... Second Semester

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Florence Owens
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1 كنا نظنك اي عيل تركتنا... لكن طيفك مل يزل يغشانا مثل الندى يسقي النبات بقطره... قد ابت ليال يف ادلىج عطشاان ما زلت أذكر حني أدخل معمال... متسك هجازا للقياس تحاكه... فأ راك جتلس هادئا حيرانا تكتب وترمس ما تراه عيانا قد كنت للطالب خري معمل... وغرست حبا يسكن الوجداان Second Semester

2 BJT Amplifiers Sec. Sem May

3 Amplifier Operation The biasing of a transistor is purely a dc operation to establish a Q-point about which variations in current and voltage can occur in response to an ac input signal. 2 May

4 AC Quantities ac and dc quantities are usually represented by capital letters with a change in the subscripts dc quantities: capital non-italic subscripts like I B, I C, I E, V C, V E, V CE V ac quantities: small rms avg V ce italic subscripts, for V CE V ce V ce V ce example, I c, I e, I b,v c, v ce and V ce 0 0 t 2 May

5 AC Quantities ac quantities may be represented in rms, average, peak, peak to peak. ac instantaneous quantities are represented by small letters with Lowercase small italic subscripts like i c, i e, i b, v c, and v ce Resistance is also identified with a small letter of small subscript when analyzed from an ac standpoint 2 May

6 Linear Amplifier A linear amplifier provides amplification of signal without any distortion exact amplified replica of the input signal. output signal is an ac source of internal resistance R s coupled to the base through C 1 load resistance R L the coupled to collector through C 2 The coupling capacitors block dc and thus prevent R S and R L from changing the dc bias voltage at the base and collector. For the amplifier shown, notice that the voltage waveform is inverted between the input V b and output V ce but has the same shape. 2 May

7 AC Load Line Operation of the linear amplifier can be illustrated using an ac load line as shown. The ac load line is different than the dc load line because a capacitor looks open to dc but effectively acts as a short to ac load resistor R L. the collector resistor R C appears to be in parallel with the 2 May

8 AC Load Line- Example - Determine the resulting peak-to-peak values of collector current and collector-to-emitter voltage from the graph. - What are the dc Q-point values Solution: collector current varying from 4 ma to 6 ma collector current has peak to-peak value of 2 ma I c = 2 ma V ce = 1 V I BQ = 50 µa I CQ = 5mA V CEQ = 1.5V 2 May

9 Transistor AC Model A transistor in an amplifier circuit can be represent by a model circuit. The transistor model circuit based on various internal transistor resistance parameters r that can represent its operation. Five resistance parameters (r-parameters) can be used for detailed analysis of a BJT circuit. For most analysis work, the simplified r-parameters give good results. 2 May

10 Transistor AC Model 2 May

11 The Common-Emitter Amplifier In the common-emitter (CE) amplifier, the input signal is applied to the base and the inverted output is taken from the collector. The emitter or ground is common to ac signals (V in and V out ) as shown. CE amplifiers has high voltage gain and high current gain. A common-emitter amplifier with voltage-divider bias and coupling capacitors C 1 and C 3 on the input and output and a bypass capacitor, C 2, from emitter to ground. Output voltage has a 180 phase difference from input voltage. 2 May

12 The Common-Emitter Amplifier DC analysis Considering CE amplifier circuit, dc analysis can be done by removing the coupling and bypass capacitors. Capacitors appear open with dc connected only we will have the voltage divider bias circuit shown. Using Thevenin equivalent for bias circuit 2 May

The Common-Emitter Amplifier AC analysis Ac equivalent circuit can be developed by considering: 1- The capacitors are replaced by effective shorts because their values are selected so that the

13 The Common-Emitter Amplifier AC analysis Ac equivalent circuit can be developed by considering: 1- The capacitors are replaced by effective shorts because their values are selected so that the capacitive resistance X C is negligible at the signal frequency and can be considered to be 0 Ω. * Note that C 2 must be large enough so that X C2 is very small compared to R E (10X C R E ) At given frequency. X C can be calculated using the relation 2- The dc source is replaced by ground. No ac voltage can be developed across it so it appears as an ac short. This is why a dc source is called an ac ground. The ac equivalent circuit for CE amplifier is 2 May

14 The Common-Emitter Amplifier AC analysis: input and output resistances The total input resistance or voltage divider bias circuit for input ac voltage is 2 May

