ADVANCED LEVEL PHYSICS

Transcription

1 AL Syllabus Electronics diode Power supplies The NPN silicon bipolar junction transistor Input, current transfer, collector, and input/putout voltage characteristics in the common emitter configuration. Current amplification factor b. Notes The diode as an uni-directional circuit element (internal mechanism not required). Half-waveband full-wave rectification. Bridge rectifier and application in a.c. measuring instruments. Full wave rectifier with storage capacitor and inductor smoothing. Qualitative treatment only. The transistor as a three-terminal device, the properties of which can be deduced, the propertied of which can be deduced from measurements at its terminals. Knowledge of internal structure not required. Knowledge of internal mechanisms not required. Determination from current transfer characteristic. Simple calculations involving base and collector currents, input and output voltages. Linear voltage amplification Single NPN transistor in the common emitter configuration. Simple biasing techniques. Derivation of voltage gain»-b R L / R B. I. Vacuum Diode Construction: Heater with filament made of tungsten Circuit for obtaining I-V characteristics Anode Vacuum To 6.3V V To 6.3V Cathode ma From S to O, V is negative, hence I = 0. I/A From O to P, as the p.d. of diode increases, the attraction of electrons by electric field increases, hence current increases. At P, since the rate of emitting electrons from cathode is dependent on the heater current (or voltage), for an unchanged current, the rate of productions of electrons is S O fixed. Higher heater voltage Lower heater voltage P Q V/V Therefore, at P, all electrons produced will be attracted to the anode as soon as they are emitted since the anode s potential is too high. An increase of diode p.d. will not increase current I since I is limited by the rate of productions of electrons. So, from P onwards, the current is saturated. CSKMS AL PHYSICS E 2004 by Leo Chung P.1

2 A higher heater potential applied will: (i) have a greater current at the same p.d. (ii) have a higher saturation current. II. Electronic Diode A. Construction B. I-V characteristics I F = Forward bias current V F = Forward bias voltage I R = Reverse bias current V R = Reverse bias voltage V B = Reverse voltage at which the diode breakdown Turn on voltage 0.6 V Resistance at forward bias = 1 / slope after turn on voltage; which is very small. Resistance at reverse bias before break down = 1 / slope between OA; which is very great. Resistance after break down It follows from the graph that the forward current I F is small until the forward voltage is about 0.6 V, thereafter very small change in V F causes a large increase in I F. The reverse current I R is negligible and remains so as the reverse voltage V R is increased. At a certain reverse voltage V B, called the breakdown voltage, the insulation of the barrier layer breaks down and I R increases suddenly and rapidly and damage may occur by overhearing. The resistance is very small, indicating that a large current can pass through it even at a median CSKMS AL PHYSICS E 2004 by Leo Chung P.2

3 voltage. C. Other Junction Diodes (a) Light emitting diode (LED) In recombining the holes and electrons of a forward biased p-n junction, a quantum of energy will be released. When this quantum is specifically chosen, many of these quanta are emitted as light (in gallium arsenide phosphide) So, LED are used as indicator lamps. The advantages of LEDs over filament lamps are their small size, reliability, long life and high operating speed. (b) Zener diode The I-V characteristic of a zener diode is similar to the figure shown in section A, in addition that the breakdown region AB is nearly perpendicular to the y-axis (V). When the zener diode is reversed bias and the p.d. across it reaches zener p.d., the diode break down and the current flows through it will be unlimited except limited by a resistor connected in series. The voltage across the diode will be maintained at V B. The zener diode is used to stablize the voltage output of a power supply. In the figure, (note that the zener diode is connected in reverse sense) the resistor R is to limit the maximum current flows through the diode. E.g. E = 10V, Zener diode: 6.2 V, 1.3 W Calculate the minimum values of R to protect the diode. E R + To drive a radio In this circuit, even when the radio does not draw current, the current passes through the diode would not exceed its maximum limit (or the diode will be burnt out). When the radio draws current, the current through the diode become less. However, the voltage across the CSKMS AL PHYSICS E 2004 by Leo Chung P.3

