Figure Zener diode. (a) Schamatic symbol (b) alternative symbol; (c) graph of current versus voltage

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1 1.1 Special Diodes PN junction diodes are the most common type of devices. They are used in power supplies to convert ac voltage to dc voltage. But this process called rectification is not all that a diode can do. Now we will discuss diodes used in other applications. This session begins with the Zener diode, which is optimized for its breakdown properties.zener diodes are very important because they are ket to voltage regulation. This session also covers optoelectronic diodes, Schottky diodes, varactors and other diodes The Zener Diode Small-signal and rectfier diodes are never intentionally operated in the breakdown region because this may damege them. A Zener diode is different;it is a silicon diode that the manufacturerhas optimized for operation in the breakdown region. The Zener diode is the backbone of voltage regulators, circuits that hold the load voltage almost constant despite large changes in line voltage and load resistance. Figure Zener diode. (a) Schamatic symbol (b) alternative symbol; (c) graph of current versus voltage I-V Graph Figure 1.46(a) shows a schematic symbol of a Zener diode; 1.46(b) is an alternative symbol. In either symbol, the lines resemble a Z, which stands for Zener. By varying the doping level of silicon diodes, a manufacturer can produce Zener diodes which breakdown voltages from about 2 to over 1000 V. These diodes can operate in any of three regions: forward, leakage and breakdown. 1

2 Figure 1.46(c) shows the I-V graph of a Zener diode. In the forward region, it starts conducting around 0.7 V, just like an ordinary silicon diode. In the leakage region between zero and breakdown, it has only a small reverse current. In a Zener diode, the breakdown has a very sharp knee, followed by an almost vertical increase in current. Note that the voltage is almost constant, approximately equal to Vz over most of the breakdown region. Data sheets usually specify the value of Vz at a particular test current IZT. Figure 1.46(c) also shows the maximum reverse current IZM. As long as the reverse current is less than IZM, the diode is operating within its safe range. If the current is greater than IZM, the diode will be destroyed. To prevent excessive reverse current, a current-limiting resistor must be used. Ideal Voltage Reference Circuit 1.47 shows a Zener voltage regulator circuit. For this circuit, the output voltage should remain constant, even when the output load resistance varies over a fairly wide range, and when the input voltage varies over a specific range. The variation in VPS may be the ripple voltage from a rectifier circuit. We determine, initially, the proper input resistance Ri. The resistance Ri limits the current through the Zener diode and drops the excess voltage between VPS and VZ. We can write Figure 1.47 A Zener diode voltage regulator circuit (1.36) which assumes that the Zener resistance is zero for the ideal diode. Solving this equation for the diode current, IZ, we get (1.37) where IL = VZ /RL, and the variables are the input voltage source VPS and the load current IL. For proper operation of this circuit, the diode must remain in the breakdown region and the power dissipation in the diode must not exceed its rated value. In other words: 2

3 1. The current in the diode is a minimum, IZ(min), when the load current is a maximum, IL(max), and the source voltage is a minimum, VPS(min). 2. The current in the diode is a maximum, IZ(max), when the load current is a minimum, IL(min), and the source voltage is a maximum, VPS(max). Inserting these two specifications into Equation (1.36), we obtain and (1.38.a) (1.38.b) Zener Resistance In the ideal Zener diode, the Zener resistance is zero. In actual Zener diodes, however, this is not the case. The result is that the output voltage will fluctuate slightly with a fluctuation in the input voltage, and will fluctuate with changes in the output load resistance. Figure 1.48 shows the equivalent circuit of the voltage regulator including the Zener resistance. Because of the Zener resistance, the output voltage will change with a change in the Zener diode current. Two figures of merit can be defined for a voltage regulator. The first is the source regulation and is a measure of the change in output voltage with a change in source voltage. The second is the load regulation and is a measure of the change in output voltage with a change in load current Photo Diode and LED Circuits Figure 1.48 A Zener diode voltage regulator circuit with a nonzero Zener resistance A photodiode converts an optical signal into an electrical current, and a light emitting diode (LED) transforms an electrical current into an optical signal. 3

4 Photodiode Circuit Figure 1.49 shows a typical photodiode circuit in which a reverse-bias voltage is applied to the photodiode. If the photon intensity is zero, the only current through the diode is the reverse-saturation current, which is normally very small. Photons striking the diode create excess electrons and holes in the space-charge region. The electric field quickly separates these excess carriers and sweeps them out of the space-charge region, thus creating a photocurrent in the reverse-bias direction. The photocurrent is Figure 1.49 A photodiode circuit. The diode is reverse biased (1.39) where η is the quantum efficiency, e is the electronic charge, is the photon flux density (#/cm 2 s), and A is the junction area. This linear relationship between photocurrent and photon flux is based on the assumption that the reverse-bias voltage across the diode is constant. This in turn means that the voltage drop across R induced by the photocurrent must be small, or that the resistance R is small. Example 1.13 Calculate the photocurrent generated in a photodiode. For the photodiode shown in Figure 1.49 assume the quantum efficiency is 1, the junction area is 10 2 cm 2, and the incident photon flux is cm 2 s 1 From Equation (1.39), the photocurrent is Comment: The incident photon flux is normally given in terms of light intensity, in lumens, foot-candles, or W/cm 2. The light intensity includes the energy of the photons, as well as the photon flux. 4

