EEE2315 ANALOGUE ELECTRONICS IV

EEE235 ANALOGUE ELECTRONICS IV Large signal amplifiers. Analysis of class A, B, AB and C amplifiers (push-pull and complementary symmetry circuits). Calculation of power gain and efficiency, estimation of distortion. Thermal variation and device parameters. Design of power amplifiers. Noise in amplifiers: noise sources, types, signal/noise ratio, noise figure temperature. Common electronic circuits. Pulse amplifiers: charge control model, rise time, fall time compensation techniques, cascaded states, radio frequency amplifier models and use of admittance. Miller effect. Analysis of broad amplifiers for various circuits configuration. Design of R.F. tuned and pulse amplifier. REFERENCE BOOKS Electrical Technology by Theraja Principles of electronics by V. K. Mehta & R. Mehta Schaum's outline of theory and problems of electronic devices and circuits by Jimmie J. Cathey Electrical and electronic principles and technology by John Bird Electronic principles by Albert Paul Rf-power-amplifiers by Miriam K RF Power Amplifiers for Wireless Communication by Steve C. Cripps High Performance Audio Power Amplifiers by Ben Duncan Electronic Devices and Amplifier Circuits with MATLAB Applications by Steven T. Karris Feedforward Amplifiers for Wideband Communication Systems by Jon Legarda Compiled by Omae M. Oteri 20

ANALOGUE ELECTRONICS IV: DC biasing: Base bias. Base bias with emitter feedback. Base bias with collector feedback. Base bias with two power supplies. Base bias with emitter and collector feedback. Voltage divider. All these connections are using the common emitter connection. They have a single transistor. BASE BIAS: Parameters that can be obtained from all the connections above. Current Voltage Gain IB VBE Ai = B = hfc IC VCE Av IE VCB hre Transconductance (g). Base bias analysis: KCL: I 2 = IC + IB VC VB VE KVL: VCC = ICRC + VCE ; VCC = IBRB + VBE ; VC = VCE = VCC ICRC VB = VBE ; VE = 0 Ai = B = I C I B AV = V CE V BE Impedance (hie) Admittance (hce) 2 Compiled by Omae M. Oteri 20

In this connection, base current is fixed while collector current varies when the current gain varies Base bias with emitter feedback: It is called feedback because the change in emitter voltage is fed back to the base emitter. KCL: IE = IC + IB VCC = ICRC + VCE + IERE VCC = IBRB + VBE + IERE VC = VCE + IERE = VCC ICRC VB = VBE +IERE = VCC IBRB VE = IERE VCC = ICRC + VCE + I C α RE IC = V CC V CE R C + R E α IE = I C α In terms of β: IC = V CC V CE α = I C I E R C + R E(+ β) β Ai = I C AV = V CE Power gain AP = I CV CE I B V BE I B V BE Has a ve feedback since it opposes the original change in collector current. 3 Compiled by Omae M. Oteri 20

Base bias with collector feedback: KVL: (IC + IB)RC + VCE = VCC VCC = (IC + IB)RC + IBRB + VBE VC = VCE = VCC (IC + IB)RC VB = VBE = VCC - (IC + IB)RC IBRB VE = 0V Exercise: Given RC = 3KΩ, RB = 6KΩ, VCC = 30V, β = 00 for base bias with emitter feedback and base bias with collector feedback. Calculate all the possible parameters. Repeat the same for base bias with RE = 0 SOLUTIONS: VBE = 0.7 VCC = IBRB + VBE + IB(β + )RE V IB = CC V BE R B + (β+)r E = 30 0.7 500+(0 x 6000) = 0.048mA IE = V CC V BE 30 0.7 R R E + B = = 4.8mA 6000+ 5000 (+ β) 0 IC = IE IB = 4.8 0.48 = 4.752mA VCC = IBRB + VBE + IERE = 0.048 x 0-3 x 5000 + 0.7 + 4.8 x 0-3 x 6000 = 28.8V 4 Compiled by Omae M. Oteri 20

VB = VBE + IERE = 28.8 + 0.7 = 29.5V VC = VCC ICRC = 30 (4.32 x 0-3 x 3000) = 7.04V VCE = VC IERE IC + IB = 30 = 0mA 3000 3000 IB =( 3000+5000 )0mA = 3.75A IC = (0 3.75) = 6.25mA VC = VCE = VCC (IB + IC)RC = 30 (0mA X 3K) = 0 Base bias with collector and emitter feedback: VCC = (IC + IB)RC + VCE + IERE VCC = (IC + IB)RC + IBRB + VBE + IERE VB = VBE + IERE = VCC IBRB (IB + IC)RC VC = VCE + IERE = VCC (IB + IC)RC VE = IERE 5 Compiled by Omae M. Oteri 20

Base bias with two power supplies: VEE = IERE + VBE + IBRB VC = VCC ICRC Voltage divider bias: VR = ( R R )VCC VR2 = 2 VC R + R 2 R + R 2 VB = VR2 = VBE + IERE Collector emitter eqn. VCC = ICRC + VCE + IERE VC = VCE + IERE = VCC ICRC VE = IERE 6 Compiled by Omae M. Oteri 20

Thevenin s Theorem: VTh = R 2 R + R 2 VCC VTh = VR2 = R 2 R + R 2 VCC RTh = R R2 = R R 2 R + R 2 Thevenin s equivalent circuit: 7 Compiled by Omae M. Oteri 20

Analogue: VBB = IBRB + VBE + IERE VCC = ICRC + VCE + IERE Example: Given RB = 5KΩ, RB2 = 4KΩ, RE = 3KΩ, VCC = 20V Determine (i) IB, IC, IE, VCE, VC where β = 00 Solution: Approximation method: VRB2 = V CCR B2 R B + R B2 = 20 X 4 5+4 = 8.89V VB = VBE + IERE but VB = VBB2 = 8.89 8.89 = 0.7 + IE x 3R IE = 8.89 0.7 = 2.73 x 3 X 0 3 0-3 A I IB = E 2.73 X 03 = = 2.7 x 0-5A (+ β) 0 IC = IE IB = (2.73 x 0-3 ) (2.70 x 0-5 ) = 2.7 x 0-3 A VCC = ICRC + VCE + IERE 20 = (2.7 x 0-3 x 4K) + VCE + (2.73 x 0-3 x 3K) VCE = 20 0.856 8.9 = V VC = VCE + IERE = + 2.73 x 3 = 9.9V Thevenin s method: VTh = VRB2 = 8.89V RTh = R BR B2 = 5 X 4 = 2.22KΩ R B + R B2 9 I VBB = IBRB + VBE + IERE = E RB + VBE + IERE (+ β) 8 Compiled by Omae M. Oteri 20