4 6.2 V? III. Rectification of A.C. Rectification Circuit A. Rectifiers radio is always at V B no matter how much current is drawn by the radio, provided that the current drawn by radio is less than the I MAX of the diode. Why we use a zener diode to provide a steady 6.2 V output instead of using a battery of Having a low resistance in forward direction & high resistance in reverse direction. Three TYPES of rectifiers: (i) (ii) (iii) Thermionic diode valve bulky, expensive & high power consumption, but it is still being used in high quality Hi-Fi circuits. Selenium rectifier not being used today, and is replaced by Semi-conductor diode cheap, wide range of operating conditions (e.g. different diodes ranging from 0.5 V to 500 V are available), reliable and small size. B. Half-wave rectification Connect up the circuit as shown below. Observe the input voltage Vin of the circuit and the output voltage Vout across the load with a double beam oscilloscope. As expected the diode conduct in forward bias; the output waveform is almost identical with the input waveform. In reverse bias, the diode does not conduct and the output voltage is zero. Thus the current through the load is always in one direction, although the magnitude varies; it is a fluctuation d.c. The diode acts as a rectifier. The process is known as half-wave rectification. Note that V in = V O + V AB, Drawbacks: (i) the d.c. is too fluctuating. (ii) half of the time there is no current passing. CSKMS AL PHYSICS E 2004 by Leo Chung P.4

5 C. Full-wave rectification by center-tap transformer + = current flow in +ve half cycle = current flow in ve half cycle the amplitude of I depends on the value of R. The transformer should have a turn ratio of 1:2, so that the voltage output is the same as V in. The current ratio, in such case, is 1:1 (Why?) Drawbacks: A step-up center-tap transformer is needed. D. Bridge full-wave rectifier Observe the p.d. across the load with a CRO. the trace is same as above Fig. Suppose in the 1st half-cycle, A is at a higher potential than B. Then current flows from A to B via D 1, R and D 2..So + is at a higher than -. In the 2nd half-cycle, B is at a higher potential than A. Then current flows from B to A via D 3, R and D 4. Again + is at a higher potential than -. Since this circuit is the simplest and cheap, bridge rectifier is the most common rectifier used in nowadays circuitry. IV. Smoothing Circuits A. Reservoir Capacitor It is a large capacitor connected in parallel with load R, its value depends on the current drawn by the load. CSKMS AL PHYSICS E 2004 by Leo Chung P.5

6 BC Charging (as V in > V C ) AB Discharging (as V C > V in ) The C 1 must be large so that the time constant RC 1 is large compared with the period of the rectified d.c. input. A large C 1 gives a better smoothing performance. However, the value of C 1 is limited by the large initial surge current (as C 1 is at 0 p.d. at starting) which may damage the rectifier circuit. How can we estimate the power output to the load? B. Capacitor-input filter A reservoir capacitor has a useful smoothing effect, but it is usually supplemented by a filter circuit. C 1 is the reservoir capacitor L is a choke (i.e. an iron-cored inductor) of about 15 H (very large inductance). C 2 is a large capacitor. Let V 1 be the voltage across the capacitor C 1. V 1 can be resolved into a steady d.c. component + an a.c. component. In order to understand the behaviour of the filter, we redraw the above circuit. CSKMS AL PHYSICS E 2004 by Leo Chung P.6

7 X L = impedance of L = ωl Hence, for d.c. component, X L = 0 While for a.c. component, X L is large. X C2 = impedance of C = 1/ωC Hence, for d.c. component, X C2 = While for a.c. component, X C2 is small compared with X L. So, when we consider the X L & X C2 as potential divider, the a.c. ripple is mainly developed across L while the d.c. component is developed across C 2 only. Hence, the C 2 output give a very steady d.c. voltage. C. Laboratory H.T.Power supply unit The H.T, power supply employs the principle of the smoothing technique discussed above. The 6.3 V output is for doing thermionic emission experiment which supply voltage for the heater. The output resistor is a variable resistor acts as a potential divider. When the output voltage is smaller, much energy is dissipated in the potential divider, and as the resistor used must have a high power rating (usually about 50 W) and have a good channel to ventilate its excess heat developed. Also, the input voltage 240 V in r.m.s. values. Hence, the 350 V secondary r.m.s. voltage has a peak voltage of V 500 V. So the H.T. power supply gives a d.c. output of 500 V at no load and nearly 400 V when 100 ma (full-load current) is supplied. CSKMS AL PHYSICS E 2004 by Leo Chung P.7

8 V. Transistor A. Construction and Operating Principle There are 2 types of transistor, namely the bipolar or junction transistor and the unipolar or field effect transistor (FET). We would consider the bipolar transistor only. I. Construction There are 2 kinds of junction transistors: n-p-n type & p-n-p type. A transistor has 3 terminals: (i) emitter E; (ii) base B; (iii) collector C. II. Operating principle Consider the n-p-n transistor: V C > V B > V E ; so CB junction is reverse biased & BE junction is forward biased. The emitter is heavily doped with electrons, while the base is highly doped with holes. When the base-emitter is about +0.6 V, electrons cross the junction from E into the base (if the biased voltage is less than +0.6 V, the flow of electron is extremely small and we say the transistor has not been turned on). The moving of emitter electrons from the external circuit constitutes the emitter current (of course in opposite direction), some electrons flow from E & recombine with the holes in B. But since the base is very thin & the p-type doping is very light, only ~ 1% of emitter electrons recombine with the holes. Most of the electrons are swept through the base due to being attracted by the positive voltage on the collector & cross the base-collector junction to become the collector current in the external circuit. The small amount of electron-hole recombination which occurs in the base gives it a momentary negative charge which is immediately compensated by battery B 1 supplying it with (positive) holes. The flow of holes to the base from the external circuit creates a small base current. This keeps the base-emitter junction CSKMS AL PHYSICS E 2004 by Leo Chung P.8

