Page 269 - Mechatronics with Experiments

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JWST499-c05
JWST499-Cetinkunt
ELECTRONIC COMPONENTS FOR MECHATRONIC SYSTEMS 255
and when it is disconnected from the supply and connected to the B side,
di(t)
0 = L ⋅ + R ⋅ i(t); t ≤ t ≤ t f (5.57)
2
dt
with the initial conditions of current being the last value during the previous state. Another
way of treating this problem is to consider the first equation with a pulse input voltage,
di(t)
V (t) = L ⋅ + R ⋅ i(t); 0 ≤ t ≤ t f (5.58)
s
dt
where V (t) is a pulse signal, that is the circuit is connected to 24 VDC (switch is on A side)
s
and 0.0 VDC (switch is on B side),
V (t) = 24 ⋅ (1(t − t ) − 1(t − t )) (5.59)
1
s
2
where 1(t) represent the unit step function, 1(t − t ) represent the unit step function shifted
1
in time by t .
1
The solution of the above differential equations can be shown to be as follows, or
®
more easily with Simulink (Figure 5.7).
i(t) = 0.0; t ≤ t ≤ t 1 (5.60)
0
3 −(t−t 1 )∕(L∕R)
i(t) = 24∕(10 × 10 ) ⋅ (1 − e ) A (5.61)
−(t−0.0001)∕0.0001
= 2.4 × (1 − e )mA; t ≤ t ≤ t (5.62)
1 2
i(t) = i(t ) ⋅ (e −(t−t 2 )∕(L∕R) )A = 2.356 ⋅ e −(t−0.0005)∕0.0001 mA; t ≤ t ≤ t (5.63)
2 2 f
Then the current across the resistor and inductor in each time period can be found by
V (t) = R ⋅ i(t) (5.64)
R
di(t)
V (t) = L ⋅ = V (t) − R ⋅ i(t) (5.65)
s
L
dt
We can confirm our physical expectation by the plots of these results. Initially, the inductor
will oppose, by generating an opposing voltage, the large rate of change in current and the
opposition will get smaller as the rate of change of current gets smaller, and as the current
reaches steady-state. When the voltage source switches suddenly to zero volts, the current
will start to decrease and the inductor will generate voltage in opposition to that change. In
other words, the inductive voltage will be negative since di(t)∕dt is negative.
For an RC circuit, the circuit relationships are, when connected to supply,
V (t) = R ⋅ i(t) + V (t) (5.66)
s
c
( t )
1
= R ⋅ i(t) + Q(t ) + i( )d ; t ≤ t ≤ t 2 (5.67)
1
1
C ∫
t 1
and when it is disconnected from the supply and connected to the B side,
( t )
1
0.0 = R ⋅ i(t) + Q(t ) + i( )d ; t ≤ t ≤ t f (5.68)
2
2
C ∫
t 2
The voltage across the capacitor at any given time (t) is the initial voltage due to the initial
charge at a given time (t ) plus the net change in the charge (integral of the current from
0
the initial time (t )topresent time(t))
0
( )
t
1
V (t) = Q(t ) + i( )d (5.69)
c
0
C ∫
t 0