180 Transient analysis
Example 8.5
Derive the waveforms of the current and the output voltage for the differ-
entiator circuit given in Fig. 8.11 when R = 2 kIl, L = 8 mH and the input
voltage pulse is 5 V in magnitude with a width of 40 I~S.
Solution
This is the same circuit as that of the previous example simply reconfigured to
make VL the output voltage. The time constant of the circuit is therefore the
same as before at 4 Ixs,
The pulse width is ten times the time constant of the circuit so that, following
the application of the pulse, the current and both element voltages will reach
their final steady state values well before it is subsequently removed.
Consider the period 0-40 IXS. The current waveform is identical to that
derived for the previous example. The output voltage Vo is now the voltage
across the inductor VL and this is given by Equation (8.6) as
V L "- V exp (-Rt/L)
After t = z(4 txs)
vL = 5 exp (-1) = 0.368 • 5 = 1.84 V
After t = 5z(20 ~s)
VL~-V=5V
Consider now the period following the removal of the pulse, i.e. after 40 Ixs.
Again the current waveform is the same as that obtained in Example 8.4. The
output voltage is given by Equation (8.11) to be vL = -V exp (-Rt'/L) where
t' = (t- 40) IXS.
At t = 40 txs, t' = 0 and VL = --V exp (0) = -V
At t = 44 txs, t' = 4 Ixs and vL = -V exp (-1) = -0.368 • 5 = 1.84 V
At t = 60 txs, t' = 20 IXS and VL = --V exp (-5) = 0
The waveforms of current and output voltage may now be drawn as in
Fig. 8.12.
i (mA)ti, Vo
- -5-
0 L--5iI20 40 ]f o "Y(,s)
Figure 8.12