15 The Common-Emitter Amplifier AC analysis: Voltage gain ac equivalent circuit for the bias circuit capacitively coupled with R L is With no load, the voltage gain for ac voltage is 2 May

16 The Common-Emitter Amplifier attenuation = V b V s = R in(tot) R in(tot) + R s The overall voltage gain of the amplifier V b / V s 2 May

17 The Common-Emitter Amplifier AC analysis: Example: for the amplifier shown below, calculate: (a) the signal voltage at the base (V b ). (b) the minimum value for the emitter bypass capacitor, C 2, if the amplifier must operate over a frequency range from 200 Hz to 10 khz. (c) Calculate the base-to-collector voltage gain of the amplifier (without and with C 2 ) if there is no load resistor. (d) If a load resistor of 5kΩ is added at V out, calculate the voltage gain (include C 2 ). (e) the overall voltage gain if C 2 and R L are included. (f) the signal voltage at the collector (V c ). 2 May

18 The Common-Emitter Amplifier 2 May

19 The Common-Emitter Amplifier 2 May

20 The Common-Emitter Amplifier AC analysis: Addition of Swamping Resistor. voltage gain is essentially depends on specially when R E is bypassed by C 2. Since changes with temperature the voltage gain becomes unstable. On the other hand, removing C 2 cause the gain to go to its lowest value. Hence we can add a swamping resistor (R E1 ) to reduce the effect of Greater gain stability can be achieved and the gain will be lower as a result. The voltage gain for the circuit shown becomes: 2 May

21 The Common-Emitter Amplifier 2 May

22 The Common-Emitter Amplifier Multisim is a good way to check your calculation. For an input of 10 mv pp, the output is 378 mv pp as shown on the oscilloscope display for the swamped CE amplifier. input output 2 May

23 The Common-collector Amplifier The common-collector amplifier (emitter-follower) has a voltage gain of approximately 1, but can have high input resistance and current gain. The input is applied to the base and taken from the emitter. +V CC C 1 R 1 V in I in C 2 V out R 2 R E R L 2 May

24 The Common-collector Amplifier The power gain is the ratio of the power delivered to the input resistance divided by the power dissipated in the load. This is approximately equal to the current gain. That is, A p A i You can also write power gain as a ratio of resistances: A p P V 2 L L L = = = 2 Pin Vin R R R 1 = RL in( tot) R in( tot) in( tot) R L A 2 v R in( tot) R L V in C 1 R 1 V CC C 2 R 2 R E V out R L 2 May

25 The Common-collector Amplifier 2 May

26 The Common-collector Amplifier By thevenizing from the base back to the source, the circuit is simplified to the form shown in Figure (b). V out = V e, I out = I e, and I in = I b 2 May

27 The Common-collector Amplifier 2 May

28 The Common-collector Amplifier: Example 2 May

29 The Common-collector Amplifier: Example Determine the total input resistance of the emitter-follower in Figure below. Also find the voltage gain, current gain, and power gain in terms of power delivered to the load, R L. Assume and that the capacitive reactances are negligible at the frequency of operation. β ac = May

30 The Common-collector Amplifier: Example 2 May

31 The Common-collector Amplifier: Example Calculate the power gain to the load for the CC amplifier using a ratio of resistances. Assume A v = 1 and β ac = 200. Use r e ' = 2Ω. V CC +15 V R in(tot) = R 1 R 2 b ac (r e ' + R E R L ) = 39 kω 220 kω 200(2 Ω+ 500Ω) = 24.9 kω R L = 1.0 kω A p Rin( tot ) 24.9 kw = = = R 1.0 kw L V in 24.9 R 1 C 1 39 kw 0.22 mf R kw C 2 V out R 3.3 mf E R L 1.0 kw 1.0 kw 2 May

32 The Common-collector Amplifier: Example The input voltage-divider in the previous example is not rock-solid but the overall power gain is good. A rock solid stiff voltage-divider is not always the best design. Can you spot the problem illustrated here? R in(tot) = R 1 R 2 β ac (r e ' + R E R L ) = 10 kω 10 kω 200(25 Ω+ 3.0 kω) = 4.96 kω R L = 10 kω V in R 1 C 1 10 kw V CC +10 V b = 200 C 2 V out A p Rin( tot ) 4.96 kw = = = R 10 kw L 0.496! The problem is the power gain is less than 1! R 2 10 kw R E 4.3 kw R L 10 kw 2 May

