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1160 ⏐⏐⏐ SYSTEM ANALYSIS: AN INTRODUCTION<br />

Recall an earlier conclusion that the larger the value of R L, the larger<br />

the loaded voltage gain. For current levels, Equation (26.12) reveals that<br />

the larger the level of R L, the less the current gain of a loaded<br />

amplifier.<br />

In the design of an amplifier, therefore, one must balance the desired<br />

voltage gain with the current gain and the resulting ac output power<br />

level.<br />

For the system of Fig. 26.17, the power delivered to the load is<br />

determined by E 2 o/R L, whereas the power delivered at the input terminals<br />

is E 2 i/R i. The power gain is therefore defined by<br />

A G � � � � � � 2<br />

and AG � A (26.13)<br />

2 Ri<br />

v�� R<br />

Expanding the conclusion,<br />

A G � (A v) � A v � � (A v)(�A i)<br />

so AG ��AvAi (26.14)<br />

Don’t be concerned about the minus sign. A v or A i will be negative<br />

to ensure that the power gain is positive, as obtained from Eq. (26.13).<br />

If we substitute A v ��A iR L/R i [from Eq. (26.10)] into Eq. (26.14),<br />

we will find<br />

AG ��AvAi ��� � �Ai Ri<br />

or AG � A (26.15)<br />

2 i � RL<br />

�<br />

R<br />

which has a format similar to that of Eq. (26.13), but now A G is given<br />

in terms of the current gain of the system.<br />

The last power gain to be defined is the following:<br />

2<br />

AGT ��� ��� 2 � � �<br />

Pg<br />

Eg/(R �Eg�� RL �<br />

g � Ri) or AGT � A (26.16)<br />

2 vT��R g �<br />

R<br />

Expanding:<br />

P L<br />

Po �<br />

Pi<br />

E 2 o/RL �<br />

EgIg E 2 o/RL �2 Ei/Ri Ri �<br />

RL<br />

E Ri �<br />

RL<br />

2 o<br />

E 2 o/R L<br />

A GT � A vT� A v T � R g<br />

�A iR L<br />

Ri �� L<br />

� Ri �<br />

R �<br />

and AGT ��AvTAiT (26.17)<br />

� E 2 i<br />

L<br />

i<br />

L<br />

Eo �<br />

Ei<br />

E o<br />

Ri �<br />

RL<br />

R g � R i

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