Question 13.7: FINDING THE TRANSIENT RESPONSE OF AN AVERAGING CIRCUIT An in...

FINDING THE TRANSIENT RESPONSE OF AN AVERAGING CIRCUIT

An input signal is to be averaged and then fed to a load resistance R_{L} = 1 GΩ. This is shown in Figure 13.27(a). The input signal is a triangular wave, as shown in Figure 13.27(b). Use PSpice to plot the transient response of input and output voltages for a duration of 0 to 33.33 msec in steps of 10 μsec.

13.27(a)
13.27(b)
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The averaging is accomplished by the RC circuit of Figure 13.27(c). This is similar to Example 13.6, except that the value is divided by time. The input frequency is f = 1 kHz.

The PSpice schematic is shown in Figure 13.28(a) with an averaging circuit. Figure 13.28(b) shows an implementation with an ABM integrator.

The listing of the circuit file is as follows:

Example 13.7 Averaging circuit

SOURCE     VIN 1 0 PULSE (0 1V 0 0.5MS 0.5MS 1NS 1MS)
CIRCUIT      RL 2 0 1G

* Calling subcircuit AVRG:
X1    1        2      0          INTG
*      Vi−  Vo+  Vo−   model name

* Subcircuit definition for AVRG:
.SUBCKT AVRG  1        3         2
* model name   Vi+    Vo+   Vo−
RI 1 0 1G
GAVR 2 4 1 0 1
C 4 2 1 IC=0V
R 4 2 1G
EOUT 3 2 VALUE = {V(4,2)/TIME}
RO 3 2 1G
.ENDS AVRG                               ; End of subcircuit definition

ANALYSIS   .TRAN 10US 4MS UIC                ; Use initial condition in transient

analysis
.PLOT TRAN V(2) V(1)                ; Prints on the output file
.OPTIONS ABSTOL = 1.00N RELTOL = 0.01 VNTOL = 0.1 ITL5 = 0
.PROBE                                        ; Graphics post-processor

     .END

The plots of the transient response for the input and output voltages are shown in Figure 13.29. Under steady-state conditions, the average value of a triangular wave remains constant at 0.5bh/T = 0.5 × 1 msec × 1 V/1 msec = 0.5 V.

13.27(c)
13.28(a)
13.28(b)
13.29

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