It has already been discussed in the previous article that a pn junction conducts current easily when forward biased and practically no current flows when it is reverse biased. This unilateral conduction characteristic of pn junction (i.e. semiconductor diode) is similar to that of a vacuum diode. Therefore, like a vacuum diode, a semiconductor diode can also accomplish the job of rectification i.e. change alternating current to direct current. However, semiconductor diodes have become more *popular as they are smaller in size, cheaper and robust and usually operate with greater efficiency. In this chapter, we shall focus our attention on the circuit performance and applications of semiconductor diodes.
A pn junction is known as a semiconductor diode or *crystal diode. The outstanding property of a crystal diode to conduct current in one direction only permits it to be used as a rectifier. A crystal diode is usually represented by the schematic symbol shown in Fig. 6.1. The arrow in the symbol indicates the direction of easier conventional current flow.
A semiconductor diode has two terminals. When it is connected in a circuit, one thing to decide is whether the diode is forward or reverse biased. There is an easy rule to ascertain it. If the external circuit is trying to push the conventional current in the direction of arrow, the diode is forward biased. On the other hand, if the conventional current is trying to flow opposite to arrowhead, the diode is reverse biased.
Putting in simple words :
(i) If arrowhead of diode symbol is positive w.r.t. bar of the symbol, the diode is forward biased.
(ii) If the arrowhead of diode symbol is negative w.r.t. bar, the diode is reverse biased.
Identification of crystal diode terminals. While using a crystal diode, it is often necessary to know which end is arrowhead and which end is bar. For this purpose, the following methods are available :
(i) Some manufacturers actually paint the symbol on the body of the diode e.g. BY127, BY114
crystal diodes manufactured by BEL [See Fig. 6.2 (I)].
(ii) Sometimes, red and blue marks are used on the body of the crystal diode. Red mark denotes
arrow whereas blue mark indicates bar e.g. OA80 crystal diode [See Fig. 6.2 (ii)].
Fig. 6.3 illustrates the rectifying action of a crystal diode. The a.c. the input voltage to be rectified, the diode and load RL are connected in series. The d.c. the output is obtained across the load as explained in the following discussion. During the positive half-cycle of a.c. the input voltage, the arrowhead becomes positive w.r.t. bar. Therefore, the diode is forward biased and conducts current in the circuit. The result is that a positive half-cycle of input voltage appears across RL as shown. However, during the negative half-cycle of input a.c. voltage, the diode becomes reverse biased because now the arrowhead is negative w.r.t. bar. Therefore, the diode does not conduct and no voltage appears across load RL. The result is that output consists of positive half-cycles of input a.c. voltage while the negative half-cycles are suppressed. In this way, crystal diode has been able to do rectification i.e. change a.c. into d.c. It may be seen that output across RL is pulsating d.c. It is interesting to see that the behavior of the diode is like a switch. When the diode is forward biased, it behaves like a closed switch and connects the a.c. supply to the load RL. However, when the diode is reverse biased, it behaves like an open switch and disconnects the a.c. supply from the load RL. This switching action of diode permits only the positive half-cycles of input a.c. voltage to appear across RL
Example 6.1. In each diode circuit of Fig. 6.4, find whether the diodes are forward or reverse biased.
(i) Refer to Fig. 6.4 (i). The conventional current coming out of battery flows in the branch circuits. In diode D1, the conventional current flows in the direction of arrowhead and hence this diode is forward biased. However, in diode D2, the conventional current flows opposite to arrowhead and hence this diode is reverse biased.
(ii) Refer to Fig. 6.4 (ii). During the positive half-cycle of input a.c. voltage, the conventional current flows in the direction of arrowhead and hence diode is forward biased. However, during the negative half-cycle of input a.c. voltage, the diode is reverse biased.
(iii) Refer to Fig. 6.4 (iii). During the positive half-cycle of input a.c. voltage, conventional current flows in the direction of the arrowhead in D1 but it flows opposite to arrowhead in D2. Therefore, during positive half-cycle, diode D1 is forward biased and diode D2 reverse biased. However, during the negative half-cycle of input a.c. voltage, diode D2 is forward biased and D1 is reverse biased.
(iv) Refer to Fig. 6.4 (iv). During the positive half-cycle of input a.c. voltage, both the diodes are
reverse biased. However, during the negative half-cycle of input a.c. voltage, both the diodes are
It has already been discussed that a forward biased diode conducts easily whereas a reverse biased
diode practically conducts no current. It means that forward resistance of a diode is quite small as
compared with its reverse resistance.
The resistance offered by the diode to forward bias is known as forward resistance. This resistance is not the same for the flow of direct current as for the changing current. Accordingly; this resistance is of two types, namely; d.c. forward resistance and a.c. forward resistance.
(i) d.c. forward resistance. It is the opposition offered by the diode to the direct current. It is measured by the ratio of d.c. the voltage across the diode to the resulting d.c. current through it. Thus, referring to the forward characteristic in Fig. 6.5, it is clear that when a forward voltage is OA, the forward current is OB.
