Semiconductors, amplifiers, attenutation


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Semiconductors, amplifiers, attenutation

  1. 1. ENGINEERING COUNCIL CERTIFICATE LEVEL ENGINEERING SCIENCE C103 TUTORIAL 19 – SEMI CONDUCTORS On completion of this tutorial you should be able to do the following. • Explain amplification and attenuation. • Express gain in deci Bels • Describe the basic operation of a diode. • Describe basic diode applications. • Describe the basic operation of a transistor. • Describe basic transistor applications. • Describe the basic operation of a thyristor. • Describe the basic operation of a triac. This tutorial cannot possibly cover the entire theory of semi-conductors and their circuits and it is not intended for students studying electronics. It should provide more than enough material for the Science Exam C103. The theory and terminology associated with semi-conductors is very large and complicated. It is impractical to expect a student to be fully knowledgeable about transistor and transistor circuits for a science exam. I notice that there is a large discrepancy between the model test question on a full blown bipolar transistor circuit and the actual exam questions which are usually based on diode rectification circuits. The student should consider whether even these notes are worth studying for the science exam (except for the diode sections). If you want a simplified introduction to transistor circuits and terminology, these notes will be useful and put you on the road to comprehending the harder stuff.
  2. 2. 1. AMPLIFIERS AND ATTENUATORS Amplifiers may amplify VOLTAGE, CURRENT or BOTH in which case it is a POWER AMPLIFIER. In this brief description only gain is discussed. There is much more theory not covered here. Amplifier gain may be expressed as a ratio or in decibels. The gain in dB is given by Power Output Gain(db) = 10log10 Power Input WORKED EXAMPLE No. 1 Calculate the power out put of an amplifier that has an input of 20 mW and a gain of 20 dB. SOLUTION Power Output Power Input Power Output 10 2 = 100 = Power Input Gain(db) = 20 = 10log10 2 = log10 Power Output Power Input Power Out = 100 x 20 mW = 2000 mW Since electric power into a resistive load is given as V 2 out Vout V2 then Gain(db) = 10log10 2 = 20log10 P = I2R = Vin R V in WORKED EXAMPLE No. 2 Calculate the gain of an amplifier with an input of 2 mV and output 10 V. SOLUTION Gain(db) = 20log10 Vout 10000 = 20log10 = 74 db Vin 2 ATTENUATORS These are the reverse of amplifiers and are used for reducing signal voltage and power. WORKED EXAMPLE No. 3 Calculate the gain of an attenuator with an input voltage of 12 V and output voltage of 2 V. SOLUTION Gain(db) = 20log10 Vout 2 = 20log10 = -15.6 db Vin 12
  3. 3. DIFFERENTIAL AMPLIFIERS These have two inputs and the difference between them is amplified. The terminal marked + is the first figure and the terminal marked – is the figure subtracted. These devices are ideal for producing the error between two analogue voltages or currents. The power supply must have plus and minus values so that the output can swing from positive to negative as required. WORKED EXAMPLE No. 4 Find the output voltage for each amplifier shown if the gain 10 dB in both cases. SOLUTION Gain(db) = 10 = 20log10 VOUT ∆VIN VOUT = 10 0.5 (2 − (5) ) = −9.5 V Gain(db) = 10 = 20log10 VOUT ∆VIN VOUT = 10 0.5 (2 − (−5) ) = 22.1 V SELF ASSESSMENT EXERCISE No. 1 1. A power amplifier has a gain of 15 db. The input is 0.01 A at 10 V. Determine the output power and the current if the voltage is 10 V. (31.62 W and 3.162 A) 2. A voltage amplifier with a gain of 25 db has an input of 0.2 V. What is the output? (3.557 V) 3. A differential voltage amplifier has a gain of 16 db. Calculate the output voltage when 0.5 V is applied to the + terminal and -0.5 V to the – terminal. (6.31 V) 4. A differential voltage amplifier has a gain of 12 db. Calculate the output voltage when -0.75 V is applied to the + terminal and -0.3 V to the – terminal. (-1.8 V)
  4. 4. 2. SEMI-CONDUCTOR THEORY SEMICONDUCTOR MATERIALS This section covers the basic physics of semiconductor materials and may be skipped if required. There are a group of natural materials that are neither good conductors nor good insulators. These are called semi conductors such as Silicon and Germanium. These are used to make a range of devices that are used in modern electronic circuits. To understand the electrical properties of these materials we need to go back to the atomic level. Electrons orbit in shells of fixed radius representing different fixed energy bands. The outer band of electrons on an atom is called the VALENCE band. The cloud of free electrons surrounding the molecules is called the CONDUCTION band. In a good conductor the electrons will leave the valence band and join the conduction band very easily and