Electronic Circuits Power Amplifiers Manar Mohaisen Office: F208 Email: manar.subhi@kut.ac.kr Department of EECE
Review of the Precedent Lecture Explain the Amplifier Operation Explain the BJT AC Models Introduce the Common-emitter Amplifier Introduce the Common-Collector Amplifier Introduce the Common-Base Amplifier Introduce Multistage Amplifiers
Class Objectives Explain the Operation Class A Power Amplifier Explain the Operation of Class B and AB Amplifiers Explain the Operation of Class C Amplifier
The Class A Power Amplifier Operation Biased amplifier is always operating in the linear region. Power amplifier provides power rather than voltage to a load. The output signal is an amplified replica of the input signal. Categories Small-signal Amplifier The ac signal moves over a small percentage of the total ac load line. Large-signal Amplifier The output ac signal approaches the limits of the ac load line.
The Class A Power Amplifier contd. Heat Dissipation Power devices must dissipate a large amount of internally generated heart. The collector terminal is the critical junction. Therefore, heat sink is usually connected to the collector.
The Class A Power Amplifier contd. Centered Q-Point AC and DC load lines intersect at the Q-point. When the Q-point is at the center of the ac load line A maximum class A signal can be obtained.
The Class A Power Amplifier contd. Non-centered Q-point Q-point closer to cutoff Q-point closer to saturation
The Class A Power Amplifier contd. Power Gain The ratio of the input power to the (load) output power. A p P = P P L is the power delivered to the load. P in is the input power delivered to the amplifier. L in Other formulas of the power gain P L Therefore, V = R A p 2 L L P in V = R V R R = = V 2 in in 2 L in A2 in 2 v R R in L L
DC Quiescent Power The power dissipation with no input signal. Output Power The Class A Power Amplifier contd. The maximum ac unclipped signal occurs for centered Q-point. Therefore, (the peak for Common-Emitter Amplifier) The maximum peak current swing is [peak value] Then, P = I V DQ CQ CEQ V = I R c(max) CQ c V Ic(max) = R CEQ P = I V out(max) crms ( ) crms ( ) c = (0.707 I )(0.707 V ) = 0.5I V c c CQ CEQ
Example 7-1 The Class A Power Amplifier contd. V I B R = 10 12V = 1.82 V 2 V R R CC + 1 2 66 V 0.7 = B = E R + R E1 E2 1.78mA R = R R β ( R + r ) ' in( tot)1 1 2 ac( Q1) E1 e( Q1) A = 56k Ω 10k Ω 200(68Ω+ 14 Ω ) = 8.4 kω = A A = ( 27.9)(1) = 27.9 vtot ( ) v1 v2 r' eq ( 1) A = 25 mv = 14 Ω I E R R R R = = = 27.9 c1 C 3 4 v1 r' + R r' + R eq ( 1) E1 eq ( 1) E1 A p R = A = ( 27.9) = 817,330 2 in( tot)1 2 8.4 k Ω vtot ( ) R L 8Ω
Efficiency The Class A Power Amplifier contd. The ratio of the output signal power supplied to a load to the total power from the DC supply. Average power supply current, I CC, equals I CQ. The supply voltage is at least 2V CEQ. Therefore, the total DC power is The maximum efficiency In Practice, P = I V = 2I V DC CC CC CQ CEQ P 0.5I V out CQ CEQ η max = = = 0.25 P 2I V DC CQ CEQ The efficiency of Class A Amplifiers is less than 10% (low efficiency). Therefore, their use is limited for small power applications (less than 1 W)
Operation of Class B Power Amplifier It is biased at cutoff. It operates in the linear region for 180 o. The Class B and Class AB Amplifiers Operation of Class AB Power Amplifier It operates slightly more than 180 o.
The Class B and Class AB Amplifiers contd. Advantages of Class B and Class AB over Class A They are more efficient than a class A amplifier. You can obtain more output power for a given input power. Disadvantage of Class B and Class AB It is difficult to implement the linear circuit.
Class B Operation The Q-point is at cutoff. The Class B and Class AB Amplifiers contd. I CQ = 0 and V CEQ = V CE(cutoff). It is brought into the linear region of operation When the input signal drives the transistor into operation.
The Class B and Class AB Amplifiers contd. Class B Push-Pull Operation Two class B amplifiers are used to amplify the entire cycle. There are two approaches to for using push-pull amplifiers Transformer coupling Two complementary symmetry transistors (npn/pnp BJTs)
Transformer Coupling Q1 conducts on the positive half cycle Q2 will be then in cutoff. Q2 conducts on the negative half cycle Q1 will be then in cutoff. The Class B and Class AB Amplifiers contd.
Complementary Symmetry Transistors The DC base voltage = 0 The input signal biases the transistors. Q1 conducts during the positive half cycle Q2 conducts during the negative half cycle. The Class B and Class AB Amplifiers contd.
