The market is flooded with cheap mobile charger circuit.every charger circuit is not same, some of them contains few extra capacitors or resistors. First you need to understand the basics of components.
What is PCB board?
A printed circuit board, or PCB, is used to mechanically support and electrically connect electronic components using conductive pathways, tracks or signal traces etched from copper sheets laminated onto a non-conductive substrate.
What is PWB Board?
What is PCBA Board?
circuit card assembly backplane assemblies
Common Components Of a Mobile Charger
2- Y type Capacitors
3 – Capacitors
4 – Common mode chock
5 – Transformer
6 – Housing
Main Functions Of the Components
Flow of current in the circuit is being controlled by the resistor.
It is mainly used to store the charges. It is of two types polarized and non-polarized,electrolytic capacitor is an example of polarized while ceramic and paper is non polarized.
It used to enlarge the signal strength or to open or close the circuit.
When the voltage achieves the breakdown point it starts working but in the reverse bias state.
It is a having two terminal named as anode and cathode. It allows the current to flow only in the forward direction while stopping the flow of current in the backward direction.
You can also build 6V DC, 9V, 12V, 15V etc by using proper transformer, capacitor and voltage regulator. The basic concept remains the same, you just need to arrange a heat sink for higher voltage and current.
This circuit mainly consists of a step down Transformer, a Full wave bridge rectifier and a 5V voltage regulator IC (7805). We can divide this circuit into four parts: (1) Step down AC voltage (2) Rectification (3) Filtration (4) Voltage Regulation.
1. Step down AC voltage
As we are converting 220V AC into a 5V DC, first we need a step-down transformer to reduce such high voltage.converting 220V AC to 9V AC. In transformer there are primary and secondary coils which step up or step down the voltage according to the no of turn in the coils.
Selection of proper transformer is very important. Because Current rating depends upon the Current requirement of Load circuit (circuit which will use the generate DC). The voltage rating should be more than the required voltage. Means if we need 5V DC, transformer should at least have a rating of 7V, because voltage regulator IC 7805 at least need 2V more i.e. 7V to provide a 5V voltage.
Rectification is the process of removing the negative part of the Alternate Current (AC), hence producing the partial DC. This can be achieved by using 4 diodes.
Diodes only allow current to flow in one direction. In first half cycle of AC diode D2 & D3 are forward biased and D1 and D4 are reversed biased, and in the second half cycle (negative half) Diode D1 and D4 are forward biased and D2 and D3 are reversed biased. This Combination converts the negative half cycle into positive.
The output after the Rectification is not a proper DC, it is oscillation output and has a very high ripple factor. We don’t need that pulsating output, for this we use Capacitor. Capacitor charge till the waveform goes to its peak and discharge into Load circuit when waveform goes low. So when output is going low, capacitor maintains the proper voltage supply into the Load circuit, hence creating the DC. Now how the value of this filter capacitor should be calculated. Here is the formulae:
C = I * t / V
C= capacitance to be calculated
I= Max output current (let’s say 500mA)
We will get wave of 100Hz frequency after converting 50Hz AC into DC, through full wave bridge rectifier. As the negative part of the pulse is converted into positive, one pulse will be counted two. So the Time period will be 1/100= .01 Second= 10ms
V = Peak voltage – voltage given to voltage regulator IC (+2 more than rated means 5+2=7)
9-0-9 is the RMS value of transforms so peak voltage is Vrms * 1.414= 9* 1.414= 12.73v
Now 1.4v will be dropped on 2 diodes (0.7 per diode) as 2 will be forward biased for half wave.
So 12.73 – 1.4 = 11.33v
When capacitor discharges into load circuit, it must provide 7v to 7805 IC to work so finally V is:
V = 11.33 – 7= 4.33v
So now C = I * t / V
C = 500mA * 10ms / 4.33 = .5 * .01 / 4.33 = 1154uF ~ 1000uF
A voltage regulator IC 7805 is used to provide a regulated 5v DC. Input voltage should be 2volts more than the rated output voltage for proper working of IC, means at least 7v is needed, although it can operate in input voltage range of 7-20V. Voltage regulators have all the circuitry inside it to provide a proper regulated DC. Capacitor of 0.01uF should be connected to the output of the 7805 to eliminate the noise, produced by transient changes in voltage.
In assembly there are two ways of assemble method is followed. One is manual and other is auto.
Step by step assembling process is different. Because of the product design. After the component assembly in the pcb board soldering is happening.
After completion of the soldering functional test is carried by a testing machines. Then it is covered by a charger housing.Finally package is done.