Power Amplifier

Learning objectives Explain and analyze the operation of class A power amplifier Explain and analyze the operation of class B and class AB power amplifier Describe the operation of class C power amplifier Calculate the efficiency of power amplifier Distinguish between dc load lines and ac load lines Signals are amplified in several stages. The initial stages are small signal amplifiers, they are designed to give good voltage gain, so they are called voltage amplifiers. At the final stage, the signal becomes large, the large-signal amplifier is called power amplifier, as it is designed for good power gain.

Classification of power amplifier The power amplifiers are classified according the conduction angle they produced. Conduction angle measures the portion of the input cycle that is reproduced at the output of a power amplifier. If the conduction angle is 360, which means that all of the input cycle is reproduced, the amplifier is called class A amplifier. The small signal amplifiers we studied previously are all class A amplifiers.

Every amplifier has a DC equivalent circuit and an AC equivalent circuit. Because of this, it has two load lines: a DC load line and an AC load line. DC load line of class A amplifier The load line is a graph of all possible combination of the current through a transistor and the voltage drop across the transistor. In the case of BJT, the DC load line is a graph of collector current IC compared to collector-emitter voltage VCE. When the transistor is saturated, it acts like a closed switch, VCE = 0. The full voltage of power supply is dropped across the resistor. When the transistor is cutoff, it acts like an open switch, IC = 0. The DC load line can be determined by finding the saturation current IC(sat.) (VCE = 0) and the cutoff voltage VCE(cutoff) (IC = 0). The Q point is determined from the DC circuit. AC load line The AC equivalent circuit is different from the DC equivalent circuit. The collector resistance is different because Rac = RC // RL, and the emitter resistor becomes zero. Therefore, the AC load line is different from DC load line.

How to determine the ac saturation point? The transistor is operated at its Q-point (ICQ, VCEQ), going from the Q-point to the saturation point, the collector-to-emitter voltage swings from VCEQ to near 0; that is, VCE = VCEQ. The corresponding change in collector current from the Q-point to saturation is therefore

IC = VCE / Rac = VCEQ / Rac The ac collector current at saturation is

Ic(sat.) = ICQ + VCEQ / Rac How to determine the ac cutoff point? Going from the Q-point to the cutoff point, the collector current swings from ICQ to near 0; that is, IC = ICQ. The change in collector-to-emitter voltage going from the Q-point to cutoff is therefore

VCE = ICRac = ICQ Rac The cutoff value of ac collector-to-emitter voltage is

Vce(cutoff) = VCEQ + ICQ Rac

When an ac signal introduces to the amplifier, the instantaneous operating point moves along the AC load line. In other words, the peak-to-peak sinusoidal current and voltage are determined by the ac load line. Relationship between AC load line and DC load line The AC load line has a lower value of cutoff voltage than the DC load line (as Rac = RC // RL < RC) and a higher value of saturation current (as Rac < RC + RE, think about it!), so the AC load line has a higher slope than the DC load line. The two load lines intersect at the Q point. Because the AC load line has a higher slope than the DC load line, the maximum peak-to-peak output is always less than the supply voltage. For instance, if the supply voltage is 10 V, the maximum peak-to-peak sinusoidal output is less than 10 V.

A class A amplifier requires both halves of a sine wave signal to be the same. Therefore, we must limit out output to the smaller of the two swings. When the class A amplifier is biased at the midpoint of the AC load line, the ac output signal has the maximum possible swing and the ac output power also becomes maximum.

AC output power The ac output power is the amount of power that is consumed by the load resistor and is measured by the rms value (average value, real reading from the power meter), Pout = Vrms Irms = Vrms2 / Rac (or Pout = Irms2Rac) In terms of peak ac out voltage Vp, Pout = Vp2 / 2Rac (as Vrms = Vp / 2) In terms of peak-to-peak ac out voltage Vpp, Pout = Vpp2 / 8Rac (as Vp = Vpp / 2)

How to calculate the rms value? By definition 2

0

1 ( )T

rmsV V t dtT=

For a sinusoidal wave V(t) = Vpcos(t + ), 2

0

1 cos( )2

Tp

rms p

VV V t dt

T = + =

similarly, 2p

rms

II =

thus Pout = Vrms Irms = (1/2)IpVp = Vp2 / 2Rac When Q-point is closer to saturation, the maximum collector-to-emitter voltage swing is VCEQ (i.e. Vp

= VCEQ), and the maximum collector current swing is VCEQ / Rac. The ac output power is Pout = VCEQ2 / 2Rac

When Q-point is closer to cutoff, the maximum collector current swing is ICQ (i.e. Ip = ICQ), and the

maximum collector-to-emitter voltage swing is ICQRac. The ac output power is Pout = (1/2)ICQ2Rac

When Q-point is centered, the maximum ac output power is

Pout = (1/2)VCEQICQ Average value and rms value The average value is an arithmetic average of all the values in a sine wave for one alternation, or half-cycle. The half-cycle is used for the average because over a full cycle the average value is zero, which is useless for comparison purposes. Root-mean-square value or rms value relates to the dc voltage and current that will produce the same heating effect.

