Physics SEMICONDUCTOR DIODE AND APPLICATION OF JUNCTION DIODE AS A RECTIFIER

### Topic Covered

color{blue}{star}SEMICONDUCTOR DIODE
color{blue}{star}APPLICATION OF JUNCTION DIODE AS A RECTIFIER

### SEMICONDUCTOR DIODE

color{blue} ✍️A semiconductor diode [Fig. 14.12(a)] is basically a p-n junction with metallic contacts provided at the ends for the application of an external voltage. It is a two terminal device.

color{blue} ✍️A p-n junction diode is symbolically represented as shown in Fig. 14.12(b).

color{blue} ✍️The direction of arrow indicates the conventional direction of current (when the diode is under forward bias). The equilibrium barrier potential can be altered by applying an external voltage V across the diode.

color{blue} ✍️The situation of p-n junction diode under equilibrium (without bias) is shown in Fig. 14.12(a) and (b).

color{brown} {bbul{"p-n junction diode under forward bias"}}
color{blue} ✍️When an external voltage V is applied across a semiconductor diode such that p-side is connected to the positive terminal of the battery and n-side to the negative terminal [Fig. 14.13(a)], it is said to be forward biased.

color{blue} ✍️The applied voltage mostly drops across the depletion region and the voltage drop across the p-side and n-side of the junction is negligible. (This is because the resistance of the depletion region – a region where there are no charges – is very high compared to the resistance of n-side and p-side.)

color{blue} ✍️The direction of the applied voltage (V ) is opposite to the built-in potential V_0. As a result, the depletion layer width decreases and the barrier height is reduced [Fig. 14.13(b)]. The effective barrier height under forward bias is (V_0 – V ).

color{blue} ✍️If the applied voltage is small, the barrier potential will be reduced only slightly below the equilibrium value, and only a small number of carriers in the material—those that happen to be in the uppermost energy levels—will possess enough energy to cross the junction.

color{blue} ✍️So the current will be small. If we increase the applied voltage significantly, the barrier height will be reduced and more number of carriers will have the required energy. Thus the current increases.

color{blue} ✍️Due to the applied voltage, electrons from n-side cross the depletion region and reach p-side (where they are minority carries). Similarly, holes from p-side cross the junction and reach the n-side (where they are minority carries). This process under forward bias is known as minority carrier injection.

color{blue} ✍️At the junction boundary, on each side, the minority carrier concentration increases significantly compared to the locations far from the junction.

color{blue} ✍️Due to this concentration gradient, the injected electrons on p-side diffuse from the junction edge of p-side to the other end of p-side. Likewise, the injected holes on n-side diffuse from the junction edge of n-side to the other end of n-side (Fig. 14.14).

color{blue} ✍️This motion of charged carriers on either side gives rise to current. The total diode forward current is sum of hole diffusion current and conventional current due to electron diffusion. The magnitude of this current is usually in mA.

color{brown}bbul("p-n junction diode under reverse bias")
color{blue} ✍️When an external voltage (V) is applied across the diode such that n-side is positive and p-side is negative, it is said to be reverse biased [Fig.14.15(a)].

color{blue} ✍️The applied voltage mostly drops across the depletion region. The direction of applied voltage is same as the direction of barrier potential. As a result, the barrier height increases and the depletion region widens due to the change in the electric field. The effective barrier height under reverse bias is (V_0 + V ), [Fig. 14.15(b)].

color{blue} ✍️This suppresses the flow of electrons from n -> p and holes from p -> n. Thus, diffusion current, decreases enormously compared to the diode under forward bias.
The electric field direction of the junction is such that if electrons on p-side or holes on n-side in their random motion come close to the junction, they will be swept to its majority zone. This drift of carriers gives rise to current.

color{blue} ✍️The drift current is of the order of a few μA. This is quite low because it is due to the motion of carriers from their minority side to their majority side across the junction.

color{blue} ✍️The drift current is also there under forward bias but it is negligible (μA) when compared with current due to injected carriers which is usually in mA.

color{blue} ✍️The diode reverse current is not very much dependent on the applied voltage. Even a small voltage is sufficient to sweep the minority carriers from one side of the junction to the other side of the junction.

color{blue} ✍️The current is not limited by the magnitude of the applied voltage but is limited due to the concentration of the minority carrier on either side of the junction.

color{blue} ✍️The current under reverse bias is essentially voltage independent upto a critical reverse bias voltage, known as breakdown voltage (V_br ). When V = V_(br,) the diode reverse current increases sharply.

color{blue} ✍️Even a slight increase in the bias voltage causes large change in the current. If the reverse current is not limited by an external circuit below the rated value (specified by the manufacturer) the p-n junction will get destroyed.

color{blue} ✍️Once it exceeds the rated value, the diode gets destroyed due to overheating. This can happen even for the diode under forward bias, if the forward current exceeds the rated value.

color{blue} ✍️The circuit arrangement for studying the V-I characteristics of a diode, (i.e., the variation of current as a function of applied voltage) are shown in Fig. 14.16(a) and (b).

color{blue} ✍️The battery is connected to the diode through a potentiometer (or reheostat) so that the applied voltage to the diode can be changed. For different values of voltages, the value of the current is noted. A graph between V and I is obtained as in Fig. 14.16(c).

color{brown} {"Note"} that in forward bias measurement, we use a milliammeter since the expected current is large (as explained in the earlier section) while a micrometer is used in reverse bias to measure the current.

color{blue} ✍️You can see in Fig. 14.16(c) that in forward bias, the current first increases very slowly, almost negligibly, till the voltage across the diode crosses a certain value. After the characteristic voltage, the diode current increases significantly (exponentially), even for a very small increase in the diode bias voltage.