33 Darlington pair The Darlington Pair: A Darlington pair is two transistors connected as shown (with common collector). The Darlington pair highly increase the input resistance better circuit (better Buffer) specially when using low-load resistance. The two transistors act as one super β transistor. Darlington transistors are available in a single package. With Darlington pair we have better buffering (barrier between input and output) in addition to high current gain 2 May

34 The Common-collector Amplifier: Example 2 May

35 The Sziklai Pair Another high β pair is the Sziklai pair (sometimes called a complementary Darlington), in which a pnp and npn transistor are connected as shown. This configuration has the advantage of only one diode drop between base and emitter. +V CC What is the relation between I E2 and I B1? V in β DC1 The DC currents are: I C1 is b DC1 x I B1 and is equal to I B2 I B1 I C1 βdc2 I E2 I E2 is approximately equal to b DC2 x I C1 R E Therefore, I E2 b DC1 b DC2 I B1 2 May

36 The Common-Base Amplifier The common-base (CB) amplifier provides high voltage gain with a maximum current gain of 1. Since it has a low input resistance. A typical common-base amplifier is shown in figure below. The base is the common terminal and is at ac ground because of capacitor C 2. The input signal is capacitively coupled to the emitter. The output is capacitively coupled from the collector to a load. 2 May

37 The Common-Base Amplifier 2 May

38 Multistage amplifiers Two or more amplifiers can be connected in a cascaded arrangement with the output of one amplifier driving the input of the next. The basic purpose of a multistage arrangement is to increase the overall voltage gain. The overall voltage gain, A ν,of cascaded amplifiers, as shown in the Figure, is the product of the individual voltage gains. 2 May

39 Multistage amplifiers: Example A certain cascaded amplifier arrangement has the following voltage gains: A v1 = 10, A v2 = 15, and A v3 = 20 What is the overall voltage gain? Also express each gain in decibels (db) and determine the total voltage gain in db. 2 May

40 Multistage amplifiers: Capacitively coupled Two-stage capacitively coupled amplifier in Figure below. Notice that both stages are identical common-emitter amplifiers with the output of the first stage capacitively coupled to the input of the second stage. Capacitive coupling prevents the dc bias of one stage from affecting that of the other but allows the ac signal to pass without attenuation because at the frequency of operation. 2 May

41 Multistage amplifiers: Capacitively coupled 2 May

42 Differential Amplifiers A differential amplifier (diff-amp) has two inputs. It amplifies the difference in the two input voltages. This circuit is widely used as the input stage to operational amplifiers. Differential-mode inputs are illustrated. +V CC V out1 V out2 R C1 R C Q 1 Q 2 2 V in1 V in2 R E V EE 2 May

43 Differential Amplifiers The same amplifier as in the last slide now is shown with common-mode inputs. Diff-amps tend to reject common-mode signals, which are usually due to noise. Ideally, the outputs are zero with common-mode inputs. +V CC V out1 V out2 R C1 R C2 1 2 V in1 1 Q 1 Q 2 2 V in2 R E V EE 2 May

44 Selected Key Terms r-parameter Commonemitter ac ground One of a set of BJT characteristic parameters that include α ac, β ac, r e ', r b ', and r c '. A BJT configuration in which the emitter is the common terminal to an ac signal. A point in a circuit that appears as a ground to ac signals only. Input resistance The resistance seen by an ac source connected to the amplifier input. 2 May

45 Output resistance Differential amplifier Selected Key Terms The ac resistance looking in at the amplifier output. A BJT configuration in which the emitter is the common terminal to an ac signal. An amplifier in which the output is a function of the difference between two input voltages. Commoncollector Commonmode A condition where two signals applied to differential inputs are of the same phase, frequency and amplitude. 2 May

46 Electric Machines Electric Machines

Electronic Devices. Floyd. Chapter 6. Ninth Edition. Electronic Devices, 9th edition Thomas L. Floyd

Electronic Devices. Floyd. Chapter 6. Ninth Edition. Electronic Devices, 9th edition Thomas L. Floyd Electronic Devices Ninth Edition Floyd Chapter 6 Agenda BJT AC Analysis Linear Amplifier AC Load Line Transistor AC Model Common Emitter Amplifier Common Collector Amplifier Common Base Amplifier Special