# Lab 3 Experiment 3.1 Common Emitter Amplifier Objective Hardware required Pre lab questions:

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1 Lab 3 Experiment 3.1 Common Emitter Amplifier Objective a. To sketch a common emitter amplifier circuit and explain the operation of the circuit. (i) (ii) b. To analyze the amplifier circuit to determine input resistance output resistance, voltage gain, current gain and power gain. c. To design a CE amplifier Circuit for the given specifications. d. To observe with an oscilloscope, the transient signal voltages of the input and output of the amplifier. e. To measure the voltage gain of the amplifier over and range of frequencies and plot the frequency response curve. f. To determine the values of lower and upper 3-dB frequencies and 3-dB bandwidth. g. To trouble shoot a non-operational CE amplifier. To make a dynamic test which will determine whether the ac amplifier is operating properly? To consider dc voltage and resistance norms at test points in the amplifier which is operating properly, and to draw inferences as to the nature of the trouble from the voltage and resistance measurements in a defective amplifier Hardware required a. Power supply - Variable regulated low voltage dc source b. Equipments - AFO, CRO, DMM c. Resistors - 50k POT d. Capacitors - e. Semiconductors - BC 107 (or equivalent) f. Miscellaneous - Breadboard and wires Pre lab questions: 1. Calculate the base bias voltage for the circuit shown below when no signal source is present, and when signal source is directly connected. 2. Calculate the transistor collector voltage for the circuit of problem 1 with C 2 present, and with R L directly connected. 3. The CE circuit shown in problem 1 has the following transistor parameters: hie =, hfe = 85, hoe = 2µs. Calculate zi, Zo & Av

2 4. For the circuit in problem 1, recalculate Zi and Av when the bypass capacitor is removed from R E 5. For a CE circuit with voltage divider bias, a bypass emitter resistor, a capacitor coupled signal source, and a capacitor-coupled load, (i) Sketch the dc equivalent circuit and write expressions for the dc voltages and currents. (ii) Sketch the ac equivalent circuit and approximate h-parameter equivalent circuit and write expressions for Zi, Zo, Av, A I and A P. (iii) Draw ac and dc load lines on the o/p characteristics of CE configuration and explain their significances. 6. The CE circuit has the following component values and parameter values: hie = 1.4k, hfe = 55, R L = 47k, Vcc = 12V, rs= 600 and Zo = 3.9k. Sketch the circuit and determine suitable resistor and capacitor values. Also, calculate Zi, Zo, Av, Ai and A P for the circuit designed Theory Amplifier is an electronic circuit that is used to raise the strength of a weak signal. The process of raising the strength of a weak signal is known as amplification. One importance requirement during amplification is that only the magnitude of the signal should increase and there should be no change in signal shape. The transistor is used for amplification. When a transistor is used as an amplifier, the first step is to choose a proper configuration in which device is to be used. Then the transistor is biased to get the desired Q-point. The signal is applied to the amplifier input and gain is achieved CE amplifier operation Consider a CE amplifier circuit as shown in fig When the capacitors are regarded as ac short circuits, it is seen that the circuit input terminals are the transistor base and emitter, and the output terminals are the collector and the emitter. So, the

3 emitter terminal is common to both input and output, and the circuit configuration is termed Common Emitter (CE) From the voltage waveforms for the CE circuit shown in Fig (a). It is seen that there is a 180o phase shift between the input and output waveforms. This can be understood by considering the effect of a +ve going input signal. When Vs increases in a +ve direction, it increase the transistor V BE. The drop across Rc, and thus reducing the level of the Vc. The changing level of Vc is capacitor-coupled to the circuit output to produce the ac o/p voltage, Vo. As Vs increases in a +ve direction, Vo goes in a ve direction. Similarly. When Vs changes in a ve direction, the resultant decrease in V BE reduces the Ic level, thereby reducing V RC, and producing a +ve going output. The circuit in Fig (a) has an input impedance (Zi) and an o/p impedance (Zo). These can cause voltage division of the circuit i/p and o/p voltages. So, for most transistor circuits Zi and Zo are important parameters. The circuit voltage amplification (Av), or voltage gain, depends on the transistor parameters and on resistor Rc and R L CE amplifier circuit elements and their functions (i) Biasing circuit The resistances R 1, R 2 and R E form the biasing and stabilization circuit. The biasing circuit must establish a proper operating point, otherwise a part of the ve half-cycle of the signal may be cut-off in the output. (ii) Input capacitor, C 1 An electrolyte capacitor C 1 is used to couple the signal to the base of the transistor. If it is not used, the signal source resistance, rs will come across R 2 and thus change the bias. C 1 allows (iii) only ac signal to flow but isolates the signal source from R 2 Emitter bypass capacitor, C E An Emitter bypass capacitor, C E is used parallel with R E to provide low reactance path to the amplified ac signal. If it is not used, then ac amplified ac signal following through R E will cause a voltage drop across it, thereby reducing the o/p voltage. (iv) Coupling capacitor, C 2 The coupling capacitor, C2 couples one stage of amplification to the next stage. If it is not used, the bias conditions of the next stage will be drastically changed due to the shunting effect of R C. This is because R C will come in parallel with the upper resistance R1 of the biasing network of the next stage, thereby altering the biasing conditions of the latter. In