9 forward biased and so maintain the larger collector current. Transistor action is the turning on (and controlling) of a large current through the high resistance (reverse biased) collector-base junction by a small current through the low-resistance (forward biased) base-emitter junction. B. The characteristic of an NPN silicon transistor in the Common Emitter configuration 1. The Input characteristic A graph showing the variation of the base current I b with the base-emitter potential difference V BE is called the input characteristic. Typical result: Circuit: R b Input Characteristic The base-emitter p.d. should be measured with an electrometer. Since the base currents are usually small and the voltmeter must take a current which is small compared with I b, so an electrometer is suitable. The base current only flows when the base is positive with respect to the emitter. The base current only flows when V BE is greater than ~0.5 V. Beyond 0.5 V, the base current rises very rapidly. The input characteristic indicates that V BE will be about 0.7 V for a large range of values of base current. This is a useful point to remember in design circuit. Note that I b is not proportional to V BE, i.e. base-emitter junction is not obey Ohm s law & does not have a constant resistance. And the value of the resistance r between base & emitter depends on the base current. CSKMS AL PHYSICS E 2004 by Leo Chung P.9

10 The resistor R b in the base circuit serves two purposes: It limits the base current to a maximum allowed value. Since excessive base current and produce effects which damage the transistor. If the resistor is omitted a very small variation in position of the potentiometer wiper would cause a very large variation in base current. In other words, the presence of R b, a much larger movement of the potentiometer wiper is required for the same current range, and this adjustment much easier. 2. The Current Transfer Characteristic A graph showing the variation of the collector current I C with the base current I b is called the current transfer characteristic. Typical result: R b Current Transfer Characteristic Notes: The p.d. between collector & emitter is fixed by the 6 V battery in the collector-emitter circuit. The battery should connected such that the collector is positive with respect to the emitter. The path of the collector current is from 6 V battery, into the collector, out from the emitter and back to the 6 V battery. I C is much larger than I B, and for most small or medium power transistors will be measured in milliamperes. In fact, I e = I B + I C, but since I C >> I B I e I C The above transistor circuit is connected in the common emitter configuration. Since the CSKMS AL PHYSICS E 2004 by Leo Chung P.10

11 emitter lead is common to both the input circuit & the output circuit. Current Amplification: The above current transfer characteristic is the key to understanding current amplification. Since the characteristic is linear over a large part of its range, I C is approximately proportional to I B. Moreover, a small variation in I B cause a change in I C which is much larger than the change in I B. The current gain of the transistor is defined as: δi β = δi where β is the slope of the current transfer characteristic. The d.c. current gain is also defined as I C / I B. Typical value of β is greater 100. C B 3. The Collector Characteristic Typical result: A graph showing the variation of the collector current I C with the collector-emitter potential difference V CE for a constant value of the base current I B is called a collector characteristic. Notes: For V CE greater than about 0.2 V, I C is almost constant for a given I B. i.e. the current flowing from collector to emitter is almost independent of the potential difference between the collector & emitter. The transistor does not obey Ohm s law. It is usual to plot a family of curves for different values of I B. These show quite clearly that I C is determined by the values of I B and the curves are approximately equally spaced. This reflects the linear current amplification property of the transistor. Collector Characteristic CSKMS AL PHYSICS E 2004 by Leo Chung P.11

12 4. The Input / Output Voltage Characteristic R L R B Input/Output Voltage Characteristic Notes: R L is called the load resistor. As I C varies, the p.d. across R L varies & so the output voltage V out varies. Hence R L converts current variations to voltage variations. Before the base current starts to flow we say that the transistor is in cut-off region. When the V in exceeds about 0.5 V, base current flows & a collector current I C = βi B flows through the load resistor. There will then be a p.d. across the load resistor so that V out falls from 6 V. This region is called linear region. When the V in exceeds about 1.1 V, further increase of V in has very little effect on V out which remains approximately constant at about 0.1 V. This transistor is then said to be saturated or bottomed. Or in the region of saturation. Details of the characteristics: The input voltage V in is always applied to the transistor via R B (note that V in is not the same as V BE ). In fact V in = V BE + I B R B The output voltage V out is the p.d. across C and E, i.e. V out = V CE is related to I C and R L according to = Experiment Note: Both V in and V out can be measured by moving coil voltmeter. According to the above figure (a), the transistor falls into one of three stages, depending on V in. CSKMS AL PHYSICS E 2004 by Leo Chung P.12

