UNIT II SMALL SIGNAL AMPLIFIER
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1 UNIT II SMALL SIGNAL AMPLIFIER HYBRID MODEL OF A CIRCUIT: A linear ckt have input and output terminal which can be analyzed by 4- parameters (ie.2 -dimensionless and 2- dimension parameters), then it is called a hybrid model. Advantages of Hybrid parameters: They are real numbers at audio frequencies. They are easy to measure They can be determined from the transistor static characteristics curves. They are convenient to use in circuit analysis and design. Electronic Devices and Circuits 15
2 Hybrid Equation: Here h 11, h 12, h 21, h 22 are called the hybrid parameters or the h-parameters V 1 = h 11 i 1 +h 12 v 2 i 2 = h 21 i 1 + h 22 v 2 By the above equation and fig 2 V 2 =0 V 1 = h 11 i 1 h 11 = v 1 / i 1 (input impedance) By the above equation and fig 2 V 2 = 0 i 1 = h 12 i 1 h 21 = i 2 /i 1 (forward current gain) By the above hybrid equation and fig 3 i1 = 0 v 1 = h 11 i 1 + h 12 v 2 v 1 = 0 + h 12 v 2 h 12 = v 1 /v 2 (reverse voltage gain) By the above hybrid equation and fig 3 i 1 = 0 i 2 = h 21 i 1 + h 22 v 2 i 2 = 0+h 22 v 2 h 22 = i 2 / v 2 (output admittance) Electronic Devices and Circuits 16
3 h Input Impedance h output Admittance h Reverse Voltage gain h Forward current gain ut impedance It has moderate value of current gain &voltage gain Hybrid equivalent for CB e It has moderate value of high current gain & low voltage gain Hybrid equivalent for CC Electronic Devices and Circuits 17
4 It is mainly used for impedance matching. It has high input & low output impedance. It act has an emitter follower circuit. CALCULATING H-PARAMETERS FOR TRANSISTOR: 1. Current Gain - Io AI= Ii Applying KCL in the output side of the ckt. Io = hfii + I = hfii + Vo ( 1/ho) = hfii + Voho Io+IoRLho = hfii = hfii - IoRLho (Vo = -IoRL) Io (1+RLho) = hfii Io/Ii = hf (1+RLho) AI = - hf (1+RLho) 2.Voltage Gain Av = Vo/Vi Apply KVL in the input side of ckt Vi = hiii + hrvo = -hi {(Io+IoRLho)/hf } +hrvo Electronic Devices and Circuits 18
5 (Io+IoRLho)/hf ] = -hi { Io(1+RLho)/hf } +hrvo [ sub Ii = = -Iohi {(1+RLho)/hf } +hrvo (Vo = -IoRL) = -Io hi (1+RLho) /hf + hr Vo = -(Vo/RL) hi (1+RLho) /hf + hr Vo = Vo(hr + hi (1+RLho) /hfrl) [ sub = Vo(hrhfRL - hi(1+rlho))/hffrl Vi/Vo = (hrhfrl - hi(1+rlho))/hfrl Av = hfrl / (hrhfrl + hi (1+RLho)) 3. Input Impedance Zi = Vi / Ii Apply KVL in the input side of the ckt Vi = hiii + hrvo = hiii +n hr(-iorl) =hiii - IohrRL = hiii - (Iihf/(1+RLho))hrRL Io = (Iihf/(1+RLho)) ] = Ii(hi - (hf/(1+rlho))hrrl [ sub = Ii(hi - (hf /1+RLho))hrRL = Ii(hi - hrrlhf/(1+rlho)) Vi/Ii = hi - hrrlhf/(1+rlho) Zi = hi - hrrlhf/(1+rlho) Electronic Devices and Circuits 19
6 AI Av - hf / (1+RLho) hfrl / (hrhfrl + hi (1+RLho)) ut Impedance 4. O u t p Considering Rs resistance Zo = Vo/Io Apply KCL Io = hfii + hovo [ sub Ii = -hrvo/(rs+hi) ] Io = -hrvohf/(rs+hi) + hovo = Vo( ho- (hrhf)/(rs+hi)) = Vo( ho(rs+hi) - hrhf)/ (Rs+hi) Vo/Io = (Rs+hi)/ ( ho(rs+hi) - hrhf) Zo = (Rs+hi)/ ( ho(rs+hi) - hrhf) 5. Overall Voltage Gain Avs = (Av.Zi)/(Zi+Rs) 6. Overall Current Gain AIS = (AI*Rs)/(ZI+Rs) CONCLUSION Electronic Devices and Circuits 20
7 Zo Avs Zi hi - hrrlhf/(1+rlho) (Rs+hi)/ ( ho(rs+hi) - hrhf) (Av.Zi)/(Zi+Rs) AIS (Ai.Rs)/(Zi+Rs) Approximate conversion formulas for hybrid parameters CC hic= hie CB hib = hie /(1+hfe) hfc= - hfb = - hfe / (1+hfe) (1+hfe) hrc= 1 hrb = ((hie.hoe) / (1+hfe))- hre hoc= hoe hob = hoe/(!+hfe) Amplifier :- 1. Depending upon the type of input signal 2. Depending upon the number of stages 3. Depending upon the type of configuration whether: Depending upon the type of configuration Common Emitter configuration (CE) Common Base configuration (CB) Common Collector configuration (CC) POWER AMPLIFIER Electronic Devices and Circuits 21
