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ABSTRACT

The aim of this lab assignment is to design a DC power supply. For this purpose we have to assemble a rectifier circuit on the output of a transformer. It also involve the calculation of different components used so the DC supple contains lesser ripples. The DC power supply has wide applications in the modern world. Every day we use Mobile charger , Laptop Charger , Ring bell, TV etc. All these equipments require DC power supply to operate.

Step 1: Process

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Object:- To design a regulated DC power supply of (+5Volt/500mA).

Rectification:- The diode is an ideal and simple device to convert AC into DC. The process is called rectification. We shall focus our attention on some performance measure of a rectifier:

Transformer A Transformer is a static piece of equipments used either for raising or lowering the voltage of an ac supply with a corresponding decrease and increase in current. It essentially consist of two windings primary and secondary, wound on a common laminated magnetic core as shown in figure. N1: no. of turns in primary coil N2: no. of turns in secondary coil If N1< N2 :- Step-up transformer N1> N2 :- Step-down transformer The following points may be noted carefully:- i. The transformer action is based on the law of electromagnetic induction. ii. There is no electrical/physical connection between the primary & secondary windings. The ac power transferred from primary to secondary through magnetic flux. iii. There is no change in frequency i.e. output power has the same frequency as the input power.

process:-

AC supply 230v/50hz >>>Transformer (step-down)>>> Diode Rectifier >>>Filter Circuit >>>Voltage Regulator

Step 2: Transformer: Selection & Checking

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Transformer

A Transformer is a static piece of equipments used either for raising or lowering the voltage of an ac supply with a corresponding decrease and increase in current. It essentially consist of two windings primary and secondary, wound on a common laminated magnetic core as shown in figure. N1: no. of turns in primary coil N2: no. of turns in secondary coil If N1< N2 :- Step-up transformer N1> N2 :- Step-down transformer The following points may be noted carefully:- i. The transformer action is based on the law of electromagnetic induction. ii. There is no electrical/physical connection between the primary & secondary windings. The ac power transferred from primary to secondary through magnetic flux. iii. There is no change in frequency i.e. output power has the same frequency as the input power. AC supply 230v/50hz Transformer (step-down) Diode Rectifier Filter Circuit Voltage Regulator Rectification Process:- F. The losses that occur in transformer are:

(a) Core losses- eddy current & hysteresis losses.

(b) Copper losses-in the resistance of a winding. Relation b/w voltages and no. of turns is: (V1/V2)=(N1/N2) Checking of Transformer:-

1. Cold check(without connecting power supply):-

(a) Insulation of Cu wire(short circuit) :- if the circuit is short than its resistance will be “0”. (b) Test for open circuit :- if the winding is break (open) from anywhere than it will show very high “infinite” resistance. (c) Insulation b/w winding and core & b/w primary and secondary windings:- these are tested using “megger”.  If megger show some value whem connect to two terminals means insulation is not proper b/w both terminals. Other wise it will show “out of limit”.

2. Hot Check(using power supply):- Rating error:- It is to verify whether output of a transformer is according to its rating(voltage and current) or not. It is identified by measuring Voutput and Ioutput using multimeter. The transformer which we have used is given bellow type:-9-0-9 ;Current rating= 500mA

Step 3: Rectification Circuit

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Full Wave Rectifier

A Full Wave Rectifier Circuit produces an output voltage or current which is purely DC or has some specified DC component. Full wave rectifiers have some fundamental advantages over their half wave rectifier counterparts. The average (DC) output voltage is higher than for half wave, the output of the full wave rectifier has much less ripple than that of the half wave rectifier producing a smoother output waveform. The Full Wave Bridge Rectifier Another type of circuit that produces the same output waveform as the full wave rectifier circuit above, is that of the Full Wave Bridge Rectifier. This type of single phase rectifier uses four individual rectifying diodes connected in a closed loop “bridge” configuration to produce the desired output. The main advantage of this bridge circuit is that it does not require a special centre tapped transformer, thereby reducing its size and cost. The single secondary winding is connected to one side of the diode bridge network and the load to the other side as shown below. The Diode Bridge Rectifier:-The four diodes labelled D1 to D4 are arranged in “series pairs” with only two diodes conducting current during each half cycle Vp-p=27.20V Output of Transformer

The Positive Half-cycle:- During the positive half cycle of the supply, diodes D1 and D2conduct in series while diodes D3 and D4 are reverse biased and the current flows through the load as shown below. The Negative Half-cycle During the negative half cycle of the supply, diodes D3 and D4 conduct in series, but diodes D1 and D2 switch “OFF” as they are now reverse biased. The current flowing through the load is the same direction as before.