13 (I) Cut-off state If V in is less than 0.7 V, the base current I b = 0 and the collector current I c = 0. As a result, the p.d. across R L is zero and the output voltage V out = 6 V. The transistor is in the cut-off state. (II) Linear region When V in increased from 0.7 V, the base current is given by = + = = β = = β = β Suppose R B = 15 KΩ, R L = 2.2 kω and b = 100. = (5) Or = + = β Clearly, V out varies linearly with V in. As Vin increases, I b increases and V out decreases. (III)Saturation state From equation (1) = As I c increases, V out decreases. When the minimum value of V out = 0.2 V is reached (Fig D2), increasing V in further would not affect the output. The transistor saturates. The minimum input voltage that causes saturation is given by = = 0.2V V in =1.1 V Thus for V in > 1.1 V, the transistor saturates. Taking into account the voltage C and E during saturation, V out = 0.2 V, the p.d. across the load resistor R L is V L = = 5.8 V and the largest collector current is The minimum base current for saturation to occur is = = = = = = µ β CSKMS AL PHYSICS E 2004 by Leo Chung P.13

14 The application of transistor If the input voltage V in changes abruptly and largely (e.g. between 0 and 5 V), the transistor switches between cut-off state (V out = 6V) and saturation state (V out = 0.2 V or 0 V). So the transistor can be used as switch or a gate in binary circuits. If V in varies within the linearly region, V out varies linearly with V in. With suitable base resistor R B and Load resistor R L, the transistor can be used to amplify a.c. (a) Linear Voltage Amplification A linear voltage amplifier The Figure above shows the circuit for amplifying an a.c. voltage DV in using a transistor. Before the a.c. is applied, the transistor is adjusted by R1 to lie in the middle of the linear region. (Such input is called optimal biasing voltage, i.e. 0.9 V in the above graphs) Thus, V out = 3.0 V. Consider the base circuit, = + = Where V BE is constant (about 0.7 V) when the transistor is conducting. Consider the collector circuit, From (1) and (2), we have = = β (2) = β (3) When a.c. of peak voltage DV in is applied through C 1, the input voltage becomes (V in + DV in ) this causes the p.d. across C and E fluctuate about V out (V out = 3.0 V). CSKMS AL PHYSICS E 2004 by Leo Chung P.14

15 The a.c. output DV out is obtained via C 2. From equation (3), Let G=DV out /DV in is called voltage gain. + = β +..(4) The output signal is 180 o out of phase with input signal. = = β a.c. component Suppose a transistor of current gain = 100 has a load resistor R L = 2.2 kand a base resistor R B = 15 k The voltage gain is 2.2 G = 100 = Thus, the amplitude of output signal is 15 times that of the input signal. To avoid distortion, the output signal s amplitude must be within 3.0 V. Thus, the peak voltage of the input a.c. must be within 3/15=0.2 V (i.e. the maximum peak input voltage is V inmax = 0.2 V d.c. component (biasing) In the absence of the a.c. input, the base current I b should flow to keep V in at the mid-way in the linear region. I b is called the biasing current. In our example, for V out = 3.0 V, I c is given by 3 3 I c = = = 1. ma R 36 L 2200 Thus, the biasing current is I b = I c /b = 13.6 ma/100 = 13.6 ma Power gain An amplifier not only provides voltage gain, but also power gain. The extra energy comes from the power supply connected to the amplifier. A transformer is not an amplifier. Although it can increase an a.c. voltage, it required a considerable current flow in the primary. Experimental Notes: (i) C 1 at the input is called blocking capacitor. It is to prevent the steady biasing current I b flowing into the a.c. source, thereby upsetting the bias condition. (ii) C 2 at the output is to block the steady p.d. (3.0 V in our example) which is necessary for biasing the transistor. This ensures that the output voltage us pure a.c.. CSKMS AL PHYSICS E 2004 by Leo Chung P.15

16 Amplifying an input a.c. by a linear voltage amplifier Transistor as Switches (1) Use with a light sensing device In the above figure, the light dependent resistor (LDR) has a high resistance (~600 kilo-ohm) in darkness and a low resistance (~9 kilo-ohm) in sunlight. Suppose the variable resistor is set at about 200 kilo-ohm. When the LDR is illuminated with sunlight, the p.d. across the LDR is about = + CSKMS AL PHYSICS E 2004 by Leo Chung P.16