8 Definition: A power amplifier is an amplifier which is capable to providing a large amount of power of the load such as loudspeaker or motor etc. *Power amplifier is a large signal amplifier *In power amplifier input signal is large (greater than 1 volt) *It will concerts DC power supply to the AC power supply. *It will generally used in the last stage of our circuit. (ie. Transmitter, receiver) *It will also used as a impedance matching circuit. Difference between voltage amplifier and power amplifier:- Voltage amplifier Power amplifier * signal level is low (mv) *signal level is high(v) *current gain is large * current gain is less (ie greater then 100) *Collector size is less *collector size is large *Emitter and base is lightly *both emitter and base are heavily doped doped *Input impedance is low than *input impedance is higher than output output impedance impedance. *Here Rc coupling for interstage *here transformer coupling is used Connection *it is called as the low or medium * it is called as the power transistor power transistor Here to determine the performance of power amplifier the following parameters are used namely circuit efficiency, distortion, and power dissipation. Classification of the power amplifier Power amplifier is classified on the basis of the transistor biasing condition and amplitude of the input signal 1. Class A (current conductor is for 360 ) it is an amplifier in which the transistor bias and amplitude of the input signal in such that the output current flows for the complete cycle (i.e., 360)of the input signal Electronic Devices and Circuits 22
9 2. Class B (current conductor is for 180 ) it is an amplifier in which the transistor bias and amplitude of the input signal in such that the output current flows for only one half cycle (i.e., 180)of the input signal 3. Class C (current conductor is for less than 180 ) it is an amplifier in which the transistor bias and amplitude of the input signal in such that the output current flows for les sthan half cycle (i.e., less than 180 )of the input signal CLASS A AMPLIFIER Class a amplifier is a common emitter amplifier as shown in the below figure The below circuits direct couple class-a power amplifier. Electronic Devices and Circuits 23
10 The only difference between the transformer coupled power amplifier and the class-a power amplifier is a input given range of volt. It is an amplifier in which the transistor bias and amplitude of the input signal in such that the output current flows for the complete cycle (i.e., 360)of the input signal The above condition achieved by locating the Q-point same where near the center of the load lines. In order obtain the max output signal the Q- point is said at the center of the load line. Electronic Devices and Circuits 24
11 In power amplifier we have to calculate 1.collector efficiency( c) 2.overall efficiency ( 0 ) c = (AC power delivered to the load)/(dc input power supplied to the transistors) 0 = (AC power delivered to thed load)/(dc input power) overall efficiency 0 = po(a.c)/pin(d.c) po(a.c) =- I^2 R L = (I Q /2)^2 R L po (a.c) = (Vp.Ip)/2 pin(d.c) = V CC I CQ 0 = (V P.I P )/(2(V CC.I CQ )) 0 = I CQ.V CEQ /(2(2*V CEQ.I CQ ) Electronic Devices and Circuits 25
12 =1/4 = 0.25 po (a.c) = (Vp.Ip)/2 0 = 25% pin(d.c) = V CEQ. I CQ collector efficiency C = (V P.I P )/(2(V CEQ.I CQ )) C = I CQ.V CEQ /(2V CEQ.I CQ ) = ½ =.50 C = 50% TRANSFORMER COUPLED AMPLIFIER:- The maximum value of overall or collector efficiency of a transformer coupled class A amplifier is 50% It is used to increase the efficiency It is used in order to match the impedance of load to the amplifier 0 = po(a.c)/pin(d.c) Electronic Devices and Circuits 26
13 = V P.I P /(2.V CC.I CQ ) = V CEQ.I CQ /(2.V CEQ.I CQ ) 0 = 50% CLASS B AMPLIFIER:- It is an amplifier in which the transistor bias and amplitude of the input signal in such that the output current flows for only one half cycle (i.e., 180)of the input signal in Class B Amplifier Q- point is operated in the cut off region since the Q- point is operated in the cut off region,so that the negative half cycle cannot be obtained only the positive half cycle is obtained. Efficiency of class B Po(a.c.) 0 = Pin (d.c) V P.I P P 0 = V CC.I P = P IN (d.c) = V CC.I dc P IN (d.c) = V CC.(I P / ) V CC.I P. 0 = V CC.I P = = 78.5% Electronic Devices and Circuits 27