DC input power The dc input power is the power delivered to the power amplifier from the dc supply and is equal to the dc supply voltage times the current drawn from the supply (i.e. Q point current). Pdc = VCC ICC The power supply produces the Q point current, and hence dc power is being dissipated whether or not any ac signal is amplified (the dc power stays constant regardless of the input signal). Example:

In the above circuit, ICC = ICQ + I1, if I1 is small when compared with ICQ, then ICC = ICQ and we have Pdc = VCCICQ Power amplifier efficiency Amplifier efficiency is the ratio of ac output power (load power) to dc supply power. It measures how much of the dc power can be turned into ac signal power = ac output power / dc input power = Pout / Pdc Ideal class A power amplifier efficiency When the Q point is at the center of the dc load line which is the same as the ac load line (a special case when the load resistor RL is removed). The peak ac power is equal to the maximum voltage swing VCEQ times the maximum current swing ICQ. Thus the peak ac output power is: Pout(peak) = VCEQICQ = 2Vrms 2Irms = 2VrmsIrms = 2Pout Since the peak ac output power is twice the ac output power Pout = (1/2)Pout(peck) = (1/2)VCEQICQ Then from Pdc = VCCICC = 2VCEQICQ We have = Pout / Pdc = (1/2)VCEQICQ / 2VCEQICQ = = 25% DC Quiescent Power: (transistor power dissipation) The power dissipation of a transistor with no signal input is PDQ = ICQVCEQ The quiescent power is the maximum power that the class A transistor must handle; therefore, their power rating should not exceed this value. Example The amplifier shown in the figure is producing a peak sine wave output of 4 V. Determine the dc input power, ac output power and the efficiency.

Solution The operating point of the amplifier IBQ = VCC / RB = 16V / 16k = 1 mA ICQ = IBQ = 100 1 mA = 100 mA VCEQ = VCC - ICQ Rload = 16V 100mA 80 = 8 V The base current is small enough to ignore, thus the dc input power is Pdc = VCCICQ = 16V 100mA = 1.6 W Since Vp = 4V, Rac = Rload, we have Pout = Vp2 / 2Rac = 16 / (2 80) = 0.1 W and = Pout / Pdc = 0.1 /1.6 = 6.25% *In this circuit, the ac load line and the dc load line coincide. (Why?) Class A amplifier has a poor efficiency. In order to increase the efficiency, the class B amplifier is adopted.

Class B amplifier

A class B amplifier has a conduction angle of 180 and is biased at cutoff so that ICQ = 0 and VCEQ = VCE(cutoff). The primary advantage of class B amplifier over class A is that it has a higher efficiency. Common collector class B amplifier

In the class B amplifier, the output is not a replica of the input. In order to solve this problem, a push-pull configuration is employed to get a sufficiently good reproduction of the input waveform. The push-pull amplifier A push-pull amplifier consists of two class B amplifiers arranged to obtain a 360 conduction angle. In the push-pull operation, a pair of complimentary transistors (NPN and PNP transistors) are used to conduct alternatively. The circuit has no biasing resistors, and only signal voltage drives the transistors into operation.

Load line of a class B push-pull amplifier Cutoff is in the middle of the load line and the two endpoints are the actual saturation points of Q1 and Q2.

Crossover distortion When there is no dc bias, input signal voltage must exceed VBE (= 0.7V) before a transistor conducts. As a result, there is a time interval when neither transistor is conducting. This is crossover distortion.

Class AB amplifier To eliminate crossover distortion, both the transistors in the push-pull arrangement must be biased slightly above cutoff when there is no signal. This can be done with a voltage divider and diode arrangement. This arrangement is sometimes called class AB amplifier.

In class AB amplifier, the two transistors have symmetrical characteristics, VA = VCC / 2, VB1 = VA + 0.7V = VCC / 2 + 0.7V Since VE1 = VE2 = VE = VCC / 2 So that VBE1 = VB1 VE = 0.7V Similarly, we have VBE2 = - 0.7V The transistors are slightly biased at 0.7V and 0.7V respectively, before the ac signal is introduced, thus the crossover distortion can be eliminated.

Maximum output power for push-pull amplifier The maximum peak output current Ic(ac sat.), and the maximum peak output voltage VCEQ. The maximum average output power therefore is Pout = Vrms(out)Irms(out) = (1/2)VCEQIc(ac sat.) = (1/4)VCCIc(ac sat.) Input power The input power comes from the power supply and is Pdc = VCCICC Since each draws current for a half cycle, the current is a half-wave signal with an average value of ICC = Ic(ac sat) /. Thus Pdc = VCCICC = VCCIc(ac sat) / Efficiency = Pout / Pdc = /4 = 0.785 = 78.5% Class C amplifier A class C amplifier has a conduction angle of class than 90 and is biased beyond cutoff.

Class C amplifiers are more efficient than either class A or push-pull class AB amplifiers. Thus, more output power can be obtained from class C operation. Because the output waveform is severely distorted, class C amplifiers are normally used as tuned amplifiers at radio frequency (RF).

Exercise Problems (Power Amplifier) 1. Find the end points of the dc load line and the quiescent current ICQ when VCEQ = 37V.

2. For the following power amplifier circuit, find the dc power, rms ac power and the efficiency.

3. Suppose the following circuit has the values: VCC = 30V, RC = 300, RL = 750. If the Q-point is in

the middle of the dc load line, what are the endpoint values of the ac load line, and what is the class A amplifier maximum voltage swing?

4. Draw the ac load line for the circuit.