IE = V BB V BE R R E + B (+ β) IB = = 8.89 0.7 = 2.7 x 0-3 A 3K + 2.22K (0) I E 2.7 X 03 = = 2.68 x 0-5 A (0) 0 IC = IE IB = (2.7 x 0-3 ) (2.68 x 0-5 ) = 2.67 x 0-3 A VCE = VCC ICRC IERE =.9 + (2.7 x 0-3 x 3K) =.9V SINGLE STAGE AMPLIFIER: It has one transistor, 5 resistors, 3 capacitors and voltage supply (VCC) Functions: RB and RB2 voltage dividers where RB reverse biases the collector. RB2 forward biases base emitter junction. RC and RE facilitates the collection of output signal. RE is a feedback resistor to regulate any changes of IC. C and C2 are coupling capacitors. C2 blocks any dc from appearing at the output and C blocks any dc from entering the amplifier. C3 is a decoupling capacitor to bypass any high frequency attenuating current to the ground so that is not fed back to the output. Analysis: DC load-line: Open all the capacitors: Leave all the dc sources intact and ground all ac sources. 9 Compiled by Omae M. Oteri 20

Collector emitter equation: VCC = ICRC + VCE + IERE VCC = ICRC + VCE + I C α RE IC = V CC V CE R C + R E α = = - V CC R C + R E α - V CE + R C + R E α V CE R C + R E α V CC R C + R E α This equation is a straight line in (dc load line) which gives different values of IC depending on load resistors, RC and RE. y = mx + c where m = - IC intercept: R C + R E α c = V VCE = 0 IC = CC = R C + R E IC (saturation). α VCE intercept: IC = 0 0 = - V CE V + CC R C + R E R α C + R E α VCE = VCC = VCE (cut-off) V CC R C + R E α x = VCE 0 Compiled by Omae M. Oteri 20

NB: The importance is to obtain the Q point of a transistor. Q point: Also called quiescent point, silent, operating point. Q point is where the transistor is designed to operate normally. For maximum swing of ac, it should be at the middle of the dc load line. VPP = VCC = 2VCEo ipp = IC(sat) = 2ICo Compiled by Omae M. Oteri 20

Saturation: VCE is from 0 to a few volts. Both junctions are forward bias. Transition is like an on switch. Active: VCE is from a few volts to around 20V. CB junction is reversed and BE is forward bias. Transistor operates normally. Breakdown: VCE more than 20Vc (depending on transition). CB is reversed but often breakdown both junctions would be conducting. Transition is not supposed to be operated in negative. Cut-off: IB = 0 Both junctions are reverse biased. Transistor acts like an off switch. Exercise: Given VCC = 25V, RC = 6K, RE = 3K, RB = 8K, RB2 = 0K β = 50. Determine the operating region of the above amplifier. What should be done to make it operate at mid-point. Use the dc load-line to explain the above. 2 Compiled by Omae M. Oteri 20

Q POINT: VBB = R 2 R + R 2 VCC = 2.2 0+ 2.2 x 0 =.8V VE = VBB VBE =.8 0.7 =.V The emitter current: IE = V E =. =.ma R E K 0 VR = x 0 = 8.20V 0+ 2.2 IR = V R = 8.2 = 0.82mA. R 0K IR2 = V R2 R 2 = V BB R 2 = 8 2.2K = 0.88mA. IB = IR IR2 = 0.82 0.88 = 0.002mA IC = IE IB =. 0.002 =.098mA Since IC almost equals IE, VC can be calculated as follows: VC = 0V (. X 0-3 X 3.6 X 0 3 ) = 6.04V The collector emitter voltage: VCE = VC VE = 6.04. = 4.94V NB: The calculation in this preliminary analysis does not depend on changes in the transistor, the collector current on the temperature. This is why the Q point of this current is stable, almost rock solid. The Q point of the circuit has a collector current of.ma and a collector emitter voltage of VCE = 4.94V. Plotting this point/ values we get. Since voltage divider bias is derived from emitter bias, the Q point is virtually immune to changes in current gain. One way to move the Q point is by varying the emitter resistor (RE). 3 Compiled by Omae M. Oteri 20

e.g. If RE changed to 2.2KΩ the collector current decreases to: IC = V C R E =. 2.2K = 0.5mA The voltage changes as follows: VC = 0V (0.5 X 3.6) = 8.2V VCE = VC VCE = 8.2. = 7.2V Therefore the new Q point will be QL and will have co-ordinates of 0.5mA and 7.V. NB: When RE is increased, the Q point is lowered and decreasing RE heightens the Q point. Q point in the middle of a load line: VCC, R, R2 and RC control the saturation current and the cut-off voltage. A change of this will change IC(Sat) and VCE (cut-off). RE set the Q point at any position along the load line. If RE is too large, Q point moves to cutoff point. If RE is too small, Q point moves to saturation. NB Voltage divider bias has a stable point. CORRECTION: Operating Pt: ICQ = 4.33mA. VCE = -4.06V IC(sat) = V CC R C + R E α VCE (cut-off) = VCC AC LOAD LINE: The line which gives different values of IC for VCE depending on load resistance (RC and RL). It cuts the x axis at VCE (cut-off). ib small signal IB large signal ib small + large signal. VCE(cut-off) = VCEQ + ICQRac It cuts the y axis at ic(sat) ic(sat) = ICQ + V CEQ R ac It has a steeper gradient than the dc load lines. The two lines meet at the Q point of the amplifier. 4 Compiled by Omae M. Oteri 20