4 short, the coupling capacitor C 2 isolates the dc of one stage from the next stage, but allows the passage of ac signal CE amplifier circuit currents (i) Base current i B = I B +ib Where I B = dc base current when no signal is applied ib = ac base when as signal is applied and i B = total base current (ii) Collector current Ic = I C ie Where I C = zero signal collector current ic = ac collector current when ac signal is applied and ic = total collector current (iii) Emitter Current i E = I E + ie Where I E = Zero signal emitter current Ie = ac emitter current when ac signal is applied and i E = total emitter current It is useful to keep in mind that I E = I B + I C and ie = ib +ic Also, I E I C and ie ic CE amplifier frequency response The voltage gain of an amplifier varies with signal frequency. It is because reactances of the capacitors in the circuit changes with signal frequency and hence affects the output voltage. The curve between voltage gain and signal frequency of an amplifier is known a frequency response. Fig shows the frequency response of a typical CE amplifier.

5 It is clear that the voltage gain drops off at low (< f L ) and high (> f H ) frequencies whereas it is uniform over mid-frequency range (f L to f H ). (i) At low frequencies (< f L ), the reactance of coupling capacitor is quite high and hence very small part of signal will pass from amplifier stage to the load. Moreover, C E cannot shunt the R E effectively because of its large reactance at low frequencies. These two factors cause a falling of voltage gain at low frequencies. (ii) At high frequencies (> f H ), the reactance of C2 is very small and it behaves as a short circuit. This increases the loading effect of amplifier stage and serves to reduce the voltage gain. Moreover, at high frequency, capacitive reactance of base-emitters junction is low which increases the base current. These reduce the current amplification factor. Due to these two reasons, the voltage gain drops off at high frequency. (iii) At mid frequencies (f L to f H ), the voltage gain of the amplifier is constant. The effect of coupling capacitor C 2 in this frequency range is such as to maintain a uniform voltage gain. Thus, as the frequency increases in this range, reactance of C C decreases which tend to increase the gain. However, at the same time, lower reactance means higher almost cancel each other, resulting in a uniform fain at mid-frequency CE amplifier analysis The first step in ac analysis of CE amplifier circuit is to draw ac equivalent circuit by reducing all dc sources to zero and shorting all the capacitors. Fig shows the ac equivalent circuit. The not step in the ac analysis is to draw h-parameter circuit by replacing the transistor in the ac equivalent circuit with its h-parameter model. Fig shows the h-parameter equivalent circuit for CE circuit.

6 The typical CE circuit performance is summarized below: Device input impedance, Z b = hie Circuit input impedance, Z i = R 1 R 2 Z b Device output impedance, Zc = 1/hoe Circuit output impedance, Zo = Re Zc Rc Circuit voltage gain, Av = -hfe/hie (Rc R L ) Circuit current gain, Ai = hfe R C R B / (R C +R L (R C + hie) Circuit power gain, A P = Av x Ai CE amplifier circuit design: Design of CE circuit normally commences with a specification of supply voltage, minimum voltage gain, frequency response, signal source impedance load impedance, stability factor and the Q-point. (draw CE amplifier circuit Fig ) Selection of I C, R C and R E From eq. 3-1-, Av = -hfe / hie (R C R L ) For satisfactory transistor operation, Ic should not be less than 500µA. A good minimum Ic to aim for is 1mA. The V CE should typically be around 3v to ensure that the transistor operates linearly and to allow a collector voltage swing of ±1v which is usually adequate for small-signal amplifier Note Rc should normally be very much larger than R L, so that R L ahs little effect on the voltage gain. Select V E = 5v for good bias stability in most circumstances. Note When V E >>V BE, V E will be only slightly affected by any variation in V BE (due to temperature change or other effects)