these electrons are free to form a current when a voltage is applied. The opposite is true of a good insulator. The band theory supposes that a fixed amount of energy is required to make an electron jump from the valence shell into the conduction band. This is called the energy gap. The energy gap is large for the molecules of a good insulator but for a good conductor it is zero. The resistance and resistivity of most conductors like copper increases with temperature but in the case of semi conductors like silicon the resistance goes down. Semi conductors are widely used to make temperature sensors (e.g. Thermistors) and a simple experiment with one of these will show that the resistance goes down quite dramatically when plunged into hot water. Natural semi conductors are called INTRINSIC. When they are modified by a manufacturing process to give them enhanced properties, they are then called EXTRINSIC. The Silicon atom has 14 electrons, two in the inner shell, eight in the second shell, and only four in the valence shell making it incomplete. This affects the way the atoms bond together in a crystalline form. There are no free electron and no conduction band. In such materials, the energy required to make an electron jump the energy gap can come from heat. This means that at temperature above absolute zero, say room temperature, some electron will jump the energy gap to become free electrons. The number of electrons in the conduction band rises with temperature and explains the negative thermal coefficient of resistance. When an electron makes this jump to the conduction band the parent atom becomes deficient and most text refers to this as a HOLE because it can be filled by another electron. When current flows in a semi conductor, the electrons can migrate from atom to atom so these HOLES appear to migrate in the opposite direction to the electron and this constitutes a current as well. These might be thought of as equivalent to a positively charged particle moving in the opposite direction to the electron. Extrinsic semi conductors are produced by doping them to enhance their conductivity. The conductivity of semiconductors like Silicon can be increased by adding small, controlled amounts of "impurities" that have roughly the same atomic size, but more or fewer valence electrons than the semimetal. This process is known as doping. An impurity with fewer valence electrons such as boron, aluminium or indium takes up space in the solid structure, but contributes fewer electrons to the valence band, thus generating an electron deficit and making the atoms more positively charged. This type of doping creates a hole in the valence band making it possible for the electrons in the valence band to move from one atom to another within this band and so increases the conductivity. Such doped semiconductors are known as p-type because the atoms are more positively charged.
  5. 5. Alternately, an impurity with more valence electrons such as phosphorus, antimony or arsenic contributes extra electrons to the band. Since the valence band is already filled by the semimetal, the extra electrons must go into the conduction band. This also improves the conductivity. Such dopes semiconductors are known as n-type because the enrichment of electrons makes it more negatively charged. SEMICONDUCTOR JUNCTION Consider what happens when a p-type and n-type material are brought together to form a junction. On their own there is an electron surplus in the n-type material and a hole surplus in the p-type. When the two pieces are brought into contact, electrons from the n-type diffuse into the p-type creating a junction zone with few charge carriers. This balancing out of electrons only occurs in the junction zone. If a voltage is applied across the junction, electrons can easily move from the ptype to the n-type but cannot flow in the reverse direction. The junction is a DIODE 3. THE DIODE The symbol for an ideal diode is shown. The ideal diode acts as a switch that conducts as soon as a voltage is applied. In a real diode, this occurs at the cut in voltage Vγ which is typically 0.6 V. Diodes may be built to handle small or large currents and applications are covered later. The two terminals are k (cathode) and a (anode) and the cathode is always connected to negative and the anode to positive. The cathode is the n material and the anode the p material. There is a small reverse flow of about 1 µA when the polarity is reversed and this is called the saturation current. DIODE MODEL We can produce a good working model for the diode by replacing the curve with a mean straight line (in reality the curve is not far from a straight line). The model is a resistor Rf in series with a voltage source Vγ as shown. The gradient of the line is 1/ Rf It must be remembered that reverse current is not allowed in the ideal model.