Step 4: Filter Circuit

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CAPACITOR FILTER

We saw in the previous section that the single phase half-wave rectifier produces an output wave every half cycle and that it was not practical to use this type of circuit to produce a steady DC supply. The full-wave bridge rectifier however, gives us a greater mean DC value (0.637 Vmax) with less superimposed ripple while the output waveform is twice that of the frequency of the input supply frequency. We can therefore increase its average DC output level even higher by connecting a suitable smoothing capacitor across the output of the bridge circuit as shown below.

Formulas to find
capacitor value:-

There are so many ways to find capacitor values . the formulas mostly used are:-

1). Q=CV

C=IL/(2.f.∆V)

OR

2). Q=CV

C=Q/∆V

C=I.td/∆V {because Q=I.t}

Now we have to find values of I(current) , td(discharging time period) and ∆V(ripple voltage) .

For current:

I = current rating of transformer

∆V(ripple voltage):

∆V= Vm-value of voltage assumed in input of regulator which

Is sufficient to give required output

td(discharging time period):-

the above waveform is sin wave so

v=Vm.Sinө

let instantaneous value of voltage v=8V

8= Vm.Sinө

Ө=sin^(-1)(8/Vm)

As at 180˚ angle; time is 10ms (because the frequency of wave is 50Hz)

So at angle Ө ; time =(10/180) Ө

5 tL

So from above figure it is clear that

td(discharging time period) = 5+Tl........................eq.(1)

now to find values of capacitor for Vm=18.2V

& RL= 18.4Ω

Q=CV

C=Q/∆V

C=I.t/∆V where ∆V= Vm-8= 18.2-8= 10.2V

& IL=1Amp

So C=1.t/10.2……………………………………………..eq.(2)

To find “td”

v=Vm.Sinө

let instantaneous value of voltage v=8V

8= Vm.Sinө

Ө=sin^(-1)(8/Vm)

sin Ө=8/18.2 Ө=26.075

As at 180˚ angle; time is 10ms (because the frequency of wave is 50Hz)

So at angle Ө ; time =(10/180) Ө

time =(10/180) 26.075 t=1.4486ms

from eq.(1)

now td =5+1.4486 td =6.4486ms

from eq..(2)

C =1*6.4486/10.2 C=632.17µƒ

But because of safety purpose we are using 2000µƒ capacitor.

C=2000µƒ

Fig. 3.3:- Image showing circuit of capacitive filter and output pulse from capacitor filter

Voltage across capacitor:-

Vp=24.88V

Vrms=24.88/(2√3)

Vrms=7.18223Volt

Till there are some ripples in the output waveform. So we have to use some IC’s like LM7805 or LM317 to obtain perfect DC wave. Now the next step is to put a voltage regulator IC in the circuit.

Step 5: Voltage Regulator IC

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Voltage Regulator

We are putting two IC’s one by one to obtain perfect DC of 5Volt.

(1). Designing of regulated DC Power Supply using adjustable Voltage Regulator LM7805:

It gives a constant direct voltage across its output terminals

(2).Designing of regulated DC Power Supply using adjustable Voltage Regulator IC-LM317:-

It gives variable Output DC voltage with change in value of R2 .

Electrical properties of LM317:-

to calculate the values of R1 & R2:-

(values from Datasheet)

IL= 10mA

Vref.= 1.2V

Iadj= 100µA

R1= Vref/IL =1.2V/10mA

R1=120Ω

For R2

Voutput= Vref(1+R2/R1)+(Iadj*R2)

5V=1.25(1+R2/120)+(100 µA *R2) R2=356.80 Ω

Step 6: Testing Line & Voltage Regulation

Your circuit must looks like that

and after that you can implement it upon PCB by various process like Drilling, Solding etc

About This Instructable

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Bio: Student of Electronics & Communication Engineering @ Techno India NJR Institute of Technology,Udaipur, Rajasthan, India
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