(ii) a.c. forward resistance. It is the opposition offered by the diode to the changing forward current. It is measured by the ratio of change in voltage across diode to the resulting change in current through it i.e.
The a.c. forward resistance is more significant as the diodes are generally used with alternating voltages. The a.c. forward resistance can be determined from the forward characteristic as shown in
Fig. 6.6. If P is the operating point at any instant, then the forward voltage is ob and forward current is one. To find the a.c. forward resistance, vary the forward voltage on both sides of the operating point equally as shown in Fig. 6.6 where ab = bc. It is clear from this figure that :
For forward voltage oa, circuit current is od.
For forward voltage oc, circuit current is of
It may be mentioned here that forward resistance of a crystal diode is very small, ranging from 1
to 25 Ω.
The resistance offered by the diode to the reverse bias is known as reverse resistance. It can be d.c. reverse resistance or a.c. reverse resistance depending upon whether the reverse bias is a direct or changing voltage. Ideally, the reverse resistance of a diode is infinite. However, in practice, the reverse resistance is not infinite because, for any value of reverse bias, there does exist a small leakage current. It may be emphasized here that reverse resistance is very large compared to the forward resistance. In germanium diodes, the ratio of reverse to forward resistance is 40000: 1 while for silicon this ratio is 1000000: 1
It is generally profitable to replace a device or system by its equivalent circuit. An equivalent circuit
of a device (e.g. crystal diode, transistor etc.) is a combination of electric elements, which when connected in a circuit, acts exactly as does the device when connected in the same circuit. Once the
device is replaced by its equivalent circuit, the resulting network can be solved by traditional circuit
analysis techniques. We shall now find the equivalent circuit of a crystal diode.
When the forward voltage VF is applied across a diode, it will not conduct till the potential barrier V0 at the junction is overcome. When the forward voltage exceeds the potential barrier voltage, the diode starts conducting as shown in Fig. 6.7 (i). The forward current If flowing through the diode causes a voltage drop in its internal resistance rf. Therefore, the forward voltage VF applied across the actual diode has to overcome :
(a) potential barrier V0
(b) internal drop If rf
For a silicon diode, V0= 0.7 V whereas for a germanium diode, V0= 0.3 V. Therefore, an approximate equivalent circuit for a crystal diode is a switch in series with a battery V0 and internal resistance rf as shown in Fig. 6.7 (ii). This approximate equivalent circuit of a diode is very helpful in studying the performance of the diode in a circuit.
For most applications, the internal resistance rf of the crystal diode can be ignored in comparison to other elements in the equivalent circuit. The equivalent circuit then reduces to the one shown in Fig. 6.8 (ii). This simplified equivalent circuit of the crystal diode is frequently used in diode-circuit analysis.
An ideal diode is one that behaves as a perfect conductor when forward biased and as a perfect insulator when reversing biased. Obviously, in such a hypothetical situation, forward resistance rf= 0 and potential barrier V0 is considered negligible. It may be mentioned here that although ideal diode is never found in practice, yet diode circuit analysis is made on this basis. Therefore, while discussing diode circuits, the diode will be assumed ideal unless and until stated otherwise.
It is desirable to sum up the various models of crystal diode equivalent circuit in the tabular form
Example 6.2. An a.c. voltage of peak value 20 V is connected in series with a silicon diode and load resistance of 500 Ω. If the forward resistance of diode is 10 Ω,
find :(i) peak current through diode (ii) peak output voltage
What will be these values if the diode is assumed to be ideal?
Peak input voltage = 20 V
Forward resistance, rf= 10 Ω
Load resistance, RL= 500 Ω
Potential barrier voltage, V0= 0.7 V
The diode will conduct during the positive half-cycles of a.c. input voltage only. The equivalent
circuit is shown in Fig. 6.9 (ii).
Comments. It is clear from the above example that output voltage is nearly the same whether the
actual diode is used or the diode is considered ideal. This is due to the fact that the input voltage is quite large as compared with V0 and voltage drop in rf. Therefore, nearly the whole input forward voltage appears across the load. For this reason, diode circuit analysis is generally made on the ideal diode basis.
Example 6.3. Find the current through the diode in the circuit shown in Fig. 6.10 (i). Assume
the diode to be ideal.
Example 6.4. Calculate the current through 48 Ω resistor in the circuit shown in Fig. 6.11 (i).
Assume the diodes to be of silicon and forward resistance of each diode is 1 Ω.
Solution. Diodes D1 and D3 are forward biased while diodes D2 and D4 are reverse biased. We can, therefore, consider the branches containing diodes D2 and D4 as “open”. Replacing diodes D1 and D3 by their equivalent circuits and making the branches containing diodes D2 and D4 open, we get
the circuit shown in Fig. 6.11 (ii). Note that for a silicon diode, the barrier voltage is 0.7 V.