7 Once V E, V CE and Ic are selected, V RC is determined as V RC = V CC V CE V E Then, R C and R E are calculated. R C = V RC / I C and R E = V E / I C Selection of bias resistors As discussed in lab-1, experiment-1.1, section 1.1, selection of voltage divider current (I 2 ) as I C /10 gives good bias stability and reasonably high input resistance. The bias resistors are calculated as R 2 = V B / I 2 and R 1 = V CC V B / I 2 Selecting R 2 = 10R E gives I 2 = I C /10. The precise level of I 2 can be calculate as I 2 = V B /R 2 and this can be used in the equation for R1. Selection of bypass capacitor, CE Basically the capacitor values are calculated at the lowest signal frequency that the circuit is required to amplify. This frequency is the lower cut-off frequency, f L. Choose X CE = hie / 1+hfe at f L for C E calculation to give the smallest value for the bypass capacitor. Selection of coupling capacitors, C 1 and C 2 The coupling capacitors C 1 and C 2 should have a negligible effect on the frequency response of the circuit. To minimize the effects of C 1 and C 2, the reactance of each coupling capacitor is selected to be approximately equal to one-tenth of the impedance in series with it at the lowest operating frequency of the circuit (f L ). X C1 = Zi + rs / 10 at f L X C3 = Zo + R L / 10 at f L Usually, R L >> Zo and often Zi >> rs, so that Zo and rs can be omitted in the above equations. Design problem (i) Design a single stage CE transistor amplifier using BC107 transistor with Vcc = 15V, V CEQ = 5V, V E = 3V, R L = 47K and f L = 100Hz. (ii) Determine Zi, Zo, Av, Ai and Ap for the CE circuit designed in problem (i). Procedure Given Vcc = 15V, V CE = 5V, V E = 3V, R L = 47k and fl = 100Hz. The data sheet of BC107 transistor shows: hie = 3k and hfe Selection of R C

8 R C << R L so that R L will have little effect on the circuit voltage gain. Select R C = R L / 10 = 47k/10 = 4.7k. (Standard value) Selection of R E R E = V E /I E V E /I C Where IC = V RC /R C = V CC V CE V E /R C = (15-5-3) V / 4.7k = 1.4mA R E = 3V/1.4mA = 2.14k (use a standard 2.2 k ) Selection of R 1 and R 2 Selection of voltage divider current I 2 as I C /10 gives good bias stability and reasonably high input resistance Selecting R 2 = 10 R E gives I 2 = Ic / 10 Ie., R 2 = 10x2-2k = 22k (standard value) and I 2 = Ic/10 = 1.4mA/10 = 140µA R 1 = Vcc-V B / I2 = 15-(V BE +V E / 140µA = 15-(0.7+5) / 140µA = 66.43k (use standard 68k ) Selection of c 1 and c 2 The coupling capacitors C1 and C2 should have negligible effect on the frequency response of the circuit. So, the reactance of each coupling capacitor is selected to be approximately equal to 1/10 th of the impedance in series with it at the lowest operating frequency for the circuit. X C1 Zi / 10 = R 1 R 2 hie /10 = 68k 22k 3k /10 = C 1 = 1/2πf L x C1 Xc2 RL/10 = 47k/10 = 4.7k C2 = 1/2πfLXC2 = 1/2πx100x4.7k = 0.34μF (use a standard 0.03μf) Selection of C E X CE = hie / 1+hfe = 3k / = C E = 1/2πfLXC E = μf (use a standard 100μf) Calculation of Zi, Zo, Av, Ai and Ap Input impedance, Zi = R 1 R 2 hie = 68k 22k 3K = Output impedance, Zo = R C = 4.7k

