Problem:

a$.$ We want to build a mobile phone charger in the lab. We are getting $230$ V $($RMS$)$ and $50$ Hz from AC mains and we want to build a rectifier circuit that gives us $5$ V DC at the output.

$(7$ Marks$)$

Hints:

Use a transformer that steps down the $230$ V $($RMS$).$ Calculate $V_{\text{p}\text{e}\text{a}\text{k}}$ or $V_{\text{m}\text{a}\text{x}}$

Calculate $V_{dc}$ that gives output ratings of voltage more than $5$ V $.$

The output after the Rectification is not a proper DC$,$ it is oscillatory and has a very high ripple factor. We don't need that pulsating output, for this we use Capacitor. Calculate capacitance using following details:

$Q=CV$

$C=I**\frac{t}{V}$

$C=$ capacitance to be calculated

$I=$ Max output current $($let$\text{'}$s say $500$ mA $)$

Frequency $=100Hz($Calculate time period$)$

b$.$ Draw complete circuit diagram and waveform at each step of the rectifier build in above section.

$(3$ Marks$)$

Step down AC voltage

As we are converting $220$ V AC into a $5$ V DC $,$ first we need a step$-$down transformer to reduce such high voltage. Here we have used $9-0-91$A step$-$down transformer, which convert $220$ VAC to $9$ VAC $.$ 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. 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.

$V_{\text{m}\text{a}\text{x}}=\sqrt[\phantom{2}]{2}{V}_{RMS}$

$V_{dc}=\frac{2(V)\text{m}\text{a}\text{x}}{}$

Rectification

Rectification is the process of removing the negative part of the AC $,$ hence producing the partial DC$.$ This can be achieved by using $2$ diodes. Diodes only allow current to flow in one direction. In first half cycle of AC diode D$1$ is forward biased and D$2$ is reversed biased, and in the second half cycle $($negative half$)$ Diode D$2$ is forward biased and D$1$ is reversed biased. This Combination converts the negative half cycle into positive.

Filtration

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:

$C=I**\frac{t}{V}$

$C=$ capacitance to be calculated

$I=$ Max output current $($let$\text{'}$s say $500$ mA $)$

$t=10ms,$

We will get wave of $100$ Hz frequency after converting $50$ Hz 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 $\frac{1}{100}=\mathrm{.}01$ Second $=10ms$