17 This is approximately the input voltage V in for the transistor. Since V in < 0.5V, the transistor is in cut-off state. There is no current though the collector terminal. The magnetic relay switch remains open. The bulb in the secondary circuit does not light up. As the LDR placed in darkness, the resistance rises to 600 kilo-ohm and its p.d. is now = + Since V in is greater than 0.7 V, a base current flows through R B. As a result a large collector current flows downward through the coil of the relay. The secondary circuit is switched on and the bulb lights up. If the bulb is switch on room light, rather than in total darkness, R should be adjusted to a smaller value. The diode in the circuit serves to protect the transistor. Large backward current would induced when the coil suddenly break. (2) Squarer Connect an a.c. low tension power supple to a basic transistor unit as shown below: Set the LT at 1 V r.m.s. (1.4peak). Observe the input signal V in and output signal V out on a double-beam CRO. The sinusoidal input has changed to a square output. The circuit of the above circuit acts a squarer. The action of a squarer can be understood from the input/output voltage characteristic. When the input voltage exceeds +1 V, the transistor is saturated and the output voltage is nearly zero (in fact about 0.2 V); when the input voltage is less than 0.5 V (including the case when Vin is negative) the transistor is cut off and the output is +6 V. In between V in = +0.5 V and V in = +1 V, V out changes linearly with V in. Note that this is a relatively small part of the V in - t graph, which can therefore be approximated as a straight line. Consequently V out is also approximately linear in the time t. Also note that V out stays longer at maximum voltage than at minimum voltage; the output waveform is not CSKMS AL PHYSICS E 2004 by Leo Chung P.17

18 symmetrical. Now increase the r.m.s. voltage of the LT gradually. We see that the output signal always changes between +6 V to ~0 V. However, the waveform becomes more and more symmetrical. VI. Operational Amplifier The operational amplifier or the op-amp does for the analogue world, like what the NAND gate foes for the digital world. It can do mathematical operations and is used as a high-gain amplifier or as a switch. A. Properties 1. A d.c. amplifier, can simplify both steady d.c. voltages and varying a.c. voltages. 2. Large voltage gain, or open-loop gain (i.e. no feedback) 10 5 in general expressed in db, e.g. a gain of 10 5 = 20 log = 100 db. 3. High input impedance, 2 megaohm to ohm. And so draws very small current from the input like a high resistance voltmeter. CSKMS AL PHYSICS E 2004 by Leo Chung P.18

19 4. Low output impedance, 75 ohm 100 ohm, so output voltage would be completely consumed by a large load. 5. High slew rate, the rate at which the output voltage can change when the input voltage is changed, at about 10 6 to 10 7 V/s. B. Uses of Op-amp 1. Voltage amplification with or without inverting the signal; 2. Finding the sum of two or more signals; 3. Finding the difference between two signals; 4. As an integrating system; 5. As a comparator: switches when a signal reaches a pre-determined level; 6. As a pulse generator; 7. As s generator of alternating voltages (a signal generator). C. Configurations 1. Common op-amp comes in d.i.l. (dual in line) IC with 8 pins in type nos. of 741 & TL Needs balanced power supplies of (-5 V, 0, +5 V) up to (-15 V, 0, +15 V) in pin 4 and 7 respectively. A dual supply is necessary since the output voltage must be able to swing both + and with respect to earth. 3. A positive input would be changed to negative output if it is fed into the inverting input or the (supply ) input in pin For the non-inverting input or the (supply + ) input in pin 3, the phase of the input signal would not be changed. 5. An op-amp amplifies the voltage difference between the inputs & not the voltage between an input and the power supply zero. Hence it is also named differential amplifier. Output at pin 6 is measured relative to the power supply zero. 6. A 741 can take any value between + 15 V and 15 V. The output of a 741 can only get to within 2 V of the supply rails. So the maximum and minimum outputs are + 13 V and 13 V with the usual supply rails. CSKMS AL PHYSICS E 2004 by Leo Chung P.19

20 D. Transfer Characteristics of an Op-amp Various voltages are fed in through the inverting and non-inverting inputs and the corresponding output voltage measured. The following graph shows the variation of the output versus the difference in the input voltages. 1. The variation is linear only within a small range of input voltages. 2. Outside this range of input, the output would be close to the supply voltage and is said to be saturated. = A V, where A 0 is the 3. The output is given by ( ) open-loop voltage gain. V out 0 2 V1 CSKMS AL PHYSICS E 2004 by Leo Chung P.20