color{blue} ✍️This voltage is called the threshold voltage or cut-in voltage (~0.2V for germanium diode and ~0.7 V for silicon diode). For the diode in reverse bias, the current is very small (~μA) and almost remains constant with change in bias. It is called reverse saturation current. However, for special cases, at very high reverse bias (break down voltage), the current suddenly increases.

color{blue} ✍️This special action of the diode is discussed later in Section 14.8. The general purpose diode are not used beyond the reverse saturation current region.

color{blue} ✍️The above discussion shows that the p-n junction diode primerly allows the flow of current only in one direction (forward bias).

color{blue} ✍️The forward bias resistance is low as compared to the reverse bias resistance. This property is used for rectification of ac voltages as discussed in the next section. For diodes, we define a quantity called dynamic resistance as the ratio of small change in voltage DeltaV to a small change in current DeltaI

color{blue}(r_d= (DeltaV)/(DeltaI))

............(14.6)
Q 3119380210

The V-I characteristic of a silicon diode is shown in the Fig. 14.17. Calculate the resistance of the diode at (a) I_D = 15 mA and (b) V_D = –10 V.
Class 12 Chapter 14 Example 4
Solution:

Considering the diode characteristics as a straight line between I = 10 mA to I = 20 mA passing through the origin, we can calculate the resistance using Ohm’s law.

(a) From the curve, at I = 20 mA, V = 0.8 V, I = 10 mA, V = 0.7 V r_(fb)= DeltaV//DeltaI = 0.1V//10 mA = 10 Omega
b) From the curve at V = –10 V, I = –1 μA,
Therefore,
r_(rb) = 10 V//1μA= 1.0 × 10^7 Omega

### APPLICATION OF JUNCTION DIODE AS A RECTIFIER

color{blue} ✍️From the V-I characteristic of a junction diode we see that it allows current to pass only when it is forward biased. So if an alternating voltage is applied across a diode the current flows only in that part of the cycle when the diode is forward biased.

color{blue} ✍️This property is used to rectify alternating voltages and the circuit used for this purpose is called a rectifier.

color{blue} ✍️If an alternating voltage is applied across a diode in series with a load, a pulsating voltage will appear across the load only during the half cycles of the ac input during which the diode is forward biased. Such rectifier circuit, as shown in Fig. 14.18, is called a "half-wave rectifier."

color{blue} ✍️The secondary of a transformer supplies the desired ac voltage across terminals A and B. When the voltage at A is positive, the diode is forward biased and it conducts. When A is negative, the diode is reverse-biased and it does not conduct.

color{blue} ✍️The reverse saturation current of a diode is negligible and can be considered equal to zero for practical purposes. (The reverse breakdown voltage of the diode must be sufficiently higher than the peak ac voltage at the secondary of the transformer to protect the diode from reverse breakdown.)

color{blue} ✍️Therefore, in the positive half-cycle of ac there is a current through the load resistor RL and we get an output voltage, as shown in Fig. 14.18(b), whereas there is no current in the negative halfcycle.

color{blue} ✍️In the next positive half-cycle, again we get the output voltage. Thus, the output voltage, though still varying, is restricted to only one direction and is said to be rectified. Since the rectified output of this circuit is only for half of the input ac wave it is called as half-wave rectifier.

color{blue} ✍️The circuit using two diodes, shown in Fig. 14.19(a), gives output rectified voltage corresponding to both the positive as well as negative half of the ac cycle. Hence, it is known as full-wave rectifier.

color{blue} ✍️Here the p-side of the two diodes are connected to the ends of the secondary of the transformer. The n-side of the diodes are connected together and the output is taken between this common point of diodes and the midpoint of the secondary of the transformer.

color{blue} ✍️So for a full-wave rectifier the secondary of the transformer is provided with a centre tapping and so it is called centre-tap transformer. As can be seen from Fig.14.19(c) the voltage rectified by each diode is only half the total secondary voltage.

color{blue} ✍️Each diode rectifies only for half the cycle, but the two do so for alternate cycles. Thus, the output between their common terminals and the centretap of the transformer becomes a full-wave rectifier output.
(Note that there is another circuit of full wave rectifier which does not need a centretap transformer but needs four diodes.) Suppose the input voltage to A with respect to the centre tap at any instant is positive. It is clear that, at that instant, voltage at B being out of phase will be negative as shown in Fig.14.19(b).

color{blue} ✍️ So, diode D_1 gets forward biased and conducts (while D_2 being reverse biased is not conducting). Hence, during this positive half cycle we get an output current (and a output voltage across the load resistor RL) as shown in Fig.14.19(c).

color{blue} ✍️In the course of the ac cycle when the voltage at A becomes negative with respect to centre tap, the voltage at B would be positive. In this part of the cycle diode D_1 would not conduct but diode D_2 would, giving an output current and output voltage (across R_L) during the negative half cycle of the input ac.

color{blue} ✍️Thus, we get output voltage during both the positive as well as the negative half of the cycle. Obviously, this is a more efficient circuit for getting rectified voltage or current than the halfwave rectifier.

color{blue} ✍️Now we shall discuss the role of capacitor in filtering. When the voltage across the capacitor is rising, it gets charged. If there is no external load, it remains charged to the peak voltage of the rectified output.

color{blue} ✍️When there is a load, it gets discharged through the load and the voltage across it begins to fall. In the next half-cycle of rectified output it again gets charged to the peak value (Fig. 14.20).

color{blue} ✍️The rate of fall of the voltage across the capacitor depends upon the inverse product of capacitor C and the effective resistance R_L used in the circuit and is called the time constant.

color{blue} ✍️To make the time constant large value of C should be large. So capacitor input filters use large capacitors. The output voltage obtained by using capacitor input filter is nearer to the peak voltage of the rectified voltage. This type of filter is most widely used in power supplies.