Example one: Given an amplifier with the following parameters VCC = 20V, RC = 2K, RE = 3K, RB = 0K, RB2 = 9K, β = 00. Obtain the dc and ac load lines of the amplifier: Figure 0. VPP small signal peak-to-peak voltage is given by: VPP = 2VCEQ VPP = 2ICQRac The small value is picked to avoid clipping due cut-off, ipp = 2ICQ or ipp = 2 V CEQ ipp peak-to-peak ac current. R ac The small value is also taken to avoid clipping due to saturation. Clipping due to saturation is given by: 5 Compiled by Omae M. Oteri 20

6 Compiled by Omae M. Oteri 20

SOLUTION fig. 0:. Obtain dc load line. Open all capacitors. VTh = V CCR B2 20 X 90 = = 9.47V R B + R B2 9 + 0 0 X 9 RTh = 0 + 9 = 4.74 KΩ VBB = IBRB + VBE + IERE IB = = IBRB + VBE + IB(β + )RE V BB V BE R B + (β+)r E = 9.47 0.7 (4.74+(0 X 3)X 0 3 ) IC = βib = 00 X 2.85 X 0-5 = 2.85mA VCE = VCC ICRC IERE = 2.85 X 0-5A = 20 (2.85 X 2) (2.85 X 0-3 + 2.85 X 0-5 ) X 3 X 0 3 = 5.66V VCEQ = 5.66V DC operating point is given by(5.66v, 2.85mA) IC(sat) = V CC = R C + R E α VCE(cut-off) = VCC = 20V 20 = 3.98mA 2 X 0 3 3 X 03 +( 00 ) X 0 7 Compiled by Omae M. Oteri 20

AC load line: Steps:. Short all capacitors. 2. Ground all the dc sources. 3. Leave all the ac sources intact. AC load line parameters: IC(sat) = ICQ + V CEQ R ac ic(sat) = 2.85 X 0-3 + Rac = R CR L = 2 X 4 R C + R L 5.66.33 X 0 3 = 7.mA 2+4 =.33KΩ VCE(cut-off) = VCEQ + ICEQRac = 5.66 + 2.85 X 0-3 X.33 X 0 3 = 9.45V 8 Compiled by Omae M. Oteri 20

VPP = 2VCEQ = 5.66 X 2 =.32V VPP = 2ICQRac = 2 x 2.85 x 0-3 x.33 = 7.58 (The smallest is chosen) IPP = 2ICQ = 2 X 2.85 = 5.70mA IPP = 2 V CEQ = 5.66 = 8.52mA R ac.3k Low power amplifier: IC 2mA. High IC>0mA will burn the transistor. Make the small signal be 20% of the large signal. I 0IB where I current through RB and RB2 Example: Given VCC = 20V, VCE = 3V, VBE = 0.3V, IC = 2mA, I = 0IB, β = 50, RC = 3KΩ Obtain the values of RB, RB2 and RE. Solution: IB = I C = 2mA = 0.04mA β 50 IE= IC + IB = 2 + 0.04 = 2.04mA VC = ICRC = 2 X 0-3 X 3 X 0 3 = 6V IERE = VCC VCE ICRC = 20 3 6 = V RE = 2.04 X 0 3 = 5.39KΩ VRB2 = VB = VBE + IERE = 0.3 + (2.04 X 5.39) =.3V RB2 = V RB2 I =.3 0 X 0.04X 0 VRB + VRB2 = VCC IRB +.3 = 20 20.3 RB = 0.4X 0 3 = 2.75KΩ 3 = 28.25KΩ 9 Compiled by Omae M. Oteri 20

EXERCISE: Given VCC = 0V, VCE = 2V, VBE = 0.7V, IC =.5mA, I = 0IB, β = 00, RC = 4KΩ. Obtain the value of RB, RB2 and RE. Draw the dc and ac load lines and show where IB = β = 2mA 50 TRANSISTOR CHARACTERISTICS: I/P characteristics: The variation in the input voltage causes variation in the input current which controls the transistor operation. Voltage changes from VgE min to VBE max while. For a CE connected amplifier, the transistor is controlled by the current dc(i) on which the ac current will be leading. The variations in the input cause variation at the output where VCE changes and IC. This can also be plotted as shown below: 20 Compiled by Omae M. Oteri 20

The small signal causes variation in IC and VCE. This gives for VCE, we have IC and for VCE2 gives IC2. The output current varies as the input current where the output is amplified depending on the transition of amplifier. AMPLIFIER CLASSIFICATION: Amplifiers can be classified according to size of the signal being handled by the amplifier (small signal or large signals). For small signals; deals with mostly voltage while large signals deal with power amplification. (This also depends on the output resistance.) For large resistance, the signal s being dealt with are small signals and vice-versa. i.e. Ro is small for large signals in the range of Ω i.e 4-8Ω. Classification according to number of transistors used. Where for one transistor is called a single stage amplifier while for several transistors is called a several stage amplifier. E.g. three stage amplifier for three transistors. Classification according to biasing of the transistor. Under this classification we have the following amplifiers: i) Class A. ii) Class B. iii) Class C. iv) Class AB 2 Compiled by Omae M. Oteri 20

Class A: Biasing is done in such a way that the operating point is at the middle of the load line. This gives maximum sing of the input and output signal - Both current and voltage. In most cases if the biasing is maintained at mid point then there is no clipping of the output signal. Output signal is 360 thence for the whole cycle to appear at the output. Class B: Biasing is done such that the operating point is operated at cut-off (50% of the input signal is used). Only 80 of the output cycle will appear meaning half of the cycle is clipped due to cutoff. They are subjected to cross-over distortion if transition from one active element is not perfect. 22 Compiled by Omae M. Oteri 20

CLASS C: Is an amplifier which allows an output of about 50 of the cycle. Slightly less than half of the signal. It is biased deep into cut-off. CLASS AB: It comprises characteristics of both A and B. Biased in a way that it produces an output between 80-360. Slightly above 80. The two active elements conduct more than half of the time as a means to reduce cross-over distortion. 23 Compiled by Omae M. Oteri 20

