CHAP. 15]

CIRCUIT ANALYSIS USING SPICE AND PSPICE

* node * node * node * node * node Rin Cin Rout Eout .ENDS Vac R1 Rf X1 .MODEL .STEP .AC .PROBE .END

1 2 3 4 5

is is is is is

395

the non-inverting input the inverting input the output the output reference (negative end of dependent source) the positive end of dependent source 1 2 10 E5 1 2 100 pF 3 5 10 k 5 4 1 2 1 E5 1 0 1 2 2 3 0 2 3 0 GAIN LIN LIN

AC 10 m 0 10 k Rgain 1 OPAMP RES(R ¼ 1Þ RES Rgain(R) 1 k 801 k 500 1000 k 1 000 000 k

200 k

The frequency response is graphed in Fig. 15-34(c). Compared with the open-loop circuit of Fig. 1533(a), the dc gain is reduced and the bandwidth is increased.

15.14 Referring to the RC circuit of Fig. 15-22, choose the height of the initial pulse such that the voltage across the capacitor reaches 10 V in 0.5 ms. Verify your answer by plotting Vc for 0 < t < 2 ms. The pulse amplitude A is computed from Að1  e1=2 Þ ¼ 10

from which

We describe the voltage source using PULSE syntax. Pulse-Step Vs R C .TRAN .PROBE .END

A ¼ 25:415 V

The source ﬁle is

response of RC, dead beat in RC/2 seconds 1 0 PULSE( 10 25:415 1:0E  6 1:0E  6 0:5 m 3 m Þ 1 2 1k 2 0 1u 1:0E  6 2:0E  3 UIC

The response shape is similar to the graph in Fig. 15-22(b). During the transition period of 0 < t < 0:5 ms, the voltage increases exponentially toward a dc steady state value of 25.415 V. However, at t ¼ 0:5 ms, when the capacitor voltage reaches 10 V, the source also has 10 V across it. The current in the resistor becomes zero and steady state is reached.

15.15 Plot the voltage across the capacitor in the circuit in Fig. 15-35(a) for R ¼ 0:01 and 4:01 . The current source is a 1 mA square pulse which lasts 1256.64 ms as shown in the i  t graph. Model the resistor as a single-parameter resistor element with a single parameter R and change the value of R from 0.01 to 4.01 in step of 4. We use the .AC command to sweep the frequency from 500 Hz to 3 kHz in 100 steps. The source ﬁle is Pulse response of RLC with variable R Is 0 1 Pulse( 0 1 m 100 u 0:01 u 0:01 u R 1 2 LOSS 1 C 1 0 2000 n IC ¼ 0 L 2 0 5m IC ¼ 0

1256:64 u 5000 u Þ

Mahmood_Nahvi_eBook_Schaum_s_Outlines_Theory_An
Mahmood_Nahvi_eBook_Schaum_s_Outlines_Theory_An