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Solutions:
a. Using the current divider rule, we have
(1/ho)h fI1 h fI1 I2 ��� ���
(1/ho) � ZL 1 � hoZL and Ai � � (26.45)
I2
h f
� � �� I 1 � h
b. Applying Kirchhoff’s voltage law to the input circuit gives us
E1 � hrE2 E1 � hiI1 � hrE2 � 0 and I1 � ��
hi
Apply Kirchhoff’s current law to the output circuit:
I2 � h fI1 � hoE2 However, I 2 ��
so � � h fI 1 � h oE 2
Substituting for I 1 gives us
� �hf� ��h E2 E1 � hrE2 � �� oE2 ZL hi
or h iE 2 ��h fZ LE 1 � h rh fZ LE 2 � h ih oZ LE 2
and E 2(h i � h rh fZ L � h ih oZ L) � �h fZ LE 1
with the result that
Av � � E2
�h fZL � � ��� (26.46)
hi(1 � hoZL) � hrh fZL E 1
E2 �
ZL
1
E2 �
ZL
EXAMPLE 26.11 For a particular transistor, hi � 1 k�, hr � 4 �
10 �4 , hf � 50, and ho � 25 ms. Determine the current and the voltage
gain if ZL is a 2-k� resistive load.
Solution:
Ai � �
� � �47.62
Av �
�
�100 � 10 100
� �����99 1.01
The minus sign simply indicates a phase shift of 180° between E2 and
E1 for the defined polarities in Fig. 26.51.
3
����
(1.050 � 10 3 ) � (0.04 � 10 3 �(50)(2 k�)
����
(1 k�)(1.050) � (4 � 10
)
�4 h f
50
�� ��
1 � hoZL 1 � (25 mS)(2 k�)
50
50
���3
�
1 � (50 � 10 ) 1.050
�h fZL ���
hi(1 � h oZL) � hrh fZL )(50)(2 k�)
oZ L
HYBRID (h) PARAMETERS ⏐⏐⏐ 1177