CLASS A: There are several types which include, voltage divider bias and all the other forms of bias we dealt with earlier; transformers coupled darlington transistor amplification. Voltage divider bias: The output signal is an exact scaled up replica of the input with no clipping. They are used to implement small signal amplifiers. Characteristics: Output signal is 360 (a whole cycle) conduction. (00% of input signal is used) Biased at the middle point of the load line. No clipping of the signal. IC max is at IC(sat). VCE max is at VCE(cut-off)/VCC. All the elements can affect the Q point. It has a lot of power wastage. Continues even when power is off. Absence of VCC will disable the amplifier. It has a single transistor. Deals with amplification of small signals and high signals. It has a limit to the input signal. Efficiency is obtained by the following formula: η = P o P in Maximum η is approximately 25%. Very simple to design. 24 Compiled by Omae M. Oteri 20

ANALYSIS OF CLASS A AMPLIFIER: Current gain: Ai = I C I B large signal. Output current Ai = Input current I C I B = β for small signals. = I C max I C min I B max I B min = 2I CQ 2I BQ = I CQ I BQ Voltage gain: AV = V CE min V CE max V BE max V BE min = 2V CEQ 2V BEQ = = V CEQ V BEQ = V CC 2V BEQ Power gain: AP = AiAV Power Analysis: Input power: Pi(dc) = IiVi = (IC + IB)VCC = IEVCC The power supplied to the amplifier for it to operate is equivalent to the input power (Pi). It depends on the VCC and IE using the above relation. This supply of power is continuous even when the small signal is not present. The element goes through a load equivalent to RC + RE. 25 Compiled by Omae M. Oteri 20

When the small signal is present, the resistance will be consuming much of this power as heat dissipation. Output power: Po = Voio Rms values: Vrms = V max 2 Vo pp = VCE max VCE min Io pp = IC max IC min Vop = (VCE max VCE min) 2 Iop = (IC max IC min) 2 2 (V CE max V CE min ) Vo rms = 2 2 io rms = (I C max I C min ) 2 (I C max I C min ) Po = io rmsvo rms = 4 2 x (V CE max V CE min ) 2 = (IC max IC min) (VCE max VCE min) 8 Efficiency: 8 (I C max I C min )(V CE max V CE min ) η = P o x 00% = P in I E V CC This is the efficiency of a simple class A amplifier. Maximum class A efficiency. 8 ηmax = (I C max I C min )(V CE max V CE min ) 8 = (2I CQ(2V CEQ ) I E V CC I E V CC 8 ηmax = (2I CQ(V CEQ ) I E V CC IE ICQ ηmax = 8 (2I CEQ(V CC ) = X 00% = 25% I CQ V CC 4 ηmax = 25% Execute: In the previous example, calculate efficiency and explain why the two are not the same. Fig. 0. Example: Given VCC = 20V, RC = 2K, RB = 0K, RB2 = 9K, RE = 3K, β = 00, RL = 4K. Obtain its efficiency: Q point parameters: ICQ = 2.85mA, VCEQ = 5.66V 26 Compiled by Omae M. Oteri 20

η = P o rms P i dc Po rms = (IC max IC min)(vce max VCE min) 8 IC min = 3.98 2.85 Pi dc = 2.87 x 0-3 x 20 = 5.74 x 0-2 3.98 X 0 3 = (3.98.72)(.32-0) X 0-3 8 = 3.98 X0-3 η = X 00 = 5.57% 5.74 X 0 2 This is the minimum efficiency before clipping starts. It can be measured by varying the values of RE and RC (changing the operating point which will shift to the centre). Application of class A amplifiers: Impedance matching. Power amplification applications. Small signal amplifications e.g. public address systems/ heat rooms. Heating rooms. Advantages: No clipping of the signal. It has no distortion. (harmonics). Easy to maintain. Simpler designs than other classes. It has no turn on time because it never shuts off completely. It experiences no cross-over distortion problems. 27 Compiled by Omae M. Oteri 20

Disadvantages: Low efficiency i.e η max = 25%. High power loss. Require heat sinking devices. Require high power input. A fault in any of the devices affect it. Easily affected by temperature. CLASS A Transformer coupled amplifier: Transformer coupling: Transformation ratio ɳ = V P V S = N P N S = I S I P Primary power Pp = IpVp Rp = V P I P But V S RL = V S I S = I P = V P I S ɳ Ps = IsVs RL in terms of Rp => R L R P = V SI P V P I S R L = x = Therefore RL = R P R P ɳ ɳ ɳ 2 ɳ 2 The transformer can be used for impedance matching where Rp = η 2 RL Example: Determine the input resistance of a transformer which will result from an RL = 20Ω. If the input current is 5mA and output current is 2mA. Solution: η = I S = 2mA = 0.4 I P 5mA Rp = η 2 RL = ( 2 )2 x 20 = 3.2Ω 5 28 Compiled by Omae M. Oteri 20

CLASS A Amplifier: This consists of a voltage divider circuit with RB and RB2 as power divider. The resistor RC is replaced with a transformer that has a resistance RL connected at the output. There is also a resistor RE connected from the emitter side. The transformer reduces losses since it is assumed to have negligible resistance at the output (primary). RL will be used to collect the output current. The output is completely isolated from the input by the use of the transformer. Very low power is lost as heat energy. C will block any dc and C2 will bypass high frequency ac to ground. Operation of the device: The input current will be changing and result to a change in the output current. When the output current is collapsing, it will induce an input current which will be equivalent to double the value of ICQ as maximum and will give a minimum of zero. This will also have an effect in the voltage, VCE. The VCE can go up to a maximum of double VCE and a minimum of zero. 29 Compiled by Omae M. Oteri 20

