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The Junction Diode
Basic Operation
The diode is fabricated of a semiconductor material, usually silicon, which is doped with two impurities. One side is doped with a donor or n-type impurity which releases electrons into the semiconductor lattice. These electrons are not bound and are free to move about. Because there is no net charge in the donor impurity, the n-type semiconductor is electrically neutral. The other side is doped with an acceptor or p-type impurity which imparts free holes into the lattice. A hole is the absence of an electron which acts as a positive charge. The p-type semiconductor is also electrically neutral because the acceptor material adds no net charge. Figure 1(a) illustrates the cross section of the diode. The junction is the dividing line between the n-type and p-type sides. Thermal energy causes the electrons and holes to move randomly. Electrons diffuse across the junction into the p-type side and holes diffuse across the junction into the n-type side. This causes a net positive charge to develop in the n-type side and a net negative charge to develop in the p-type side. These charges set up an electric field across the junction which is directed from the n-type side to the p-type side. The electric field opposes further diffusion of the electrons and holes. The region in which the electric field exists is called the depletion region. There are no free electrons or holes in this region because the electric field sweeps them out.
Figure 1: (a) Diode cross section. (b) Reverse biased diode. (c) Forward biased diode.
Figure 1(b) shows the diode with a battery connected across it. The polarity of the battery is such that it reinforces the electric field across the junction causing the depletion region to widen. The positive terminal pulls electrons in the n-type side away from the junction. The negative terminal pulls holes in the p-type side away from the junction. No current can flow. The diode is said to be reverse biased. Figure 1(c) shows the diode with the battery polarity reversed. The battery now tends to cancel out the electric field in the depletion region, causing its width to decrease. The positive terminal forces holes toward the junction. The negative terminal forces electrons toward the junction. A current flows which increases rapidly if the applied voltage is increased. The diode is said to be forward biased.
i − v Characteristics
Figure 2(a) shows the circuit symbol for the diode. The arrow part of the symbol points in the direction of current flow when the diode is forward biased. The upper terminal is called the anode. The lower terminal is called the cathode. These names come from vacuum tube diodes. The theoretical equation for the diode current is
iD = IS
exp
μ vD nVT
where IS is the saturation current, n is the emission coefficient, and VT is the thermal voltage. The emission coefficient accounts for recombinations of electrons and holes in the depletion region, which tend to decrease the current. For discrete diodes, it has the value n ' 2. For integrated circuit diodes, it has the value n ' 1. The reason it is different for the two cases is because an integrated circuit diode is fabricated as a bipolar transistor with the collector connected to the base. The impurity doping in transistors is done so as to minimize recombinations. Thus, n ' 1 when recombinations can be neglected. A typical plot of iD versus vD is given in Fig. 2(b). For vD ≤ 0 .6 V, the current is very small. For vD > 0 .6 V, the current increases rapidly with vD. The voltage at which the diode appears to begin conducting is called the cutin voltage. This is approximately 0 .6 V for the plot in Fig. 2(b).
Figure 2: (a) Diode symbol. (b) Typical current versus voltage.
The thermal voltage is given by VT = kT q
where k is Boltzmann’s constant, T is the Kelvin temperature, and q is the electronic charge. At T = 290 K, the thermal voltage has the value VT = 0.025 V. The default value for T in SPICE is T = 300 K. In this case, the thermal voltage has the value VT = 0.02585 V. This value should be used in any hand calculations that are to be compared to SPICE simulations. Because VT increases with T , the equation for iD seems to imply that increasing T decreases iD. However, IS increases rapidly with temperature which causes iD to increase with T. A rule of thumb that is often quoted for silicon diodes is that if iD is held constant, vD decreases by about 2 mV for each degree C as T increases.
Example 1 Calculate the amount by which vD would have to be increased to double the diode current. Assume exp (vD/nVT ) À 1 and VT = 0.025 V.
Solution. We have 2 =
IS [exp (vD 2 /nVT ) − 1] IS [exp (vD 1 /nVT ) − 1]
' exp
μ vD 2 − vD 1 nVT
Solution for ∆v = vD 2 − vD 1 yields
∆v = nVT ln 2 = 17 .3 mV for n = 1 = 34 .7 mV for n = 2
This example illustrates how fast the diode current increases with increasing diode voltage.