Crossover Distortion Transistors are off for the range ]-V BE, V BE [ The Class B and Class AB Amplifiers contd.
The Class B and Class AB Amplifiers contd. Biasing the Push-Pull Amplifier for Class AB Operation The transistors are conducting even without the input signal. The collector current is given by I CQ V = CC 0.7 R 1
AC Operation Q1: AC cutoff at VCC with V 0.7 I = CC CQ R The AC saturation current V I = CC csat ( ) R The Class B and Class AB Amplifiers contd. 1 L
The Class B and Class AB Amplifiers contd. AC Load Line for Q1.
The Class B and Class AB Amplifiers contd. Efficiency When there is no signal, transistors have a very small current. The power dissipation is small. Therefore, the efficiency is higher than that of class A amplifiers. Example 7-3 The ideal maximum out voltage V out(peak) = V CC = 20 V The maximum peak current I 20 I CC = = out( peak) c( sat) RL 150 V 133mA
Single Supply Push-Pull Amplifier The Class B and Class AB Amplifiers contd. V V V CC out( peak) CEQ 2 I VCEQ V = I = = R 2R out( peak) c( sat) L CC L
Maximum Output Power The Class B and Class AB Amplifiers contd. P = I V out out( rms) out( rms) I = 0.707I = 0.707I out( rms) out( peak) c( sat) P = 0.5I V out csat ( ) CC VCEQ = VCC /2 V = 0.707 V = 0.707 V out( rms) out( peak) CEQ P = I V DC CC CC Each transistor working on half cycle, then: I P CC DC = = I I csat ( ) π V π csat ( ) CC η max P 0.25I V out csat ( ) CC = = = 0.25π = 0.79 P I V / π DC csat ( ) CC
Input Resistance The Class B and Class AB Amplifiers contd. R = β r + R R R in ' ac( e ) E 1 2
Darlington Class AB Amplifier When the load is relatively small, Darlington pair is used The input resistance is increased The Class B and Class AB Amplifiers contd. R = β r + R R R in ' ac( e ) L 1 2 = β β ( r + R ) R R ' ac1 ac2 e L 1 2
The Class C Amplifier Operation Conduction occurs for much less than 180 o. More efficient than either class A and class B/AB It is biased below cutoff with a negative V BB supply.
The Class C Amplifier contd.
The Class C Amplifier contd. Operation more The input peak exceeds V BB + V BE for conduction. For entire ac load line operation I c is ideally I c(sat) and V ce is V ce(sat).
The Class C Amplifier contd. Power Dissipation Maximum Power dissipation (during the on period) P = I V D( on) c( sat) ce( sat) The average power dissipation over the period T P = t T P on D( avg) D( on)
The Class C Amplifier contd. Tuned Operation The output is not a replica of the output, Therefore, a resistive load is of no value in liner applications. A parallel resonant circuit is thus used. The resonant frequency f r = 1 2π LC
The Class C Amplifier contd. Operation Transistor on Capacitor charges Transistor off Capacitor discharges Inductor charges Capacitor charges in opposite direction. Capacitor discharges. Inductor charges Capacitor charges. Transistor on Due to input pulse.
The Class C Amplifier contd. Operation If The transistor remains off However, due to the input signal, I c turns on and off i.e., transistor is on periodically. As a result, the output voltage does decay to zero. The Class C Amplifier As a Frequency Multiplier f r = 1 = 2 f 2π LC in
The Class C Amplifier contd. Efficiency The output voltage has a peak of approximately (2V CC ), therefore 2 (0.707 V ) 2 P Vrms CC out = = R R Rc is the equivalent parallel resistance of the collector tank circuit. Has a low value. The (Total) Input Power Therefore, the efficiency is T c P = P + P out η = out = T out D( avg) P P P P + P Therefore, efficiency is close to 1 (100 %) for P out >> P D(avg)! out c D( avg) P = I V P D( on) c( sat) ce( sat) = t T P on D( avg) D( on)
The Class C Amplifier contd. Example 7-8 V CC = 24 V, Rc = 100 Ω, P D(avg) = 4 mw P out = (0.707 V ) 2 V R = R = 2 rms c CC c 2.88 W Pout P η = = out = 2.88W 1 P P + P 2.88W + 4 mw T out D( avg)
The Class C Amplifier contd. Clamper Bias Conduction happens for a short time.
The Class C Amplifier contd. Example 7-9 V = (1.414)(1V) = 1.4 V s( p) The base is clamped at ( 0.7) = 0.7 V dc V s ( p ) The signal at the base has A negative peak of V s ( 0.7) 1.4 0.7 2.1 V ( p + = = ) A positive peak of +0.7 V Output Voltage V 2 30 @ 1 pp = V = V fr = = 411kHz CC 2π LC
Discussion & Notes K K A K A A A A K K K K A K A K K A K A K A