21 E. Feedback The process whereby all or part of the output generated by a system is fed back to the input of that system. For positive feedback, it results in continued or even greater output. For negative feedback, it will tend to cancel or reduce the effect which produced it. In general, there are 3 distinct ways in which an op-amp may be used. They may be used in open-loop mode (i.e. without any feedback), with positive feedback and with negative feedback. The last method is the most common way in the use of an op-amp. F. Open-Loop Mode When an op-amp is used as a processing unit on its own, it acts as a comparator. ii. It looks at the signals coming into its two inputs. iii. It decides which one has the higher voltage. iv. It pushes the output to one of the supply rails. e.g. if non-inverting input > inverting input, the output is + 14 V or + 15 V. if inverting input > non-inverting input, the output is 14 V or 15 V. CSKMS AL PHYSICS E 2004 by Leo Chung P.21

22 Consider the following two circuits in which the op-amp is used as a switch. For the left circuit, if the thermistor is cold, (i.e. Y is low) and X> Y, the op-amp is in non-inverting mode, V out becomes + and the green LED is lights. If the thermistor is hot, Y will increase until Y > X, the op-amp is in inverting mode and V out becomes which makes the red LED lights. The temperature at which the LEDs swap over is controlled by the potentiometer. The op-amp acts as a switch and the system is a hot/cold detector. For the right circuit, the Led only lights if V in lies in the range + 5V to + 10V. When V in lies outside this range, the outputs of the two op-amps have the same + or values of the supply rails. The LED can only come on when the outputs are different. G. Negative Feedback the inverting amplifier *** Saturation V out V in As can be seen in the above circuit, a fraction of V out is fed back to the inverting input through R f (f = feedback). This lowers the input and the output as they are in anti-phase. The gain with negative feedback is called closed-loop gain would be less than the open-loop gain. Saturation CSKMS AL PHYSICS E 2004 by Leo Chung P.22

23 i. Gain is predictable, independent of the characteristics of op-amp. ii. iii. iv. Greater stability. Less output distortion, or more linear amplification. Increase in the bandwidth of frequency response. i. Negligible current drawn in each input due to large input resistance. ii. The inverting input is close to zero or earth potential (called virtual earth). As the gain of the op-amp is so large that p.d. between the two inputs is small (non-inverting input is earthed). Let V l be + ve and so V o is ve, Using Ohm s law, i 0 V = I ( ) O 2R ( ii) Vl 0 = I 1Rl f Hence, the closed-loop gain is given by, V R O f A0 = = ( Q I 2 I 1 ) V R l l i. Hence the closed-loop gain depends only on the value of the resistors. ii. When R f = Rin, the closed-loop gain = - 1, i.e. the circuit acts as an inverter (the NOT iii. gate). If the input is an a.c. source, the gain of the amplifier decreases as the frequency of the input signal decreases. The use of negative feedback increases the range of frequencies over which the gain is constant, i.e. the bandwidth is increased. CSKMS AL PHYSICS E 2004 by Leo Chung P.23

24 H. Others Applications of Op-amp The summing amplifier uses an op-amp to add 2 or more voltages and produces an output that is proportional to their sum. As negligible current flows into the inverting input, thus current in R f equals the sum of the currents in R 1, R 2 and R 3. The non-inverting input is at zero potential. Therefore, V1 V2 V + + R R R 1 when V = R out f R 1 = R2 = R3 = R f, V out = ( V + V + V ) The input V in is supplied to the non-inverting (+) terminal of the op-amp. By the same assumptions in inverting amplifier, we have, V V in out = I f = I R f in thus the gain ( R + R ) f in V ( R + R ) out f in A 0 = = = 1+ Vin Rin R R f in V + R f 10 k V in R in V out V - CSKMS AL PHYSICS E 2004 by Leo Chung P.24

25 If the output is directly all fed to the inverting (-) terminal, then R f = 0, and so the gain is 1. Thus V out is very close to V in. The circuit is called a voltage follower. It has very high input impedance and a low output impedance (due to the op-amp). It is used as the input stage of an analogue voltmeter for giving a high input impedance. M.C.Questions << Diode & power supplies >> I 48 Which of the following statements about a thermionic diode is/are correct? (1) A thermionic diode can be used to convert a.c. into d.c. (2) Increasing the filament current will increase the saturation current. (3) The current flow to the anode increases with the anode voltage within a certain range. A. (1), (2) and (3) B. (1) and (2) only C. (2) and (3) only D. (1) only E. (3) only I 29 For the circuit shown above, which of the graphs shown below best represents the variation of current I with time t? CSKMS AL PHYSICS E 2004 by Leo Chung P.25