POWER ANALYSIS: Pi dc = VCCICQ VCE pp = VCE max VCE min = 2VCEQ 0 = 2VCEQ = 2VCC ic pp = ic max ic min = 2ICQ 0 = 2ICQ VCE rms = 2V CC 2 2 = V CC 2 ic rms = 2I CQ 2 2 = I CQ 2 Po rms = ic rms X VCE rms = (I CQ)(V CEQ ) 2 η = P o rms X 00% = (I CQ)(V CC ) x P i dc 2 V CC I CQ X 00% = 50% Example: Determine the effect of introducing a transformer at the output stage of the following amplifier. NB β = 00. VTh = V CCR B2 0 X 6 = = 5V R B + R B2 2 RTh = 6 X 6 = 3KΩ 2 VBB = IBRB + VBE + (β + )RE IB = V BB V BE = 5 0.7 R B 3 X 0.43 X 00 ICQ = IE X α = 0 3 =.43 =.46 VCEQ = VCC ICRC IERE = 0 (.4mA X 3K) = 5.76V ηmax = 8 (I C max I C min )(V CE max V CE min ) I E V CC Pdc = VCCICQ = 0 X.46 X 0-3 =.46 X 0-2 VCE pp = (0 5.76) x 2 = 8.48V ic pp = 2ICQ = 2 x.46 = 2.832 30 Compiled by Omae M. Oteri 20

Po rms = 8 (IC max IC min) (VCE max VCE min) = (8.48) 2.832) = 3mW 8 η = P o rms 3 X 0 X 00% = 3 P i dc.46 X 0 2 X 00% = 2.2% When 2VCEQ>VCC you take (VCC - VCEQ) X 2 to get VCE pp. Introducing a transformer of turn ratio 5: we have RP = η 2 RL = 5 2 X 4 = 00Ω Calculate the power delivered to the speaker and compare it with the power delivered to speaker with no transformer. Solution: RP = η 2 RL = 00Ω VRB2 = V CCR B2 0 X 6 = = 5V R B + R B2 6+6 IERE = VRB2 VBE IE = V CC V BE R E IEQ = 5 0.7 3 X 0 3 =.433mA VCC = ICRC + VEQ + IEQRE VCEQ = VCC ICRC IEQRE = 0.42 X 0-3 X 00.43 X 0-3 X 3 X 0-3 = 45.568V ICQ = IEQ X α = 00 X.43 =.42mA 0 Pi dc = VCCICQ = 0 X.42 X 0-3 =.42 X 0-2 W Po rms = Io rmsvrms VCE pp = VCE max VCE min = 2VCEQ 0 = 2X 5.57 =.4V ic pp = 2ICQ ICP = ICQ =.42mA Po rms = i p X V CE p 5.57 X.42 X 0 3 = = 3.95mW 2 2 2 η = P o rms P i dc = 3.95 X 0 3.42 X 0 2 X 00% = 27.82% 3 Compiled by Omae M. Oteri 20

CLASS B Amplifiers: Characterestics: This is where the output is 80 of the input signal. There are several types i.e Single transition Push and pull. Complementary push and pull. It experiences large amount of distortion. Characteristics: Q point is at cut-off. Output signal is 80. Half of the signal is clipped out. ic max is at IC (sat). VCE max = VCC All elements affect its operation. Reduced power wastage since, it s only on when there is an input signal. Q point is not stable. Can be affected by temperature. +ve variation of VCC cause ve variation of ic. Absence of VCC will distort it. Can have more than one transistor e.g. push and pull. Has a limit to the input signal. Maximum efficiency is approximately 78.5%. η = P o P in. It is complex to design because of the use of transformers. 32 Compiled by Omae M. Oteri 20

It has distortion i.e output is not exact replica of the input. Single transistor amplifier: Operation: This is operated in such a way that the Q point is at VCC. The input signal should be large enough to bias the transistor. When the input is +ve going, the transistor is forward biased at the BE junction for Si>0 for Ge>0.3V. The change in the input reflected at the output as ic which varies according to output resistance and supply voltage VCC. The maximum current ic can fo up to ic(sat). The transistor comes on only when there is an input signal which is positive going. When the input voltage/ current goes ve, the transistor goes to cut-off and cannot conduct. The output will be zero i.e. for current ic and voltage VCE. 33 Compiled by Omae M. Oteri 20

No ve signal appears at output; only positive cycle appears. The max voltage can go up to VCE min = 0 for VCC point at which bias starts. When imax is ic(sat) then we have VCE min = 0. When VC min = 0 then VCE = VCC. POWER ANALYSIS: Since output signal is only half cycle. Then icp = ic max ic min but ic min = 0, icp = ic max. VCEP = VCE max VCE min but VCE max = VCC VCEP = VCC VCE min. Output power: Po rms = io rmsvceo rms where io rms = I C max 2 2 since o/p signal is half cycle. Vo rms = V CC V CE min 2 2 NB: The o/p signal is half cycle and therefore half rms value. Po rms = io rmsvo rms = I C max 2 2 X V CC V CE min 2 2 = [ic max(vcc VCE min)] 8 Input power: The i/p power will be obtained considering dc values i.e. Idc and VDC. VDC = VCC Idc = I C max π This is because when the i/p is +ve; that is what makes the transistor switch on. The o/p IC varies according to variation of the input i.e ib. Idc = I C max π Pi dc = IdcVDC = Idc = I C max VCC Efficiency η = P o rms P i dc X 00% = π 8 {i C max (V CC V CE min )} i C max V π CC = π 8 (V CC V CE min V CC ) x 00% = π 8 ( - V CE min V CC ) x 00% Maximum efficiency occurs when ic max = ic (sat) and VCE min = 0. η = π 8 ( - 0 V CC ) x 00% = 39.27% Example: Given an amplifier with VCC = 0V, RB = 5K, RB2 = 5K, RC = RE = 0Ω β = 00, VBE = 0.7V. Solution: VB = VBR2 = V CCR B2 R B + R B2 = RTh = 5 X 5 0 = 2.5V VB = VBE + IERE 0 X 5 0 = 5V 34 Compiled by Omae M. Oteri 20