5 LED Circuit A light-emitting diode (LED) is the inverse of a photodiode; that is, a current is converted into an optical signal. If the diode is forward biased, electrons and holes are injected across the space-charge region, where they become excess minority carriers. These excess minority carriers diffuse into the neutral n- and p-regions, where they recombine with majority carriers, and the recombination can result in the emission of a photon. LEDs are fabricated from compound semiconductor materials, such as gallium arsenide or gallium arsenide phosphide. These materials are direct-band gap semiconductors. Because these materials have higher band gap energies than silicon, the forward-bias junction voltage is larger than that in silicon-based diodes. It is common practice to use a seven-segment LED for the numeric readout of digital instruments, such as a digital voltmeter. The seven-segment display is sketched in Figure Each segment is an LED normally controlled by IC logic gates. Figure 1.51 shows one possible circuit connection, known as a common-anode display. In this circuit, the anodes of all LEDs are connected to a 5 V source and the inputs are controlled by logic gates. If VI 1 is high, for example, D1 is off and there is no light output. When VI 1 goes low, D1 becomes forward biased and produces a light output. Figure 1.50 Seven-segment LED display Example 1.14 Determine the value of R required to limit the current in the circuit in Figure 1.51 when the input is in the low state. Assume that a diode current of 10 ma produces the desired light output, and that the corresponding forward-bias voltage drop is 1.7 V. Figure 1.51 Control circuit for the sevensegment LED display 5

6 If VI = 0.2 V in the low state, then the diode current is The resistance R is then determined as Comment: Typical LED current-limiting resistor values are in the range of 300 to 350 Ω. Figure 1.52 Optoisolator using an LED and a photodiode One application of LEDs and photodiodes is in optoisolators, in which the input signal is electrically decoupled from the output (Figure 1.52). An input signal applied to the LED generates light, which is subsequently detected by the photodiode. The photodiode then converts the light back to an electrical signal. There is no electrical feedback or interaction between the output and input portions of the circuit The Schottky-Barrier Diode (SBD) The Schottky-barrier diode (SBD) is formed by bringing metal into contact with a moderately doped n-type semiconductor material. The resulting metal- semiconductor junction behaves like a diode, conducting current in one direction (from the metal anode to the semiconductor cathode) and acting as an open-circuit in the other, and is known as the Schottky-barrier diode or simply the Schottky diode. In fact, the current-voltage characteristic of the SBD is remarkably similar to that of a pn-junction diode, with two important exceptions: 6

7 1. In the SBD, current is conducted by majority carriers (electrons). Thus the SBD does not exhibit the minority-carrier charge-storage effects found in forward-biased pn junctions. As a result, Schottky diodes can be switched from on to off, and vice versa, much faster than is possible with pn-junction diodes. 2. The forward voltage drop of a conducting SBD is lower than that of a pnjunction diode. For example, an SBD made of silicon exhibits a forward voltage drop of 0.3 V to 0.5 V, compared to the 0.6 V to 0.8 V found in silicon /wz-junction diodes. SBDs can also be made of gallium arsenide (GaAs) and, in fact, play an important role in the design of GaAs circuits. Gallium-arsenide SBDs exhibit forward voltage drops of about 0.7 V. Apart from GaAs circuits, Schottky diodes find application in the design of a special form of bipolar-transistor logic circuits, known as Schottky- TTL, where TTL stands for transistor- transistor logic. Before leaving the subject of Schottky-barrier diodes, it is important to note that not every metal-semiconductor contact is a diode. In fact, metal is commonly deposited on the semiconductor surface in order to make terminals for the semiconductor devices and to connect different devices in an integratedcircuit chip. Such metal-semiconductor contacts are known as ohmic contacts to distinguish them from the rectifying contacts that result in SBDs. Ohmic contacts are usually made by depositing metal on very heavily doped (and thus low-resistivity) semiconductor regions Varactor Diode The Varactor (also called the voltage-variable capacitance, varicap, epicap and tuning diode) is widely used in television receivers, FM receivers and other communications equipment because it can be used for electronic tuning. Basic Idea In Figure 1.53(a) the depletion layer is between the p region and n region. The p and n regions are like the plates of a capacitor and the depletion layer is like the dielectric. When a diode is reverse biased, the width of the depletion layer increases with the reverse voltage. Since the depletion layer gets wider with more reverse voltage, the Figure.1.53 Varactor. (a) Doped regions arelike capacitor plates separated by a dielectric (b) ac equivalent circuit 7