  6. 6. DIODE APPLICATIONS HALF WAVE RECTIFICATION A diode is a solid state device that only conducts in one direction. When used with a.c. it prevents electricity flowing over one half of each cycle so it may be used to produce direct current from alternating current, a process called rectification. A single diode may be used as a half wave rectifier as shown. Because reverse conduction is not possible, only the positive half cycles appear at the output. The output is always positive but is far from being pure direct current. The circuit usually has a load resistor and smoothing capacitor added as shown. This produces the effect shown on the output. During the conduction period the capacitor charges and usually this is quick enough for the voltage to follow the sinusoidal curve. After the peak the conduction stops and the voltage starts to fall so the capacitor releases its charge and the decay is exponential rather than sinusoidal and the voltage has not fallen to zero before the next conduction point is reached. This produces d.c. with a ripple and may be considered as d.c. plus an a.c. component. The exact wave form depends on the values of the resistor and capacitor and the affect of changing them is like this. Graph A shows a typical half wave rectified voltage. During the time the voltage is rising, the capacitor charges and when the voltage starts to fall, the capacitor discharges. The result is DC plus an AC ripple. Graph B shows the effect of increasing the load resistance. A higher resistance means less current and so the voltage drop across R is reduced and the output voltage increases. Graph D shows the effect of increasing R. The voltage drop across R is increased so the output voltage is reduced. Graph C shows the effect of increasing C. A larger capacitor stores more charge so the voltage falls at a slower rate and the result is less AC ripple.
  7. 7. FULL WAVE RECTIFICATION The diagram shows a bridge rectifier. The a.c. is applied as shown and if the current paths are traced over the entire cycle, it will be seen that the current and voltage at the output terminals is always positive. In affect the negative half cycles are inverted. A more realistic circuit is shown with a smoothing capacitor and resistors. Graph A shows a typical full wave rectified voltage. During the time the voltage is rising, the capacitor charges and when the voltage starts to fall, the capacitor discharges. The result is DC plus an AC ripple. Graph B shows the effect of increasing the load resistance. A higher resistance means less current and so the voltage drop across R is reduced and the output voltage increases. Graph D shows the effect of increasing R. The voltage drop across R is increased so the output voltage is reduced. Graph C shows the effect of increasing C. A larger capacitor stores more charge so the voltage falls at a slower rate and the result is less AC ripple.
  8. 8. 4 THE BIPOLAR JUNCTION TRANSISTOR There are many web sites full of information on transistors. I found the following most helpful. If we use two junctions we have a TRANSISTOR. You can have n-p-n or p-n-p transistors. By injecting or removing electrons at the junction, the junction zones can be made to conduct and so pass a much larger current and producing a gain. The symbols for the transistor are like this. The three terminals are called the emitter, collector and base. For the n-p-n type, the base is normally kept positive relative to the emitter with no signal input (Typically 0.7 V). It might be easier to understand the bipolar transistor if we first study an ideal model. TRANSISTOR MODEL The model for an n-p-n transistor is shown. The diamond symbol is for a current controlled source. The base current Ib is controlled by the source resistor Rs and this produces a current in the controlled device of Ic = β Ib. β is the current gain. The reasoning goes like this. When the current is zero, the controller is an open circuit so Vce = Vcc. When the current is very large, the device is a short circuit so Vce = and Ic= Isc= short circuit current. The algebra goes like this. The source voltage is Vs = Ib Rs Ib= Vs/Rs Hence Vce = Vcc – Ic Rc Ic= (Vcc – Vce)/Rc Ic Rc= (Vcc – Vce) βIb Rc= ( Vcc – Vce) β(Vs/Rs) Rc= ( Vcc – Vce) Vce = Vcc - β Vs Rc/Rs If we plot Ic against Vce we get the graph shown with a gradient of -1/Rc and Isc = Vcc/Rc In reality, these devices are not ideal and neither Isc nor Vcc reach zero due to effects like the saturation resistance of the device and the leakage current. Never the less, this is quite a good model. The next diagram shows a typical set of characteristic curves for a n-p-n transistor in common emitter configuration. The base current Ib is kept constant and various collector voltages are applied. The collector current Ic is measured and plotted. We see that for any base current there is a much larger collector current. The ratio Ic/Ib is denoted β and is the current gain. This is pretty well a constant so long as the transistor is not in the saturation region.