IE = V B V BE 5 0.7 = = R E 0 Example: Given VCE min = 2V and VCC = 0V. Calculate η of the amplifier. Solution: η = π ( - 2 ) x 00% = 3.42% 8 0 Guiding parameters: ICQ = 0 and VCEQ = VCC in designing of the amplifier. OPERATION: CLASS B PUSH AND PULL: Po rms = io rmsvo rms To obtain io rms ic pp = ic max ic min and icp = i C max i C min 2 Considering one transistor: Ip = ic max Therefore for two transistors icp = ic max io rms = i cp = i c max 2 2 Similarly: Vo rms = V CE V CE min 2 2 Po rms = io rmsvo rms = i c max 2 Input power: = V CC V CE min 2 2 Vdc = VCC and Idc = 2i c max π Pi dc = 2i c max X V CC π Efficiency: η = P o rms P i dc V CC V CE min 2 2 = ic max (VCC VCE min) 8 = i c max (VCC VCE min)/ 2i c maxv CC = π ( - V CE min ) 4 π 4 V CC ηmax = π 4 ( - 0 V CC ) X 00 = 78.54% 35 Compiled by Omae M. Oteri 20

Example: Given VCE min = 3V and VCC = 20V. Calculate the efficiency of this amplifier: SOLUTION: η = π ( - 3 ) X 00 = 66.76% 4 20 COMPLEMENTARY CLASS B: Push and pull amplifier: Because the disadvantage of class B push and pull amplifier i.e. requirement of input transformation to slip the input signal and output transformer to complete the o/p signal (both being centre-tapped). There is need to eliminate being the following. (i) Using the complementary amplifier without transformer. (ii) Using class AB amplifier. Using complementary transistor: Construction: It is constructed using basic components i.e capacitors and resistors to bias the two transistors Q and Q2. Q is an n-p-n transistor while Q2 is p-n-p transistor. The doping of the transistors is the same in terms of the amount of dopant but opposite terms of P type and n-type materials. The second transistor forms a segment of the first transistor. C compress the input signal to the amplifier. C2 compress the o/p signal to the o/p load RL. RB biases the transistor when ON> VCC supply voltage to Q VCC2 supply voltage to Q2. 36 Compiled by Omae M. Oteri 20

Operation: When i/p signal is +ve going, it will forward bias Q and reverse bias Q2 and vice-versa. Since Q, is forward biased, change in i/p signal will have a corresponding effect on ic which will also change. The o/p signal due to the above effect will be +ve because it s being picked from the emitter. This will be the current ie. Advantages compared with transformer couple: It eliminates the use of input and output transformer. It utilizes only one input signal. Excellent efficiency. Less power losses hysteresis. Easier to analyze. Low cost. Easier to fabricate. Reduced space occupied. Disadvantages: Must use accurately doped transistors. Prone to noise effects coupling capacitor allowing some dc. Advantages of class B amplifier compared to class A: Lower power losses. When no input signal is present, the amplifier is off. High efficiency of about 78.5% Disadvantages: Are complex. Require 2 transistors for push and pull. Occupy large space/ are bulky. Input voltage must be large enough to turn on the transistor. They are expensive due to use of transformers. Experience high levels of cross-over distortion. Cross-over distortion: 37 Compiled by Omae M. Oteri 20

This is a situation which results when the transistors take time to switch one. i.e. the two transistors. It is called cross-over distortion because it occurs when the operation is transferred to the other transistor. It crosses over from one transistor to the other. Chances of not getting a replica of the i/p are high. CLASS AB amplifier: To solve the problem of distortion, the class B amplifier are biased with a small voltage just enough to switch the transistor. For silicon, the bias voltage is.4v and Ge = 0.6V for both transistors. This will solve the problem and since there is a small bias, this brings in the property of class A amplifier. This kind of amplifier is called class AB amplifier. i.e. one that combines both properties of A and B amplifier. There are two types: Resistor biasing. Diode basing. This will require only one voltage supply. GENERAL CHARACTERISTICS: Conducts between 80 and 360 of the cycle. Requires a small pre-bias, slightly above cut-off. Signal can be clipped. Combines both properties of class A and B amplifier. Eliminates the cross-over distortion. Efficiency will be between 30 80%. Less than half the signal is clipped. Absence of VCC disables the amplifier. 38 Compiled by Omae M. Oteri 20

DIODE PRE-BIASING: Construction. Constructed using 2 capacitors, 2 resistors, 2 diodes, 2 transistors and a voltage source. The two diodes are the ones that produce the bias voltage of.4v for Si and 0.6V for Ge. D drops 0.7V and D2 drops 0.7V to give.4v. OPERATION: When the input signal is +ve going, it rides on the small dc bias to produce an output signal which is full of half +ve cycle. 39 Compiled by Omae M. Oteri 20

When the input signal is ve going some of it will reduce the dc bias to zero before the transistor is cut-off. A bit of the ve signal o/p will be clipped due to cut-off. This can range from 80 to 360 i.e. class B to class A. The above is for transistor Q When the ve signal goes more negative, it will switch ON. Q2 which will conduct the signal cut-off due to transistor Q. POWER ANALYSIS: The input power is obtained by Pi dc. The output power is given by Po rms. Pi dc = VCCIdc There are two dc components. ICQ and I C max π Idc = ICQ + I C (sat) I CQ π Pi dc = VCC {ICQ + I C (sat) I CQ π io rms = i c max 2 Similarly: Vo rms = V CE max 2 Po rms = i c max 2 Efficiency: X V CE max 2 η = P o rms P i dc X 00% = Using pre-biasing: } = ic maxvce max 2 2 i C maxv CE max V CC (I CQ + I C sat I CQ ) π 40 Compiled by Omae M. Oteri 20

The operation is the same as that of diode biasing. The difference is that this uses resistors for biasing. The voltage across the resistor must be at least.4v. CLASS C AMPLIFIER: Characteristics: It conducts between 20 and 50 of the cycles. i.e. less than 50%. Clips more than half of the o/p cycle. Output not an exact replica of input. Efficiency is from 80% to 90%. Consists of components equivalent to those of class A amplifier. Consists of components equivalent to those of class A amplifier. It is biased deep into cut-off. Distortion at the output is high. POWER analysis: Po rms = io rmsvo rms where io rms = i c max 2 Similarly: Vo rms = V CE max 2 4 Compiled by Omae M. Oteri 20