8 Capacitance becomes smaller. It s as though you moved apart the plate of a capacitor. The key idea is that capacitance is controlled by reverse voltage. Equivalent Circuit and symbol Figure 1.53(b) shows the ac equivalent circuit for a reverse-biased diode. In other words, as far as an ac concerned, the varactor acts the same as a variable capacitance. Figure 1.53(c) shows the schematic symbol for a varactor. The inclusion of a capacitor in series with the diode is a remainder that a varactor is a device that has been optimized for its variablecapacitance properties. Capacitance Decreases at Higher Reverse Voltage Figure (c) Schematic symbol (d) capacitance versus reverse voltage Figure 1.53(d) shows how the capacitance varies with reverse voltage. This graph shows that the capacitance gets smaller when the reverse voltage gets larger.the really important idea here is that reverse dc voltage control capacitance. How is abvaractor used? It is connected in parallel with an inductor to form a oarallel resonant circuit. This circuit has only one frequency at which maximum impedance occurs. This frequency is called the resonant frequency. If the dc reverse voltage to the varactor is changed, the resonant frequency is also changed. This is the principle behind electronic tuning of a radio station, a TV channeland so on. 8

9 Exercise problems 1. (a)calculate Vbi for a GaAs pn junction at T = 300 K for Na = cm 3 and Nd = cm 3. (b) Repeat part (a) for a Germanium pn junction with the same doping concentrations. (Ans. (a) Vbi = 1.23 V, (b) Vbi = V) 2. Calculate the junction capacitance of a pn junction. Consider a silicon pn junction at T = 300 K, with doping concentrations of Na = 1016 cm 3 and Nd = cm 3. Assume that ni = cm 3 and let Cjo = 0.5 pf. Calculate the junction capacitance at VR = 1V and VR = 5V. (Ans. For VR = 1V, Cj=0.312 pf, For VR = 5V, Cj= pf) 3. A silicon pn junction at T = 300 K is doped at Nd = cm 3 and Na = cm 3. The junction capacitance is to be Cj = 0.8pF when a reverse bias voltage of VR = 5V is applied. Find the zero-biased junction capacitance Cjo. (Ans. Cjo = 2.21 pf) 4. Consider the circuit in Figure shown below. Let VPS = 4 V, R = 4 kω, and IS = A. Determine VD and ID, using the ideal diode equation and the iteration method. (Ans. VD = V, ID = ma) Figure for problem 4. A simple diode circuit 5. Determine the diode voltage and current in the circuit shown in Figure for problem 4 using a piecewise linear model. Also determine the power dissipated in the diode. (Ans. VD=0.622V, ID=2.19mA, PD=1.36mW) 6. Consider the diode and circuit in problem 4, Determine VD and ID, using the graphical technique. (Ans. VD 0.54V, ID 0.87 ma) 7. Determine the diode voltage and current in the circuit shown in figure for problem 4 using a piecewise linear model. Assume piecewise linear diode parameters of Vγ=0.6 V and rf = 10Ω.(Ans. ID=2.19 ma, VD=0.622 V) 9

10 8. The power supply (input) voltage in the circuit of Figure for problem 4 is VPS = 10 V and the diode cut-in voltage is Vγ = 0.7 V (assume rf = 0). The power dissipated in the diode is to be no more than 1.05 mw. Determine the maximum diode current and the minimum value of R to meet the power specification. (Ans. ID = 1.5 ma, R = 6.2 kω) 9. Determine diode voltages. The reverse saturation currents of a pn junction diode and a Schottky diode are IS = A and 10 8 A, respectively. Determine the forward-bias voltages required to produce 1 ma in each diode. (Ans.0.539v for PN junction diode and 0.299V for Schottky diode) 10. A pn junction diode and a Schottky diode both have forward-bias currents of 1.2 ma. The reverse-saturation current of the pn junction diode is IS = A. The difference in forward-bias voltages is V. Determine the reverse-saturation current of the Schottky diode. (Ans. IS = A) 11. Consider a simple constant-voltage reference circuit and design the value of resistance required to limit the current in this circuit. Consider the circuit shown in Figure below. Assume that the Zener diode breakdown voltage is VZ = 5.6 V and the Zener resistance is rz = 0. The current in the diode is to be limited to 3 ma. (Ans. R=1.47 KΩ) Figure for problem 11 Simple circuit containing a Zener diode in which the Zener diode is biased in the breakdown region 12. Consider the circuit shown in Figure for problem 11. Determine the value of resistance R required to limit the power dissipated in the Zener diode to 10 mw. (Ans. R = 2.46 kω) 13. A Zener diode has an equivalent series resistance of 20Ω. If the voltage across the Zener diode is 5.20 V at IZ = 1 ma, determine the voltage across the diode at IZ = 10 ma. (Ans. VZ = 5.38 V) 10

11 14. Determine the value of resistance R required to limit the current in the circuit shown in Figure below to I = 15 ma. Assume Vγ = 1.7 V, rf = 15 Ω, and VI = 0.2 V in the low state. (Ans. R = 192 Ω) Figure for problem 14 Control circuit for the sevensegment LED display 15. Consider the circuit in Figure below. Let Vγ = 0. (a) Plot vo versus vi over the range 10 vi +10 V. (b) Plot i1 over the same input voltage range as part (a). Figure for problem For the circuit in Figure below, (a) plot vo versus vi for 0 vi 15 V. Assume Vγ = 0.7V. Indicate all breakpoints. (b) Plot id over the same range of input voltage. Figure for problem 16 11