26 I 30 The above figure shows a half-wave rectifier with a smoothing circuit. The time constant of C and R should be A. small compared with the time of one cycle. B. equal to half the time of one cycle. C. equal to the time of one cycle. D. equal to twice the time of one cycle. E. large compared with the time of one cycle I 42 Which of the following circuits can give a half-wave rectification waveform on a C.R.O.? CSKMS AL PHYSICS E 2004 by Leo Chung P.26

27 I 42 In the above a.c. smoothing circuit, the ripple on the current passing through the load R can be reduced by increasing (1) the load resistance R (2) the capacitance C (3) the a.c. supply frequency A. (1), (2) and (3) B. (1) and (2) only C. (2) and (3) only D. (1) only E. (3) only CSKMS AL PHYSICS E 2004 by Leo Chung P.27

28 I 37 Which of the following graphs best represents the variation with time t of the current I through the segment AB in the above circuit? IIA 29(AL)/1995 IIA 15(AS) A 'black box' containing two unknown components is connected to a cell, a resistor and an ammeter as shown. A current flows steadily no matter which way the box's terminals are connected to he cell and the same ammeter reading is obtained. The two components in the 'black box' could be A. two diodes in series. B. two diodes in parallel. C. two capacitors in parallel. D. a diode and resistor in parallel. E. a diode and a capacitor in parallel. CSKMS AL PHYSICS E 2004 by Leo Chung P.28

29 IIA 35(AL) When a CRO is connected to a circuit, the trace obtained is shown above. Which of the following circuits has the CRO been connected? << NPN silicon bipolar junction transistor >> (AL only) I 30 The diagram shows a transistor circuit with two similar light bulbs L1 and L2. The bulb L2 lights up brightly, but L1 does not glow at all. Which of the following could be a possible reason for this? A. The filament of L1 is burnt out. B. The cell E1 should be connected the other way round. C. The cell E2 should be connected the other way round. D. The collector current is very much less than the emitter current. E. The base current is very much less than the emitter current. CSKMS AL PHYSICS E 2004 by Leo Chung P.29

30 I 32 In the transistor circuit shown above, the voltage VAE, VPE, VCE and VBE were measured and tabulated as follows : VAE VPE VCE VBE 3.6 V 2.6 V 2.5 V 0.6 V What are the values for the base current IB and the collector current IC? I B / ma I C / ma A B C D E I 25 A transistor is used for current amplification in the common emitter configuration. The measured currents through the emitter, collector and base of the transistor are ie, ic and ib respectively. What are the possible values of ic/ie and ie/ib as obtained from the measurement of the various currents? i c / i i / i A B. 1 1 C D E I 43 e e b CSKMS AL PHYSICS E 2004 by Leo Chung P.30

31 In the above circuit, the sliding contact S is moved between X and Y to give different voltages across ce. Which of the following graphs best represents the variation of the collector current Ic with the voltage Vce across the emitter and the collector? I 44 In the above circuit, the reading of the voltmeter is zero. What should be the voltage applied at Input 1 and Input 2 respectively? Input 1 Input 2 A. 0 V 0 V B. 0 V 6 V C. 3 V 3 V D. 6 V 0 V E. 6 V 6 V CSKMS AL PHYSICS E 2004 by Leo Chung P.31

32 I 43 A NPN transistor is operated as a linear voltage amplifier and the output voltage is displayed on a CRO screen as shown : What changes will occur when RL is slightly increased? Vdc Vpp A. decreases increases B. decreases decreases C. increases increases D. increases decreases E. no change increases I 45 In the above circuit, if the current amplification factor b = 130, what is the output voltage Vout? A. 0 V B. 3.0 V C. 4.1 V D. 4.9 V E. 6 V I 40 A CRO is used to display the current transfer characteristic curve of a transistor. How should points P, Q and R be connected to the X-input, Y-input and the earth terminal of the CRO? CSKMS AL PHYSICS E 2004 by Leo Chung P.32

33 X-input Y-input Earth A. P Q R B. P R Q C. Q P R D. Q R P E. R P Q I 39 The circuit above shows an NPN transistor and two resistor 10 kw and 1 kw connected to a 5 V d.c. supply. The current gain of the transistor is 100. What is the value of the collector current? A. 5 µa B. 500 µa C. 5 ma D. 50 ma E. 500 ma I 43 An n-p-n transistor is used in the above circuit as a pulse shaper or a squarer. When a sinusoidal voltage whose magnitude varies between +2 V and -2 V is applied to the input, what will be the output voltage? CSKMS AL PHYSICS E 2004 by Leo Chung P.33