  9. 9. If we superimpose the straight line from the ideal model we have the load line. The actual line position depends on Rc. BIAS VOLTAGE When a n-p-n transistor is used to amplify signals with large variation, we need to stop the base going negative so a positive voltage of about 0.7 V is arranged at the base. This may be produced by a voltage divider as shown. Before we go on, there is a little trick that you need to know. Consider the voltage divider made from R1 and R2 to produce a required voltage Vo from a supply Vs as shown. We can replace the circuit with a single resistance Rb in series with Vs to get the same result where R1R 2 RB = R1 + R 2 WORKED EXAMPLE No. 5 For the circuit shown, calculate IB, VCE and IC. The current gain β is 150 and VBE is 0.7 V Draw the small signal model and calculate the voltage gain vo/vi in dB. Take hFE = 150 and hie = 6.5 kΩ. SOLUTION VBB is the voltage produced by the divider network if the current is equal in both resistors. VBB = VCC x 11/121 = 1.09 V This is not the actual voltage at the base because a current is drawn by the base. Now replace the 110 k and 11 k resistors with a single one in series with VBB as shown earlier. 110 x 11 RB = = 10 k 110 + 11 Next draw the equivalent circuit using the ideal model. The ideal model separates the source from the controlled current and allows independent calculations.
  10. 10. The voltage drop across the 10 kΩ resistor is 1.09 – 0.7 – IE (1.2k) = 0.39 – IE (1.2k) IB ={0.39 – IE (1.2k)}/10k IC = βIB IE = (β+1) IB 0.39 − I B (β + 1)(1.2) 10I B + I B (β + 1)(1.2 ) = 0.39 10 0.39 0.39 0.39 IB = = = 0.00204 mA I B { + 1.2(β + 1)} = 0.39 I B = 10 {10 + 1.2(β + 1)} 10 + (1.2 x 151) 191.2k IC = 150 x 0.00204 = 0.306 mA IE = 0.306 + 0.00204 = 0.30804mA IB = Voltage drop across the 1.2 k resistor is 0.30804 x 1.2 = 0.37 V Voltage drop across the 1.8 k resistor is 0.306 x 1.8 = 0.554 V VCE = 12 – 0.554 - 0.37 = 11.08 V For the small signal model we do away with the bias resistors and put an input impedance instead. SELF ASSESSMENT EXERCISE No.2 Calculate the values of VB, VC, VE, IB and IC in the circuit shown. Assume that VBE = 0.7 V and the current gain β = 200. VB = 3.97 V VC = 7.36 V VE = 3.27 V IB = 0.00493 mA IC = 0.987 mA
  11. 11. SMALL SIGNAL MODEL The diagram shows a widely accepted model for a bipolar junction transistor with small signals. Explanations for all the component symbols are not given here. The unlabelled resistors and capacitors represent components of the model that are only effective at higher frequencies and so are not present for low frequency and d.c. models. rb is usually small and ignored and ro is very large so a simplified model is shown. The model is mainly used for small amplitude a.c. signals. In this case we are only interested in the gain of the a.c. component so we need not include d.c. components such as the bias voltage or indeed the supply voltage. The external resistors set the operating point and quiescent current and the a.c. produces variations about this point. If a resistor has a capacitor in parallel with it, the a.c. component will experience very small impedance and so the component is effectively ignored for calculating a.c. gain although the resistor plays an important roll in fixing the operating point of the transistor. ho and hi are the output and input impedance respectively. hox is often very small and ignored so in order to produce an output voltage vo a load resistor is needed in the external circuit. The model shown is the h model in which parameters are designated h and these are used by manufacturers in their data sheets. Confusingly, text books use many other symbols without relating them to manufacturer's symbols so beware. The diamond symbol represents a current controlled device and the current is related to the current in the base circuit. In this case hfx = β. Most text books confusingly show the arrows in the circuit in the opposite direction to that used here. The arrows used here represent conventional current flow. There are three ways that the transistor can be configured as an amplifier. These are described in the next section and they are the common Emitter, the common Collector (also called emitter – follower) and the common Base. In the model, the letter x should be replaced with e, c or b to denote which mode is in use. Small letters are used when the current varies with time; that is for a.c. applications. For d.c. applications, capital letters are used. Here are some definitions. vce - voltage difference measured between the collector and emitter. vbe - voltage difference measured between the base and emitter. vbc - voltage difference measured between the base and collector. hi – the input impedance when vce = 0 hf – the d.c forward current gain. ho – the output impedance with ib = 0. (This is often shown as an admittance 1/ho in parallel) hr – the reverse transfer voltage ratio with ib = 0 and this is usually ignored in models.
