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# AE_Lecture10_Supplementary

Module by: Bijay_Kumar Sharma. E-mail the author

Summary: AE_Lecture10_Supplement gives the power analysis of Class A _ CE Amplifier . Its poor power conversion efficiency does not justify its applications for Power Amplification applications.

AE_Lecture10_SupplementaryNotes.

Class A Direct Coupled Resistive Load Power Analysis.

Here we are analyzing Class A_ CE Amplifier with resistive load connected as RC .

Under no signal condition:

The dc power drawn from the battery is = Pbattery = ICEQVCC

For maximum symmetrical swing, Q point is chosen to give :

ICEQ = VCC/(2RC) and VCEQ = VCC/2 ;

Therefore Pbattery = VCC2/(2RC) ;

Power dissipated in the device = Pdissip = ICEQVCEQ = VCC2/(4 RC);

Power dissipated in the collector resistance = PRC = (ICEQ)2. RC = VCC2/(4 RC) ;

Pbattery = Pdissip + PRC ;

This means power drawn from the battery is equally dissipated across the collector resistance and the device .

When maximum signal is applied at the input, we get maximum output signal.

It will be noticed in Figure 2 that when maximum current flows in BJT there is minimum voltage across BJT and when there is maximum voltage across BJT then there is minimum current through BJT. Both these situations minimize the device dissipation. The saved portion of device dissipation is converted into signal power.

The signal voltage across the load i.e. RC is = (VCC/2)Sin(ωt)

Therefore signal power developed across the load = Vrms2/RC ;

Vrms = ( amplitude of the sinusoidal voltage swing) = (VCC/2)/√2 ;

Signal Power = Psignal = Vrms2/RC = ((VCC/2)/√2)2/RC = (VCC)2/(8RC) ;

Power Conversion of efficiency = η = Psignal/ Pbattery = 2/8 = 1 / 4 =25%;

Figure 4 gives the static and dynamic load line in case of ClassA CE transformer coupled resisitive load. Since transformer primary coil offers 0Ω under dc condition hence the static load line has a slope of = -1/0 = ∞. Therefore Static load line is a vertical line parallel to Y axis and cutting X axis at VCC.

Q point is so chosen that we obtain maximum symmetrical swing. For this the vertical from Q must bisect the X axis intercept of the Dynamic Load Line.

Hence slope of OQ = negative slope of the dynamic load Line= 1/RL

where RL’ = n2 RL and n= turns ratio of primary to secondary..

Therefore Q point is : VCEQ = VCC and ICQ= VCC/ RL’ .

Under no signal condition the power drawn from the battery =

Pbattery = VCC× ICQ =( VCC)2/ RL’= this power is totally dissipated in the device. There is no resistive dissipation as the dc ohmic resistance is zero in collector circuit.

Signal Power delivered to the load as seen on the primary side is = Vrms2/RL’ ;

But Vrms = amplitude of maximum sinusoidal swing/ √2 ;

As seen from the graph in Figure 4 the amplitude of maximum sinusoidal swing= VCC;

Hence Psignal = (Vrms= VCC/√2 )2/RL’ = ( VCC)2/ (2RL’);

Therefore η = 1/2 = 50% . Almost hundred percent improvement over the direct coupled load but still not permissible for Power Amplifiers.Here half of the device power dissipation under no signal condition is being transformed into signal power under signal condition. Thus we achieve 50% power conversion efficiency.

So we go for Class B mode of operation where we will achieve 80% power conversion efficiency.

In RF applications we will go for Class C mode of operation which will provide 99% power conversion efficiency.

Here the question arises why do we not go for Class C mode of operation in Audio Frequency Range.

This is because Class C suffers from very severe harmonic distortion which can be removed only by tuned circuits and tuned circuits are not practical at audio range.

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