Application of Junction Diode as a Rectifier

Answers & Solutions

Work the questions above first, then check yourself here. Each answer is set out the way it should be written in the paper.

1. 1.↑ back to the question

   The blank is RECTIFY. A junction diode conducts current only when forward biased and blocks current when reverse biased — this one-directional conduction is exactly what allows it to convert (rectify) an alternating voltage into a unidirectional (pulsating DC) voltage.

2. 2.↑ back to the question

   A rectifier is based on the fact that a p-n junction conducts current only when it is forward biased, and offers a very high resistance (does not conduct any appreciable current) when it is reverse biased — i.e. the p-n junction is able to conduct current in one direction only. It is precisely this asymmetric, one-way conduction property — low resistance in forward bias, very high resistance in reverse bias — that makes a p-n junction diode suitable for converting an alternating (bidirectional) voltage into a unidirectional (pulsating DC) voltage, i.e. for rectification.

3. 3.↑ back to the question

   The question is a real board question. Its answer and working are ours — this paper was published without an official answer key.

   Two differences

   Point	Half wave rectifier	Full wave rectifier (centre-tap)
   Part of the input used	only one half of each input cycle is rectified; during the other half the diode is reverse biased and the output is zero	both halves of each input cycle are rectified, the two diodes conducting in alternate half cycles
   Output pulses and ripple frequency	one pulse per input cycle, so the output frequency equals the input frequency, $\nu_{out} = \nu_{in}$ ($50$ Hz in gives $50$ Hz out)	two pulses per input cycle, so the output frequency is double the input, $\nu_{out} = 2\nu_{in}$ ($50$ Hz in gives $100$ Hz out)
   Number of diodes and transformer	one diode; an ordinary secondary	two diodes; a centre-tapped secondary (or four diodes in a bridge)
   Efficiency and ripple	lower rectification efficiency, large ripple, harder to smooth	roughly twice the efficiency, smaller ripple, easier to smooth with a capacitor filter

   Output waveforms

   The input and output waveforms of the two rectifiers are shown alongside — the half wave output first, with its gap in every alternate half cycle, and the full wave output below it, in which the second diode fills that gap.

   Answer

   The two differences worth writing for the two marks are:

   • i. A half wave rectifier rectifies only one half of each input cycle, so its output has a flat zero gap after every pulse; a full wave rectifier rectifies both halves, so its output has no gaps.
   • ii. Consequently the ripple frequency of a half wave rectifier equals the input frequency, $\nu_{out} = \nu_{in}$, whereas that of a full wave rectifier is twice the input frequency, $\nu_{out} = 2\nu_{in}$.

   
4. 4.↑ back to the question

   Circuit: the secondary winding of a step-down transformer (whose primary is connected to the AC mains) has one end (A) connected through a diode $D$ to a load resistor $R_L$, and the other end (B) connected directly to the other end of $R_L$; the output voltage $V_o$ is taken across $R_L$. During the positive half cycle of the input AC, end A of the secondary winding becomes positive and end B becomes negative. Thus diode $D$ becomes forward biased and conducts, so current flows (from A, through the diode and $R_L$, to B) during this half cycle. During the negative half cycle of the input AC, end A becomes negative and end B becomes positive. Diode $D$ is now reverse biased and does not conduct, so no current flows in the circuit during this half cycle. Since current flows through $R_L$ only during the positive half cycle, and always in the same direction (A to B), the output across $R_L$ is a series of one-directional pulses (occurring once every full input cycle) — this pulsating, unidirectional output is the rectified (DC) output of the half-wave rectifier.

5. 5.↑ back to the question

   (i) V-I characteristic: for forward bias, the curve for Ge rises steeply past about 0.3 V and for Si past about 0.7 V (the current stays near zero before this and then increases very rapidly); for reverse bias, only a small, nearly constant current (of order a few $\mu$A) flows over a wide range of reverse voltage, until breakdown, beyond which the current suddenly increases sharply. (ii) Threshold voltage: in forward bias, the diode current first increases only very slowly with applied voltage, until the voltage across the diode crosses a certain value — beyond this, the current increases significantly even for a very small further increase in voltage. This voltage is called the threshold (or cut-in) voltage. Breakdown voltage: a very small current flows through a p-n junction when it is reverse biased, due to minority charge carriers. If the reverse bias voltage is continuously increased, at a certain reverse voltage the current through the junction increases abruptly. This reverse voltage is known as the breakdown voltage. (iii) Since a junction diode conducts current in forward bias but does not conduct (blocks current) in reverse bias, it is used to rectify AC voltages — this one-way (unidirectional) conduction property is what makes it suitable for rectification.

