to login/signup page.

message box displaying circuit saved message

signup or login to your account

Step 5 : Login to your account and head over to your dashboard.

head over to your dashboard

You’ll see the name of the experiment/circuit which you have saved. From here you can carry on your experiment or share your work with your friends.

Note : You can try this feature when you’re not login to your account. In case if you are already logged in to your account the experiment will be directly saved in your dashboard. 

Check out the Lab to give it a try.

AUTO WIRE CUTTING

I would to introduce a new feature that will make connecting components super easy, as users will have to create one wire only, and it will get splitted into wires to connect each component.

Here is a video for the new feature: