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The CE h-parameter transistor model (with hre hoe 0 was applied to the CC ampli er in Section 6.2. Taking iB and vEC as independent variables, develop a CC h-parameter model which allows for more accurate representation of the transistor than the circuit of Fig. 6-2(c).

Table 5-1. Frequency Response of the Low-pass Filter !, rad/s f , Hz jv2 =v1 j 0 0 1 10 1.59 1 100 15.9 0.995 103 159 0.707 104 1:59 103 0.1 105 15:9 103 0.01

13.1C

CC characteristics are not commonly given by transistor manufacturers, but they would be plots of iB vs. vBC with vEC as parameter (input characteristics) and plots of iE vs. vEC with iB as parameter (output or emitter characteristics). With iB and vEC as independent variables, we have vBC f1 iB ; vEC iE f2 iB ; vEC 1 2

The circuit of Fig. 5-30 compares the voltage v1 with a reference level vo . Since the open-loop gain is very large, the op amp output v2 is either at Vcc (if v1 > vo ) or at Vcc (if v1 < vo ). This is shown by v2 Vcc sgn v1 vo where sgn stands for sign of. For vo 0, we have Vcc v1 > 0 v2 Vcc sgn v1 Vcc v1 < 0

POSTULATE 1:

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Next we apply the chain rule to form the total di erentials of (1) and (2), assuming that vbc vBC % dvBC , and similarly for ie : vbc vBC % dvBC @vBC ib @vBC vec @iB Q @vEC Q @i @i ie iE % diE E ib E vec @i @v

Fig. 5-30 EXAMPLE 5.23 In Fig. 5-30, let Vcc 5 V, vo 0, and v1 sin !t. Find v2 . For 0 < t < =!, v1 sin !t > 0 For =! < t < 2 =!, v1 sin !t < 0 v2 5 V One cycle of v2 is v2 5 V

13.1D

The output v2 is a square pulse which switches between 5 V and 5 V with period of 2 =!. given by 5V 0 < t < =! v2 5 V =! < t < 2 =!

3 4

The sum of the lengths of two sides of a triangle is greater than the length of the third side. (Corollary: The length of the longest side of a triangle is less than the sum of the lengths of the other two sides and greater than their difference.)

EXAMPLE 5.24 The circuit of Fig. 5-31 is a parallel analog-to-digital converter. The Vcc and Vcc connections are omitted for simplicity. Let Vcc 5 V, vo 4 V, and vi t (V) for 0 < t < 4 s. Find outputs v3 ; v2 ; and v1 . Interpret the answer. The op amps have no feedback, and they function as comparators. The outputs with values at 5 or 5 V are given in Table 5-2.

PRINCIPLE 1:

Finally, we de ne @vBC % vBC @iB Q iB Q @v vBC Reverse voltage ratio hrc BC % @vEC Q vEC Q @i i Forward current gain hfc E % E @iB Q iB Q @i iE Output admittance hoc E % @v v Input resistance hic

Table 5-2 time, s 0<t<1 1<t<2 2<t<3 3<t<4 input, V 0 < vi 1 < vi 2 < vi 3 < vi <1 <2 <3 <4 v3 v3 v3 v3 5 5 5 5 outputs, V v2 v2 v2 v2 5 5 5 5 v1 v1 v1 v1 5 5 5 5

Fig. 13-2

The binary sequences fv3 ; v2 ; v1 g in Table 5-2 uniquely specify the input voltage in discrete domain. However, in their present form they are not the binary numbers representing input amplitudes. Yet, by using a coder we could transform the above sequences into the binary numbers corresponding to the values of analog inputs.

5 6 7 8

PRINCIPLE 2:

CHAP. 5]

A circuit that satis es (3) and (4) with de nitions (5) to (8) is displayed by Fig. 6-16.

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