Idc = I C max π Pi dc = VCC I C max π 8 Efficiency η = i C maxv CE max I V C max = π V CE max CC 8 V CC π The class C amplifier has two modes of operation: tuned and unturned. RANGES OF FREQUENCY: Audio amplifier operates in the range of 20 Hz to 20KHz. Radio frequency amplifiers (RF) e.g. AM radios frequencies between 535 to 605KHZ. FM radios between 88 08MHz. AMPLIFIER GAIN: Variations with frequency: Av mid Mid frequency voltage. 2 Av mid 3dB frequency voltage gain. f lowest cut-off frequency. f2 upper cut-off frequency. f = f2 f (Bandwidth). The above rep a graph of voltage gain versus frequency for class a A amplifier (Audio amplifier). This gives the bandwidth of operation of the amplifier as f = f2 f. All the signal frequencies in range are amplified almost equally. The first cut-off frequency f is located at 3dB downward from the mid frequency gain. The same applies for f2. For power it can be said to be Ap mid. 2 For voltage gain it can be referred to as 2 Av mid. The frequency below f will not be amplified. The same case applies to signals with frequencies above f2. 42 Compiled by Omae M. Oteri 20

The amplifier gain at low frequencies is affected by large capacitances C, C2 and C3. At high frequencies it is affected by small capacitances Cbc, Cbe and Cs where: Cbc collector base capacitance. Cbe base emitter stray capacitance. Cs due to wiring used in the amplifier. At low frequencies using the expression: Xc = 2πfc Since as f->0, the reactance increases reducing the gain. C, C2 and C3 will be like open while Cbc, Cbe, and Cs will be like shorts. This is what affects the amplifier gain in these frequencies. At mid frequencies, the capacitors C, C2 and C3 will behave as short circuits. Both categories of capacitors do not affect the amplifier gain. This gives the BW (Band width). f2 f where there is uniform gain. At high frequencies, i.e. beyond f2, the capacitors C, C2 and C3 behave as short circuits but Cbe, Cbc and Cs will behave as open circuits. The capacitors Cbe, Cbc and Cs will affect the gain in this frequency range which starts reducing from f2 onwards. Calculation of cut-off frequency: Using the above circuit, considering the capacitor C its cut-off frequency can be obtained by taking the resistor along its path. This gives the following circuit. 43 Compiled by Omae M. Oteri 20

Cut-off due to C: fc cut-off = 2πC R eg From the equivalent circuit Reg = Rs + RB RB2 (βre + βre) Reg Rs + RB RB2 βre since βre<< βre Taking Rs = 2KΩ, RB = 5KΩ, RB2 = 6KΩ, RC = 3KΩ, RE = 4KΩ, RL = 8KΩ, β = 00, C = 47µF and Vcc = 20V. SOLUTION: Reg = Rs + RB RB2 βre = 2 X 0 3 + 5 X 0 3 6 X 0 3 00 X 4 X 0 3 Reg = 4.7KΩ fc = 2πC R eg = 2π X 47 X 0 6 X 4.7 X 0 3 = 0.79Hz For C2 = Equivalent circuit will be: fc2 cut-off = 2πC 2 R eg where Reg = RL + RC Using the above parameters: Reg = 3K + 8K = K fc2 cut-off = =.45Hz. 2π X 0 X 0 6 X X 0 3 NB. For the lower cut-off frequency, we take the smaller value which is 0.79Hz. MILLER EFFECT: From the CE single stage amplifier, the transistor has capacitance given by Cbc and Cbe. These are represented in the circuit below. The effect of these capacitors is referred to as miller effect; more so with the bc capacitor. When the bc capacitor is taken from the input, it has a feedback effect due to the connection between the input and output i.e b and c. 44 Compiled by Omae M. Oteri 20

Because of the feedback, the effective collector base capacitance is given by (A + )Cbc where A is the gain of the transistor. When a small signal Vs is applied to the input of the transistor with gain A, the charge in Vc will be given by AiVi. Usually, the total charge for cb voltage ic VCB = Vi + AiVi VCB = Vi ( + Ai) When Vi applied has an effective charge of Q = CV Q at the i/p = Cbc X VCB Q = Cbc + ( + A)Vi = ( + A)CbcVi The effective base collector capacitance is given by Cat base = ( + A)Cbc -> Miller effect. When one looks into the circuit, the capacitors will give an effective capacitance of ( + A)Cbc Cbe Ceg = ( + A)Cbc Cbe = ( + A)Cbc + Cbe Example: Given an amplifier with gain of A = 200, Cbc = 0pF and Cbe = 20pF. Calculate the cut-off frequency f2 if the equivalent resistance of the amplifier is 0KΩ. SOLUTION: Ceg = ( + A)Cbc + Cbe = ( + 200)0 X 0-2 + 20 X 0-2 = 2030 X 0-2 F f2 = = 2πC eg R eg 2πC 2 R eg 2π X 2030 X 0 2 X 0 X 0 3 = 7840Hz. For the example above, if the wiring capacitance is 40pF: a) Calculate the new cut-off frequency CW = 40pF = 40 X 0-2 F Ceg new = CW + Ceg = 40 X 0-3 + 2030 X 0-2 f2 new = 2πC eg new R eg = = 2070 X 0-2 F 2π X 2070 X 0 2 X 0 X 0 3 = 7688.6Hz 45 Compiled by Omae M. Oteri 20

TRANSISTOR TUNED AMPLIFIER: A tuned amplifier is constructed differently from a normal audio amplifier. The output resistance RC is replaced with a capacitor. Because of the capacitor and inductor, the combination resonates at the same frequency called fr (resonant frequency). At this frequency, there is maximum impedance. The operation is in such a way that when the input signal reaches the frequency at which the amplifier resonates, it will be allowed to pass and as a result get amplified since it appears at the output of the transistor. It also has a bandwidth of operation though small while compared with those of audio amplifiers. These amplifiers are not suitable for audio amplifiers since they have a narrow bandwidth. These amplifiers deal with radio frequency signals RF. For audio amplifiers, the bandwidth is around BW = f2 f = 20KHz 20Hz For RF amplifier which has a bandwidth of 40KHz will be given by f = 780KHz and f2 = 800KHz if the centre of the frequency is 500KHZ. FEATURES: Resonant frequency. Impedance at resonance. 46 Compiled by Omae M. Oteri 20