6. 6.↑ back to the question

   A full wave rectifier consists of two diodes $D_1$ and $D_2$ connected to the two ends (A and B) of the secondary winding of a centre-tapped step-down transformer, with the load resistor $R_L$ connected between the (common) output point X (where both diodes' other terminals join) and the centre tap of the transformer, which is taken as the reference (ground/Y) point. During the positive half cycle of the input AC, end A of the secondary becomes positive and end B becomes negative (relative to the centre tap). So diode $D_1$ (connected to A) becomes forward biased and conducts, while diode $D_2$ (connected to B) becomes reverse biased and does not conduct — current flows through $D_1$ and through $R_L$, from X to Y. During the negative half cycle, end A becomes negative and end B becomes positive. Now diode $D_1$ is reverse biased (does not conduct), while diode $D_2$ becomes forward biased and conducts — current flows through $D_2$ and through $R_L$, again from X to Y. So during BOTH halves of the input cycle, current flows through $R_L$ in the same direction — this gives a pulsating DC output at twice the frequency of the input: the input waveform $V_{in}$ is a full sine wave (positive and negative lobes alternating), while the output waveform $V_{out}$ across $R_L$ consists of a series of positive-only humps, one for every half-cycle of the input (so twice as many humps as the number of input cycles).

7. 7.↑ back to the question

   (i) Two p-n junction diodes $D_1,D_2$ are connected to the two ends of a centre-tapped secondary winding, with load $R_L$ between the common diode output and the centre tap. During the positive half cycle of the input AC, $D_1$ is forward biased (conducts) and $D_2$ is reverse biased (does not conduct); during the negative half cycle, $D_2$ is forward biased (conducts) and $D_1$ is reverse biased. In both halves, current flows through $R_L$ in the SAME direction, producing a pulsating DC (unidirectional) output whose ripple frequency is double the input frequency. (ii) V-I characteristics: negligible forward current until the threshold voltage is crossed, then a steep rise; negligible, nearly voltage-independent reverse current until breakdown, then a sharp rise. Because a junction diode conducts appreciably in only ONE direction (forward) and blocks current almost entirely in the other (reverse) direction, if an alternating voltage is applied across it, current will flow only during the half-cycle for which the diode is forward biased — this asymmetric conduction is exactly what is used to rectify (convert to unidirectional) an AC voltage. (iii) Carbon, silicon and germanium all belong to the same group (they each have 4 valence electrons and the same diamond lattice structure), but their 4 bonding (valence) electrons sit at different distances from the nucleus: in carbon they occupy the second orbit (closest to the nucleus, most tightly bound), while in silicon they occupy the third orbit (farther out, less tightly bound). Because carbon's valence electrons are held much more tightly (a much larger energy is needed to free an electron for conduction), essentially no electrons become free at ordinary temperatures — carbon behaves as an insulator. In silicon, the outer valence electrons are more loosely bound (smaller energy gap), so a small but significant number of electrons can be thermally freed at room temperature — silicon behaves as an (intrinsic) semiconductor.

8. 8.↑ back to the question

   The two important processes are diffusion and drift. Full-wave rectifier circuit and working: two p-n junction diodes $D_1,D_2$ are connected to the two ends A, B of a centre-tapped secondary winding of a step-down transformer, with the load resistor $R_L$ connected between the common diode-output node X and the centre tap Y. During the positive half cycle of the input AC, end A becomes positive and end B negative; diode $D_1$ becomes forward biased (conducts) while $D_2$ becomes reverse biased (does not conduct) — current flows through $D_1$ and $R_L$ from X to Y. During the negative half cycle, end A becomes negative and end B positive; now $D_1$ is reverse biased and $D_2$ is forward biased (conducts) — current flows through $D_2$ and $R_L$, again from X to Y. Since current flows through $R_L$ in the SAME direction during both halves of the input cycle, a pulsating DC (unidirectional) output results, at double the frequency of the input: the input waveform $V_{in}$ is a full alternating sine wave, while the output waveform $V_{out}$ across $R_L$ is a series of positive-only humps occurring twice as often (one per half-cycle of the input). Characteristic property suitable for rectification: since junction diodes conduct in forward bias but do not conduct in reverse bias, this is used to rectify ac voltages — the diode allows current to flow through it in only one direction, converting the bidirectional AC into unidirectional (pulsating DC) current.

9. 9.↑ back to the question

   Circuit: it consists of a diode $D$ connected in series with a load resistor $R_L$ across the secondary winding of a step-down transformer, whose primary is connected to the AC supply. The secondary's ends are labelled A and B; diode $D$ connects A to the output node X, and $R_L$ is connected between X and Y (Y being the other end, B, of the secondary). During the positive half cycle of the input AC, end A of the secondary winding becomes positive and end B becomes negative. Thus diode $D$ becomes forward biased and conducts the current through it — so current in the circuit flows from A to B, through the load resistor $R_L$. During the negative half cycle of the input AC, end A of the secondary winding becomes negative and end B becomes positive. Thus diode $D$ becomes reverse biased and does not conduct any current — so no current flows in the circuit during this half. Since electric current through the load $R_L$ flows only during the positive half cycle, and in one direction only (from A to B), a unidirectional pulsating output (DC) is obtained across $R_L$: the input waveform $V_{in}$ is a full sine wave, while the output waveform $V_{out}$ consists of only the positive humps of that sine wave, with a flat (zero) region during each negative half-cycle of the input — occurring at the same frequency as the input.