14 CLASS B PUSH PULL AMPLIFIER:- As we seen in the Class B power amplifier it is operated in the cut off region,so that the negative half cycle cannot be obtained only the positive half cycle is obtained In order to avoid this contain we are using two transistors connected in the push pull amplifier in this arrangement one of the transistors conducts during one half cycle and the other conducts during the second half cycle. In transformer coupled class _B push pull amplifier It has two center tapped transformer t 1 and t 2 and two identical transistors q1,q2. Here the transformer t1 is called the input transformer or it is called the phase spliter. It is required to produce two signal voltage,which are 180 out of phase with each other Here the transformer is called the output transformer. During the positive half cycle Q2 transistor conducting and the Q1 transistor is in OFF During the negative half cycle Q1 transistor conducting and the Q2 transistor is in OFF Advantages of Class-B push pull Amplifier:- 1.It has high efficiency(i.e 78.5%) Electronic Devices and Circuits 28
15 2.use of push pull amplifier will eliminates even order harmonics in the a.c output signal. Disadvantage:- 1.It is bulky and expensive. 2.It is difficult for tapping. EFFICIENCY :- Po(a.c.) 0 = Pin (d.c) V P.I P P 0 = V CC.I P = P IN (d.c) = V CC.I dc P IN (d.c) = V CC.(2I P / ) V CC.I P. 0 = V CC.I P = = 78.5% TRANSFORMERLESS CLASS B PUSH PULL AMPLIFIER:- It is used used in order to avoid the disadvantages of of the Class-B push pull amplifier In this figure the input center tapped transformer is removed by an input driver called the phase splitter or phase inverter. It consists of an NPN transistor with equal collector and emitter resistances Since the load is directly connected to the amplifier there will not be any problem of impedance mismatch between the output impedance of the circuit and the load resistance. Electronic Devices and Circuits 29
16 COMPLEMENTARY SYMMETRY CLASS B PUSH PULL AMPLIFIER:- In this amplifier the term complementary means that the circuit uses two identical transistors i.e. one NPN and the is PNP. Electronic Devices and Circuits 30
17 Both these transistors are have identical input and output characteristics In this amplifier the term symmetry means that the biasing resistances are equal. Here r1&r2 is act as the biasing resistors During the positive half cycle Q1 transistor will conduct and Q2 will not conduct. During the negative half cycle Q2 transistor will conduct and Q1 will not conduct because it is PNP transistor In this method we do not use the phase spliter FET AMPLIFIER In fet amplifier: - * It has high input impedance * Fet amplifier is easy to fabricate * It has less noisy Difference between BJT and FET BJT:- 1. Both majority &minority carriers are present. 2. It act has a current controlling device 3. Ic Ib FET:- 1. It has only majority charge carriers. 2. It act has a voltage controlling device. 3. Io VGS SMALL SIGNAL MODEL OF FET Electronic Devices and Circuits 31
18 Common Source Amplifier (CS) 2. Ac equivalent ckt:- Input is gate terminal Output is drain termial Source is earth VDD = o; Electronic Devices and Circuits 32
19 3.Replace the fet by small signal model Av = Vgs Vo Vo = - id* rd gm Vgs *rd Id = Rd+rd -gm Vgs rdrd vo= (rd+rd) Av = Av= gm rd *Rd *Vgs (rd+rd)*vgs -gm rd *Rd (rd+rd) zo ( without source resistance) = rd zo (with source resistance) = Rd!! rd Common Source With Unbypassed Source Resistor:- Electronic Devices and Circuits 33
20 1.Ac equivalent ckt:- 2.Replace the FET by its small signal model From fig:- In loop 2 Electronic Devices and Circuits 34
21 In loop 1 idrd +(id - m Vgs)rd+id Rs = 0 Vgs = Vin idrs Thevenins equivalent circuits for above small signal model vo = - id *rd id = vi rd+rd+( +1)Rs - vi vo = *Rd rd+rd+( +1)Rs Av = vo Vi Av = Rd rd+rd+( +1)Rs Common Drain Amplifier (Source Follower) Electronic Devices and Circuits 35
22 1.Ac equivalent ckt:- 2.Replace the FET by its small signal model From fig 2:- Loop 2:- Loop 1 :- Idrd + id Rs +(id gm vgs)rd = 0 Vgs = vi id Rs Electronic Devices and Circuits 36
23 Thevenins equivalent circuits for above small signal model id = vi id = rd+rd +1 vi rd+rd+( +1)Rs Rs vi vo = *Rs rd+rd+( +1)Rs Av = Av = Av = vo Vi vi Rs Vi(rd+Rd+( +1)Rs) Rs rd+rd+( +1)Rs Electronic Devices and Circuits 37
24 general Av =1 If Rs is large Av = if the is large Av = 1 Electronic Devices and Circuits 38
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