The current at resonance. Quality factor. The inductor has small resistance due to the wire resistance. The circuit will be given as: The phasor diagram for the circuit along side will be given by: RESONANCE: At resonance, the capacitor current is equivalent to the insulator current. IC = ILsin QL IL = V Z L ZL2 = R2 + XL2 X C = X L Z L 2 XCXL = ω C ω L sin QL = X L Z L (2πfL)2 + R2 = L C and XCXL = ZL2 = R2 + XL2 L C = ZL2 = R2 + XL 2πfL = L C R2 fγ = R2 2π LC L 2 This expression gives the resonant frequency. If R is very small as expected with inductors, the above expression will reduce to: fγ = 2π LC 47 Compiled by Omae M. Oteri 20

Resonant impedance (Zγ): At resonance, the circuit current is given by: I = ILcos QL V = V X R Z γ Z L Z L Zγ = Z L 2 R Zγ = Z L 2 = ω L x = L R ω C R CR Resonant current: I = V Z γ = V L CR Quality factor: = VCR L Q = X L R Quality factor: usually indicates the selectivity of an RF amplifier. The higher the Q, the better the selectivity and therefore the better the amplifier to selectively amplify a specific frequency. When the quality factor is high, the graph will have a sharper response as below. Example: C = 0μF, L = 2mH, R = 20Ω. Calculate resonant frequency, resonant impedance, Iγ and Q factor. Solution: fγ = = 2π LC 2π 2 X 0 3 X 0 X 0 Zγ = L = 2 X 0 3 = 0Ω CR 0 X 0 6 X 20 Iγ = V = 20 = 2A Z γ 0 Q = X L = 2πfL R R = 2π X 25.4 X 2 X 0 3 20 6 = 25.4Hz = 0.707 Advantages of tuned amplifiers: Low power lost since the LC circuit dissipates very low power. High selectivity which is important when amplifying small based signals. Low VCC since there is low voltage dropped on the capacitor, inductor circuit. Low voltage is required to bias the amplifier. 48 Compiled by Omae M. Oteri 20

Disadvantages: Affected by temperature. Mostly applied in high frequency range i.e. not applied for radio frequency. Narrow bandwidth. For them to be applied in low frequency the LC should be very large. Relationship between bandwidth and Q -factor : Q = f r where fr resonant frequency. BW The lower the BW, the higher the Q factor and therefore the higher the selectivity. Factors which affect the Q factor: Bandwidth Q BW Capacitance Q C Inductance Q L DC EQUIVALENT CIRCUIT: The inductor has some resistance R which affects the dc current ie, IC VCC = ICR + VCE + IERE 49 Compiled by Omae M. Oteri 20

VRB2 = VBE + IERE = V CCR B2 R B + R B2 AC EQUIVALENT CIRCUIT: At resonance: There is maximum impedance. The output current will not pass through LC but will go through RL. This gives a maximum of output voltage. XL + XC. Below resonance: f < fr and the resonant circuit behaves like a capacitor i.e. with capacitance XC>XL. Above resonance: f>fr : The resonant circuit behaves like an inductor I.e. it has inductance XL>XC NB. For below and above resonance, there is current which flows through the resonant circuit. This reduces the current passing through RL. Effectively, the output voltage and hence the gain of the amplifier is reduced. CLASS C TUNED AMPLIFIER: 50 Compiled by Omae M. Oteri 20

The amplifier conducts less than 80 of the waveform. This makes a class C amplifier. It operates in such a way that it is biased deep into cut-off. The output current IC is given by the diagram below. When the input voltage goes beyond +0.7V, the BE diode conducts. Anything above 0.7V will come out. If the input voltage is 2V, then only.3v will come out. This will form a pause which affects the output current IC according to amplification factor. The pulse at the output will be produced by extra.3v at the output. This will be fed to the resonant circuit which if it is only one pulse; it will produce a dying sinusoid. If they are several pulses, then they produce a pure sinusoid at the output due to oscillation. The circuit is also called a pulse amplifier. Resonance: At resonance, the signal is ve going and the BE diode is reverse biased hence there is no conduction. But there will be some output since the capacitor stores some charge hence the output will be a sinusoid. DC load: When there is no ac input, the transistor is off and therefore no current flows through. If there is an input ac signal when it is more than 0.7V, the transistor conducts the resistance along the path ic. AC load: The resistance of the coil can be used to determine the equivalent parallel resistance of the circuit. This parallel resistance will be in parallel with RL and is given by RP = Q coil XL The equivalent circuit is as below: 5 Compiled by Omae M. Oteri 20

The load will be given by Rac = Rp RL Rac = R PR L R P + R L \ Maximum ac power: Since the RE is replace with an inductor and capacitor. Almost all the VCC will be appearing at the collector side because of the inductor. The ac signal can change from 0 to VCC as maximum sing i.e. VCE max = VCC and ic max = V cc R ac Pmax = V CE max X i C max 2 = V cc 2 R ac DC input power is given by: Pidc = Pmax + Po Where Po power dissipated in the transistor amplifier. η = P o P X 00% = o max X 00% P i dc P o max + P o Since Po for class C amplifier is very small, then it can be negligible. PD 0 η = P o P i dc X 00% = 00% Example: Given an amplifier with R = 0Ω, L = mh, C = μf, RL= 0KΩ. Calculate fγ, Rp and Rac. Solution: fγ = = = 5033Hz 2π LC 2π X 0 3 X X 0 6 Rcoil = X L = 2πfL = 2π X 5033 X X 0 3 = 3.6 R R 0 Rp = QcXL = 3.6 X 2π X 5033 X 0-3 = 00Ω Rac = R PR L 00 X 0 X 03 = = 99Ω R P + R L 00+(0 X 0 3 ) 52 Compiled by Omae M. Oteri 20