Example 6.5. Determine the current I in the circuit shown in Fig. 6.12 (i). Assume the diodes to
be of silicon and forward resistance of diodes to be zero.
Solution. The conditions of the problem suggest that diode D1 is forward biased and diode D2 is
reverse biased. We can, therefore, consider the branch containing diode D2 as open as shown in
Fig. 6.12 (ii). Further, diode D1 can be replaced by its simplified equivalent circuit.
Example 6.6. Find the voltage VA in the circuit shown in Fig. 6.13 (i). Use simplified model
Solution. It appears that when the applied voltage is switched on, both the diodes will turn “on”. But that is not so. When voltage is applied, germanium diode (V0= 0.3 V) will turn on first and a level of 0.3V is maintained across the parallel circuit. The silicon diode never gets the opportunity to have 0.7 V across it and, therefore, remains in an open-circuit state as shown in Fig. 6.13 (ii).
VA = 20 − 0.3 = 19.7 V
Example 6.7. Find VQ and ID in the network shown in Fig. 6.14 (i). Use a simplified model.
Solution. Replace the diodes by their simplified models. The resulting circuit will be as shown in
Fig. 6.14 (ii). By symmetry, current in each branch is ID so that current in-branch CD is 2ID. Applying Kirchhoff’s voltage law to the closed-circuit ABCDA, we have,
Example 6.8. Determine current through each diode in the circuit shown in Fig. 6.15 (i). Use a simplified model. Assume diodes to be similar.
Solution. The applied voltage forward biases each diode so that they conduct current in the same
direction. Fig. 6.15 (ii) shows the equivalent circuit using a simplified model. Referring to Fig. 6.15 (ii),
Comments. Note the use of placing the diodes in parallel. If the current rating of each diode is 20
20 mA and a single diode is used in this circuit, a current of 28.6 mA would flow through the diode,
thus damaging the device. By placing them in parallel, the current is limited to a safe value of 14.3 mA for the same terminal voltage.
Example 6.9. Determine the currents I1, I2, and I3 for the network shown in Fig. 6.16(i). Use a simplified model for the diodes.
Solution. An inspection of the circuit shown in Fig. 6.16 (i) shows that both diodes D1 and D2 are forward biased. Using simplified model for the diodes, the circuit shown in Fig. 6.16 (i) becomes
the one shown in Fig. 6.16 (ii). The voltage across R2 (= 3.3 k Ω) is 0.7V.
Example 6.10. Determine if the diode (ideal) in Fig. 6.17 (i) is forward biased or reverse biased.
Solution. Let us assume that diode in Fig. 6.17 (i) is OFF i.e. it is reverse biased. The circuit
then becomes as shown in Fig. 6.17 (ii). Referring to Fig. 6.17 (ii), we have,
Example 6.11. Determine the state of the diode for the circuit shown in Fig. 6.18 (i) and find ID and VD. Assume simplified model for the diode.
Solution. Let us assume that the diode is ON. Therefore, we can replace the diode with a 0.7V
battery as shown in Fig. 6.18 (ii). Referring to Fig. 6.18 (ii), we have,
Since the diode current is negative, the diode must be OFF and the true value of diode current is ID = 0 mA. Our initial assumption was wrong. In order to analyze the circuit properly, we should replace the diode in Fig. 6.18 (i) with an open circuit as shown in Fig. 6.19. The voltage VD across the diode is
We know that 0.7V is required to turn ON the diode. Since VD
is only 0.4V, the answer confirms
that the diode is OFF
While discussing the diode circuits, the reader will generally come across the following terms :
It is the current flowing through a forward-biased diode. Every diode has a maximum value of forwarding current which it can safely carry. If this value is exceeded, the diode
may be destroyed due to excessive heat. For this reason, the manufacturers’ datasheet specifies the maximum forward current that a diode can handle safely.
It is the maximum reverse voltage that a diode can withstand without destroying the junction. If the reverse voltage across a diode exceeds this value, the reverse current increases sharply and breaks down the junction due to excessive heat. Peak inverse voltage is extremely important when diode is used as a rectifier. In rectifier service, it has to be ensured that reverse voltage across the diode does not exceed its PIV during the negative half-cycle of input a.c. voltage. As a matter of fact, PIV consideration is generally the deciding factor in diode rectifier circuits. The peak inverse voltage may be between 10V and 10 kV depending upon the type of diode.
It is the current that flows through a reverse-biased diode. This current is due to minority carriers. Under normal operating voltages, the reverse current is quite small. Its value is extremely small (< 1μ A) for silicon diodes but it is appreciable (j 100 μA) for germanium diodes. It may be noted that the reverse current is usually very small as compared with the forward current. For example, the forward current for a typical diode might range up to 100 mA while the reverse current might be only a few μA—a ratio of many thousands between forward and reverse currents.
Reference: Principles Of Electronics By V K Mehta And Rohit Mehta
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