  12. 12. 5. BASIC CIRCUITS FOR AMPLIFIERS The diagram shows the three transistor configurations mentioned previously. n – p – n TYPE The current in the emitter, collector and base are denoted Ie, Ic and Ib respectively. The ratio Ic/Ib is the current gain of the transistor and it is denoted with the symbol β or hfx. The emitter current Ie = Ib + Ic. The ratio Ie/Ib is β +1 The ratio Ic/Ie is less than unity and is denoted with the symbol α (alpha). If you play around with the algebra you will find that β = α/(1-α) COMMON EMITTER The Input Signal is applied to the base. This changes the base current and a larger change is produced in the collector current by a ratio β. For example if β = 100 and ∆Ib = 20 µA, the Ic = 100 x 20 = 2000 µA or 2 mA. The output voltage is V – Ic RL COMMON COLLECTOR The only difference is that the output is developed at the emitter. The emitter current Ie = Ib + Ic. For example if β = 100 and ∆Ib = 20 µA, the Ic = 100 x 20 = 2000 µA and Ie = 2020 µA. In affect the current gain β is 101 when applied to the emitter. The output voltage is Ie RL COMMON BASE The Input Signal is applied to the emitter. A change in voltage at the emitter produces a change in base current and hence a gain in the emitter and collector current. p-n-p TYPE There is no difference in the theory; it is just a case of reversed polarity. The circuits are shown below.
  13. 13. Before we do the next example here is another trick you need to know that is an extension of the last one called Thevenin's theorem. The two circuits shown are equivalent and the voltage VT and resistance RT are given by: VR R 1R 2 VT = s 2 and R T = R1 + R 2 R1 + R 2 This may be verified by equating for i in both circuits. WORKED EXAMPLE No. 6 The circuit shows an amplifier for small a.c. signals. Draw the small signal equivalent model and determine the ratio of the output and input voltages in dB. hFE = 200 and hie = 625 Ω. hoe is negligible. The impedance of all capacitors is very small and may be ignored. SOLUTION First replace Rs and RB with RT and VT VR R s R B 0.6 x 220 10 x 220 = 9.973 V = 0.598 kΩ VT = s B = RT = R B + R s 220 + 0.6 R s + R B 220.6 Note these are almost 10 V and 600Ω because RB>>Rs Since R has a capacitor across it, it plays no affect on the a.c. gain and is left out of the model. Next draw the small signal model with the external circuit. Although VT and the supply volts are shown in the circuit, they plays no role in the a.c. gain and could be omitted as shown. vo = RLic = RLib hfe = 0.500 x 200 ib vo = 100 ib mA The variation in ib = (vs)/(0.598 + 0.625) mA The variation in vo = 100 vs/(1.223) vo 100v s In dB the voltage gain is 20 log 81.8 = 38.26 dB = = 81.8 v s 1.223v s
  14. 14. WORKED EXAMPLE No. 6 Repeat the last example with C omitted and R = 2.8 kΩ The simplified circuit must now include R as this will affect the a.c. component. vo = RLic = RLib hfe = 0.500 x 200 ib vo = 100 ib mA The variation in ib = (vs)/(0.598 + 0.625 + 2.8) mA The variation in vo = 100 vs/(4.203) vo 100v s = = 24.86 In dB the voltage gain is 20 log 24.86 = 27.9 dB v s 4.203v s SELF ASSESSMENT EXERCISE No. 3 1. For the same circuit as in S.A.E. No. 1, determine the voltage gain in dB Ignore the impedance of the coupling capacitor. Take hfe = 200 and hie = 6.5 kΩ Assume hoe and hre are negligible. Answer 33 dB 2. Repeat the problem with a capacitor placed across the 3.3 kΩ resistor with negligible impedance. Answer 29.3 dB
  15. 15. 6. OTHER DEVICES THYRISTOR A thyristor is like a diode but it has a third terminal called the gate. The diode only conducts in the forward direction when the voltage at the gate reaches a critical value. This means that if a pulse is applied at the gate the diode only conducts from that moment on. It switches off automatically on the negative half cycles so the timing pulse is required once every cycle. The circuit diagram shows how a variable voltage is produced with the potentiometer P1 that is turned into ½ wave pulses by the diode B. This controls the trigger point on the conduction ½ cycle of the thyristor C and controls the power to the load. TRIACS Triacs are two thyristors in parallel and this enables the ac waveform to be switched on during both halves of the cycle. This gives good power control for applications like ac motors and light dimmers. You can get a full description and animated demonstration of a triac at the following web link. The circuit diagram below shows how Triac may be used to control an AC motor or other load. The triac is triggered by signal at the gate and the timing of this signal is changed by altering the phase at the gate with respect to the load. Adjusting the potentiometer changes the phase angle.