9 Voltage gain, Av = -hfe / hie (Rc R L ) = / 3K (4.7k 47k) = Current gain, A I = hfe Rc R B / Rc + R L ) (Rc+hie) = 190x4.7k x (68k 22k) (4.7k+47k) (4.7k + 3k) Power gain, Ap = Av x A I = Trouble shooting a CE amplifier When you are faced with having to trouble shoot a circuit, the first thing you need is a schematic with the proper dc and signal voltages labeled. You must know what the correct voltages in the circuit should be before you can identify an incorrect voltage. Instability After the circuit has been constructed, the power supply should be set to the appropriate voltage and then connected and switched on. An oscilloscope should be connected to monitor the output of the amplifier to check that the circuit is not oscillating. If the circuit is oscilloscope the oscillations must be stopped before proceeding further. Amplifier instability can be the result of incorrect design or poor circuit layout. It can also be caused by feedback along the conductors from the power supply to the circuit. To stabilize an unstable amplifier, commence by connecting a 0-01μf decoupling capacitor from the +ve supply line to ground. Where a plus minus supply is used, connect capacitor from each supply line to ground. If the circuit is still unstable, small shunt capacitors should be connected from the transistor collect terminal to ground, or between collector and base

10 Fig CE amplifier circuit with decoupling capacitors and small shunt capacitors to eliminate oscillations in an unstable DC Voltage measurements Once it is established that the circuit is stable the next step is to measure the dc voltage levels at all transistor terminals. A digital multimeter (DMM) should be used IF the dc vulgates are not satisfactory; they must be corrected before proceeding further. The measured dc voltage levels should atleast show that (i) The EB junction of the transistor should be forward biased. The VBE may vary from 0.65 to 0.75V (for silicon transistor). (ii) The V CB may range from approximately half the supply voltage to almost the full battery voltage. Inferences from DC voltage measurements DC voltage readings are used to draw inferences of proper or improper junctioning in transistor circuits. To shown this, we will assume certain abnormal voltages in the circuit of Fig and analyze the possible causes of these voltages. 1. Vc = Vcc = +15V. Possible troubles could be (a) an open in the emitter circuit, (b) an open in the base circuit, (c) base-emitter short circuit, (d) base short-circuited to ground. 2. Vc = or. Possible troubles include (a) open collector circuit, and (b) collector short-circuited to ground. 3. V E = V C collector to emitter short-circuited

11 4. V E =OV. Possible troubles include (a) there is no current following in the emitter, or (b) the emitter is short-circuited to ground. Dynamic test When satisfactory dc levels are established throughout the circuit, dynamic test may proceed. Suppose that there is no signal at the o/p terminals of the amplifier for a specified signal input. The dynamic signal-tracing method can be used to determine what is wrong with the circuit. The procedure is as follows. A sine-wave signal no larger than the amplifier can handle is injected into the input terminals of the amplifier and observed at these terminals with an oscilloscope. If the observed signal is normal, the oscilloscope probe is moved to point B (base) of the amplifier. (Refer Fig.3-1-7) The sine-wave signal at this point should be approximately the sameas at the input terminals if the amplifier input is normal. If there is no signal at the base, two possible reasons exist. The first is that the capacitor C1 is open. The second is that the base terminal is short circuited to ground. An open capacitor may readily be found by connecting a 0-1μf capacitor across C1 and observing with oscilloscope the output signal. If an O/P signal appears, this indicates that capacitor is open. If no signal appears at o/p terminals, the oscilloscope probe is connected to the base. No signal indicates a short circuit in the base circuit. We may also determine if capacitor C2 is open by signal tracing. Assume that the input circuit, including C1, is found to be operating correctly, but that there is no signal at the o/p terminals of the amplifier. The oscilloscope probe is then connected directly to the collector of transistor. If normal signal appears at the collector, but non exists at the o/p terminals, we know that C2 is open. Resistance measurements Resistance measurements in transistor circuits, always made with power turned off, are helpful in determining defective components. The resistance measured at the base B to G should be 22k. The resistance measured at the emitter E to G should be 2-2k. The resistance from collector C to G would be the sum of Rc, R 1 and R 2 ; in this case 94.7k. These value, then, are the standard or norm for the circuit of Fig