34 I 30 The graphs show the characteristics for a transistor operating in the common emitter mode. I C is the collector current, I B is the base current and V CE is the potential difference between the collector and emitter. The current gain for this transistor is A. 20 B. 50 C. 80 D. 100 E I 31 The graph show s the transfer characteristic of an electronic device. The input is a sinusoidal voltage with a peak value of 11.5 V and a mean value of zero. Which one of the following waveforms best represents the variation of the output voltage with time? A. B. C. D. E. CSKMS AL PHYSICS E 2004 by Leo Chung P.34

35 I-40 The above diagrams show an NPN transistor consist and its input/output voltage characteristic. What is the current amplification factor of the transistor? A. 10 B. 30 C. 60 D. 75 E IIA-31(AL) In the above transistor circuit, the voltage across the base and the emitter is 0.5 V when the transistor works. The current amplification factor of the transistor is 80. What input voltage (V in ) would give an output voltage (V out ) of 4 V? A V B V C V D V E V << Operational amplifier >> (AL only) I 32 In the circuit above, what is the voltage amplification? A with inversion B with inversion C. 50 with inversion D. 50 without inversion E without inversion CSKMS AL PHYSICS E 2004 by Leo Chung P.35

36 I-41 Two electrical signals V 1 and V 2 are fed into an operational amplifier. The variations of V 1 and V 2 with time are shown above. Which of the following graphs represents the variation of the output V out with time? A. B. C. D. E I-35(AL) Which of the following statements about an operational amplifier is/are correct? (1) It amplifies the difference between the voltages at its two inputs. (2) For d.c., the open loop voltage gain is of the order (3) For a.c., the open loop voltage gain decreases with increasing frequency. A. (1) only B. (3) only C. (1) and (2) only D. (2) and (3) only E. (1),(2) and (3) CSKMS AL PHYSICS E 2004 by Leo Chung P.36

37 IIA-36(AL) The above figure shows an operational amplifier circuit which uses a ±15 V supply (not shown). If the input potential is V, what is the potential at point X? A. -5 V B. -15 V C V D. +5 V E. +15 V IIA-43(AL) An operational amplifier is connected as shown below with input voltage V i = +2 V. What is the output voltage V o? A. -6 V B. -4 V C. +4 V D. +6 V E. +12 V IIA-32(AL) An input voltage(v in ) of 2.0 V is applied to an ideal operational amplifier connected as shown. The current flowing through the 8 kω resistor is A ma from X to Y B ma from X to Y C ma from Y to X D. 1.0 ma from X to Y E. 1.0 ma from Y to X CSKMS AL PHYSICS E 2004 by Leo Chung P.37

38 IIA-32(AL) A circuit is required for doubling the amplitude of a sinusoidal alternating voltage. Which of the following circuits might be suitable? (1) (2) (3) A. (1) only B. (3) only C. (1) and (2) only D. (2) and (3) only E. (1), (2) and (3) IIA-42(AL) In the above operational amplifier circuit, the output voltage V out is A. 5 V B. 2.5 V C. 0 V D V E. +5 V CSKMS AL PHYSICS E 2004 by Leo Chung P.38

39 Structured Type Questions I-9(AL) Figure 9.1 Shows a common-emitter transistor circuit and also its input/output voltage characterisics. (a) (i) Find the voltage V c. (ii) What is the voltage between the base and the emitter when V i = 1.5 V. (iii) Find the voltage gain and the current amplification factor of this transistor circuit. (b) In order to amplify the rectified a.c. signal shown in Figure 9.2 two resistors and two capacitors are added to the circuit as shown in Figure 9.3. Figure 9.2 Figure 9.3 (i) (ii) State the function of the capacitors in the circuit What is the maximum value of V p such that the signal can be amplified without chopping off the peaks? (iii) Draw the corresponding output signal in the space provided if V P = 0.2 V CSKMS AL PHYSICS E 2004 by Leo Chung P.39

40 IA-4(AL) (a) Figure 4.1 shows an operational amplifier circuit. Figure 4.1 A graph of output voltage V out plotted against input voltage V in is shown below. (i) What is the resistance of the resistor R f? (ii) With reference to the graph, explain the function of the above circuit. (b) Figure 4.2 Figure 4.2 shows the circuit of a comparator. The LED lights up when the input voltage V in is less than 4.5 V. Find the resistance of R. (c) A stable power supply is essential for the operation of computer parts, which usually work at 5 V d.c. A fluctuation in the supply voltage V in of more than 10% is certainly not tolerable, so it is therefore important to keep a regular check on it. Figure 4.3 shows a warning device designed for such a purpose. (i) State the potential at P and at Q. Briefly explain the operation of the circuit. (ii) Calculate the resistance of the resistors R 1 and R 2. CSKMS AL PHYSICS E 2004 by Leo Chung P.40