P-N Junction As A Rectifier

Half-wave Rectifier

Look at the diagram given above. An alternating voltage is applied across a junction diode which is connected to a load in a series connection. In this case, only during those half cycles of the AC input when the diode is forward biased, will a voltage appear across the load. This type of circuit, which rectifies only one-half of the input current is a Half-wave Rectifier.

The alternating current is supplied at points A and B. During the alternating cycle, when the voltage at point A is positive, the diode is forward-biased. This will happen when the diode conducts. On the other hand, the diode is reverse-biased when the voltage at point A is negative, and it doesn’t conduct. Generally, for all practical purposes, the reverse saturation current can be considered zero since it is negligible.

Hence, we will get output voltage only through one-half of the input cycle. Also, there will be no current available in the other half. Hence, the output still varies between positive to zero but the negative cycle is cut off and the output voltage is said to be rectified.

Full-wave Rectifier-

Look at the figure given above. In the circuit given above, two junction diodes are connected to a load. In this circuit, both positive and negative halves of the AC cycle will come out. Hence, it is a Full-wave Rectifier. In this circuit, the p-sides of both the diodes are connected to the input while the n-sides are connected together and connected to the load. To complete the circuit load is connected to the mid-point of the transformer. Since this mid-point of connection is also called the Center tap and because of this, the transformer is called a Center tap transformer.

Here two diodes are connected, one diode rectifies the voltage for one half of the cycle while the other diode rectifies it for the other half. Therefore, the output between the centre tap of the transformer and their common terminals becomes a full-wave rectifier output. Let’s see how this works-

If the voltage at point A is positive, then that at point B is negative. In this case, the diode D1 is forward-biased while D2 is negatively biased. So, D1 conducts while D2 blocks the current. So, during the positive half of the input AC cycle, we will get the output current. Afterwards, the voltage at point A becomes negative and that at point B becomes positive. In this case, D2 conducts while D1 blocks the current. So, we will get an output current in the negative half of the input AC cycle too. So this circuit rectifies both the halves of the input voltage, that's why it is called Full-wave Rectifier. But, one thing should be noted the output is pulsating and not steady. So to derive a steady DC output there is a need for a capacitor across the output terminals (parallel to the load).

Role of a Capacitor

The role of the capacitor is to filter out the AC ripple and provide pure DC output. Let us discuss how it works:

We can see in the circuit given above, a capacitor is connected parallel to the load. The capacitor gets charged when the voltage across the capacitor rises and It discharges only when a load is connected to it and the voltage across it falls. As the AC cycle changes and the second diode kicks in, the capacitor charges again to its peak value and then again discharged due to the presence of the load.

One should note that the rate of discharge depends on the inverse product of Capacitor and Resistance (or load). To increase the discharge time and get a steady DC output, we should connect large capacitors. The idea is to obtain an output voltage close to the peak voltage of the rectified current.

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Solved Example Based on P-N Junction as a Rectifier

Example 1: A half-wave rectifier has an input voltage of $220 \mathrm{~V}_{\mathrm{rms}}$ If the step-down transformer has a turns ratio of $4: 1$, what is the peak load voltage? Ignore the diode drop.

1) 154.7 V
2) 310 V
3) 53 V
4) 88 V

Solution:

We need to convert the RMS voltage to peak voltage using the relation

$\begin{aligned} & V_p=\sqrt{2} \times V_{\mathrm{rms}} \\ & V_p=\sqrt{2} \times V_{\mathrm{in}} \\ & V_p=\sqrt{2} \times 220, \mathrm{~V} \\ & V_p=220 \times 1.414 \\ & V_p=310.48, \mathrm{~V}_{\text {(approximately) }}\end{aligned}$

Low-voltage rectifiers require a step-down transformer to reduce the strength of AC voltage.

$\begin{aligned} & \frac{V_1}{V_2}=\frac{N_1}{N_2} \\ & \frac{N_1}{N_2}=\frac{4}{1}\end{aligned}$

$\mathrm{V}_p$ at the output of the stepdown transformer:
$
\mathrm{V}_p=\frac{210.48}{4} \approx 53 \mathrm{~V}
$
Hence, the answer is option (3).