12 The most obvious resistor defects are opens, which can be spotted very easily. For example, if the resistance measured from E to G is infinite ( ), then either R E is open, or the connective wiring is open. So, the resistance checks and continuity checks will reveal where the defect is Experiment 1. DC voltage measurements 1.1 Assemble the dc equivalent of the CE amplifier circuit you have designed, as shown in Fig Use the BC 107 transistor. 1.2 Measure Q point and other transistor terminal voltages and currents as per the procedure given in experiment tabulate the readings in Table Compare the measure values with the calculated values. 2 Transient voltage measurements 2.1 Feed 100mV (peak-topeak) sinusoidal signal at 1KHz frequency as the input signal Vs to the CE circuit shown in Fig Observe the input and output voltages simultaneously on a CRO. Note down the amplitude, frequency and phase difference between the two voltages in the table Compute the gain of the amplifier circuit and compare it with the calculated value. 2.3 Plot on a linear graph the transient voltage of the i/p & o/p of the amplifier circuit 3. Frequency response curve measurements 3.1 In the above assembled circuit, keep the magnitude of the source same, ie., 100mv and decrease the frequency from 1KHz and measure voltage gain of the amplifier at each frequency. Now increase the frequency from 1 KHz to 1MHz and measure the voltage gain of the amplifier at each frequency. Take atleast 5 readings on either side of the 1 KHz frequency. Tabulate the reading in Table Plot on a semi log graph sheet the frequency response (voltage gain Vs frequency) curve using the above measure ments. 3.3 From the plob, determine the values of (a) Mid band voltage gain, Av(mid), (b) Lower cut-off frequency,(c) upper cut-off frequency and (d) Bandwidth. 4. Resistance measurements 4.1 In the above assembled circuit, keep the magnitude and frequency of the source same, ie., 100mv 1KHz frequency.

13 4.2 Connect a potentiometer Rin (variable resistance) in series with the circuit input terminal and the signal source, as shown in Fig Connect a two-channel CRO to simultaneously monitor the i/p and o/p signal voltage waveforms. 4.4 Adjust the POT until a new o/p signal Vo, equal to one-half the original measured value of Vo is obtained. Now, remove Fig Experimental CE circuit to measure input and output impedances Rin from the circuit and measure its resistance using Dmn The measured value in ohms equals the input impedance, Zi 4.5 To measure the o/p impedance Zo of the amplifier, connect a potentiometer Rout to the o/p circuit. 4.6 Adjust the POT until a new o/p signal Vo, equal to one-hal the original measured value of Vo is obtained. Now, remove Rout from the circuit and measure its resistance using Dmn The measured value in ohms equals the o.p impedance, Zo. Tabulate the readings in Table Post Lab questions 1) Check your understanding by answering these questions. (a) The voltage gain of the amplifier in Fg is 50. when C E is opened the gain of the amplifier should (increase, decrease, remain the same) (b) The circuit of Fig must amplify sine-wave signals in the frequency range 20 to 20,000 Hz. The design value of R E is The highest value X C of C E which will act a good bypass for this amplifier is at Hz. The value of C E is μf. (c) In an audio amplifier the collector-to-base must be (forward, reverse) biased. (d) The ac signal voltage measured at the base of a C E amplifier is 50mr. The o/p signal voltage measured at the collector is 2.5v. The voltage gain of the amplifier is (e) The rms voltage measured at the collector of the amplifier in Fig is 4.6v without load. When a 250 load is connected across the o/p, the rms voltage measured at the collector load is 2.3v. The o/p impedance of the circuit is. (f) A CE amplifier has a gain of 50, an i/p impedance of 1000, and an o/p impedance of 200. The power gain of this amplifier is. (g) The decibel power gain of the amplifier in question 1 (f) is d B.

14 (h) A sine wave injected into the base of the transistor Fig results in a normal o/p. When the generator lead are moved to the i/p, no signal appears in the o/p. The most probable cause of trouble is a. (i) In the circuit shown in Fig , a technician measures 10v at the collector. Vcc = 12v. The collector current is. (j) If Rc in Fig is open, a resistance check from C to G will measure. (k) If the emitter is short circuited to base in Fig , a resistance check from the base to ground will show approximately. (l) For the conditions in question (k), the voltage at the collector will measure v (m) If R 1 in Fig is open, the voltage measured from E to G will be v (n) How do coupling capacitors C 1 and C 2 affect the frequency response? Why? 2) What is the effect on the amplifier performance of omitting R E? 3) What is the effect on i/p impedance of removing bypass capacitor C E? 4) (a) What is the phase relationship between the input and o/p signals of a C E amplifier? (b) Was this relationship confirmed by the results of your experiments? Explain how. 5) Is the o/p impedance of a CE amplifier a fixed quantity? Confirm your answer by referring specifically to any substance tiating data in this experiment. 6) Classify amplifiers 7) List the methods for coupling amplifiers in cascade.

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