Page 186 - Schaum's Outline of Theory and Problems of Electric Circuits

HIGHER-ORDER CIRCUITS AND COMPLEX FREQUENCY
CHAP. 8]
8.11 HIGHER-ORDER ACTIVE CIRCUITS 175

Application of circuit laws to circuits which contain op amps and several storage elements produces,
in general, several first-order differential equations which may be solved simultaneously or be reduced to
a higher-order input-output equation. A convenient tool for developing the equations is the complex
frequency s (and generalized impedance in the s-domain) as used throughout Sections 8.5 to 8.10.
Again, we assume ideal op amps (see Section 7.16). The method is illustrated in the following examples.
EXAMPLE 8.13 Find HðsÞ¼ V 2 =V 1 in the circuit of Fig. 8-41 and show that the circuit becomes a noninverting
integrator if and only if R 1 C 1 ¼ R 2 C 2 .
Apply voltage division, in the phasor domain, to the input and feedback paths to find the voltages at the
terminals of the op amp.
1
At terminal A: V A ¼ V
1 þ R 1 C 1 s 1
R 2 C 2 s
At terminal B: V B ¼ V 2
1 þ R 2 C 2 s
But V A ¼ V B . Therefore,
1 þ R 2 C 2 s
V 2
¼
V 1 ð1 þ R 1 C 1 sÞR 2 C 2 s
Only if R 1 C 1 ¼ R 2 C 2 ¼ RC do we get an integrator with a gain of 1=RC
ð t
V 2 1 1
¼ ; v 2 ¼ v 1 dt
V 1 RCs RC 1
EXAMPLE 8.14 The circuit of Fig. 8-42 is called an equal-component Sallen-Key circuit. Find HðsÞ¼ V 2 =V 1
and convert it to a differential equation.
Write KCL at nodes A and B.
V A V 1 V A V B
At node A: þ þðV A V 2 ÞCs ¼ 0
R R
V B V A
At node B: þ V B Cs ¼ 0
R
Let 1 þ R 2 =R 1 ¼ k, then V 2 ¼ kV B . Eliminating V A and V B between the above equations we get
k
V 2
¼ 2 2 2
V 1 R C s þð3 kÞRCs þ 1
2
2
R C 2 d v 2 þð3 kÞRC dv 2 þ v 2 ¼ kv 1
dt 2 dt
EXAMPLE 8.15 In the circuit of Fig. 8-42 assume R ¼ 2 k
, C ¼ 10 nF, and R 2 ¼ R 1 . Find v 2 if v 1 ¼ uðtÞ.
By substituting the element values in HðsÞ found in Example 8.14 we obtain
2
V 2
¼
s þ 2 10 s þ 1
4 10 10 2 5
V 1
2
d v 2 4 dv 2 8 9
þ 5 10 þ 25 10 v 2 ¼ 5 10 v 1
dt 2 dt
The response of the preceding equation for t > 0to v 1 ¼ uðtÞ is
v 2 ¼ 2 þ e t ð2 cos !t 2:31 sin !tÞ¼ 2 þ 3:055e t cos ð!t þ 130:98Þ
where ¼ 25 000 and ! ¼ 21 651 rad/s.
EXAMPLE 8.16 Find conditions in the circuit of Fig. 8-42 for sustained oscillations in v 2 ðtÞ (with zero input) and
find the frequency of oscillations.
In Example 8.14 we obtained
k
V 2
¼ 2 2 2
V 1 R C s þð3 kÞRCs þ 1