Example 2: A full-wave $\mathbf{p}$ - $\mathbf{n}$ diode rectifier uses a load resistor of $1500 \Omega$ No filter is used. The forward bias resistance of the diode is $10 \Omega$ The efficiency of the rectifier is:
1) $81.2 \%$
2) $40.6 \%$
3) $80.6 \%$
4) $40.2 \%$

Solution:

Here, $\mathrm{r}_{\mathrm{i}}=10 \Omega, \mathrm{R}_{\mathrm{L}}=1500 \Omega$
The efficiency of the full wave rectifier is
$
\eta=\frac{\mathrm{P}_{\mathrm{dc}}}{\mathrm{P}_{\mathrm{Bc}}}=\frac{\left(2 \mathrm{I}_{\mathrm{m}} / \pi\right)^2 \mathrm{R}_{\mathrm{L}}}{\left(\mathrm{I}_{\mathrm{m}} / \sqrt{2}\right)^2\left(\mathrm{r}_{\mathrm{f}}+\mathrm{R}_{\mathrm{L}}\right)}=\frac{0.812 \mathrm{R}_{\mathrm{L}}}{\mathrm{r}_{\mathrm{f}}+\mathrm{R}_{\mathrm{L}}}=\frac{0.812 \times 1500}{10+1500}=0.806=80.6 \%
$

Hence, the answer is option (3).

Example 3: For a given electric voltage signal dc value is

1) 6.28 V
2) 3.14 V
3) 4 V
4) 0 V

Solution:

$\mathrm{V}_{\mathrm{dc}}=\mathrm{V}_{\mathrm{ac}}=\frac{2 \mathrm{~V}_0}{\pi}=\frac{2 \times 6.28}{3.14}=4 \mathrm{~V}$

Hence, the answer is option (3).

Example 4: A P-N junction D shown in the figure can act as a rectifier.

An alternating current source $(V)$ is connected in the circuit. The current $(I)$ in the resistor $(R)$ can be shown by

1)

2)

3)

4)

Solution:

The frequency of output is the same as input a.c.

This circuit represents a wave rectifier

During half-time, the diode is forward biased and hence current will flow through it and for the next half-time cycle, the diode is in reverse half biased and hence no current will flow through it output is

Hence, the answer is option (3).

Example 5: Match the rectifier types in List I with their corresponding descriptions in List II.

1)(a) - (iii), (b) - (ii), (c) - (iv), (d) - (i)

2)(a) - (i), (b) - (ii), (c) - (iv), (d) - (iii)

3)(a) - (i), (b) - (iii), (c) - (ii), (d) - (iv)

4)(a) - (iii), (b) - (ii), (c) - (iv), (d) - (i)

Solution:

(a) Diodes: Diodes are electronic components that allow current to flow in one direction while blocking it in the opposite direction.

(b) Capacitor: A capacitor is an electronic component that stores electrical energy and releases it when needed. In rectifiers, capacitors are often used to smooth the output voltage by reducing voltage ripple. They store charge during the peak voltage periods and discharge during the lower voltage periods, resulting in a more stable DC output.

(c) Transformer: A transformer is an electrical device that can step up or step down the AC voltage. It consists of two coils, primary and secondary, that are magnetically coupled. Transformers are commonly used in rectifier circuits to step down or step up the input AC voltage to a desired level for rectification.

(d) Rectifier: A rectifier converts Alternating Current into Direct Current.

Now, let's match the rectifier types in List I with their corresponding descriptions in List II based on the above explanations:

(a) Diodes - (iii) Blocks the reverse current flow

(b) Capacitor - (ii) Smoothes the output voltage

(c) Transformer - (iv) Steps down or steps up the AC voltage

(d) Rectifier - (i) Converts AC voltage into DC voltage

Therefore, the correct match is:

(a) Diodes - (iii)

(b) Capacitor - (ii)

(c) Transformer - (iv)

(d) Rectifier - (i)

Summary

A P-N junction includes the junction of a semiconductor material of p-type and that of n-type. When joined together, they form a junction with special electrical properties. It allows current to flow in one direction, forward bias but blocks it in the opposite direction, reverse bias. This property of one-way conductivity makes it an effective rectifier that can convert AC into DC. This P-N junction is very critical in power supplies, to ensure that a steady and reliable DC voltage reaches the electronic devices. Knowledge about how this works comes in quite handy during the design and use of electronic circuits and power systems.