Reverse Breakdown
If the diode is reverse biased and the voltage is increased, a point will be reached when the diode enters reverse breakdown and a current will flow. The voltage at which this occurs is called the zener voltage. Fig. 3 illustrates the variation of diode current with voltage. The reverse breakdown voltage is labeled −VZ and the cutin voltage is labeled Vγ. In diode applications, the reverse breakdown voltage must be greater than the maximum applied reverse voltage to prevent diode failure. As an example, diodes used to rectify 120 V rms ac line voltage must have a reverse breakdown voltage greater than 120
2 = 170 V. Diodes which are fabricated to have a specific value of VZ are called zener diodes. These diodes are used as voltage reference diodes. The zener or reverse breakdown voltage is a function of the n and p type doping levels in the diode. The higher the doping, the lower the value of VZ. For VZ less than approximately 4 V, the breakdown is due to zener breakdown. For VZ greater than approximately 5 V, the breakdown is due to avalanche breakdown.
Figure 5: (a) i − v characteristics with tangent line at (VD, ID). (b) Linear diode model.
The equivalent circuit which models this equation is shown in Fig. 5(b). Because vD = VD when iD = ID, it follows that the voltage VDO is given by
VDO = VD − ID rd
Example 3 A diode is biased at ID = 1 mA. It is given that IS = 3. 6 × 10 −^9 A, n = 2, and VT = 25.9 mV. Solve for rd and VDO in the linear model.
Solution. VD = nVT ln (ID/IS ) = 0.649 V, rd = nVT /ID = 51. 8 Ω, VDO = VD − IDrd = 0.598 V.
Small-Signal Model
Because the diode equation for iD as a function of vD is non-linear, the tools of linear circuit analysis cannot be applied, in general, to circuits containing diodes. However, if the diode current is known for a particular voltage, linear circuit analysis can be used to predict the change in current for a given change in voltage, provided the change is not very large. Such an approach is called a small-signal analysis. Let the diode voltage and current be written
vD = VD + vd iD = ID + id
where VD and ID are dc bias values and vd and id are small-signal changes about the bias values. Let the diode equation be denoted by iD = f (vD). We can write
ID + id = f (VD + vd) ' f (VD) + f 0 (VD) vd
where the approximation is a first-order Taylor series expansion about the point (VD, ID). Because ID = f (VD), we can solve for id to obtain
id = f 0 (VD) vd = d dVD
IS
exp
μ VD nVT
× vd =
IS
VT
exp
μ VD nVT
× vd =
ID + IS
nVT
× vd
The small-signal resistance is defined as the ratio of vd to id and is given by
rd = nVT ID + IS
nVT ID
This is the same rd as in the linear model of the diode in Fig. 5(b). This equation says that the diode small-signal resistance is inversely proportional to the current through it. Each time the current is doubled, the resistance is halved. It follows from the linear diode model that rd can be interpreted graphically as the reciprocal of the slope of the iD versus vD curve at the point (VD, ID).
Figure 6: (a) Diode attenuator circuit. (b) Small-signal circuit for calculating vo.
Figure 7: Plot of voltage gain versus control current.
Example 4 Figure 6(a) shows a diode attenuator circuit. The input voltage vi is a small-signal differential voltage represented by two series sources with the common terminal grounded. The current IQ is a dc control current. Solve for the small-signal gain vo/vi. Assume the diodes are identical, n = 2, and VT = 0.025 V.
Solution. For the dc solution, we set vi = 0. In this case, vo = 0. For identical diodes, the current IQ splits equally between the two. It follows the dc voltages across the diodes cancel in calculating VO so that VO = 0. The small-signal resistance of each diode is rd 1 = rd 2 = 2nVT /IQ = 0. 1 /IQ. The small-signal circuit is shown in Fig. 6(b). The current IQ does not appear in this circuit because it is a dc source which becomes an open circuit when zeroed. It follows by voltage division that
vo vi
rd 1 + rd 2 2 R + rd 1 + rd 2
2 × 0. 1 /IQ
2 R + 2 × 0. 1 /IQ
1 + 10IQR
Thus the small-signal gain of the circuit can be varied by varying the dc current IQ. A typical plot of the gain versus current is shown in Fig. 7.
